TCU Organic Chemistry: Sergei Dzyuba
Chemistry Discussion Board:
TCU Organic Chemistry: Sergei Dzyuba
|By Sergei on Saturday, August 19, 2017 - 07:44 am: Edit|
Welcome to CHEM 30123/O-Chem lecture!
Feel free to post questions related to O-Chem.
Have a great semester!
Hi Dr. Sergei,
I am wondering if there is any way to find the answers to the homework problems at the end of each chapter in the textbook? Thanks
|By Sergei on Wednesday, August 23, 2017 - 08:22 pm: Edit|
There is a study guide that comes with the book (the bookstore usually orders those with the book).
There might be an online version of the study guide as well.
For your reference
|By arielburden on Tuesday, August 29, 2017 - 05:19 pm: Edit|
on page 24 of the inquisition workbook, it states that ethene is sp hybridized on the answer website, yet further down on the page it states "the carbons in ethene are flat because their sp2 hybridization". I am confused about which is correct, because I thought it was sp2 but the answer page had sp. Thank you!
|By Sergei on Tuesday, August 29, 2017 - 08:02 pm: Edit|
hybridization of C in ethene is sp2 (there must be a typo on the website)
Also, carbons are not flat, the atoms in ethene are located in a plane, bc of the 120 angles that are required for the sp2 hybridized carbons
Also, please look at this problem/issue from a different perspective:
lets assume that there is no typo, in that case how would the molecule (C2H4) look like, i.e., if C was sp-hybridized?
If you try to draw the molecule (or MO) assuming sp hybridization, it will be obvious that something is not working; thus, the inability to provide a reasonable structure, MO, etc will put into question sp hybridization (something similar to what we have done in the lecture on some occasions), and therefore, a different hybridization should be proposed.
|By TD on Thursday, August 31, 2017 - 06:49 am: Edit|
Dear Dr. Dzyuba,
When we rank the acidity levels of HF, Hal, HBr, HI, we talked in terms of strength of conjugate base (or stability) and conclusively strength of acid (or reactivity).
I just want to clarify, if a weak conjugate base holds electron tightly to itself, and a strong conjugate base readily loses electron.
Also, is it correct that we should rank these molecules in regard to EN, because they are in different periods.
Last question, what does it mean when I- is a better Lewis base than an F-. Is it because the further away the distance from nucleus, the more "favorable" the lone pairs of electrons? Or it's the stability, as from the above example, that matters?
|By TD on Thursday, August 31, 2017 - 07:13 am: Edit|
Can you give examples of hard acid, hard base, soft acid and soft base, as well as the favorable bond formations btw hard-hard, and soft-soft.
one example I know is H+ and F- ( hard acid-hard base) form bonds faster than Ag+ (soft acid) and F- (hard base). Other is Ag+ and I- (soft acid-soft base) forms faster bond than H+ and I- (hard acid-soft base).
Also, when you gave examples about H+, Li+, Na+, K+, and Rb+, and Cs+ to rank which is harder and softer, do we deal primarily with size and delocalization and not thing else ( like EN, stability, reactivity, etc.)?
|By Sergei on Thursday, August 31, 2017 - 08:04 am: Edit|
TD on Thursday, August 31, 2017 - 06:49 am:
TD on Thursday, August 31, 2017 - 07:13 am:
In general, to assess (rank) the acidity of acid(s), we should be looking at the stability of the conjugate base.
In other words, by removing the appropriate proton (at the moment we only looked at molecules that have one type of proton), and then evaluate how localized/delocalized the formed electron pair would be.
Delocalized electron pair (or delocalized charge, in general) leads to decreased reactivity, and thus increased stability.
Higher reactivity implies that the conjugate base will be readily donating the electrons, whereas lower reactivity (weaker base) will not be reacting as much/fast (b/c electron pair is more delocalized, and thus less available to react with an acid).
Please consider all the concepts that we covered at this point.
Since we introduced hard-soft acid-base theory, we cannot say any more that I- is a better base than F-, without specifying towards what.
(to clarify your statement/question: the fact that the electrons are close/far from nucleus doesn't make them more or less "favorable" (also, favorable for what?)
Again, we have to consider all the concepts, and once we moved away from Bronsted acidity (into Lewis), everything becomes relative.
So, the electron pair on I- is farther from the nucleus (as opposed to F-, and other Hal-), and therefore it cannot be held as tightly (compared to F-), the large size of I also implies that that electron pair is more delocalized (as compared to other Hal-), which makes I- a softer base (softest amount Hal-), thus I- should be more reactive, but towards soft(er) acids.
The example that we covered in lecture/the ones that you are giving in the post are the demonstrating that.
In regard to H+, Li+, Na+, etc:
just because we moved from the electron pair to a vacant orbital, we should not be looking for different ways to assess stability/reactivity/etc. Also, recall that when we gave definitions of hard/soft acids/bases, all the parameters are interchangeable/connected (large/small radius indicates the position of HOMO/LUMO, for example)
We will be going over more examples on Friday/next week.
For now, you could try to evaluate hardness/softness of the following species:
Cu+ vs Cu2+, (CH3)3N vs (CH3)3P, H2S vs H2O, Hg2+ vs Al3+
|By TD on Thursday, August 31, 2017 - 08:52 am: Edit|
Dear Dr. Dzyuba,
Cu+ is larger than Cu2+ so its softer.
(CH3)3-N is smaller than (CH)3-P so its harder.
H2S is larger than H2O so its softer.
Hg2+ is larger than Al3+ so its softer.
Should I consider the size of molecule/atom or should I consider the delocalization between the atoms sharing bond, i.e. C and H, C and P; or H and S, H and O in the above example? If so, how?
|By Sergei on Thursday, August 31, 2017 - 09:45 am: Edit|
You got harder/softer correct, but for somewhat wrong reasons.
Why would Cu+ be larger than Cu2+; i.e., Cu is Cu?
If it's the same atom, the size should be the same, but the distribution of charge would change if you have 1+ or 2+.
Seems like you are trying to apply one parameter to everything.
Recall charge localization/delocalization, and the whole discussion about how elements in the same period could be treated over the elements in different periods.
Also, when considering a molecule, decide which atom/center is responsible for specific properties.
For example, in the case of H2S vs H2O:
S is larger than O (3rd period vs 2nd period), thus the electron pairs will be more delocalized on S in H2S as compared to O in H2O.
The electron pair is not going to be delocalized over Hs in both cases. Therefore the size of H2S/H2O (as a molecule) is not really relevant.
Keep in mind: it's not just the size or delocalization or EN - it's always going to be a combination of factors/parameters, depending on the situation.
We will continue on this topic tomorrow.
|By Sofia on Thursday, August 31, 2017 - 10:16 am: Edit|
Hi Dr Dzyuba,
We talked about the inductive effect on an atom being related to electronegativity. Does that mean that the inductive effect on an sp3 carbon will be weaker than the inductive effect on an sp2 carbon?
Also, if we have two sp3 carbons, for example in methane and ethane, which carbon will have a stronger inductive effect?
|By Taylor Gray on Thursday, September 07, 2017 - 07:37 pm: Edit|
I have a quick question in regards to Lewis acids and Bronsted acids. I ran into a bit of a stump when asked if HCl could be considered a Lewis acid. I definitely understood how it could be a Bronsted acid but I don't understand how it could be a Lewis acid. During tutoring it was explained that it was because when it disassociated it has a proton, which has an empty orbital. I understand that but I don't understand how or when I would have to look at the molecules behavior when it disassociates. Is it because it is a strong base?
|By Sergei on Thursday, September 07, 2017 - 09:06 pm: Edit|
Sofia on Thursday, August 31, 2017 - 10:16 am:
Apologies for overlooking the post:
Csp2 is more electronegative than Csp3, so in regard to +I effect, Csp2 would be donating less than Csp3.
In regard to electron-withdrawing ability, Csp2 would be have a stronger -I effect than Csp3
By Taylor Gray on Thursday, September 07, 2017 - 07:37 pm:
H+ is a Lewis acid (it's also a Bronsted acid), because it's an empty orbital. This is a definition of the Lewis acidity - anything with an empty orbital is a Lewis acid.
Recall our whole approach that we have been using: it does not matter if/when/how a bond is broken (or in this case how/when the dissociation takes place), lets assume that it already happened.
In the case of HCl, H+ is a Lewis acid, Cl- is a Lewis base.
But bc Cl– is stable (non-reactive), it's not going to donate it's electron pair, this means that the empty orbital of H+ is readily available, and hence you have the associated acidity.
In other words, even though the dissociation of HCl give a lewis acid and lewis base, the Lewis base is non-reactive (charge is delocalized, noble gas configuration, size, electronegativity), which means that only H+ is the reactive species; hence it's a Lewis acid.
|By Sergei on Friday, September 08, 2017 - 07:25 am: Edit|
In regard to " if we have two sp3 carbons, for example in methane and ethane, which carbon will have a stronger inductive effect?"
I am assuming that you are referring to methyl/Me (CH3) and ethyl/Et (CH3CH2) case:
more carbons with Hs, more donation; also, recall the examples we went through in class.
|By YourfavoriteOchemStudent on Friday, September 08, 2017 - 05:42 pm: Edit|
Hey Dr. Dzyuba
What is the name of your dog? I am actually curious
|By YourfavoriteOchemStudent on Friday, September 08, 2017 - 06:21 pm: Edit|
|By Sergei on Monday, September 11, 2017 - 08:03 am: Edit|
Quiz1 is graded (it's in the box, in the Chem Office):
Key was e-mailed to you.
|By TD on Monday, September 11, 2017 - 08:41 am: Edit|
So I don't understand 2 concepts:
1) CH4 is not a Lewis acid, I agree, but it shouldn't be Lewis base as well, because the e- pairs are contributed both by H and C. so it's a shared pair. I have looked at the examples in the book and all of them need the lewis base to have a lone pair, not a shared pair between two different atoms.
2) Why F can't be hybridize but not O and N?
|By anon on Monday, September 11, 2017 - 05:13 pm: Edit|
This explains why methane can be a Lewis Base (in very rare cases)
|By Sergei on Monday, September 11, 2017 - 05:38 pm: Edit|
TD on Monday, September 11, 2017 - 08:41 am:
1. Lewis base is anything with a pair of electrons.
Lone pair is not a prerequisite.
We'll be looking at alkenes and alkynes next, and the pi-bond is the Lewis basic center.
2. Please recall our whole discussion on the hybridization when we were making the MO diagrams, and used the bond angles to define the hybridization.
For C=O and CN - the bond/directionality is defined, because the double bond(s) controls the geometry/angle (also there is no free rotation around the double/triple bond). In the case of F (or any other halogen), the directionality/angle could not be defined (there is a free rotation around single bond).
anon on Monday, September 11, 2017 - 05:13 pm:
Just to clarify: the role (acid/base) could change based on the reaction conditions, however, without any indication what the conditions/reagents are - we should follow the definition, and look at what's available, i.e., a vacant orbital or an electron pair.
In methane only electron pair is present, hence it's a Lewis base.
it's a very weak base (again, based on MO diagrams, we know that sigma electrons are the lowest in energy, hence most stable, least reactive; the only thing that this implies, it requires a very strong acid or what's known as super acid).
|By Esteban on Thursday, September 14, 2017 - 05:31 pm: Edit|
On the book on Ch. 2 #50 (b), would the correct name be 3-isopropylhexane or 3-ethyl-2-methylhexane?
|By Sergei on Thursday, September 14, 2017 - 08:24 pm: Edit|
because following one of the rules for branched alkanes, larger number of branched points should be chosen.
The way we looked at it in the lecture: given a choice, smaller/simpler substituents should be used
|By Sergei on Sunday, September 17, 2017 - 03:53 pm: Edit|
Quiz 2 is graded (in the box in the Chem Office):
|By AD on Monday, September 18, 2017 - 02:13 pm: Edit|
How do we determine if a neutral atom, such as H3PO4, is a lewis acid or lewis base?
|By Sergei on Monday, September 18, 2017 - 03:17 pm: Edit|
Because of the ionic nature of the bond, you can think of this acid as the one that contains a Lewis acidic center (H+) and a Lewis basic center (H2PO4-).
Consider how strong the LA vs LB could be: LB (large delocalized) vs LA (small, localized).
H3PO4 should act as a Lewis acid, bc of the higher reactivity of the Lewis acidic center (H+).
Just to be sure: H3PO4 is not an atom
|By anon on Monday, September 18, 2017 - 05:59 pm: Edit|
Will we still be tested over topics from ch 4 and 6 that we didn't discuss in class?
|By anon on Monday, September 18, 2017 - 06:05 pm: Edit|
Hi Dr. Dzyuba,
In the molecule 1-methyl-1,2,3-triazole, why would protonation occur selectively at the Nitrogen on the third position?
|By AA on Monday, September 18, 2017 - 06:19 pm: Edit|
What does the H+ work up do at the end of the alkene reaction with H2O, H2SO4, and Hg2+?
|By anon on Monday, September 18, 2017 - 06:22 pm: Edit|
in lecture today you added NaOH on top of an arrow and HOOH on the bottom of an arrow in the reaction that produced the cyclic pentanol from the cyclopentane with BH2 attached... where does the HOOH go? How does this make an alcohol and get rid of BH2?
Also, when describing that H- is a base AND nucleophile, you drew an arrow going from H- to a carbon attached to Hg+ as a substituent in a cyclopentanol. Why does H- attack the Carbon, and not the Hg+?
|By Sergei on Monday, September 18, 2017 - 09:29 pm: Edit|
anon on Monday, September 18, 2017 - 05:59 pm:
only the topics covered in class will be on the exam.
Chapter 6 will not be the exam.
Some material from chapter 4 will be, as it relates to the addition to C=C (and alkynes).
anon on Monday, September 18, 2017 - 06:05 pm:
Draw the conjugate acids with H being on N2 and N3 (N1 is out, since its electron pair is in conjugation, we haven't covered aromatic compounds yet, but if you consider the knowledge that we have in regard to alkenes, and how the electron pair/bond should be located relative to each other to get the delocalization, it will be clear that only the electron pairs on N2 and N3 are available for protonation).
Then draw the resonance structures, and evaluate the stability of those structures.
It should be clear why protonation is taking place on N3.
AA on Monday, September 18, 2017 - 06:19 pm:
Just to clarify:
H2SO4 is the equivalent reaction to Hg2+; in other words, either or.
HgSO4 was in parenthesis (under H2SO4) to indicate that these conditions could be used independently (almost).
However, when we should consider that Hg is large (as large as Br), the mercurial intermediate should be possible.
That 3-memebered carbocation bc all bonds are "locked") should not be undergoing rearrangements (where as in the case of H2SO4, H+ is the LA, and the carbocation could rearrange, if it give a more stable carbocation)
H+ work-up was in regard to the fact that at the end of the reaction between the alkene, water and HgSO4, there will be a C-Hg bond present. To get rid of this bond, and replace it with C-H (since this reaction is a way of synthesizing alcohol from alkenes, and alcohols do not have C-Hg bond), we need to use NaBH4, which will not only replace C-Hg with C-H, but because H– is a base (in addition to being Nu), the OH will get deprotonated.
In other words, the use of NaBH4 implies basic conditions (OH of the alcohol can get deprotonated, and since the reaction is in water, the H of H2O (or MeOH) will be deprotonated as well, giving OH– (or MeO–), hence basic conditions).
Since the product should be neutral, an acidic work-up is required.
anon on Monday, September 18, 2017 - 06:22 pm:
H2O2 will create a Nu (upon deprotonation by NaOH), i.e., HOO–, which will attack the vacant orbital of BH2. following that there will be a rearrangement step (which will not be covered this semester), which essentially leads to the electrons of the C-B bond attacking the O of the B-O-OH, which leads to the loss of OH– and the formation of the C-O-BH2, which will follow by the attack of HO– on the BH2.
In the nutshell: the rearrangement of the C-B-O-OH is responsible for the formation of C-O bond
The mechanism of demercuration is a bit more complicated, it's a multistep process, and very likely involves some radical chemistry, which we are not covering this semester. In that case, you would be correct to bring H– s to Hg+ (but then several steps would be required to form C-H bond, including some rearrangements).
At this point, with what we know, the formal reaction mechanism (although not entirely correct) was shown to highlight a difference between H– being Nu and Base.
|By AA on Tuesday, September 19, 2017 - 07:45 am: Edit|
So just to make sure, out of all the reactions that the textbook has (and doesn't), we should only know the boration, oxymercuration, halohydrin formation, bromonium formation, hydride (or CH3) shift as well as the common sense ones like the addition of HX, H2O, and X2? I feel like part d of question 11 on the review is something about KMnO4 and acidic solutions which we didn't cover..
|By AA on Tuesday, September 19, 2017 - 07:50 am: Edit|
Also in response to your response about my question about acidic workup, is the H+ coming from H2SO4 in the environment? So is the point of having H2SO4 only for the acidic workup step?
|By BH3 on Tuesday, September 19, 2017 - 10:24 am: Edit|
I'm confused on why in the major product in the boration process, BH2 goes to the branch with more hydrogens and H goes to the branch with a CH3. Don't we always add hydrogen where there's more hydrogen? Also, is BH3 addition the only case as far as what we've covered so far that is anti Markinikovs?
|By Sergei on Tuesday, September 19, 2017 - 01:04 pm: Edit|
AA on Tuesday, September 19, 2017 - 07:45 am:
I will list the reactions that will not be on the exam shortly
AA on Tuesday, September 19, 2017 - 07:50 am:
In the case of H2SO4/water reaction: basic work-up would be required, because the reaction is done under acidic conditions (thus there is an excess of H+ that would have to be neutralized).
You questions involved also HgSO4 as a Lewis acid.
In that case, there is an additional step: NaBH4, and that changes the reaction media from acidic to basic; hence acidic work-up would be required to get neutral alcohol.
BH3 on Tuesday, September 19, 2017 - 10:24 am: Yes, this is the case of anti-Markovnikov additions.
However, we should not be governed by names when deciding which product to draw, but rather by very fundamental principles: reactions proceed through the most stable intermediates/most favorable transitions states.
IN hydroboration, sterics plays a role, hence the observed regioselectivity.
|By Sergei on Tuesday, September 19, 2017 - 01:05 pm: Edit|
Quiz 2 is graded:
the key will be e-mailed shortly
|By Sergei on Tuesday, September 19, 2017 - 01:10 pm: Edit|
Topics that WILL NOT be on tomorrow's exam:
Chapter 6 (chirality on sp3 carbon)
Oxidation of alkenes (epoxidation, hydroxylation, C=C cleavage, and epoxide ring opening)
Hydrogenation of alkenes/alkynes
Reactions of acetylide anons with alkyl halides
|By JP on Tuesday, September 19, 2017 - 01:26 pm: Edit|
Have you determined which reactions will and won't be on the exam yet?
|By JP on Tuesday, September 19, 2017 - 01:26 pm: Edit|
Nevermind so sorry just saw that you already answered that!!!
|By skromer on Saturday, September 23, 2017 - 02:27 pm: Edit|
Hi Dr. Dzyuba,
Yesterday in class when we were evaluating which compounds can be considered aromatic, we looked at the cyclic structure of C4H4O. You said it is considered aromatic because it is planar, conjugated, and the electrons = 4n+2. I am confused how this compound is considered planar if the Oxygen is sp3 hybridized. How does that not cause the ring to lose its planarity? Thanks
|By Sergei on Saturday, September 23, 2017 - 06:50 pm: Edit|
Hi skromer on Saturday, September 23, 2017 - 02:27 pm:
The atoms would do whatever is needed to optimize/lower the overall energy
sp3 - 109, sp2 - 120 - thus, it's not that much to become planar, and aromaticity is very very energetically favorable.
