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!!!