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?