In other words, sp3-like O adopts sp2 hybridization to attain aromaticity
|By Anon on Monday, September 25, 2017 - 04:45 pm: Edit|
During aromatic substitution, we produce a carbocation. Why do we have to show resonance before elimination? Thank you.
|By Sergei on Tuesday, September 26, 2017 - 10:03 am: Edit|
the formed carbocation is delocalized (that's why we drew resonance structures); those resonance structures show where the positive charge could be located once the electrophile is added.
It will be important once we start looking at the reactions on the di-sibstituted benzenes, for example.
|By Sergei on Wednesday, September 27, 2017 - 08:47 am: Edit|
Exam 1 is graded:
The key will be e-mailed later today
|By anon on Wednesday, September 27, 2017 - 04:12 pm: Edit|
For the exam, for the second reaction in Q6:
-Why HgSO4 is preferred? I read online that oximercuration produces in the more substituted C since the positive charge is more heavily at the tertiary C: http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch14/ch14-4-6.html?
-Also, why the boronation with NaOH and iPrOOH not correct mechanistically?
|By Sergei on Wednesday, September 27, 2017 - 05:10 pm: Edit|
more substituted product would be the major, what was shown was the minor.
Recall our discussion on more accessible versus more electrophilic sites (when we looked at 3-membered rings): because we have the mercurium 3-membered ring, both reaction sites should be considered.
There are no other possibilities to make this compound (based on what we covered thus far).
if we consider the mechanism, the O that is becoming the alcohol is coming from the peroxide, but it's not the one closest to iPr, i.e., (iPr(O)OH), but rather the iPrO(O)H. , and iPrO– is the leaving group. Thus, iPrOOH will not work.
|By AA on Saturday, September 30, 2017 - 11:54 am: Edit|
Hi Dr. Dzuyba,
I'm assuming you are not following the schedule on the syllabus anymore, so I was wondering which section are you going to be teaching on Monday?
|By Sergei on Saturday, September 30, 2017 - 12:08 pm: Edit|
Finish electrophilic aromatic substitution, and start on nucleophilic substitution, which will incorporate the chirality (Ch6).
Thus, I think we are somewhat on schedule (maybe the dates on which specific chapters are covered would move).
The only major modification thus far is chapter 6
|By AA on Saturday, September 30, 2017 - 01:02 pm: Edit|
Thank you! Also, Why is there a -R effect in the case of nitrobenzene? Why can't we move the electrons on the double bond of NO2 and into the ring?
|By Sergei on Saturday, September 30, 2017 - 01:47 pm: Edit|
There is no way to move the electrons on O towards the aromatic ring, you will have to break sigma bond (which is not allowed; also there are no electrons on N, hence N+)
In other words, try to do it, and you will see that it's not going to work, hence +R effect is not possible
On the other hand, moving electrons from Ar-ring to NO2 gives a resonance structure that is not violating any rules.
|By AA on Saturday, September 30, 2017 - 06:02 pm: Edit|
In order to produce a nitronium ion, how do we do the hydride shift after H2SO4 gives a H+ to O-? I thought hydride shift is when the + on carbocation changes its position with a H nearby, but here there is no + to replace with H..
|By Sergei on Saturday, September 30, 2017 - 06:52 pm: Edit|
There is no hydride (H: ) shift in HNO3/H2SO4!!
All the steps that we outlined were proton (H+) transfer steps, i.e., protonation/deprotonation.
We are dealing with acid-base chemistry here.
|By Sergei on Monday, October 02, 2017 - 08:31 am: Edit|
Quiz4 is graded:
|By alexm on Monday, October 02, 2017 - 10:37 am: Edit|
When is the creation of a aromatic ring considered to be favorable/when should we make a ring like the question on Quiz 4?
|By Sergei on Monday, October 02, 2017 - 01:59 pm: Edit|
Just to clarify: there is a difference between an aromatic ring and a cyclic structure.
In regard to aromatic ring/aromaticity:
if the starting material is aromatic, the product should be aromatic as well; hence whatever is required to restore the aromaticity should be explored (since the intermediates may or may not be aromatic).
There are very few reactions when the aromatic structure is not retained (e.g. p172, hydrogenation)
Cyclic structures (as the one on the Quiz4):
if interamolecular processes are possible and are favorable, then you should consider the formation of the cyclic structure.
In that case, if you analyze the problem from the stand point of what is a Nu, what is E, the Nu and E are in the same molecule (and it is going to proceed via a 5-membered TS), hence an intramolecular reaction should lead to the major product.
|By abjackson on Monday, October 02, 2017 - 05:03 pm: Edit|
I am a little confused by something you discussed in lecture today. In the Sn1 reaction, two products are possible, and both are "chiral." But then later you said that since the products exist as a racemic mixture, the chirality of the starting material is lost through the achiral intermediate. How can this be true if the racemic mixture is made up of two chiral compounds? Aren't they chiral even if they don't rotate light?
Also, why is the concerted mechanism referred to as "Sn2"? What does the 2 mean? Thank you!
|By Sergei on Monday, October 02, 2017 - 05:53 pm: Edit|
Both compounds are chiral, meaning that each compound has a chiral center. But because these two enantiomers are forming in 1 to 1 ratio, the resulting mixture is not optically active.
Please recall one of the earlier notes about the distinction between chirality and optical activity.
Achiral intermediate leads to the two enantiomers in equal amounts; hence there is no optical activity.
In SN1: the concentration of Nu had not impact on the rate of the reaction, because, carbocation had to form first, before it can react with Nu.
Thus, the rate of the reaction depends on only 1 compound, and as a result 1 (1 - unimolecular).
In SN2, the concentration of Nu plays a role, thus two compounds are present in the transition state (one of the last reactions that we put), because there are two compounds in the transition state, it had to be reflected in the naming/nomenclature (2 - bimolecular).
|By anon on Monday, October 02, 2017 - 07:02 pm: Edit|
why does nitrobenzene not undergo Friedel Crafts acetylation?
|By Sergei on Monday, October 02, 2017 - 07:39 pm: Edit|
Just not as readily as benzene (and by a lot).
NO2 is a very EWG, hence nucleophilicity of the Ph group is greatly diminished.
Thus, heating, long(er) reaction times will be required, compared to benzene, for example.
|By arielburden on Tuesday, October 03, 2017 - 08:35 pm: Edit|
Hi Dr. Dzyuba,
in a reduction process, how would we know whether to use Zn/Hg and HCl or 1) H2NNH2 2) NaOH/delta?
|By Sergei on Tuesday, October 03, 2017 - 08:40 pm: Edit|
Either of those conditions will work.
If you see one or the other, you should think reduction of the C=O to CH2 (conversely, if you see a ketone as the starting material and the product is CH2, you can propose one of those options as the reagents)
|By anon on Wednesday, October 04, 2017 - 05:12 pm: Edit|
Hi Dr. Dzyuba,
just to clarify I have this correct.... the things we look at when determining if it will be SN1 or SN2 are: the solvent, substrate, and reagent. If it is a primary or secondary carbon, it will most likely proceed through SN2, and tertiary carbons will go through SN1 due to steric crowding. If the carbocation produced would be primary and cannot be immediately delocalized by resonance, SN1 will not take place.
We use polar solvents for SN1 reactions and aprotic for SN2. Why?
|By AB on Wednesday, October 04, 2017 - 05:15 pm: Edit|
How do we know whether the reaction will be a SN1 or E1? I know you said higher temps favor Elimination reactions, but is this all we can go off of? Besides how hard/soft the nucleophile is to tell if it is acting as a nucleophile or base?
|By Sergei on Wednesday, October 04, 2017 - 09:06 pm: Edit|
anon on Wednesday, October 04, 2017 - 05:12 pm:
Yes, for most parts:
1. "If the carbocation produced would be primary and cannot be immediately delocalized by resonance" - allylic/benzylic carbocations are not primary that have immediate delocalization; they are distinct from primary.
2. Since charges are involved in these types of reactions, solvents must be polar.
Carbocation formation, requires the breaking of the bond, and leads to the formation of ions; ionic species require polar media that can stabilize these charges. Protic solvents (water, alcohols, acids) will provide possibilities for H-bonding with various charged species, hence they will delocalize the formed charges better than solvents that do not have H-bonding capabilities.
In SN1 - the formed carbocation is a very reactive electrophile, and hence the reactivity of Nu is not very important (in other words whether it's surrounded by solvent molecules, H-bonded to solvent or not; more generally - solvated by solvent or not).
In SN2 - there is only a partial charge on the C, hence reactivity of Nu is important; if we use protic solvent, it will H-bond with Nu, and decrease Nu's reactivity (i.e., Nu will be solvated/surrounded by the molecules of the solvent).
This will be discussed on Friday.
AB on Wednesday, October 04, 2017 - 05:15 pm:
These always compete, hence you will always have SN1-product and E1-product.
Temperature is only one of the factors. If you are given two sets of conditions which differ only in regard to temperature, then - yes: the higher temperature will lead to E1-product, which will be the major one.
The issues is more complex than just temperature; solvent also plays a role. For example: water (as solvent) can act not only as a base but also as a nucleophile.
Overall, SN1/E1 is a mess, and a single product should never be expected; which one is major, and which one is minor will be determined by the conditions (T, Nu/B), and it's not going to be a straightforward assessment.
We will be looking at more examples in the next few lectures.
|By AA on Friday, October 06, 2017 - 06:18 pm: Edit|
I was wondering what is the difference between dextorotory and R configuration in enantiomers?
|By Sergei on Friday, October 06, 2017 - 06:48 pm: Edit|
R (same as S) is nomenclature
dextro (same as levo) is a physical property (rotation of polarized light to the right or to the left)
the two should not necessarily correlate, meaning that it is possible to gave R-(+) and R-(–), same as S-(+) and S-(–).
Another important thing to keep in mind: if one enantiomer is R-(+) the other has to be S-(–)
|By AA on Friday, October 06, 2017 - 07:54 pm: Edit|
Do we need to know the reaction of (-) to (+) malic acid?
|By Sergei on Saturday, October 07, 2017 - 09:47 am: Edit|
We will cover the synthesis of RCl from ROH next week.
|By Sergei on Saturday, October 07, 2017 - 09:58 am: Edit|
Quiz 5 is graded:
The office opens on Monday morning.
|By Anon on Saturday, October 07, 2017 - 11:27 am: Edit|
Hi Dr. Dzyuba,
I was wondering how we know where we stand in the class? I know that the final groupings for grades don't happen until after the final, but do you give us a rough estimate at some point during the semester? Such as, if you were to make the grades right now, an A would be scores between x and y, a B would be between scores y and z, etc...
|By Sergei on Saturday, October 07, 2017 - 05:01 pm: Edit|
In general: above/below ave, close to top, etc should give you some idea. After each exam, the overall ave/max/min are posted
Just as a VERY VERY VERY rough estimate, per your request, assuming the semester ends today, which it is not; there are still 2 exams, at least 5 quizzes, and the final. If you are within +/–10pts from the grade cutoff, assume that you are at the cutoff:
For those who have excused absences, these numbers would translate to percentages; %-wise of the highest score (which is 203 at the moment):
|By AA on Sunday, October 08, 2017 - 11:08 pm: Edit|
So does that mean that these percentages are always like this, so like even after the second exam, is 88 percent of the whatever the highest grade is and above considered an A?
|By Sergei on Monday, October 09, 2017 - 07:08 am: Edit|
The percentages are based on the gaps; the gaps change after each quiz/exam. The grades are not based no fixed percentages (please check the syllabus).
Therefore, the above distribution is only valid for this point of the semester (ie., E1+Q1-5).
After Exam 2, the percentages will be whatever the gaps between the scores will be.
|By Sergei on Friday, October 13, 2017 - 07:54 am: Edit|
Quiz6 is graded:
|By AA on Friday, October 13, 2017 - 06:35 pm: Edit|
I just watched this video on why SN1 is good in protic and SN2 bad in protic solvents. It said something about the partial negative charge on the solvent resulting from the H bond, will stabilize the carbocation in SN1 and the negative charge of the nucleophile is weakened by the Hs in the solvent in SN2. I don't understand how it is discussing Nu vs solvent in SN2 and substrate vs solt in SN1? It makes no logical sense..
|By Sergei on Saturday, October 14, 2017 - 08:06 am: Edit|
lets sort it out:
1. protic solvent (H-Sol) will hydrogen bond to anything with a negative charge/electron pair using H, hence the Nu will be hydrogen bonded to the solvent, and Nu's electron density (i.e., negative charge/electron pair ) will be decreased, hence it will be less nucleophilic.
2 . If protic solvent/H-Sol could hydrogen bond, Sol-part is electron rich (partial negative charge), hence it can donate it's electron density to the carbocation, and thus stabilize the carbocation
3. H-Sol must be polar, and polar media can support charged species (carbocation, LG)
Thus, the H-Sol solvent will weaken the nucleophilicity of the Nu using H of H-Sol (which is not good for Sn2, because the electrophilicity of the E is not very high), and the H-Sol could help in the stabilization of not only the carbocation (Sol part of H-Sol), but also the LG (H of H-sol part will provide H-bonding, since LG has a negative charge ) in Sn1. Carbocation + LG (right part of the equation of the first step of Sn1) is kind of a salt.
Hope it clarifies the issue, but if anything please feel free to follow up.
|By AA on Sunday, October 15, 2017 - 07:33 pm: Edit|
If a reactant molecule doesn't have dashes or wedges (in a E2 reaction for instance), should we not even think about stereo-chemistry at all?
|By Sergei on Tuesday, October 17, 2017 - 10:19 pm: Edit|
If the stereochemistry of a chiral center is not defined, it's probably not important, i.e., that chiral center does not participate in the reaction, both stereoisomers could react, etc.
but, if the stereocenter is specified, it does not mean that it's should be involved in the reaction.
In other words, always consider what is the particular situation (conditions, question).
|By Sergei on Thursday, October 19, 2017 - 08:56 am: Edit|
Quiz 7 is graded:
|By AA on Saturday, October 21, 2017 - 10:00 am: Edit|
In the Inquisition book, there is this question about drawing ortho-M2C6H4 and the answer is just like ortho-methyltoluene. Why?
|By AA on Saturday, October 21, 2017 - 10:11 am: Edit|
Isn't phenylcyclopropane a faulty name? Shouldn't it be cyclopropanebenzene instead since the number of carbons in the phenyl is higher than cyclopropane so it shouldn't be a branch?
|By AA on Saturday, October 21, 2017 - 11:01 am: Edit|
Why is cyclohept-2,4,6-triene with a positive charge on carbon number 7, considered an aromatic compound? I thought it's not conjugated?
|By AA on Saturday, October 21, 2017 - 11:57 am: Edit|
Do we need to know the name of all the benzene derivatives in the textbook, or only the ones you mentioned in lecture? Like do we need to know styrene?
|By AA on Saturday, October 21, 2017 - 12:51 pm: Edit|
Can we say that if a compound follows Huckel's rule and all of its carbons are sp2, it is %100 with no doubt an aromatic compound, and if it does not have those two conditions, it is %100 non-aromatic?
|By Anon on Saturday, October 21, 2017 - 03:03 pm: Edit|
Hi Dr. Dzyuba,
I am still having trouble distinguishing between whether it will be an sn1, sn2, e1, or e2. My struggle is I don't know where to start a problem. I know it is all relative and depends on the reagent, substrate, and the solvent. Once I figure out if it's substitution or elimination, then I can finish a problem. So I guess my question is when we come to a problem, what is a good place to start to determine which pathway?
|By AA on Saturday, October 21, 2017 - 07:13 pm: Edit|
When we are doing the mechanism for bromination for instance, are you okay if we do Br+ and Br- in the beginning out of nowhere, or do you explicitly want us to start the arrow on the double bond to the partially positive Br and then another arrow on the bond between the two Br's to the partially negative Br to form Br-?
|By AA on Saturday, October 21, 2017 - 07:24 pm: Edit|
In that step in nitration when the water leaves, can we say that O- is the nucleophile?
|By Sergei on Saturday, October 21, 2017 - 07:58 pm: Edit|
AA on Saturday, October 21, 2017 - 10:00 am:
ortho dimethyl benzene - 2 methyl groups should be 1,2 or ortho to each other.
If toluene's used as the parent name, then the substituent ortho to the methyl group of toluene will give the 1,2-dimethylbenzene
AA on Saturday, October 21, 2017 - 10:11 am:
either cyclopropylbenzene or phenylcyclopropane are used
(cyclopropanebenzene should NOT be used)
either cyclopropyl is a substituent or phenyl is the substituent.
since both of the names are giving only one structure, you can use either one of them.
AA on Saturday, October 21, 2017 - 11:01 am:
Draw the resonance structures; the positive charge could move from one carbon to the other (it's 1 sigma bond away from the pi-system).
There are 6e (4n+2, with n=1), and 7-membered ring is flat.
Thus, all conditions for aromaticity are fulfilled.
AA on Saturday, October 21, 2017 - 11:57 am:
Names in the book are a fair game.
Keep in mind that trivial names and IUPAC names are used interchangeably (so when you don't remember/don't know the trivial name, IUPAC naming is a way to go)
AA on Saturday, October 21, 2017 - 12:51 pm:
Not quite; sp2 is not the requirement; the orbital overlap, and the complete conjugation (thus planar structure) are.
Anon on Saturday, October 21, 2017 - 03:03 pm:
There is no good/bad point to start (e.g. solvent, substrate, etc); you just have to consider all of them (or most of them), instead of just one.
For example, tertiary halide will not undergo Sn2 (i.e., the rate of the reaction is extremely slow), no matter what the solvent is, bc it too hindered.
Sn vs E: consider whether negative charge species are likely to act as bases or nucleophiles (hard/soft, etc)
E1 vs E2 is easy: E1 is under acidic conditions, E2 is under basic
E1 and Sn1 always compete; at higher temperatures E1 will dominate.
When you are working on a problem, try to look at it from every angle, in other words, if Sn2 takes place, what will happen, and what are the factors that will favor it; if E2, that's what should be expected. Think in terms of reasonable possibilities, write them all out, and then propose why one should be favored. If a clear decision is not possible, the mixtures should be expected.
We will have a review lecture on these issues before the exam.
AA on Saturday, October 21, 2017 - 07:13 pm:
in bromination you don't have Br+ and Br-, those are all deltas.
Br2 is getting polarized by neighboring Br2 molecules, but Br+ is never generated
AA on Saturday, October 21, 2017 - 07:24 pm:
in the formation of nitronium ion (NO2+)?
Yes, if you want to (it's typically not presented as such)
|By StevenK on Saturday, October 21, 2017 - 08:40 pm: Edit|
Will the material you cover in lecture on Monday and Wednesday be included on Friday's exam? Also are we having a review session this week? Thanks
|By David on Saturday, October 21, 2017 - 11:56 pm: Edit|
We haven't learned the birch reduction. Correct?
|By Sergei on Sunday, October 22, 2017 - 09:21 am: Edit|
StevenK on Saturday, October 21, 2017 - 08:40 pm:
Yes and Yes.
Unless, we move faster than expected, we'll start alcohols on Wed; in this case whatever is related to alcohols will not be on the exam.
David on Saturday, October 21, 2017 - 11:56 pm:
This is correct.
|By AA on Sunday, October 22, 2017 - 11:09 am: Edit|
But whenever we are deciding between which one should be the main chain and which one the branch, don't we count the number of carbons and the one with more carbons would be the main chain? So how is phenylcyclopropane a correct name if the branch has more carbons?
Regarding aromaticity, isn't being planar the same as being sp2 (since it would be 120 degrees)?
|By Sergei on Sunday, October 22, 2017 - 11:44 am: Edit|
You are correct, following IUPAC nomenclature, it should be cyclopropylbenzene.
however, there are plenty of examples in literature that use phenylcyclopropane (sometime it's almost like a trivial name, sometimes it's due to the fact how it was made synthetically).
Important thing is that both names give the same structure.
not necessarily, not every cyclic structure with sp2 atoms will retain planarity. For example: cyclodec-1,3,5,7,9-pentaene.
5e-pairs, so 4n+2=10; everything is in conjugation, but the ring is too big/flexible to retain planarity.
I understand that these are rare cases, which we would not be encountering often, but still it's better to use more general terms.
Importantly, recall the cases of pyrrole and furan - those N and O don't look like sp2, but they are forced into the planar form (although in some textbooks they are presented as if they are forced to become sp2).
Carbanions are not quite sp2 either, but in cyclopendienyl anion, that CH– is forced into planar configuration.
|By AA on Sunday, October 22, 2017 - 12:20 pm: Edit|
When we make a carbocation for either EAS or E1, do we care more about where the + is and we do hydride shift according to that (more substituted better), or do we care more about where the resulting double bond will be and decide according to that? Or will it turn out to be the same anyways?
|By AA on Sunday, October 22, 2017 - 01:16 pm: Edit|
In friedel-craft acylation, do we always use the resonance structure with C being +? Do we never use the one with the triple bond and + on O?
|By AA on Sunday, October 22, 2017 - 02:16 pm: Edit|
I don't understand how halogens can have +R effect, how can they make a double bond?
|By sarahK on Sunday, October 22, 2017 - 03:20 pm: Edit|
Hi Dr. Dzyuba,
What are the general guidelines for determining if a step is reversible or not. I know that in general acid/base steps are reversible and nucleophile/electrophile steps less so. Are there other ways to tell? Thanks
|By AA on Sunday, October 22, 2017 - 05:29 pm: Edit|
Are R and S only applicable in the case of chiral carbons, or any other atoms as well?
|By AA on Sunday, October 22, 2017 - 07:58 pm: Edit|
Why is OH- more localized than OR-? Doesn't the R group donate electrons to O by the I effect, making it more localized?
|By Sergei on Sunday, October 22, 2017 - 10:47 pm: Edit|
AA on Sunday, October 22, 2017 - 12:20 pm:
carbocations will do anything possible to get rid of the positive charge, and it's not just either/or, but rather and/and.
Also, you should consider that even if elimination is possible, and the alkene is forming, would that alkene react with a LA, and in those instances it might be that it would turn out to be same anyways, exactly as you said.
AA on Sunday, October 22, 2017 - 01:16 pm:
the nucleophile will come to C, this is the electrophilic center.
O with 3 bonds and positive charge is not an electrophile.
AA on Sunday, October 22, 2017 - 02:16 pm:
consider how bonds are made: 2 electrons and 2 orbitals (please look through the MO diagrams that we drew in the beginning of the semester).
Hal has 2 electrons, if there is a C with a plus charge, i.e., empty orbital (one sigma bond away), then those electrons could be shared.
It's also the resonance structures that we are talking about.
sarahK on Sunday, October 22, 2017 - 03:20 pm:
If the LG is as nucleophilic as the Nu (recall the issue with RI reacting with NaI, for example)
AA on Sunday, October 22, 2017 - 05:29 pm:
For any chiral center (atom), not just carbon.
R1S(O)R2 or R1R2R3N
AA on Sunday, October 22, 2017 - 07:58 pm:
If R is an alkyl group, then RO– is more localized than HO–
exactly as you said - due to +I effect of R
|By Sergei on Monday, October 23, 2017 - 08:43 am: Edit|
Quiz 8 is graded:
|By elyse on Monday, October 23, 2017 - 03:19 pm: Edit|
When we were comparing -O--R and -S--R, we said -O is harder relative to -S because -S delocalizes the charge better making it softer.
My question is: Why does this make -S--R only able to go through SN2 and not E2, while the -O--R is able to go through both mechanisms?
|By AA on Monday, October 23, 2017 - 06:57 pm: Edit|
When doing KMnO4 oxidation, would an alkene or alkyne also act like an alkane chain and become COOH no matter the complexity?
|By Dave on Monday, October 23, 2017 - 07:27 pm: Edit|
We haven't learned aldol condensation-elimination, correct?
|By Sergei on Monday, October 23, 2017 - 11:14 pm: Edit|
elyse on Monday, October 23, 2017 - 03:19 pm:
S is below O, hence it's larger, and hence it could delocalize the charge better; in other words, S is softer than O.
Softness is associated with nucleophilicity, while hardness is related to basicity.
With RS– the substitution (SN2) is the major (if not the only pathway).
AA on Monday, October 23, 2017 - 06:57 pm:
Alkenes will react differently depending on the conditions (basic, acidic, neutral). We will look at this issue after the exam 2.
Just to be sure: alkanes are not oxidized by KMnO4!!
Only benzylic C with a H (the benzylic CH) could be oxidized to CO2H (CH which is a sigma bond away from the aromatic ring.
Dave on Monday, October 23, 2017 - 07:27 pm:
Not yet; this is related to carbonyls.
However, E1cb is a part of of the aldol reaction (addition of the neonate to the carbonyl, followed by elimination, which takes place by E1cb).
|By AA on Tuesday, October 24, 2017 - 12:47 am: Edit|
I feel like I can always do e1cb reaction on anything that I like. Can you please explain to me what the limiting factor is for doing a e1cb reaction, or when are we not allowed to do a e1cb reaction?
|By sarahK on Tuesday, October 24, 2017 - 01:03 pm: Edit|
In the "SN-2 like" Epoxide ring opening under acidic conditions, why does the nucleophile attack the secondary carbon over the primary? Isn't primary better for Sn2?
|By Anon on Tuesday, October 24, 2017 - 01:06 pm: Edit|
Hi Dr. Dzyuba,
I was wondering when the review session is going to be? I just got confused with everyone asking to move it.
|By Sergei on Tuesday, October 24, 2017 - 01:57 pm: Edit|
AA on Tuesday, October 24, 2017 - 12:47 am:
E1cb requires an EWG that could delocalize the negative charge (thus it cannot be on any substrate), and that delocalization has to be through -R effect (most efficient) or –I (CF3 or C2F2n+1, there are not that many groups that could delocalize C– via induction)
sarahK on Tuesday, October 24, 2017 - 01:03 pm:
Under acidic conditions, both Sn1 and Sn2 are possible, and depending on the stability of the carbocation, one may dominate the other.
Secondary carbocation is more stable than primary, thus under acidic conditions, Sn1-like mechanism would be more favorable. Sn2-like is still possible (but to a lesser degree, since the nucleophile is now more solvated, and hence less reactive), hence the product from this mechanism is also forming, but it's not likely to be a major product.
The important thing is that under acidic conditions both mechanisms are possible, but under basic conditions, only SN2.
Anon on Tuesday, October 24, 2017 - 01:06 pm:
It will be tomorrow (Wednesday) for sure.
I will let you know about the exact time (6-8pm or 7-9 or 6-9pm) as soon as possible
|By AA on Tuesday, October 24, 2017 - 05:26 pm: Edit|
Do we need to know alkene reactions for the exam?
|By Sergei on Tuesday, October 24, 2017 - 05:35 pm: Edit|
Everything upto this point is a fair game (for example, material for the exam1 could be/will be incorporated into the question that is related to material of exam2; similar to some examples given in lectures).
|By AA on Tuesday, October 24, 2017 - 06:31 pm: Edit|
Whenever we have acetone, how do we know whether the purpose is to ppt NaX or is it meant to be an aprotic solvent?
|By AMenke on Tuesday, October 24, 2017 - 09:38 pm: Edit|
How do we know when the E1 reactions is favored over the SN1 reactions. IE: which reaction is going to produce the major product
|By Sergei on Tuesday, October 24, 2017 - 11:10 pm: Edit|
AA on Tuesday, October 24, 2017 - 06:31 pm:
Acetone is an aprotic solvent regardless of the reaction.
NaI/acetone and alkyl halide(Cl or Br) is specific for the Finkelstein reaction. Here, it is an aprotic solvent, in which NaI is soluble, but NaCl and NaBr are not soluble (hence they precipitate).
If acetone is the solvent and/or reagent under other circumstances, it has nothing to do with SN2/Finkelstein reaction.
AMenke on Tuesday, October 24, 2017 - 09:38 pm:
higher temperature (if the reaction is done under reflux) will favor E1 product; hence lower temperatures favor SN1
|By AA on Wednesday, October 25, 2017 - 11:53 pm: Edit|
On question 3 on the review exam, upon adding HBr, do you want us to add H to the less substituted and Br to the more substituted right away, or should we add H to the less substituted, resulting in a positive charge on the more substituted part of the previously double bond, then do a hydride shift resulting in a benzylic carbocation and then add Br to the carbocation?
|By anonnn on Thursday, October 26, 2017 - 07:47 am: Edit|
How do we know when a Nu/B can be only a Nu, only a B, or both? Does it have to do with hard/softness?
|By Sergei on Thursday, October 26, 2017 - 09:27 am: Edit|
AA on Wednesday, October 25, 2017 - 11:53 pm:
to be honest, it's not about what I want you to do, it's what the reagents, reactants and conditions allow you to do.
Reactions proceed via most stable intermediates, it is not energetically favorable to make something unstable then do another step, to get to more stable species, if those species could be obtained directly.
Please don't confuse it with Friedel-Crafts reaction - the formation of carbocation in that case is predetermined by the location of C-Hal, hence the carbocation can only form where the Hal is; then it could rearrange to a more stable intermediate (if the structure allows it).
In other words, there is no option of forming a more stable intermediate without going through a less stable one first.
anonnn on Thursday, October 26, 2017 - 07:47 am:
Ina "vacuum", yes - with hardness/softness, but in reality with all other parameters: size, localization of the charge, solvation, neighboring environment, and the substrate.
|By Sergei on Thursday, October 26, 2017 - 02:50 pm: Edit|
Quiz 9 is graded:
|By KD on Thursday, October 26, 2017 - 04:38 pm: Edit|
Hi Dr. Dyzuba,
Would you be able to post a rough estimate of the class cut offs after exam 2 is graded so that we have somewhat of anode of where stand in the class?
Please and thank you!
|By Sergei on Thursday, October 26, 2017 - 05:01 pm: Edit|
|By AB on Thursday, October 26, 2017 - 06:34 pm: Edit|
Hi Dr. Dzyuba,
If you have a tertiary alkyl halide, but a charged Nu/B in polar protic solvent, would the reaction that takes place be SN1 or E2?
|By AA on Thursday, October 26, 2017 - 06:43 pm: Edit|
How do you name a compound if we have multiple double bonds along with multiple chiral centers? So like (3Z-5E)(2R-6S)blahblahblah?
|By AA on Thursday, October 26, 2017 - 06:47 pm: Edit|
Sorry, also on the first exam, there is a cyclic structure we had to name that had a double bond. Should we never specify E-Z for a double bond in a cyclic structure even if it's not symmetric with regards to the type and number of substituents?
|By AA on Thursday, October 26, 2017 - 08:35 pm: Edit|
On the first exam, when we were adding the HCl, we first added H and then we moved the carbocation to a more stable position. Why do we not do the same in that question on the review were we add HBr that I asked you about? Like why do we do that there but not here?
|By Sergei on Thursday, October 26, 2017 - 10:41 pm: Edit|
AB on Thursday, October 26, 2017 - 06:34 pm:
Charged base will typically mean basic conditions, hence SN1 is not possible.
AA on Thursday, October 26, 2017 - 06:43 pm:
just like that (except for blah blah-part): stereochemistry must be specified, and the absolute stereochemistry of the chiral center would have to be specified as well.
AA on Thursday, October 26, 2017 - 06:47 pm:
It depends on the size of the cycle.
In that case E/Z was not possible, due to the "small" ring size.
But cyclooctene can exist as cis and trans.
AA on Thursday, October 26, 2017 - 08:35 pm:
H+ will react with C=C to form a carbocation, if the carbocation has a chance to rearrange to a more stable one, it will be a favorable thing to do.
On the practice set: you don't want to form primary carbocation, then rearrange it to tertiary.
In this case, tertiary is forming from the start (please write split resonance structures of C=C to be sure that you should be forming tertiary carbocation).
in your other question: Friedel-Crafts - there the formation of the carbocation is somewhat different, as the position of Hal determines where the carbocation should form first.
|By AA on Sunday, October 29, 2017 - 11:37 am: Edit|
I'm so nervous about this exam grade.... Please go easy on us
|By AA on Sunday, October 29, 2017 - 08:22 pm: Edit|
Hi Dr. Dzyuba,
I was wondering when we might expect to have the tests graded? I do realize there are a lot to go through. I am just thinking about when we might want to start considering whether or not to drop the class, since we don't know the actual grade distribution until after the final. We can look at where we generally are with the averages, but what if we have been borderline or slightly under for example?
Thanks so much!
|By AA on Sunday, October 29, 2017 - 08:25 pm: Edit|
Can you please make your own username instead of using mine, especially when posting something weird like this?
|By sarahk on Thursday, November 02, 2017 - 05:17 am: Edit|
Hi Dr. Dzyuba,
In the example you gave in class yesterday to show how a tertiary alcohol will only react with KMnO4 if the conditions are acidic...after water left and there was a hydride shift, you shifted electrons off of the methoxy oxygen and onto the ring and then onto the carbocation. Why would oxygen just give up its electrons? How is its charge restored? Thanks
|By sarahk on Thursday, November 02, 2017 - 05:40 am: Edit|
Also, after allylic or propagylic alcohols are oxidized to aldehydes, what do they immediately decompose to?
|By Sergei on Thursday, November 02, 2017 - 08:24 am: Edit|
In regard to the oxygen of MeOH-group - that was a resonance structure, just to show the extra stabilization of the positive charge, which could explain/justify the rearrangement (hydride shift) step, in principle (however, tertiary carbocations are among the most stable ones and do not rearrange to another tertiary, for example; but in some rare cases, where rearrangement provides extra stabilization, like the one shown in lecture, i.e., +R effect of a very strong EDG, or producing an aromatic structure), the rearrangement could be justified.
in many cases the originally formed aldehydes/ketones would undergo further oxidation to carboxylic acids (especially in the case of propargyl alcohols), but not due to MnO2 treatment, but rather due to the oxidation with O2/air via several intermediates, which we are not covering this semester.
The oxidation of an allylic OH would lead to C=O, which gives the enone/enal (depending on whether it was secondary or primary allylic C-OH). These products would be stable in the cases of substitute allylic alcohols; but in the case of non-substituted alcohol the product is quite volatile and hard to handle.
However, there are other ways to make enones (aldol reaction, for example), which are a bit more practical, bc of the accessibility of starting materials.
These are a bit of the practical issues, for the lecture course, the important thing is that MnO2 will oxidize allylic/benzylic/propargylic alcohols to the corresponding aldehydes (or ketones), but the aldehydes will not be over oxidized to the carboxylic acid (by MnO2!).
|By kacio on Thursday, November 02, 2017 - 09:01 am: Edit|
Hi Dr. Dzuyba!
I just wanted to clarify some details and make sure I'm understanding the products of epoxidation correctly. For the trans alkene, a racemic mixture is created, right? I wrote down that the product has no plane of symmetry and achiral centers, but I don't quite understand why it has achiral centers. For the cis alkene, a meso compound is created because there is a plane of symmetry. I also wrote down that the product has chiral centers, which, again, I don't quite understand why. Also, is a racemic mixture also created after the epoxidation of a cis alkene?
Thank you for your help!!
|By kacio on Thursday, November 02, 2017 - 09:05 am: Edit|
Also, I just realized that I accidentally misspelled your name. I am so sorry!!
|By AA on Thursday, November 02, 2017 - 11:12 pm: Edit|
For benzylic alcohol, allyl alcohol, and propagyl alcohol which we used MnO2 in order to convert them to aldehydes, could we have used DMP/PCP as well?
|By Sergei on Friday, November 03, 2017 - 08:52 am: Edit|
Ave – 66
MAX – 130
MIN – 5
Grade distribution (still +/– 10-20 pts from the cut-offs should be treated as the same; for those with excused absences - %s are in parentheses):
A: >338 (70%)
B: >260 (54%)
C: >203 (42%)
D: > 110 (23%)
F: <110 (<23%)
|By NH on Friday, November 03, 2017 - 10:53 am: Edit|
Hi Dr. Dzyuba,
If we have excused absences, are the percentages strict cutoffs? Like for the B range, 20 points within that would be 240, which would be around a 50%. Would that mean that the cutoff for the B could (theoretically) be 50%?
|By AA on Friday, November 03, 2017 - 11:12 am: Edit|
I was wondering can you also post how many people are in the A/B/C/D/F range as well if that's not trouble for you? Thank you!
|By Sergei on Friday, November 03, 2017 - 12:07 pm: Edit|
kacio on Thursday, November 02, 2017 - 09:01 am:
Yes, in that case (both R are the same), trans alkene is going to give two stereoisomers (enantiomers), there are two chiral centers (a chiral center is the opposite of an achiral center)
The symmetry of the alkene is going to be transferred into the epoxide.
because two groups on alkene were the same, we had an issue with the plane of symmetry.
In that particular case: cis and trans alkenes (both R groups are the same):
cis: the two R groups will be on the same side (either up/up or down/down), hence there is a plane of symmetry.
trans: the two R groups are on the opposite sides (up/down of the 3-membered ring), hence there is no plane of symmetry.
In both cases (cis/trans) there are two chiral centers upon epoxidation of the C=C bond, because each C of the epoxide (C-O) has 4 different substituents (i.e., R, H, O and C, which is connected to R, H O)
It's because of the plane of symmetry the two stereoisomers are now the same, hence it's the meso compound.
This is exactly the same reasoning for any meso compound: compound with more than one chiral center, and a plane of symmetry.
Hope it clarifies the issue; if anything let me know.
kacio on Thursday, November 02, 2017 - 09:05 am:
AA on Thursday, November 02, 2017 - 11:12 pm:
PCC (Pyridinium ChloroChromate)
|By Sergei on Friday, November 03, 2017 - 01:54 pm: Edit|
NH on Friday, November 03, 2017 - 10:53 am:
For excused absences - please use % only.
+/– 10 pts from the cut-off should be converted to %s as well.
so lets say the total possible # of points (2exams - 300pts, 9 quizzes - 180pts) is 480
203 is the bottom of the C (which means that 213 and 193 is the range; in percentages - 44 to 40%)
In other words, 10 pts is a 2% at this stage of the semester.
AA on Friday, November 03, 2017 - 11:12 am:
At this point of the semester: 2 exams, 9 quizzes.
|By AA on Saturday, November 04, 2017 - 09:46 am: Edit|
Do we need to know the mechanisms for how the alcohols are oxidized with KMnO4/H2CrO4/etc?
|By Sergei on Saturday, November 04, 2017 - 05:47 pm: Edit|
No, the mechanism is out of scope for this semester
|By AA on Saturday, November 04, 2017 - 06:25 pm: Edit|
In general for the class, do we need to know what POCl3 do to an alcohol?
|By AA on Saturday, November 04, 2017 - 07:09 pm: Edit|
|By Sergei on Saturday, November 04, 2017 - 08:42 pm: Edit|
We didn't cover this reaction specifically in the class, but you would be expected to be able to postulate the outcome and the mechanism of the reaction, based on what we have learned about alcohols (in terms of their nucleophilicity/bacisity), and the similarities between POCl3 and PCl3 (or PBr3).
In general, try to think about the reactions form the stand point of acid/base, nucleophilic/electrophile types of interactions, rather than simply relying on memorizing a set of reactions.
POCl3 will lead to dehydration of alcohols in the presence of the base (pyridine or triethylamine).
|By IBC on Sunday, November 05, 2017 - 10:07 pm: Edit|
In ether cleavage where the ether has more than one electrophilic center for I- to attack, will it preferentially attack the least-sterically hindered one?
|By Sergei on Monday, November 06, 2017 - 06:32 am: Edit|
Yes, for any process that goes via SN2 mechanism sterics will be one of the major factors.
However, if increasing steric constrains are associated with the formation of a more stable carbocation, SN1 could also start to contribute; this would affect the major/minor percentages.
|By VinC on Monday, November 06, 2017 - 08:11 pm: Edit|
In discussing haloform formation in class today, you mentioned that if the base present cannot also act as a nucleophile then the reaction will not go all the way to haloform. I am wondering if the counter halogen ion (Br-, Cl-, I- etc) would be able to act as a nucleophile...or will these only ever act as base? Thanks
|By Sergei on Monday, November 06, 2017 - 09:56 pm: Edit|
Hal– is a viable Nu in general; but in this case - C=O is fairly favorable and we are gonna get O–, which is not favorable; thus, Hal would not act as a Nu.
But, even if Hal– is added, you would like to restore the carbonyl.
Going through possible LG - Hal will be the best option, which means that it would be a reversible reaction.
Hal– as a base -this would mean that the conjugate acid is HHal, but we are working under basic conditions, hence HHal is not possible.
Thus, Hal– will not be a base either
|By AA on Tuesday, November 07, 2017 - 08:51 am: Edit|
In the haloform reaction, why does the double bond have a higher affinity for the partially positive Br rather than the O- ?
|By AA on Tuesday, November 07, 2017 - 09:17 am: Edit|
Sorry also another clarification, does the OH- Nu attack the carbon under the double bond rather than the carbon bonded to the three bromines because the carbon under the double bond is more partially positive due to the R effect but the other carbon only has the I effect acting on it?
|By AA on Tuesday, November 07, 2017 - 09:22 am: Edit|
I'm so sorry for asking so many questions, but I was also wondering why does the CBr3- prefer to act like a base, deprotonating the H on the carboxylic acid rather than acting like a nucleophile attacking the carbonyl carbon? Isn't C much softer than H?
|By Sergei on Tuesday, November 07, 2017 - 10:31 am: Edit|
AA on Tuesday, November 07, 2017 - 08:51 am:
hard-soft; C=C is a softer nucleophile, while O– is a harder nucleophile.
Br2 (or I2) are soft electrophiles.
AA on Tuesday, November 07, 2017 - 09:17 am:
No need to say sorry.
C=O's C is more accessible, since it's planar, while the alpa-C with R' and Hal is more hindered.
In addition, HO– is a harder Nu rather than softer. C=O is a harder electrophilic center than C-Br, bc Br is softer than O.
Also, you can use the argument of C=O being more electrophilic due to -R/–I effects of O, as opposed to -I of Br (although in this case, we would come back to hard/soft, since stronger -I/-R implies that =C is going to be more localized, "bigger", hence harder).
AA on Tuesday, November 07, 2017 - 09:22 am:
No need to say sorry.
Questions are good.
Any nucleophile is also a base, and in the presence of an acidic H, Nu will act as a base; this situation is pretty much the same for Grignard, for example. The acidity of RCOOH is pretty high (pKa ca.5), hence the deprotonation will be the only viable process. Also with 3 Hal (–CBr3), C– is a bulky Nu, and thus would not be attacking the electrophilic center (but this is kind of an irrelevant argument, due to the presence of the acidic H).
|By Sergei on Tuesday, November 07, 2017 - 01:52 pm: Edit|
Quiz 10 is graded:
|By AA on Tuesday, November 07, 2017 - 03:55 pm: Edit|
What exactly will you be teaching tomorrow?
|By Sergei on Tuesday, November 07, 2017 - 04:26 pm: Edit|
imine and acetal formation
|By lauren on Tuesday, November 07, 2017 - 08:30 pm: Edit|
In a reaction between chroman and HI would the O be protonated and then the I- act as a nucleophile, attacking on the less hindered carbon and forming O- on the benzene ring (in conjugation which then is protonated to an alcohol) and I on the end of the carbon chain? This is what makes sense to me but is not correct on the answer key of the sheet i am looking at. Can you explain why?
|By Sergei on Tuesday, November 07, 2017 - 09:08 pm: Edit|
The electron pair of O will react with H+, only after that I– will attack the CH2 (of the Ar-O(H+)-CH2); it's not because it's the least hindered, the other carbon is sp2/Ar, cannot be attacked by a Nu in the SN2 process.
Once I– attacks, the O-CH2 bond breaks (concerted process, SN2), the OH is already there, no protonation at that stage, i.e., no O– followed by protonation.
Hope this clarifies it, but if anything, please repost.
|By Lauren on Wednesday, November 08, 2017 - 08:36 am: Edit|
Thanks for the explanation. So the alcohol will end up on the benzene ring and the -I will form and alkyl halide?
|By Sergei on Wednesday, November 08, 2017 - 01:42 pm: Edit|
|By sarahk on Wednesday, November 08, 2017 - 02:16 pm: Edit|
Hi Dr. Dzyuba,
Can hydrate formation of aldehydes and ketones act the same way as acetal formation? As in, does it protect from nucleophilic attack? Thx
|By AA on Wednesday, November 08, 2017 - 04:36 pm: Edit|
I was wondering what else will you be teaching before the exam? I see that there's only one concept left from chapter 9 and that is conjugate nucleophilic addition reactions. Can you please let us know the titles of the other things you are planning on teaching before the exam? Thank you
|By AA on Wednesday, November 08, 2017 - 05:01 pm: Edit|
For the acetal formation, some sources use H+ to protonate the alcohol and then they use that as the electrophilic center but in the lecture we just used a general H+ source. Is it not significant which method we use?
Also some sources, instead of doing a proton shift, they just protonate the O twice so that water would leave as the LG. Why are there so many different ways? Do we just do the one you told us?
|By Sergei on Wednesday, November 08, 2017 - 05:05 pm: Edit|
Chapters 9-11 (chapters 10 and 11 are all related to carbonyl-type chemistry, except polymers)
Thus, everything that is covered through next Wednesday Nov15 is a fair game for exam3.
Whatever is not covered form those chapters before exam3 will be discussed after the exam 3.
|By AA on Wednesday, November 08, 2017 - 06:53 pm: Edit|
When doing a LiAlH4 reduction on a carboxylic acid, we end up with a molecule with Oil on it. Can O- act as a Nu at this point and do stuff?
|By sarahk on Wednesday, November 08, 2017 - 07:26 pm: Edit|
Hi Dr. Dzyuba,
Can hydrate formation of aldehydes and ketones act the same way as acetal formation? As in, does it protect from nucleophilic attack?
Also, could you clarify when to use formyl- versus carbaldehyde- ?
|By Sergei on Wednesday, November 08, 2017 - 10:05 pm: Edit|
sarahk on Wednesday, November 08, 2017 - 02:16 pm:
sarahk on Wednesday, November 08, 2017 - 07:26 pm:
Apologies for missing the earlier post:
Hydral formation removes the electrophilic center, but Nu still can act as a base, and in the case of hydral (same as in hemiacetal) there is an acidic H, which could be deprotonated. Hence the synthetic utility of hydrals and hemiacetals is not great.
Formyl vs carbaldehyde:
formyl is the prefix, i.e., CHO is a substituent, meaning that there is a higher priority group present (ester, nitrile, etc, those that we haven't covered yet at length)
carbaldehyde is the suffix (CHO is the highest priority group, and it's at the end of the name) when CHO is attached to the ring.
AA on Wednesday, November 08, 2017 - 05:01 pm:
The important thing is that you need a Lewis acid (H+ or general case of a Lewis acid)
O of the C=O is a Lewis basic center; hence it will react with a Lewis acidic center (H+ or LA).
I'm not quite sure what do you mean by "protonate the O twice" but as long as the two Hs are not added in one step, it's probably quite similar to what we did anyway.
Initial protonation of the C=O, will follow by the attack of the ROH, then proton migration (not quite the shift, and the word "shift" should not be used, to avoid confusion with "hydride shift", for example).
A lot of those steps are semantics, meaning that sometime the order of protonation/reprotonation does not matter.
Recall today's case of the hemiacetal (could be done in a different number of steps, from different intermediates); it's kind of a similar issue to H2SO4/HNO3 - which O of the HNO3 is getting protonated first, which H+ to move to which O, etc - the important things is that at some point you need to generate a good LG.
Same with the acetal, imine, etc formation.
|By Sergei on Wednesday, November 08, 2017 - 10:10 pm: Edit|
Hi AA on Wednesday, November 08, 2017 - 06:53 pm:
LiO– will be a LG.
The reason for using it as a LG is a somewhat covalent character of the bond between O and Li, and they are both hard; hence LiO– is likely to act a base than a Nu.
In addition, H– is a better/stronger Nu than LiO–; thus LiO– will not do anything, but pick up a H+ during the work-up step, for example.
|By AA on Thursday, November 09, 2017 - 04:50 pm: Edit|
What concepts will you exactly be teaching tomorrow (Friday)?
|By Sergei on Thursday, November 09, 2017 - 05:46 pm: Edit|
aldol and related reactions
|By kacio on Thursday, November 09, 2017 - 08:12 pm: Edit|
Hi Dr. Dzyuba!
I was wondering if you will be sending out the practice set for exam #3 soon?
|By AA on Thursday, November 09, 2017 - 08:37 pm: Edit|
Will the review be on Wednesday night?
|By Anon on Friday, November 10, 2017 - 11:44 am: Edit|
Hi Dr. Dzyuba,
Today in class you said that for naming we could have the ending be ane as opposed to -yl. On test 2 for naming, the third one was supposed to be cyclopentyl magnesium chloride. Would it then have also been correct to say cyclopentane in that case too? If so, would we get the points back?
|By Sergei on Friday, November 10, 2017 - 01:48 pm: Edit|
kacio on Thursday, November 09, 2017 - 08:12 pm:
yes; I'll e-mail it by Monday
AA on Thursday, November 09, 2017 - 08:37 pm:
Anon on Friday, November 10, 2017 - 11:44 am:
Unfortunately (in regard to the points back) it would not be correct to say cyclopentane magnesium chloride.
Carbaldehyde by itself does not make sense, i.e.,it requires something to go with it; hence cyclopentane carbaldehyde will work.
On the other hand, magnesium chloride is MgCl2, so cyclopentane magnesium chloride leaves room for more than one possibility, and that's why we need to specify that cyclopentyl part is "attached" to the MgCl part, i.e., they are the same compound.
|By AA on Friday, November 10, 2017 - 05:58 pm: Edit|
So I'm still pretty uncomfortable with the H bond in the Cannizzaro reaction attacking the carbonyl carbon. Why did you say that's possible to do?
|By AA on Friday, November 10, 2017 - 06:00 pm: Edit|
Also, do you have a rough estimate on what exact topics you are planning on covering on Monday AND on Wednesday?
|By AA on Friday, November 10, 2017 - 06:40 pm: Edit|
I was reading about the cannizzaro reaction online, and in there, they didn't do the deprotonation of the hydroxy group by OH-. Also their final products were completely different, and included one alcohol. Which reaction should I study?
|By Sergei on Friday, November 10, 2017 - 08:30 pm: Edit|
AA on Friday, November 10, 2017 - 05:58 pm:
Too much of electron density on the same carbon, plus the restoring carbonyl is thermodynamically advantageous.
Also, it's not the H-bond, it's the hydride that attacks the carbonyl.
If it helps with the comfort level: think about it as an intermolecular hydride shift, which is facilitated by the formation of the carbonyl.
AA on Friday, November 10, 2017 - 06:00 pm:
carbonyl related chemistry (chapters 9-11)
AA on Friday, November 10, 2017 - 06:40 pm:
Cannot really comment on what people post on line, you can show me what you have or post a link here and we'll try to rationalize it.
But this reaction does require the nucleophilic addition of HO– to the aldehyde, followed by deprotonation of the OH by another molecule of the base. Since one aldehyde is reduced, the other molecule of the aldehyde has to get "oxidized" and hence you are getting both alcohol and carboxylic acid from the same aldehyde.
|By AA on Friday, November 10, 2017 - 08:56 pm: Edit|
I will let you know about that online course in a bit.
Also, when doing aldol condensation, do we stop at the aldol product or do we attack an alpha H with another OH- like you demonstrated in class today? I was doing this problem where an aldehyde was treated with a strong base but they stopped the reaction at the aldol product but I stopped at the very end where we get rid of OH- and make a double bond. How do I know which one is right?
|By AA on Saturday, November 11, 2017 - 01:02 pm: Edit|
So I didn't understand why aldol condensation and crossed aldol condensation are considered different. Isn't crossed aldol condensation basically the same as aldol condensation but between two different molecules both of which having alpha hydrogens?
|By Sergei on Saturday, November 11, 2017 - 02:16 pm: Edit|
AA on Friday, November 10, 2017 - 08:56 pm:
everything what we covered thus far (substrates and the conditions) should lead to the aldol condensation product, ie., enone.
We will look at the cases when the aldol product will form (i.e., no further elimination to enone).
Again, cannot really tell you why certain steps are proposed or not proposed without seeing the complete problem.
AA on Saturday, November 11, 2017 - 01:02 pm:
From the fundamental point of view these two reactions should not be considered different. We didn't look at them as two different processes either.
Cross-aldol: at least one of the starting materials should have alpha hydrogen. If both components have alpha hydrogens, 2 enolates should be expected (under basic conditions, for example).
|By AA on Sunday, November 12, 2017 - 08:58 am: Edit|
In KMnO4/OH- reaction, if the double bond has a E conformation, (so maybe groups of X and Y being trans to each other.) After we do the oxidation, will X and Y have to be in Z conformation because we have to have our OHs be cis?
|By AA on Sunday, November 12, 2017 - 10:17 am: Edit|
In the reaction of alcohol with TsCl, why doesn't Cl then act as a nucleophile attacking the carbon by the oxygen, expelling the whole OTs group altogether, like it does in SoCl2?
|By AA on Sunday, November 12, 2017 - 10:50 am: Edit|
Just to clarify, will benzylic, allylic, and propagyl alcohols go to carboxylic acid in the presence of a strong oxidizing agent like KMnO4?
|By Sergei on Sunday, November 12, 2017 - 09:50 pm: Edit|
AA on Sunday, November 12, 2017 - 08:58 am:
You cannot really be talking about E/Z in regard to the product, since there is no C=C
AA on Sunday, November 12, 2017 - 10:17 am:
the base (either pyridine or Et3N) deprotonates H, and at that stage, Cl– is no longer good enough of a Nu to remove OTs (unlike in the case of SOCl2); Et3NH+ requires a counter ion, and since the reaction is done in non-polar solvent (CH2Cl2, for example), salts are not soluble , and Et3NHCl precipitates out, hence the concentration of Cl– is negligible.
AA on Sunday, November 12, 2017 - 10:50 am:
The CH2OH of benzylic, allylic and propargylic will get oxidized to the carboxylic acid; however, the C=C and the alkyne will also undergo the oxidation
Thus, benzyl alcohol will get oxidized to benzoic acid; for other cases, the unsaturated group will not be present in the product.
|By anon on Sunday, November 12, 2017 - 10:22 pm: Edit|
under what circumstances would you choose to protect a carbonyl group with TMS over acetal formation? I'm slightly confused on the differences between the two and when to use which.
|By Sergei on Sunday, November 12, 2017 - 10:34 pm: Edit|
TMS is "protecting" enol from undergoing a tautomerization to ketone, so that the aldol product could be isolated (on the contrary to the enone if TMS for example is not used); but TMS-protected enol is still a Nu.
Acetal is not a Nu, and it does not have an electrophilic center; hence by converting ketone/aldehyde to a ketal/acetal, other electrophile centers in the molecule could be treated with nucleophilic reagents.
We will address this issue tomorrow (Monday).
|By BLV on Monday, November 13, 2017 - 06:41 am: Edit|
On the syllabus, it says we do not have class on Monday, November 20. I just wanted to confirm this.
|By Sergei on Monday, November 13, 2017 - 06:45 am: Edit|
That is correct.
|By Sergei on Monday, November 13, 2017 - 08:18 am: Edit|
Quiz 11 is graded:
|By Anon on Monday, November 13, 2017 - 05:14 pm: Edit|
You mentioned that chapters 9-11 will be on the exam on the syllabus. Are we going to be learning all that we haven't learned from them on Wednesday? If not all, can you let us know what topics are you planning on teaching so that we can prepare ourselves for the lecture on Wednesday? Thanks.
|By AA on Monday, November 13, 2017 - 08:26 pm: Edit|
How should we have approached the problem today from the quiz - how were we supposed to get the -HCO2?
|By Sergei on Monday, November 13, 2017 - 09:47 pm: Edit|
Anon on Monday, November 13, 2017 - 05:14 pm:
Whatever we cover by the end of Wednesday's lecture will be on the exam. What we don't cover will be covered after the break (per syllabus).
Wednesday: we will look at the relative reactivities of all carbonyl compounds, and ester synthesis, most likely.
AA on Monday, November 13, 2017 - 08:26 pm:
I'll send the key soon; but there were the following reactions:
aldol, haloform, Grignard, and two oxidations: one with KMnO4/H+ and the other KMnO4 only (to avoid the elimination of H2O).
|By AA on Tuesday, November 14, 2017 - 08:43 am: Edit|
Do we need to be familiar with furan reactions?
|By Sergei on Tuesday, November 14, 2017 - 08:47 am: Edit|
Not sure what you mean by fur reactions? Please clarify.
Furan as an aromatic compound - yes (electrophilic aromatic substitution).
|By Sergei on Tuesday, November 14, 2017 - 08:48 am: Edit|
Quiz12 is graded.
|By AA on Tuesday, November 14, 2017 - 10:57 am: Edit|
I mean like making acetal from furan.
|By Sergei on Tuesday, November 14, 2017 - 11:10 am: Edit|
you mean dihydropyran, which is used for protection of alcohols (giving an acetal like product), right?
And this process goes through the same mechanism as the acetal formation.
Yes, you are expected to provide a reasonable mechanism that explains this transformation.
|By aa on Tuesday, November 14, 2017 - 01:53 pm: Edit|
I have a question about making an enone with the aldol condensation process. In order to form the double bond, do you protonate the O- twice (thus producing water, a good leaving group), or do you protonate the O- once and perform an E1cb reaction? Or do both work?
|By Sergei on Tuesday, November 14, 2017 - 02:39 pm: Edit|
under acidic conditions, you have HO on the aldol product, and protonation gives –OH2+, hence H2O is the LG
under basic conditions, the O– of the aldol will get protonated to give the OH, and the HO– will be a LG via E1cb mechanism
In either case, there is no "protonate twice" type of the situation for the same oxygen. Please look through the mechanism - there are various protonation/reprotonation steps; but under acid conditions, there is no O–, and under basic conditions, there is no OH2+
|By Anon on Tuesday, November 14, 2017 - 06:06 pm: Edit|
Should we know dies alder reaction?
|By Anon on Tuesday, November 14, 2017 - 06:41 pm: Edit|
Does KET happen in basic conditions as well?
|By Sergei on Tuesday, November 14, 2017 - 07:56 pm: Edit|
Anon on Tuesday, November 14, 2017 - 06:06 pm:
(this reaction is not a part of the course at this time)
Anon on Tuesday, November 14, 2017 - 06:41 pm:
technically not, since under basic conditions one has an enolate, not an enol (ie., the proton is gone/deprotonated, rather than being moved from one atom to the other)
|By AA on Tuesday, November 14, 2017 - 10:17 pm: Edit|
Do we need to know the common names or only IUPAC naming for this exam?
|By Sergei on Wednesday, November 15, 2017 - 06:55 am: Edit|
you should be able to recognize both trivial and IUPAC naming, if the name of the compound is given.
if you asked to name a compound (i.e., the structure is given) - you can use either one.
|By anon on Wednesday, November 15, 2017 - 03:16 pm: Edit|
I just wanted to confirm topics for the exam - the material we covered that corresponds to what is in Chapter 10 of the textbook was mostly just nomenclature of carboxylic acids and their derivatives and nitrile reactions, right? Thanks!
|By Sergei on Wednesday, November 15, 2017 - 04:51 pm: Edit|
Plus the relative reactivity of all carbonyl-containing compounds
|By TCUfrog on Wednesday, November 15, 2017 - 10:20 pm: Edit|
Can you let us know what in chapter 11 WON'T be on the exam? Just for clarification, chapters 8, 9, & 10 in their entirety will be on the exam, correct?
|By frog on Wednesday, November 15, 2017 - 10:57 pm: Edit|
Hi Dr. Dzyuba,
if you want to reduce a ketone directly next to an amine group (amide function group) to get rid of the ketone completely and maintain the amine group as a secondary amine could you use wolf kishner reaction or would that not work because of the basic conditions that would deprotonate the nitrogen? Would LiAlH4 or NaBH4 be the only option?
|By Sergei on Thursday, November 16, 2017 - 12:09 am: Edit|
TCUfrog on Wednesday, November 15, 2017 - 10:20 pm:
material that WON'T be on the exam3:
CHAPTER11: malonic acid synthesis, condensation of esters/Claisen condensation
CHAPTER 10: ester synthesis/hydrolysis, amide synthesis/hydrolysis, synthesis of acid halides (although if you think about these reactions in terms of mechanisms, and correlated to what we have already covered, you should be able not only provide the products, but also the mechanism)
The rest of the material is a fair game
frog on Wednesday, November 15, 2017 - 10:57 pm:
No, amides would not react with hydrazine(s) in the same way as the ketones
It's more plausible that RC(O)-NR1R2 bond of the amide will get broken after the tetrahedral intermediate will form, in order to restore the carbonyl, and R1R2N– is a decent leaving group under basic conditions, and RC(O)NHNH2 will be the other product
|By Anonymous on Thursday, November 16, 2017 - 08:25 am: Edit|
Hi Dr. Dzyuba,
I was wondering if you were planning on posting the relative grade distribution after test 3 is graded like you have done for the other tests?
|By Sergei on Thursday, November 16, 2017 - 08:40 am: Edit|
|By anon on Thursday, November 16, 2017 - 11:45 am: Edit|
So just to clarify, we will need to know the nomenclature for alcohols, aldehydes, ketones, ether, ester, nitrile, amide, CA, anhydrides, amines, thiols, and thioethers, correct? There are some other ones in the table i the beginning of the textbook, but we shouldn't know those right?
|By anon on Thursday, November 16, 2017 - 11:55 am: Edit|
For nitriles, there are two ways of naming them, but you only want us to know the one with "cyano"? So like saying cyanocyclopentane as opposed to cyclopentanecarbonnitrile. Am I correct about this?
|By Sergei on Thursday, November 16, 2017 - 02:20 pm: Edit|
anon on Thursday, November 16, 2017 - 11:45 am:
all functional groups that we have covered thus far (ie., since August) is a fair game
The groups we haven't covered: amines, phosphates (from the table); also, the nomenclature of imines, sulfoxides, and sulfones will not be on the exam. You only need to be able to recognize what those groups are, reactivity, etc, but you would not be asked to give a name for the compound that has those groups.
anon on Thursday, November 16, 2017 - 11:55 am:
cyano is the prefix (should be used when higher priority groups are present)
-carbonitrile (-onitrile) or -nitrile are the suffix/ending
cyclopentanecarbonitile is the correct one (and it's consistent with all other namings of the cyclic systems, please look through the subchapter on naming carboxylic acids and derivatives).
|By laurenc on Thursday, November 16, 2017 - 02:22 pm: Edit|
hi dr. dzyuba!
If I am trying to do a nucleophilic addition of a carbon onto another molecules carbon adjacent to an amine group, how would I go about that?
for example: if i have Phenylacetylene and want it to connect to the alpha carbon of the amine in N-METHYLBENZYLAMINE
could i deprotonate the carbon of the alkyne and have a chlorine on the carbon of attack in n-methylbenzylamine and do a SN2 process to replace the chlorine with the alkyne group?
I hope that made sense! Thanks!
|By anon on Thursday, November 16, 2017 - 05:07 pm: Edit|
Do we always try to have all of the functional groups which we studies so far in the main chain even if the main chain turns out to be shorter as a result?
|By anon on Thursday, November 16, 2017 - 05:11 pm: Edit|
1- Nitriles are always on the end of the molecule so we won't need to number them, correct?
2- To what extent do we need to know the common names? So like all the ones that are mentioned in the textbook even if we have never encountered them?
3- I'm a little confused on how much of a chain to say when it comes with other names like with amide, so like do I say prop amide, or do I say propanamide?
4- What is exactly a carboxamide? It seems to me like it's just amide?
|By Sergei on Thursday, November 16, 2017 - 05:46 pm: Edit|
laurenc on Thursday, November 16, 2017 - 02:22 pm:
if you are proposing a nucleophilic addition, make sure you look at all possible electrophilic centers (not only sp3, but also sp2)
In addition, consider the functional group transformations, i.e., how amines could be obtained...
anon on Thursday, November 16, 2017 - 05:07 pm:
identify the highest priority group, and choose the longest chain that contains that functional group; that chain might not be the longest in the molecule (you are correct - the chain with the highest priority group might be shorter as a result)
anon on Thursday, November 16, 2017 - 05:11 pm:
1. yes (highest priority group should have the number 1, and hence no need to put 1)
2. trivial names/common names of the complex natural products and names of the drugs that are described in the textbook are not going to be on the exam. If such names are given, a structure would be provided as well.
3. remove 'e' from the alkane, keep the rest
4. carboxamide - amide is a part of the more complex structure or a ring, e.g. cyclohexanecarboxamide (cyclohexyl group connected to C(O)NH2)
|By AA on Saturday, November 18, 2017 - 01:39 pm: Edit|
I know we don’t have class on Monday, but what will be covering in class next lecture? I know we didn’t completely cover chapter 11 so i didn’t know if we were going to finish or keep going.
|By anon on Saturday, November 18, 2017 - 07:41 pm: Edit|
How many pages is the final exam?
|By Sergei on Saturday, November 18, 2017 - 08:59 pm: Edit|
AA on Saturday, November 18, 2017 - 01:39 pm:
ester, amides and everything that relates to them.
After that - amines and carbohydrates (as stated in the syllabus).
anon on Saturday, November 18, 2017 - 07:41 pm:
|By ZZ on Monday, November 20, 2017 - 08:39 am: Edit|
Why is acetonitrile, CH3CN, and doesn't have a C-O double bond, whereas all of the "aceto's we've been learning so far had that carbonyl group?
|By Sergei on Monday, November 20, 2017 - 09:10 am: Edit|
By nomenclature rules, nitriles could be viewed as derivatives of carboxylic acids
CO2H is replaced with CN
Hence oic acid(or ic acid) is replaced with onitrile; acetic acid - acetonitrile
Similarly: benzonitrile from benzoic acid
In some cases these are also stated to be common names although they do follow the rules; I think our text book also lists onitrile as an ending), and in some instantces it might say that “o” is added to sound better
In all those cases, “o” in the name does not mean oxygen in the compound
|By ZZ on Monday, November 20, 2017 - 10:00 am: Edit|
I quite don't understand why phenol is less acidic than acetic acid. I was thinking the CB of phenol will be destabilized by the benzene ring because the O- would start giving electrons to the ring. In the case of acetic acid, I was thinking O- would still be +R but not as much as phenol. Why am I wrong?
|By ZZ on Monday, November 20, 2017 - 10:45 am: Edit|
How many pages was the final exam last year?
|By ZZ on Tuesday, November 21, 2017 - 07:49 am: Edit|
Will the third exam be graded before we leave for thanksgiving break?
|By Sergei on Wednesday, November 22, 2017 - 08:40 am: Edit|
Monday, November 20, 2017 - 10:00 am:
in the CB of acetic acid, when you draw the resonance structures, the negative charge will always be on the most electronegative atom, oxygen.
In the phenolate, the negative would have to be delocalized onto carbon, which is not as good as having the negative charge on oxygen.
Thus, even though in both cases there is a delocalization of the negative charge, in the case of the acetic acid's CB it's a better delocalization, hence higher stability, and as a result higher acidity of the corresponding acid.
ZZ on Monday, November 20, 2017 - 10:45 am:
ZZ on Tuesday, November 21, 2017 - 07:49 am:
No; as soon as it's graded, you will get the e-mail with the key and the stats will be posted here.
However, it will not be before Wednesday of next week, since there are a few people taking it on Monday/Tuesday.
|By ZZ on Wednesday, November 22, 2017 - 11:27 am: Edit|
Do we ever care about the number of arrows that we will draw for the resonance structures? So like let's say if the CB of compound A is a 10 membered ring and aromatic in a way that we can resonance stabilize the charge all over the ring. Let's also say compound B is acetic acid. Will acetic acid still be the more acidic one because the charge is distributed between 2 Os as opposed to like 10 Cs?
|By ZZ on Wednesday, November 22, 2017 - 11:56 am: Edit|
Also if the conjugate base of an acid is neutral, then what charge do we look at when deciding on the strength?
|By zz on Wednesday, November 22, 2017 - 01:33 pm: Edit|
I'm just generally confused about the fact that the more the percentage of s, the more stable the CB. Should this sp/sp2/etc carbon be located in a specific distance from the acidic hydrogen, or is it saying that just in general if you have a sp carbon somewhere in your molecule, that molecule is more acidic?
|By zz on Wednesday, November 22, 2017 - 03:43 pm: Edit|
There's this online resource which says that CHF3 is a weaker acid than CHBr3. Should it not be the reverse?
|By Sergei on Wednesday, November 22, 2017 - 05:45 pm: Edit|
ZZ on Wednesday, November 22, 2017 - 11:27 am:
The number of resonance structures is important, and could be used to decide which structure/compound would be more stable, etc.
HOWEVER, they have to be comparable: carbocation to carbocation, etc.
In the previous example, delocalizing the negative charge over 2 oxygens is always going to be preferred to delocalizing the negative charge over O and 2-3 carbons.
ZZ on Wednesday, November 22, 2017 - 11:56 am:
the one on the acid.
For neutral compounds you can use split resonance structures, (recall EAS, and drawing those for aromatic compounds with EDG and EWG)
zz on Wednesday, November 22, 2017 - 01:33 pm:
for the consideration of the C-H acidity using sp/sp2/sp3: it's the C-H, one sigma bond; the carbon will get the negative charge, and hybridization will be used to evaluate the stability.
zz on Wednesday, November 22, 2017 - 03:43 pm:
it's helpful to know the solvent (also whether it's experimental data or calculated), etc.
Without this information (or assuming that it's irrelevant), here is what could be your line of reasoning:
electronegativity of F being higher than electronegativity of Br should suggest that F3C– should be more delocalized (stronger –effect of F vs –I effect of Br). However, as you recall the case of HHal, electronegativity alone fails to explain pKa (mostly because it only considers effects through sigma bonds - this is valid for elements of the same period, but for elements in different period that is not the case). Thus, we need to take into account something else, beyond EN (as we did with HHal).
1. Hard/soft: F is hard, H is hard, while Br is softer, hence C in F3C– is going to be harder than C in Br3C–, hence a stronger interaction with harder H.
2. Bond length: C-F vs C-Br; both Hal have 3 non-paired pairs of electrons, with C– (an electron pair on C) there is going to be more repulsion in C–F than in C–Br, because C-F is shorter (the electron pairs will bump into each other).
Thus, it's a combination of electronic and "steric" effect; EN is going to be offset by the repulsion of the electron pairs in the case of F3CH, whereas Br's EN while being smaller than that of F, it is not going to be offset by the repulsion of the electron pairs as much
|By zz on Wednesday, November 22, 2017 - 06:29 pm: Edit|
Can we instead of using cis/trans always use R/S? Don't they convey the same message?
|By zzzzzzzzzzzz on Wednesday, November 22, 2017 - 08:26 pm: Edit|
If there's a trans double bond, does the LA prefer to add to a certain side? (both sides are substituted equally)
|By zZzZzZzzZ on Wednesday, November 22, 2017 - 09:52 pm: Edit|
Can we do hydrogenation on a carbonyl?
|By zzzzzzzzzzzzzzzzzzzzzz on Thursday, November 23, 2017 - 06:30 pm: Edit|
Do we need to know the formation of trialkylborane for the hydroboration reaction? Or should we just know that the first step H and BH2 are added and the second step is the formation of alcohol?
|By z on Thursday, November 23, 2017 - 06:36 pm: Edit|
We shouldn't know any reactions involving ozone, right?
|By Sergei on Friday, November 24, 2017 - 07:17 am: Edit|
zz on Wednesday, November 22, 2017 - 06:29 pm:
No, they don't convey the same message; cis/trans are not related to chirality
R/S is for chiral compounds (point chirality, compounds with a chiral center).
priority rules are the same, but the way you assign them, and the structural elements are completely different.
zzzzzzzzzzzz on Wednesday, November 22, 2017 - 08:26 pm
C=C has a plane of symmetry, both sides are the same, thus, top/bottom addition/approach is the same.
To have a differential approach something must be chiral: LA, environment or R-groups on the alkene (but this is out of the scope of this course)
zZzZzZzzZ on Wednesday, November 22, 2017 - 09:52 pm:
Yes - but this is also outside of the scope
zzzzzzzzzzzzzzzzzzzzzz on Thursday, November 23, 2017 - 06:30 pm:
Do you mean the addition of the borane (R2B–H) to the double bond?
If yes, then yes.
It's the second step (insertion/rearrangement type) that we skipped in this course that you are not responsible for on the exam/quizzes.
z on Thursday, November 23, 2017 - 06:36 pm:
No, you really should.
But it's a part of cycloaddition reactions that are out of scope for this semester, so this material will not be on the exam/quizzes.
|By KH on Saturday, November 25, 2017 - 11:36 am: Edit|
Hi Dr. Dzyuba,
Finals are coming up in the next few weeks. There is obviously a lot we have covered so far in the semester. I was wondering if you had any study tips? It doesn't make sense to go about memorizing everything, because there is just too much. When we come to a potential problem on the final, how should we know where to start?
Thanks for your help
|By d on Sunday, November 26, 2017 - 12:34 pm: Edit|
Will you be teaching about amines or carbohydrates tomorrow?
|By Sergei on Sunday, November 26, 2017 - 01:43 pm: Edit|
KH on Saturday, November 25, 2017 - 11:36 am:
In no specific order:
1. functional group transformations are a must!
Memorizing these will be very helpful!
Make sure you know how each functional group is being converted into another; make small notes as to the specifics of each case, especially as it relates to the mechanism (e.g. knowing that it's SN2 for example, will indicate that all rules that apply to SN2 should be considered)
2. Don't memorize the mechanisms.
You saw on numerous occasions how some steps could be altered or the sequence of steps could be altered, depending on conditions, etc.
Try to rationalize the mechanism, i.e., what could be a base, nucleophile, acid, electrophile. If multiple options are available, explore all options, and make a choice as to which one would likely to be the major, considering basic/fundamental principles.
3. If you started drawing arrows (in the arrow-pushing mechanism) stick with them, and follow through with the bond forming/breaking steps, draw the compounds that are consistent with the arrow movements.
4. Try to remove the guess work out of the problem, instead consider multiple scenarios that are reasonable based on the given structures, conditions, etc.
5. If you stumbled on a problem that you don't know how to approach, start with the basics: assign each compound a role (acid, electrophile, base, nucleophile, solvent, catalyst; a single compound could have multiple roles), identify one compound that would have a single role, and try to match it with it's counter part (i.e., if something is only a base, you need to look for interactions only with those compounds that you identify as bases (even though they might have other roles).
If you can't identify a compound that could serve only one role, you would have to consider multiple options, but they would be fundamentally the same as the those discussed above.
6. Try to look at the material (all reactions) that we covered from the mechanistic and fundamental stand points of view.
there are a lot of similarities (e.g. the whole carbonyl chemistry is the same, mechanistically).
d on Sunday, November 26, 2017 - 12:34 pm:
No, we have to finish the carbonyl chemistry (ester formation, hydrolysis, and related reactions).
|By anon on Tuesday, November 28, 2017 - 01:27 pm: Edit|
in doing hydrolysis of an ester under basic conditions, you said choose reagents/solvents wisely. Which reagents should be used together and which should not? For instance, why can EtOH not be used with NaOH but water can be?
|By Sergei on Tuesday, November 28, 2017 - 01:36 pm: Edit|
Hydrolysis implies that the ester will be converted to the carboxylic acid (not another ester).
A combination of EtOH and NaOH gives two nucleophiles: HO– and EtO–; HO– will lead to hydrolysis, EtO– to transesterification.
In principle, this is not a huge problem, since the new ester will also hydrolyze.. eventually (since if EtOH is used as a solvent, there will always be more of EtO– than HO– (as such the rate of the hydrolysis reaction will be slow); but hydrolysis is irreversible, while transesterification is reversible, which would give the acid in the end.
In other words, "choosing wisely" means that there is no need to introduce an unnecessary reaction, if the goal is to produce the acid.
With NaOH/H2O - the only nucleophile is HO–, hence only hydrolysis will be taking place.
|By AN on Tuesday, November 28, 2017 - 01:37 pm: Edit|
How do we know when to stop deprotonating the B keto-ester? or when to keep going to get the two negative charges on both alpha carbons?
Along with this question, if we deprotonate both alpha carbons and add an alkyl halide, which negative charge would attack the alkyl halide?
Sorry if that was confusing. Thanks!
|By Sergei on Tuesday, November 28, 2017 - 02:41 pm: Edit|
Under normal conditions (i.e., no large excess of the base), the most acidic H will be deprotonated.
You would not be deprotonating multiple CHs, and the only negative charge will be delocalized over the two carbonyls.
If you do use a large excess of the base, then multiple CHs will be deprotonated (probably 2 max, since putting 3 negative charges will result in highly insoluble material, which will precipitate out of the solution).
The more reactive negative charge (i.e., the least delocalized) will react first (and if only 1 eq of the electrophile is added, it will be the end of it) - exactly as was shown in the lecture.
|By anon on Wednesday, November 29, 2017 - 04:33 pm: Edit|
Does solvation of a base (with a protic solvent) decrease basicity? Is that what the description of Et3N with water was?
|By Sergei on Wednesday, November 29, 2017 - 07:31 pm: Edit|
the steric hindrance of the three R groups (Et3N - R=Et) precludes the "proper" solvation, that would be described as solvent stabilization of the conjugate acid Et3NH+.
Since the conjugate acid of the Et3N is not as stabilized by the solvent as the conjugate acid of Et2NH, the Et3N is less basic than Et2NH.
The thing to keep in mind - the differences are fairly small for aliphatic amines; thus all of these amines are strong bases (comparative to aniline, pyridine, etc).
|By Sergei on Thursday, November 30, 2017 - 08:04 am: Edit|
Exam 3 is graded:
the key and the grade distribution will be e-mailed shortly
|By aa on Thursday, November 30, 2017 - 10:13 am: Edit|
Can you post the grading distribution please?
|By Sergei on Thursday, November 30, 2017 - 01:12 pm: Edit|
Quiz13 is graded:
TENTATIVE GRADE DISTRIBUTION
(3 exams, 13 quizzes; for those with excused absences - % are given in parentheses):
A: >480 (67%)
B: >386 (54%)
C: >265 (37%)
D: >110 (15%)
F: <100 (14%)
|By Kat on Thursday, November 30, 2017 - 01:56 pm: Edit|
I apologize if this has already been asked, but will any quizzes be dropped? Since we've taken so many?
|By T on Thursday, November 30, 2017 - 03:00 pm: Edit|
Do you mind posting how many people are in each range?
|By Sergei on Thursday, November 30, 2017 - 04:00 pm: Edit|
Kat on Thursday, November 30, 2017 - 01:56 pm:
No, no quizzes will be dropped.
Considering our grading system, dropping a score will not have a significant effect on the grade distribution/gaps
T on Thursday, November 30, 2017 - 03:00 pm:
with 3 students in between the grades.
|By C1 on Thursday, November 30, 2017 - 06:11 pm: Edit|
In the synthesis of an Amide, we used DMAP as well as DCC. I look-up that DCC refers to dicyclohexylcarbodiimide. Why is DCC considered an imide without the presence of any acyl groups?
|By anon on Thursday, November 30, 2017 - 06:15 pm: Edit|
Do the ranges for the cut offs remain the same? I just see that C to D is a big difference. How do we know what still counts as a C and what starts to count as a D?
|By AA on Thursday, November 30, 2017 - 06:18 pm: Edit|
Hi Dr. Dzyuba,
Do you think there is still hope for us? Is it still possible to pass this amazing but hard class?
|By DMAP on Thursday, November 30, 2017 - 06:38 pm: Edit|
So if one has about 460 points, is he or she in the B+ range?
|By SN2 on Sunday, December 03, 2017 - 10:23 am: Edit|
If there's a chiral center in a molecule, do we only have to specify the R/S stereochemistry? What if cis/trans is also applicable? Do we also mention those?
|By Confused on Sunday, December 03, 2017 - 01:21 pm: Edit|
Going off of the previous post, how do we know when to use and name cis/trans, R/S, or E/Z? When explaining it in class I completely understand but of course, during the exams, I have no idea when to use what! I'm sure this is probably a dumb question and probably easily solved, so I'm sorry!
Thanks Dr. Dzyuba!
|By The original AA on Sunday, December 03, 2017 - 01:26 pm: Edit|
Can we have two double bonds besides each other in a cyclic structure?
|By The Original AA on Sunday, December 03, 2017 - 04:25 pm: Edit|
We haven't learned proton NMR shifts right?
|By The Original AA on Sunday, December 03, 2017 - 05:07 pm: Edit|
How about shielding and disshielding protons//NMR spectroscopy//hyperconjunction?
|By Sergei on Sunday, December 03, 2017 - 11:06 pm: Edit|
For some reasons some of my previous answers to your posts don't show up. Apologies for the inconvenience.
Here it goes:
C1 on Thursday, November 30, 2017 - 06:11 pm:
The key is "carbodiimide"; it's not the same as "imide".
anon on Thursday, November 30, 2017 - 06:15 pm:
based on the posted cut-offs, if you score is in between 480 and 386 - you are in the B range, if your score is in between 265 and 385 - you are in the C range, etc.
AA on Thursday, November 30, 2017 - 06:18 pm
Absolutely; hope never dies, just feels a little bit of discomfort, from time to time.
And, the final is comprehensive, you will see the type of questions that would be similar to those you've seen the whole semester, so...
DMAP on Thursday, November 30, 2017 - 06:38 pm:
460 is a B right now, because it's below 480; instead of +/–, he or she should think that it's only a quiz away from a A range.
+/– would be considered after the final, but these would be exceptions rather than the rules (since it will be only a very few students).
SN2 on Sunday, December 03, 2017 - 10:23 am:
All stereocenters should be assigned.
In the case of a chiral alkene for example, R/S and E/Z should be used.
The main thing is - give the name that unambiguously gives only one structure
Confused on Sunday, December 03, 2017 - 01:21 pm:
R/S - if there a chiral center
E/Z for di-/tri-/tetra-substituted alkenes
cis/trans - di-substituted alkenes; di-substituted cyclic alkanes, for example.
The original AA on Sunday, December 03, 2017 - 01:26 pm:
do you mean an allene?
In principle, yes.
The ring size would probably have to be >8 not too generate too much strain. But allenes, in general are fairly reactive species; thus, putting them in the ring (even large one) would increase the reactivity.
The Original AA on Sunday, December 03, 2017 - 04:25 pm
Yes, no NMR this semester.
You will have it in the spring O-Chem lab.
The Original AA on Sunday, December 03, 2017 - 05:07 pm:
Nothing related to NMR in this course.
Hyperconjugation - does not relate to NMR; but it is related to stabilization of the carbocation by the neighboring sigma bond (C-H or C-C).
In principle, it's used to show that Me3C+ is more stable than MeCH2+, for example.
We used +I effect to get to the same conclusion.
We are not covering it, to avoid confusion with conjugation (+/– R effect).
For everything what we are dealing with this semester, inductive and resonance effects will be sufficient.
|By SB on Tuesday, December 05, 2017 - 11:09 am: Edit|
Hi Dr. Dzyuba,
Can an ester group be added to an aromatic ring via a mechanism analogous to Friedel-Crafts acetylation (using an ester chloride instead of an acid chloride)? If so, would be it be different than acetylation mechanism? Thanks!
|By Sergei on Tuesday, December 05, 2017 - 12:45 pm: Edit|
electrophilic aromatic substitution requires a carbocation.
Ester is an EWG, which will destabilize the formation of the carbocation.
|By Sergei on Tuesday, December 05, 2017 - 12:46 pm: Edit|
Quiz 14 is graded.
|By anon on Tuesday, December 05, 2017 - 04:04 pm: Edit|
On quiz 2, number 2, the name of the structure given on the key was 4-ethyl-1,2-dimethylcyclohexane. Why would the carbon with the ethyl group not be number one? Isn't that the highest priority substituent?
|By SAB on Tuesday, December 05, 2017 - 04:51 pm: Edit|
In class, we performed the hemiacetal formation on an aldose, which would have an H and a OH group, and you said the newly formed OH group would have a squiggly line for now. If we performed the hemiacetal formation on a ketose, how would we know whether the CH2OH group (the C1 group on the other side of the ketone) or the newly formed OH group points upwards or downwards?
|By Sergei on Tuesday, December 05, 2017 - 07:21 pm: Edit|
anon on Tuesday, December 05, 2017 - 04:04 pm:
methyl and ethyl are both alkyl groups, hence they are the same priority groups.
Per rule, the substituents must get the lowest numbers possible (also noted in the key), which is 1,2,4 (in other words - the sum of substituents numbers must be the smallest).
If you give the ethyl group #1, then the substituents will get: 1,3,4 - 2 is lower than 3 (or if we go by the sum-rule - 7 vs 8), hence it should be 1,2,4, and the methyl group gets 1.
SAB on Tuesday, December 05, 2017 - 04:51 pm:
It will not be a squiggly line after tomorrow's lecture.
Ketose or aldose would behave the same way (a mixture of the compounds will form, one with H up (equatorial) and one with H down (axial).
|By x on Wednesday, December 06, 2017 - 12:04 pm: Edit|
Why in oxymercuration, the H2O/MeOH decides to attack the more substituted par of the double bond?
|By Sergei on Wednesday, December 06, 2017 - 12:08 pm: Edit|
The reaction is done under "acidic" conditions, which favors the carbocation formation. Hence a more stable intermediate is the more substituted carbocation; i.e., this is a SN1-like process.
Mercurium intermediate prevents the carbocation rearrangement.
|By x on Wednesday, December 06, 2017 - 12:12 pm: Edit|
What is the mechanism for the first step of hydroboration? I can't find it anywhere. Is that just that the double bond would attack BH3 and then a base removes an H from the resulting structure to have BH2 on the carbon?
|By Sergei on Wednesday, December 06, 2017 - 12:19 pm: Edit|
The mechanism for the 1st step is the notes.
We looked at it from the stand point of the acid base interaction, while noting that it is a concerted process:
the vacant orbital of BH3 (LA) interacts with the C=C (LB), and the breaking of the C=C and B-H bond is taking place at the same time as the formation of the C-H and C-B.
This reaction could be analyzed from a simple "–" to "+" (i.e., acid-base chemistry).
However, because the stereochemistry is preserved, it's a concerted process, no intermediates, and it is governed by the steric-control, hence C-B is forming on the least substituted carbon.
|By x on Wednesday, December 06, 2017 - 12:26 pm: Edit|
I was wondering if we have a trans alkene and we have KMnO4 and base, would the stereochemistry of the diol still be the same for the two OH groups?
|By x on Wednesday, December 06, 2017 - 12:30 pm: Edit|
Do we need to understand the mechanism for KMnO4/H+ to alkene? If so, what happens after the concerted process with KMnO4, like how does H+ exactly break the C-C bond?
|By Sergei on Wednesday, December 06, 2017 - 12:33 pm: Edit|
it's a concerted process; the two OH groups will be "cis" to each other.
However, there is a free rotation (assuming it's not a cyclic alkene), and R-groups and the OHs might be represented with somewhat different orientations in the diol.
use Newman projections for both the product and the starting alkene, and then redraw the product using the dashed/solid wedge bonds for the newly formed chiral centers.
|By Sergei on Wednesday, December 06, 2017 - 12:41 pm: Edit|
Hi x on Wednesday, December 06, 2017 - 12:30 pm:
H+ is not breaking the bond per se (it's responsible for the formation of Mn–OH), it's still a concerted process for the breaking of the C-C bond; the process is facile under acidic conditions.
Under basic conditions, HO– is acting as a Nu towards Mn (LA, has vacant orbitals), hence the Mn–OC bond is getting broken, leading to the diol.
|By x on Wednesday, December 06, 2017 - 12:56 pm: Edit|
If you like ask a question to "do this thing in three steps," would the workup be considered in one of the steps or is it just assumed?
|By Sergei on Wednesday, December 06, 2017 - 12:57 pm: Edit|
Work-up should always be assumed, i.e., it's not a separate reaction step.
|By x on Wednesday, December 06, 2017 - 01:05 pm: Edit|
Can we use NaBH4 on a CA twice, and then do an acidic workup to get a diol?
|By x on Wednesday, December 06, 2017 - 01:10 pm: Edit|
I don't understand why LiAlH4 is stronger than NaBH4..Isn't the H in NaBH4 softer so it would have a better interaction with the partial positive carbon?
|By Sergei on Wednesday, December 06, 2017 - 01:15 pm: Edit|
x on Wednesday, December 06, 2017 - 01:05 pm:
What is CA?
If you have two aldehydes or two ketones i the same molecule (or more than 2), both (all) will be reduced by NaBH4
x on Wednesday, December 06, 2017 - 01:10 pm:
Consider H–/BH3 vs H–/AlH3; Al is harder than B, hence H– is less coordinated to the LA-center (Al vs B), and therefore more available, i.e., more reactive.
|By x on Wednesday, December 06, 2017 - 01:20 pm: Edit|
CA = Carboxylic acid; I was mostly thinking why can't we use NaBH4/LiAlH4 on a CA and protonate the H first, and then attack it again and push the =O up making a O-, so that would make two O- in the structure, and then do a H+ workup and make a diol. Can we do that?
What do you mean by "less coordinated to the LA-center"?
|By Sergei on Wednesday, December 06, 2017 - 01:36 pm: Edit|
CA: it's an acid, hence the H– will first react with CO2H, and form a carboxylate, which is not very electrophliic.
If you look through the notes, we made a special case for the O-Li bond (partially covalent bond), which made LiO– a reasonable leaving group.
O-Li - hard hard type of interaction; O-Na not so much, since Na is a bit softer than Li.
Thus, with LiAlH4 the reduction takes place, while with NaBH4 it does not. In addition, H– in NaBH4 is not as reactive as H– in LiAlH4, as we discussed in previous posts.
"less coordinated to the LA-center":
in LiAlH4 - H– + AlH3 = AlH4– - the equilibrium is likely to be shifted to the left, since H– and Al are not matching up in softness (at least not as much as in H– and B).
Thus, in NaBH4, H– + BH3 = BH4– the equilibrium is likely to be to the right; i.e., H– is in the complex with LA, rather than freely available (more coordinated to the LA-center).
This is however, is not a straightforward issue, since applying HSAB theory to thermodynamics is a bit tricky.
But within what we know, and what we intend to use it for, it's probably sufficient.
|By x on Wednesday, December 06, 2017 - 02:03 pm: Edit|
So I still don't understand why you made a big deal out of benzylic/allylic/propagyl alcohols needing a MnO2 to go to an aldehyde. I mean all types of alcohols need a mild oxidizing agent to go to aldehyde not just those three right? Why did we emphasize on those three?
|By Sergei on Wednesday, December 06, 2017 - 02:13 pm: Edit|
MnO2 is only going to oxidize allyllic/benzylic types of the alcohols; nothing else (if you have a primary/secondary alcohol and a benzylic alcohol in the same molecule, using MnO2 will allow to oxidize benzylic without affecting the primary/secondary alcohol).
Selectivity, in regard to functional group transformation, is kind of a big deal (although, I didn't really mean to make it bigger than it actually is).
|By x on Wednesday, December 06, 2017 - 02:33 pm: Edit|
Oh so we can't use PCC/DMP on benzylic/allylic alcohols?
|By x on Wednesday, December 06, 2017 - 02:44 pm: Edit|
I have something weird in my notes I wanted to check it with you: so all KMnO4, CrO3, H2O2, ROOH, tBuOOH, Fe2+, transition metals, they all can be used to oxidize alcohols?
|By x on Wednesday, December 06, 2017 - 02:49 pm: Edit|
In ether cleavage of methyl phenyl ether using HI, can we not use SN1 at all?
|By anon on Wednesday, December 06, 2017 - 03:50 pm: Edit|
today in lecture, I wrote in my notes that you said that the imine formation can only take place under acidic conditions. However, I remember in the review for the third exam you going over how to do it in basic conditions. Can we do it in basic conditions or not?
|By x on Wednesday, December 06, 2017 - 03:54 pm: Edit|
Can we ever do acetal formation in basic conditions/
|By x on Wednesday, December 06, 2017 - 04:16 pm: Edit|
The last step of aldol condensation where OH leaves is irreversible, right?
|By Sergei on Wednesday, December 06, 2017 - 04:38 pm: Edit|
x on Wednesday, December 06, 2017 - 02:33 pm:
Yes, you can.
PCC and DMP will also oxidize primary and tertiary alcohols as well (in other words, PCC/DMP does not differentiate between benzylic/allylic and the other alcohols; MnO2 does).
x on Wednesday, December 06, 2017 - 02:44 pm:
KMnO4 or CrO3 (acid) or H2CrO4 or H2Cr2O7 or a transition metal such as Fe, for example) with a peroxide (either H2O2 or ROOH, where R is tBu, for example) would be acceptable oxidizing agents.
x on Wednesday, December 06, 2017 - 02:49 pm:
This is NOT a SN1 process, it's SN2-like, where I– acts as a Nu that attacks the CH3 electrophilic center.
anon on Wednesday, December 06, 2017 - 03:50 pm:
Imine formation, in general, could happen under both acidic and basic conditions (although it's much more preferred under acidic, as it avoids generation of a not very favorable LG, N–).
Today's lecture was on sugars, and imine formation was not on just a ketone/aldehyde, but on a hydroxy aldehyde/ketone (aldose/ketose).
Under basic conditions, the deprotonation of the alpha OH is possible, and it will hinder the subsequent steps.
x on Wednesday, December 06, 2017 - 03:54 pm:
No (would not have a good LG for the few steps that require a LG).
x on Wednesday, December 06, 2017 - 04:16 pm:
Yes (enone is conjugated, more stable product, and it is a soft E, HO– is a hard Nu) .
|By pcp on Thursday, December 07, 2017 - 08:35 am: Edit|
Are you not making a review sheet out of the things after exam 3 and send it to us?
|By x on Thursday, December 07, 2017 - 09:09 am: Edit|
In the textbook, there's this questions about drawing fisher for (R)-2-chlorobutane and then in the answer it only draws the fisher for the chiral center. I thought fisher is for the whole molecule isn't it?
|By x on Thursday, December 07, 2017 - 09:21 am: Edit|
Can we say that all D sugars have R stereochemistry in the very lowest chiral center?
|By x on Thursday, December 07, 2017 - 09:26 am: Edit|
So the textbook does something weird: it asks you to draw L-Fructose from D-Fructose and then instead of only changing the stereochemistry in the lowest carbon, it flipped everything..Why?? This is so weird
|By Sergei on Thursday, December 07, 2017 - 09:54 am: Edit|
pcp on Thursday, December 07, 2017 - 08:35 am:
yes, you should have it by tomorrow morning.
x on Thursday, December 07, 2017 - 09:09 am:
Yes, Fischer projection is for the whole molecule, but the most important thing is the chiral center, i.e., the rest of the molecule is not going to matter (recall/check how CH2OH and CHO of the glucose for example is drawn)
x on Thursday, December 07, 2017 - 09:21 am:
It's better not to mix D/L and R/S, and there is no need for that. Use one or the other, but not both.
x on Thursday, December 07, 2017 - 09:26 am:
Recall the discussion on the enantiomers: every stereo center has to be inverted when going from R to S (and vice versa).
D and L are enantiomers, hence all stereo centers must be inverted (please check the notes too).
|By x on Thursday, December 07, 2017 - 09:55 am: Edit|
Does Fischer-indole have to do with fischer projections at all?
|By Sergei on Thursday, December 07, 2017 - 11:04 am: Edit|
Only in regard to Fischer (same Fischer, i.e., Emil Fischer, in both cases)
Fischer indole synthesis relates to the synthesis of indole by Fischer; has no relationship to drawing Fischer projections for sugars (more generally, chiral centers).
|By HC on Thursday, December 07, 2017 - 01:45 pm: Edit|
Hi Dr. Dzyuba,
Should go back and study the MO diagrams we did at the beginning of the semester?
|By x on Thursday, December 07, 2017 - 03:17 pm: Edit|
In aldol condensation, do we always have a hydrogen in the middle for the second step of the mechanism? If we happen to not have a hydrogen in the middle, can we not proceed with the E1cb like reaction anymore?
Also, in aldol condensation, if we have two carbonyl containing compounds, could either act as the attacker? Or should we start thinking about which one would have a more stable alpha hydrogen?
|By x on Thursday, December 07, 2017 - 03:27 pm: Edit|
So cannizzaro has to be done with at least one aldehyde molecule right? Because if we have like a ketone with two tBu groups on both sides, (so on alpha hydrogens), then there would be no H to act as a Nu in the second step, right?
|By x on Thursday, December 07, 2017 - 03:35 pm: Edit|
When using TMSCl, why did we use O Nu to attack TMS instead of the C Nu? I was thinking TMS since it has Si, it's probably more closer to C in terms of softness, no?
|By x on Thursday, December 07, 2017 - 03:49 pm: Edit|
In TMSCl reaction with a ketone, the only purpose is that "there's no base in the solution where we do the reaction for TMS" and that's why there's no OH- available to do the deprotonation of alpha H. But that kind of sounds silly I think because we could just dump the whole thing into a concentrated acid, and get rid of all bases, right? Why do we necessarily have to do something complicated like this?
Also, just to check for the mechanism, the reason the O- grabs the Si from OTMS+ is that we just don't want a positive charge on oxygen because it's bad and so we do this to get rid of it? Also, as a result of this, as opposed to regular cases of aldol, in this reaction the carbonyl in the final product is in fact for the attacker molecule not form the molecule being attacked, right?
|By Sergei on Thursday, December 07, 2017 - 04:22 pm: Edit|
HC on Thursday, December 07, 2017 - 01:45 pm:
review them so that it's clear which electrons are likely to be more reactive and how many bonds could be drawn (that was the main purpose for discussing them in the first place)
x on Thursday, December 07, 2017 - 03:17 pm:
if you don't have an alpha H after the aldol reaction, then there is nothing to deprotonate, hence no E1cb.
if both carbonyls could undergo enolization (enol or enolate) then you have two nucleophiles. Depending on the stability of the enolates, you can decide which aldol will be the major which one is going to minor.
x on Thursday, December 07, 2017 - 03:27 pm:
Yes, Cannizzaro reaction is only considered when you have an aldehyde with no alpha-H, so that the enolate formation is not possible, i.e., HO– can only act as a Nu initially.
x on Thursday, December 07, 2017 - 03:35 pm:
Hard-soft issue; Si is almost like H (just not acidic)
x on Thursday, December 07, 2017 - 03:49 pm:
Aldol condensation happens under both acidic and basic conditions. In other words, enone will form either in acid or base.
if we want to preserve the aldol (i.e., OH in the product, and not have the enone) then we must use protected enolate (to have neutral conditions).
TMS is getting transferred from one O to the other due to proximity (6-membered ring)
Aldol reaction is the same, regardless of the acid, base or TMS-based process. It's the elimination (E1cb) that is prevented with TMS (i.e., no enone).
|By x on Thursday, December 07, 2017 - 07:17 pm: Edit|
If we have an acid halide, and a conjugated system of carbonyl, would Nu attack the alkene first, or the carbonyl carbon?
|By x on Thursday, December 07, 2017 - 09:05 pm: Edit|
In my notes I have something that is a molecule with two ketones, and one of them gets tautemerized and one of them doesn't. In the margin I have written "something random someone asked" so there's really no context to give you, but do you know how we could tautemerize one carbonyl and not the other?
|By x on Thursday, December 07, 2017 - 10:12 pm: Edit|
Can you explain the special thing with F in terms of both electrophilic aromatic substitution and in terms of acid halide reactivity?
|By anon on Friday, December 08, 2017 - 09:43 am: Edit|
Will you have regular office hours during finals week?
|By Sergei on Friday, December 08, 2017 - 10:01 am: Edit|
x on Thursday, December 07, 2017 - 07:17 pm:
it's a tough call... most likely a mixture will form (in other words - both possibilities will happen at a similar rate).
Assuming that Nu is fairly soft (RS–), enone is likely to be attacked first due to hard-soft (acid chloride is relatively hard, due to O and Cl).
However, acid chlorides are very reactive electrophiles, in many cases hard-soft issue becomes a non-factor, especially if a Nu is fairly reactive.
x on Thursday, December 07, 2017 - 09:05 pm:
if you are referring to beta-diketones, where one has 2 alpha-CHs, the one between the two C=O is the most acidic.
The most acidic H will be the one to be moved in the tautomerization process.
In general, remove the alpha CH, and evaluate the stability of the conjugate base, the more stable conjugate base will mean a more acidic CH.
x on Thursday, December 07, 2017 - 10:12 pm:
F: +R ~ –I
+R effect is fairly strong when there is a 2p-2p overlap (e.g. C-F)
–I effect is fairly strong because of the electronegativity.
The two opposing effects (when they are both applicable to a given group/compound, i.e., +R effect is possible due to the proximity of the F: and the [i-system for example) cancel each other out, and it appears that F is an exception (only because the perception is that since F is the most EN atom, it must have an effect; of course in those cases, +R is rurally considered).
|By X on Friday, December 08, 2017 - 12:37 pm: Edit|
So going back to beta diketones, the structure I have is symmetrical with two carbons between the ketones and thus two possible places in between the ketone from which we could deprotonate H. But we only choose one to do this with so I still don't understand because the two sides are exactly the same.
In regards to F, so if we have an acid fluoride then the carbon by a carbonyl is not very positive because it's like not having anything there so it doesn't help?
|By Sergei on Friday, December 08, 2017 - 05:53 pm: Edit|
if there are two carbons between the ketones - these are not beta-diketones, these are gamma diketones (these two C=O will behave indepedently of each other). Both CHs seem to be the same, thus it does not matter which one is deprotonated.
Deprotonating both at the same time would required an excess of the base, and I think this was not the case for that example; hence only one was deprotonated to form the enolate.
Feel free to specify it further, if it's not clear why certain steps were done.
F: acid fluorides are still electrophilic, and they will undergo nucleophilic addition, but not as readily as it would be expected based on the EN of F.
Based on the +R/–I effect RC(O)F would be less reactive then RC(O)Cl
Acid fluorides are outside of the scope for this course (the synthesis of these compounds requires special reagents, etc).
|By abc on Friday, December 08, 2017 - 06:36 pm: Edit|
before exam 2, you gave us a reaction with a secondary amine with CH3I, then Ag2O/H2O to produce AgI precipitate, an alkene, and N(CH3)3 gas.
What exactly is this mechanism used for... what purpose does this reaction serve and when would we use it?
|By EB on Friday, December 08, 2017 - 07:23 pm: Edit|
Hi Dr. Dzyuba,
I had a question regarding the anomeric effect. If the substituent attached to C1 is -OH and not -OR is the axial position still more favored than the equatorial position?
|By Sergei on Friday, December 08, 2017 - 07:42 pm: Edit|
abc on Friday, December 08, 2017 - 06:36 pm:
In general, elimination leading to alkenes gives a more substituted alkene (Zaytsev elimination). E2 mechanism.
Formation of less substituted alkenes (Hofmann elimination) requires special substrates (or bigger bases).
In that particular case, an amine was converted into the trimethyl ammonium group (I– is the counter ion; this is SN2 process). This is a very large LG (larger than I), and hence the base (HO–, in this case) will be approaching the least sterically hindered (the least substituted) beta-CH, thus giving the less substituted alkene.
Mechanistically it's still E2.
EB on Friday, December 08, 2017 - 07:23 pm:
No, we made a special note in lecture not to include the OH in this discussion.
The anomeric effect is a bit of a complicated issue for this course, we only talked about it because there are some examples of alpha/beta percentages of some sugars in the textbook that might seem odd without the knowledge of this effect.
|By anon on Friday, December 08, 2017 - 08:19 pm: Edit|
Are the grade distributions the same as before?
|By C1 on Friday, December 08, 2017 - 08:29 pm: Edit|
Since the Final counts as 300 points, should we expect the test to be just like a regular test, just double the length?
|By Sergei on Saturday, December 09, 2017 - 09:05 am: Edit|
anon on Friday, December 08, 2017 - 08:19 pm:
Roughly (you can increase each by 10-15pts to account for 2 quizzes; but at this point 10-15 pts should not have any effect on how to approach the final)
C1 on Friday, December 08, 2017 - 08:29 pm:
|By x on Saturday, December 09, 2017 - 09:47 am: Edit|
On exam 3, first question, on the key you have a second method of getting to the product in which you add EtMg before you oxidize. I am not sure why this works because the carbon located by the OH is not the most partially positive center, so I feel like Et would be added to one of the more partially positive carbons..
|By x on Saturday, December 09, 2017 - 10:09 am: Edit|
On exam 3, how does thee enone thing go to the structure with the CN?
|By x on Saturday, December 09, 2017 - 11:45 am: Edit|
In imine-enamine reaction, do we always have a primary amine which reacts with the ketone, or can it be secondary/tertiary amine as well?
|By x on Saturday, December 09, 2017 - 12:52 pm: Edit|
Actually let me rephrase that: so in exam 3 review, there was this questions that had NH2OH instead of RNH2 but the mechanism was the same as for amine formation. So my questions is, is imine formation always with a primary amine (RNH2) or is the selection of N-containing compounds wider than only primary amine?
Also, I just tried doing it with RR'NH, and the mechanism as almost the sam except I ended up with two R groups on the N of course, so is this a right move?
|By ABC on Saturday, December 09, 2017 - 01:21 pm: Edit|
Hi Dr. Dzyuba,
Is there any way you could post/email out the blank versions of exams 1, 2 and 3 so we can practice the problems without looking at the answers? Would be much appreciated.
|By ABC on Saturday, December 09, 2017 - 01:30 pm: Edit|
Also why is the last page of the last review so hard?
|By x on Saturday, December 09, 2017 - 01:38 pm: Edit|
So I've been thinking let's say you give us a molecule that goes to another molecule and we need to write out the reagents, can we use MgBr? Like can we literally make our starting material a grignard? I'm asking because if so, I don't really know if we can make it on any carbon that we want, or is there a place that it has a higher affinity for? So like does MgBr want to attack like the most positive center or the most negative center for instance?
|By x on Saturday, December 09, 2017 - 02:17 pm: Edit|
What happens if we add Br2 and H+ to a ketone?
|By x on Saturday, December 09, 2017 - 03:14 pm: Edit|
Do you think it’d be actually possible for someone with a D to get a B or an A if they do well on the final? Asking for a friend.
I know anything is possible but do you think it’s realistic?
|By ddd on Saturday, December 09, 2017 - 03:16 pm: Edit|
How do you recommend that we approach studying for this final? Looking over notes? Reading all the chapters? Practice problems? Memorizing? Please give us some more of your insight!
|By x on Saturday, December 09, 2017 - 05:04 pm: Edit|
Hi Dr. Dzyuba,
How do you come up with the grade distribution averages? Do you just add what the average was to each of the distributions?
|By x on Saturday, December 09, 2017 - 05:08 pm: Edit|
If there's an H on a O but let's say it's in a molecule of alcohol, would the base still deprotonate that H first even though it's an alcoholic H not an acidic H?
|By x on Sunday, December 10, 2017 - 08:11 am: Edit|
In EAS, when we add like a tBu for instance with friended craft, does it add it to only one place, or does it add it to the two para positions?
|By Sergei on Sunday, December 10, 2017 - 09:57 am: Edit|
x on Saturday, December 09, 2017 - 09:47 am:
EtMgBr is added to the carbonyl (the only electrophilic center in the molecule is the aldehyde).
x on Saturday, December 09, 2017 - 10:09 am:
Please check the key.
x on Saturday, December 09, 2017 - 11:45 am:
tertiary amines are not reacting with ketones/aldehydes: a) they are more basic than nucelophlic, b) there is no H that is suitable for migration, and the formation of a suitable leaving group, etc.
Please check the mechanism that we put in lectures for the imine formation to confirm that the reaction is not possible with tertiary amines
x on Saturday, December 09, 2017 - 12:52 pm:
NH2OH could be treated as NH2R (recall hydrazine and phenyl hydrazine, which we will also treated like NH2R for reactivity purposes).
Secondary amines will give iminium salts (N+ to keep up with the bonding).
ABC on Saturday, December 09, 2017 - 01:21 pm:
It would not be as helpful as you might think, unless you analyze the problem from very basic principles, and treat each suitable problem as a practice set problem - answer is not important, but suggesting reasonable answers and identify unreasonable (and why) ones.
But sure, I'll send the copies.
ABC on Saturday, December 09, 2017 - 01:30 pm:
Think in terms what is an electrophile/nucleophile, acid/base and the answer will present itself.
The answer is not the goal, the goal for practice set problems is to explore possibilities, and have a reason (beyond "we always do it this way", "it's in the book", etc) that is based on the fundamental principles that we have been using the whole semester.
x on Saturday, December 09, 2017 - 01:38 pm:
You can use a Grignard as a starting material or a reagent, as long as it's appropriate, i.e., "not every molecule"
bromo alcohols or promo acids cannot be converted to Grignard, for example.
x on Saturday, December 09, 2017 - 02:17 pm:
H+ is an acid, hence you are looking for a base: options are O: and Br:
Acid bases reactions usually give electrophile/nucleophile pair.
Also, what is Br2 - E or Nu?; most likely it will act as an E (Br– is a noble gas configuration, stable/good LG), if it's an electrophile then you need a nucleophile...
Ketone is an electrophile, but in the presence of an acid, something could happen, right?
x on Saturday, December 09, 2017 - 03:14 pm:
D to A is not realistic, because, for that to happen all students in the A range should either not come to the exam or score in single digits (unlikely to happen), while the student in the D range should get close to a perfect score.
It's not impossible for a D student to get a perfect score, but that fact by itself might not be sufficient.
However, because the final is comprehensive, I will be considering both performance during the semester and that on the final, so there is always a reason to do as best as possible.
ddd on Saturday, December 09, 2017 - 03:16 pm:
most important tips were posted here, please check previous posts.
In the nutshell:
1. functional group interconversion, this is must! Memorization of the key elements (specific points related to mechanism, stereochemistry, structure, etc) will help.
2. in case of the question that requires writing a mechanism or of the functional group transformation is not recognized: assign specific roles for all reactant/reagents, and think in terms of acid-base and electrophile-nucleophile interactions; remove the guess work out, write appropriate arrows, and form intermediates that are consistent with arrows. You want to be sure that your arrow-movement is consistent with the structures that you are proposing.
3. While revising the material: make connections to acid-base/electrophile-nucleophile issues, and try to look at each chapter as a part of the whole course, i.e., not an isolated piece of material.
x on Saturday, December 09, 2017 - 05:04 pm:
I don't come up with anything.
The grade distributions are based on gaps in scores (I look for big, meaningful gaps (>10-15pts) starting from the top, since the highest score is an A).
x on Saturday, December 09, 2017 - 05:08 pm:
Base will always deprotonate the most acidic H, HO is very easy to deprotonate because the conjugate base is stable (O is EN, and having a negative charge on EN atom is favorable).
Is it possible that CH would be deprotonated before OH - yes, but that requires special structural features, which means that C– must be more stable than O–.
x on Sunday, December 10, 2017 - 08:11 am:
Friedel-Crafts alkylation will always give polysubstituted product, because of the higher reactivity of the product (i.e., 1st substitution, for substituents with donating effects) as compared to the starting material.
Large substituents, such as t-Bu, iPr, etc will try to minimize the steric repulsion, and para substituted product (from benzene, as the starting material) will be expected to be the major product, with t-Bu for example.
|By Q on Sunday, December 10, 2017 - 02:45 pm: Edit|
What proportion of the final can we expect to be the new material we've covered since exam 3?
|By x on Sunday, December 10, 2017 - 03:28 pm: Edit|
So when doing something let's say adding a grignard, we always need to see where it would go first, and that might not be where we want it to be. Is this the same with adding a base to deprotonate H? I feel like on the review, some of the places we need to deprotonate the H to get the carbanion are not actually the first place that would get deprotonated..
|By x on Sunday, December 10, 2017 - 03:38 pm: Edit|
I just have a general question. When we have a Br2 and we make a bridge at the double bond, and then we have a Nu attacking the carbon besides the bridge, would the Nu attack the more or less substituted carbon?
|By x on Sunday, December 10, 2017 - 03:55 pm: Edit|
Should we know dies alders reaction?
|By x on Sunday, December 10, 2017 - 04:15 pm: Edit|
Do we need to know Beckman rearrangement?
|By x on Sunday, December 10, 2017 - 06:27 pm: Edit|
What does NaNH2 in liquid NH3 do? Does it just take a triple bond and make it single bond just like Lindler's?
|By x on Sunday, December 10, 2017 - 06:31 pm: Edit|
If we have CH3Br and we want to get to CH2=CH2, how can we do it?
|By Sergei on Sunday, December 10, 2017 - 08:41 pm: Edit|
Q on Sunday, December 10, 2017 - 02:45 pm:
The material from the last part of the semester will be integrated into the questions (similar to what I was showing in lectures).
Maybe there will be 1-2 stand alone questions that will be related to sugars.
x on Sunday, December 10, 2017 - 03:28 pm:
In general, the reactions proceed not where we want them to proceed, but where the appropriate distribution of charges are.
In case of the Grignard reagents: they are nucleophiles, but also bases; thus, if there is an acidic H, it will get deprotonated, and it could be the end of the Grignard.
When deciding on which H is more acidic – deprotonate, and evaluate the stability of the conjugate base.
x on Sunday, December 10, 2017 - 03:38 pm:
Br– will be a suitable Nu as well. So unless you have another suitable Nu (for example ROH, if the reaction is done in an alcohol), dibromide would be the product.
Bromonium ion is not really a carbocation (i.e., no rearrangements), but there is an appreciable charge component, and most of the positive charge will be carried by the more substituted carbon
x on Sunday, December 10, 2017 - 03:55 pm:
No, this reaction is out of scope for the 1st semester.
x on Sunday, December 10, 2017 - 04:15 pm:
No, this reaction is out of scope for the 1st semester.
x on Sunday, December 10, 2017 - 06:27 pm:
No; NaNH2 is a Nu and a B; although in most cases it will be a B only, since Na is hard, and NH3 is a protic solvent (kind of a similar idea to EtONa in EtOH, which is more of a B than Nu).
It will not be do hydrogenation (unlike hydrogenation with Lindlar’s catalyst).
x on Sunday, December 10, 2017 - 06:31 pm:
You need to make a C-C bond, i.e., add one more C to what you have.
Consider what functional groups can give you an alkene, and how to get to that functional group(s) from alkyl halide.
|By x on Sunday, December 10, 2017 - 09:59 pm: Edit|
Is it only SN2 where the substrate can't be sp2? that doesn't apply to SN1 right?
|By x on Sunday, December 10, 2017 - 10:02 pm: Edit|
x on Sunday, December 10, 2017 - 06:31 pm:
So like can I use MeMgBr and then NaH?
|By Sergei on Sunday, December 10, 2017 - 10:12 pm: Edit|
x on Sunday, December 10, 2017 - 09:59 pm:
SN2 reaction cannot be done on sp2C - yes.
SN1 - carbocation is an sp2C.
x on Sunday, December 10, 2017 - 10:02 pm:
If you are using MeMgBr, what is the electrophile?
Why NaH? It's a base, so what would be an acidic H and a suitable LG.
once you decide on electrophile the subsequent step should be clear.
|By LewisAcid on Sunday, December 10, 2017 - 11:10 pm: Edit|
dr. dzyuba, I am still struggling a little with what to assume with solvents. I know what to do with -OH or an acid as a solvent but I get a little confused when it is something like "NaOH/MeOH". In those conditions are we to protonate or deprotonate? Are we able to make carbocations or does everything proceed through SN2? Can a poor leaving group leave like -OH can in basic conditions? I was just wondering if you could help clear up what these conditions actually mean a little bit.
|By x on Sunday, December 10, 2017 - 11:13 pm: Edit|
So PCC oxidizes everything, right? So can we instead of thinking what oxidizes what every time, just always use PCC?
|By Sergei on Monday, December 11, 2017 - 07:36 am: Edit|
LewisAcid on Sunday, December 10, 2017 - 11:10 pm:
Just to clarify: HO– and acid are not solvents.
NaOH/MeOH - this is basic media; HO– is a base, MeO– is a base as well.
Thus, there should not be any H+ (nor C+).
All protons are present in the form of conjugate acids (H2O, MeOH)
In NaOH/MeOH - E2 is probably a dominating pathway (depending on the substrate).
Bad/good LG are relative terms, what is a good LG under one set of conditions, might be a bad LG under the other set of conditions.
So, in basic media HO– is a good LG, because the media is "negatively charged"
Recall E1cb. This is the case where "bad LG" are removed, because of the media and because of the initial substrate (carboanion)
x on Sunday, December 10, 2017 - 11:13 pm:
PCC is a mild oxidant, only converts primary and secondary alcohols to aldehydes and ketones, respectively.
|By on Monday, December 11, 2017 - 12:38 pm: Edit|
I understand the mechanism of the Cannizzaro reaction; however, I don't know that I understand the point of it. What's the reasoning/purpose of doing it, and when is it best to perform one?
|By anon on Monday, December 11, 2017 - 01:25 pm: Edit|
Will the percentages of the grade cutoffs ever increase? For instance, would a B ever jump from 54% to 57%?
|By Sergei on Monday, December 11, 2017 - 01:37 pm: Edit|
:-) on Monday, December 11, 2017 - 12:38 pm:
For our purposes - it's an illustration of electrophile/nucleophile, acid/base paradigms.
is it the best way to actually make acids and alcohols from an aldehyde - no. But it's an interesting, not very trivial and rather peculiar reaction.
You should be aware of the possibility of HO– acting as a Nu in a nonreversible manner, when the conditions are right.
What is the point of any reaction ?
1. Provide us with new knowledge, provide an explanation of why certain events happen, and help us understand how to optimize them
2. Provide new materials
This reaction satisfies those criteria.
anon on Monday, December 11, 2017 - 01:25 pm:
It could happen (in fact it did happen for A, C, D, F just looking at the past 2 exams; it's a coincidence that B didn't change)
Again, I'm looking at the actual scores and big gaps between them for determining the cutoffs, not percentages. The percentages are mostly used to help with EAs.
|By Laurenc on Monday, December 11, 2017 - 04:58 pm: Edit|
hi dr. dzyuba,
I was looking at haloform reaction and was wondering if aldol condensation is a side product of that reaction? Both could occur under those conditions correct? Why is haloform more favorable, because it is non reversible?
Thanks so much for your help!
Heres a chem joke for everyone studying
What do you call an acid with an attitude?
-an amino acid!!
|By Sergei on Monday, December 11, 2017 - 06:55 pm: Edit|
Although aldol is possible, it is not going to be appreciable, because
Br2/I2 are soft E (softer than a carbonyl), hence alpha-functionalization will be more favorable.
After the first C-Hal is formed, the subsequent steps will be more favorable due to increased acidity.
|By x on Tuesday, December 12, 2017 - 08:05 am: Edit|
I have in my notes somewhere that the conjugate base, F- is more stable than OH- more than NH2- more than CH3- but I don't understand because isn't F harder than O harder than N harder than C? So it would be more localized and less stable?
|By x on Tuesday, December 12, 2017 - 08:33 am: Edit|
How do we have to know that N is bigger than S? Size decreases as we go from left to right and it increases as we go form top to bottom. N and S are across form each other meaning the two effects kind of neutralize each other, so how do we know?
|By x on Tuesday, December 12, 2017 - 08:36 am: Edit|
On quiz 1, hybridization of one of the carbons is not stated. It wasn't on purpose right? So like that carbon can be assigned a hybridization state but we just happened to not discuss it right?
|By Sergei on Tuesday, December 12, 2017 - 08:41 am: Edit|
Hard and soft is not related to stability, but rather in selecting the reactivity pattern (hard reacts faster with harder center, than with the softer).
F is the most EN, hence it will hold on to the electron pair the strongest, thus F– is the least likely to give up the electron pair.
C is the least EN in this set, hence will be the most likely to give up the electron pair, thus it's the most reactive.
|By x on Tuesday, December 12, 2017 - 08:48 am: Edit|
In drawing resonance structures, could we push two arrows that are totally unrelated? So like on quiz 2 question 1 second one, could we push the electrons on the double bond to the right, and then push the electrons on the oxygen to the right as well and make a resonance structure?
|By x on Tuesday, December 12, 2017 - 08:50 am: Edit|
By Sergei on Tuesday, December 12, 2017 - 08:41 am
But didn't we say more localized more reactive? And F- is more localized than like CH3-
|By x on Tuesday, December 12, 2017 - 08:52 am: Edit|
Just to make sure, cis and trans's rules are different from R/S right? For cis and trans we only care about big/small groups but for R/S we have all those rules about priority right?
|By x on Tuesday, December 12, 2017 - 08:58 am: Edit|
Which one of these is right? 2-cyclopropylpropane OR Dimethylcyclopropane?
|By Sergei on Tuesday, December 12, 2017 - 09:00 am: Edit|
x on Tuesday, December 12, 2017 - 08:33 am:
N is not bigger than S
N is in 2nd period, S is in the 3rd...
N vs S in regard to Nu/B, for example:
N is smaller/harder, hence more basic (although could be a Nu), S is larger/softer - hence more Nu (almost never a base).
x on Tuesday, December 12, 2017 - 08:36 am:
in C=C both Cs are sp2 (same type of bonds); it could be assigned, I probably should have put that on the structure. However, the note says which atoms could not be assigned.
|By x on Tuesday, December 12, 2017 - 09:05 am: Edit|
Just to make sure, hydrogenation an an alkene or alkyne results in an alkane. Hydrogenation in a cyclic compound results in several acyclic compounds depending on the number of places we can break the cycle, right?
|By Sergei on Tuesday, December 12, 2017 - 10:07 am: Edit|
x on Tuesday, December 12, 2017 - 08:48 am:
yes, as long as you have the correct charges as a results of that
x on Tuesday, December 12, 2017 - 08:50 am:
You cannot apply one principle to all situations, you have to consider the specifics (i.e., it's not one factor/property problem)
F– vs C–: both are in the same period, hence EN is an important factor. F– is more EN, hence it's holding on to electrons tighter than C, thus they are less available to react; which means that C– is more reactive than F–.
C– is a strong B and Nu (stronger than F), bc the electron pair is freely available, low EN of C. However, C is also a softer Nu than F (recall - charge to surface area ratio).
Also keep in mind, that hard/soft has nothing to do with strong/weak.
x on Tuesday, December 12, 2017 - 08:52 am:
Priority rules are the same!
R/S for any chiral compound, E/Z for alkenes (i.e., not for chiral centers)
x on Tuesday, December 12, 2017 - 08:58 am:
those are two different compounds (and dimethyl one exists as several constitutional isomers).
if you meant: 2-cyclopropylpropane vs iso-propylcyclopropane, then
either one would be ok (the one with cyclopropane is probably preferred by IUPAC); in general, for cycloalkane with substituents, whichever is easier should be used (either cycloalkyl as a prefix, or cycloalkane as a suffix)
|By Sergei on Tuesday, December 12, 2017 - 10:13 am: Edit|
x on Tuesday, December 12, 2017 - 09:05 am:
complete/total hydrogenation (the ones that involved H2/Pt or other transition metal given in the book or notes) will give the alkane.
Special catalysts (e.g. Lindlar's cat) will give the alkene from the alkyne's hydrogenation, i.e., partial hydrogenation.
Hydrogenation of cyclic alkenes is only limited to cyclopropane (the most common/easiest) and cyclobutane (possible, but a bit harder); any other (e.g., cyclopentane, cyclohexane, etc) will not react
|By x on Tuesday, December 12, 2017 - 10:13 am: Edit|
Do we need to memorize substituents like vinyl, propagyl, etc?
|By x on Tuesday, December 12, 2017 - 10:42 am: Edit|
On quiz 2, question 2, if you hadn't specified that the substituents should be in the axial position, we could have placed CH3 and H groups in any arbitrary order, right?
|By Qq on Tuesday, December 12, 2017 - 11:04 am: Edit|
On exam1 question 1, could we have another structure where after the attack of the molecule to its own carbocation, we could do a hydride shift to a secondary carbon on its left followed by a methyl shift of one of the methyl groups on the left to the right and have two more new structures as well?
|By x on Tuesday, December 12, 2017 - 11:28 am: Edit|
Looking back at exam 1 question 5, if we want to asses the basicity by not being acidic therefore conjugate base being unstable (so draw the conjugate base of the compound and say the least stable one is the most basic one), it seems to not work. Is this a wrong approach for assessing basicity?
|By x on Tuesday, December 12, 2017 - 11:45 am: Edit|
So I have two questions about exam 1 #6 second structure:
1- Could we have used mcPBA to make an epoxide and then use magnesium bromo isopropane as a grignard reagent to make the product instead?
2- Also is the steric hindrance you mentioned because of the fact that it is a cyclic structure? I just don't see a lot of other things bumping into Nu, hindering it from attacking the bridge...
|By Sergei on Tuesday, December 12, 2017 - 11:50 am: Edit|
x on Tuesday, December 12, 2017 - 10:13 am:
yes (also allyl, benzyl)
x on Tuesday, December 12, 2017 - 10:42 am:
Yes, but that would have been not really good, as it would have created too many options for the Newman projection
Qq on Tuesday, December 12, 2017 - 11:04 am:
No, you are generating a tertiary carbocation (after the alkene attacks the carbocation), hence shifting to a less stable secondary would not be possible.
x on Tuesday, December 12, 2017 - 11:28 am:
it's a wrong approach for that problem.
The basicity for each of those compounds is different due to the difference of the specific electron pairs: non-bonding electrons, pi-electrons and sigma electrons.
Also, if the basicity being assessed - you should be considering the conjugate acid (not the conjugate base). in other words, protonate every compound/electron pair and see which is going to be less/more stable, and then relate that to the basicity of the original bases.
|By x on Tuesday, December 12, 2017 - 11:58 am: Edit|
Qq on Tuesday, December 12, 2017 - 11:04 am
I know that the carbocation would be less stable in a secondary position than in a tertiary position, but also, one of the products resulting form what I proposed, would be a tetra substituted alkene which is more stable than if we hadn't done the less favorable hydride shift. So I think my question is do we look at the stability of the carbocation only when assessing the major product, or do we look at the stability of the resulting alkene?
|By Sergei on Tuesday, December 12, 2017 - 12:00 pm: Edit|
Hi x on Tuesday, December 12, 2017 - 11:45 am:
1. Yes, but at that time we hadn't covered the epoxidation
2. there are 2 carbons on the mercurium ion: one is primary, the other is tertiary; on one C there are 2Hs, on the other there are 2Cs.
The minor product is given, since under these conditions, the reaction is under SN1-like control, hence more stable carbocation is the preferred pathway that would give the more substituted product.
However, at that time, this was the only route we could have taken to get to this product.
|By Sergei on Tuesday, December 12, 2017 - 12:09 pm: Edit|
Hi x on Tuesday, December 12, 2017 - 11:58 am:
the alkene is forming from the carbocation, i.e., the formation of the carbocation comes first.
If the carbocation cannot form, then it does not matter how stable the alkene would might have been.
Reactions proceed via the most stable intermediates. Also,recall that you cannot be forcing the reaction to go uphill in energy over and over again, with the idea that in the end it will form the a more stable product.
for carbocations, in general: whatever is possible to form will form initially if there are no options, followed by rearrangement - if that leads to more stable carbocation (not less or equally stable, and then to more stable)
If there are options - only the more stable one will form (and not less stable, followed by rearrangement to a more stable).
Rearrangements of the carbocations are chemical reactions, hence they still require Ea.
|By x on Tuesday, December 12, 2017 - 12:21 pm: Edit|
Can we ever make a phenol using benzene and water in acidic environment? Or is H+ never a good enough LA for aromatic compounds?
|By Sergei on Tuesday, December 12, 2017 - 12:27 pm: Edit|
No, that would not work
But you can propose a multistep synthesis that will feature diazonium salt or a sulfonic acid as one of the intermediates en route to phenol
|By x on Tuesday, December 12, 2017 - 12:31 pm: Edit|
bromobenzene or phenyl bromide?
|By laurenc on Tuesday, December 12, 2017 - 12:51 pm: Edit|
Hi Dr. Dzyuba,
In regards to carbonyl isomers, I know we said that generally ketones are the more favorable isomer when compared to enols and that enones are more favorable than ketone-alcohols. What about beta-ketones vs forming a ketone next to a enol?
My thinking is that since ketones are more stable than enols then the di-ketone isomer is more favorable but by forming ketone-enols beta to each other, it is more resonance/conjugation. Which one will be the major product when form this through like Claisen's reaction for example?
Thanks for your help!
|By x on Tuesday, December 12, 2017 - 12:52 pm: Edit|
Going back to this question:
By x on Tuesday, December 12, 2017 - 08:48 am
Whenever we are trying to compare stability, we look at how many electronegative atoms for instance are sharing that neg charge. Does that count the ones that we can't do with one arrow pushing process as well?
So like comparing H2SO4 and HNO3, which of these would be correct?
a) negative charge is distributed between 3 oxygens in H2SO4 and 2 oxygens in HNO3.
b) negative charge is distributed between 4 oxygens in H2SO4 and 3 oxygens in HNO3.
|By x on Tuesday, December 12, 2017 - 01:06 pm: Edit|
As far as we've learned, we never attack a positive atom like N, S, etc, with the exception of electrophilic aromatic substitution, right?
|By x on Tuesday, December 12, 2017 - 01:59 pm: Edit|
I have some questions regarding this table of benzene reactivity in the presence of different substituents in the textbook:
1- Why does F as a substituent make the ring more reactive than other halogens? Isn't a stronger EWG and so it would take the negativity out of the ring making the ring less reactive than let's say Br does?
2- Also, I know that ortho position is bad because of hindrance and repulsion, but if the EDG has +I, then wouldn't the ortho position be more nucleophilic and therefore a better position to attack?
|By x on Tuesday, December 12, 2017 - 02:00 pm: Edit|
Could we use either clemmenson or wolf-kishner for reduction of things, or are they used for specific things?
|By x on Tuesday, December 12, 2017 - 02:03 pm: Edit|
It's never possible to have 1, 3, 5 tribromobenzene right? Because once we put the first Br, the other ones only get added to ortho and para positions and never to meta positions, correct?
|By x on Tuesday, December 12, 2017 - 02:12 pm: Edit|
Actually never mind the last question. We can take other routes of course what was I thinking
|By x on Tuesday, December 12, 2017 - 03:13 pm: Edit|
Why doesn't Cl- come back to attack the carbon in TsCl reaction with an alcohol, making an alkyl chloride?
|By x on Tuesday, December 12, 2017 - 03:31 pm: Edit|
So two enantiomers do have the same physical properties but two diasteromers don't?
|By x on Tuesday, December 12, 2017 - 03:41 pm: Edit|
In E2, should the H and LG be BOTH anti AND axial?
|By Sergei on Tuesday, December 12, 2017 - 03:57 pm: Edit|
x on Tuesday, December 12, 2017 - 12:31 pm:
Bromobenzene is preferred (IUPAC).
laurenc on Tuesday, December 12, 2017 - 12:51 pm:
beta-diketones are a bit different from beta-keto-esters.
With beta-diketones enol is favored, due to formation of an extra H-bond, 6-membered ring.
With keto-esters the "enol" form is not a much favored (esters are not great H-bonding groups, worse than ketones), but enol form could be proposed (and would be taken as a correct answer).
x on Tuesday, December 12, 2017 - 12:52 pm:
If you are trying to compare the conjugate bases, make sure you have that negative charge, i.e., proper comparison is between NO3– and HSO4–
Write all charges, and see over how many atoms the negative charge could be delocalized.
In the case of NO3–, the "–" can only be delocalized over 2 oxygens
x on Tuesday, December 12, 2017 - 01:06 pm:
But NO2+ is not an exception; if you draw resonance structures and/or arrow-pushing mechanism, there is no violation of basic rules (electrons have somewhere to go, bonding is ok, etc)
x on Tuesday, December 12, 2017 - 01:59 pm:
1. F - +R-effect is about the same as –I-effect. For other halogens, –I-effect is a bit stronger than the +R-effect for reasons that we have discussed in the lecture (2p-2p overlap vs 2p-3p/4p/5p)
Again, once comparing atoms/groups from different periods, EN cannot be considered as the primary factor, all components should be considered.
2. +I or +R effect, it's the steric bulk that is hindering the approach of the electrophile. Me-group only has +I effect but it's still bulky (it occupies quite a bit of space, tetrahedral, 3D arrangement)
x on Tuesday, December 12, 2017 - 02:00 pm:
Either one of them could be used for the reduction of the C=O to CH2
x on Tuesday, December 12, 2017 - 02:03 pm:
You cannot make 1,3,5-tribromobenzene from bromobenzene.
You have to go through a somewhat longer route (i.e., diazonium salt)
x on Tuesday, December 12, 2017 - 02:12 pm:
|By Sergei on Tuesday, December 12, 2017 - 04:05 pm: Edit|
x on Tuesday, December 12, 2017 - 03:13 pm:
There is a base (Et3N, pyridine) that is converted to the conjugate acid, and Cl– is the counter ion, the resulting salt precipitates out.
The base deprotonates OH+ from the intermediate, hence the LG is going to be O–, and Cl– is not good enough of a Nu for that.
In the case of SOCl2, the LG is neutral, hence Cl– works as Nu.
x on Tuesday, December 12, 2017 - 03:31 pm:
x on Tuesday, December 12, 2017 - 03:41 pm:
beta-H and LG must be anti
In cyclic compounds (cyclohexane) they happen to be axial
|By NervousWreck on Tuesday, December 12, 2017 - 06:30 pm: Edit|
I'm just here to say that I'm very very very scared for tomorrow. I'm sure there are others out there that are just as nervous for tomorrow as I am so I just wanted to say it and wish everyone good luck! We can do it guys!
Dr. Dyzuba, thanks for being a great professor this semester! This has been one heck of a class. I hope to pass so I don't have to retake it... no offense!
|By aa on Tuesday, December 12, 2017 - 06:40 pm: Edit|
is there any way to make a grignard add only once to a carbonyl or is it always going to add multiple times?
|By x on Tuesday, December 12, 2017 - 06:56 pm: Edit|
So if we add HNO3 and H2SO4 to a benzene, does that cover the 1, 3, 5, positions with NO2 or does it stop after adding one?
|By x on Tuesday, December 12, 2017 - 07:26 pm: Edit|
If we are in an acidic environment but we want to do SN2, do we still attack the less hindered site of epoxide?
|By x on Tuesday, December 12, 2017 - 08:11 pm: Edit|
Have we learned how to make secondary and tertiary amines?
|By Sergei on Wednesday, December 13, 2017 - 05:08 am: Edit|
NervousWreck on Tuesday, December 12, 2017 - 06:30 pm:
Thank you and good luck!
aa on Tuesday, December 12, 2017 - 06:40 pm:
Grignard addition to ketones/aldehydes and nitriles happen only once (although for different reasons).
in the case of the ester: the intermediate is more reactive than the starting materials, hence the second addition.
x on Tuesday, December 12, 2017 - 06:56 pm:
You might expect/propose dinitration at most; the 3rd nitration is not very likely, since 2 strong EWGs will deactivate the ring towards the electrophilic attack.
Although, trinitrobenzene is possible to make via nitration, it would typically require some EDG which would be subsequently removed.
|By Sergei on Wednesday, December 13, 2017 - 05:09 am: Edit|
x on Tuesday, December 12, 2017 - 07:26 pm:
x on Tuesday, December 12, 2017 - 08:11 pm:
|By HC on Wednesday, December 13, 2017 - 11:04 am: Edit|
Hi Dr. Dzyuba,
Do you send out an email with the grades once the tests are graded? Or rather, how do we know what we received in the class?
|By Done on Wednesday, December 13, 2017 - 01:12 pm: Edit|
Follow up question from the previous one, will we get to know what we made on the final exam? Or will we just find out our final grade for the class?
|By Sergei on Thursday, December 14, 2017 - 08:42 am: Edit|
Hi HC and Done,
As soon as I'm done grading (not going to be today), I'll e-mail the results, post the grades distributions here and submit the grades to the registrar
|By RIPochemI on Tuesday, January 30, 2018 - 02:04 pm: Edit|
Hi Dr. Dzyuba,
We miss you.
|By anon on Sunday, February 18, 2018 - 10:09 pm: Edit|
Hi Dr. Dzyuba,
I was wondering what was your reasoning behind including people who retook the class into our class curve? Some of us are still very much confused about that.
|By Sergei on Monday, February 19, 2018 - 10:23 am: Edit|
There was never a curve.
The grading was done based on gaps in the scores (as we discussed many time in class, on this board and it is also in the syllabus).
Everyone is graded based on their performance in the current class, not anything else.