TCU Organic Chemistry: Sergei Dzyuba

Chemistry Discussion Board: TCU Organic Chemistry: Sergei Dzyuba
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By Sergei on Monday, August 01, 2022 - 07:54 am:  Edit

GREETINGS!

Welcome to O-Chem I lecture (CHEM-30123)!
Please post all O-Chem related questions here.

Have a great semester!

By anonymous on Tuesday, August 23, 2022 - 09:47 am:  Edit

In Monday's lecture when you drew the electron orbital diagrams for C, "C+" and "C-", how did you determine how many bonds the atom could form?

By Sergei on Tuesday, August 23, 2022 - 10:38 am:  Edit

Hi anonymous,

neutral carbon has 4 orbitals that matter (i.e., out shell,2nd level): 1 of 2s and 3 of 2p, and there are 4 electrons total on those 4 orbitals;
2s2 2p2

If we are removing one electron, neutral carbon will become C+.
One empty orbital has to be reserved to indicate/specify that there is a "+" charge. Thus, 3 orbitals remain (also, at this point it is not important which ones, just the fact that one is reserved for "+", and 3 are left).
Because 3 orbitals lefts, and now we have 3 electrons (because one was removed) - the maximum number of bonds that could be drawn to "C+" is 3.
Recall that to make a covalent bond, we need two electrons per orbital, thus, C will provide 3 electrons and 3 orbitals (again not important which orbitals and electrons exactly, at this time) to form a bond to something else.


For "C–" the reasoning is similar, except we are going to add an electron to neutral C. Adding an electron implies that there will be an electron pair that does not participate in the bond formation, i.e., we have to have/"reserve" an orbital with 2 electrons on it to indicate that we have "C–" (similarly we had an empty orbital to indicate "C+").
Once we have one orbital and 2 electrons reserved for "C–" designation, how many orbitals will be left? 3, right? since the total is 4, and one is used to show that it's "C–".
On those 3 orbitals, we have 3 electrons (again, at this time, it's not important which electrons are on which orbitals). Therefore, with 3 orbitals and 3 electrons, how many bonds can be formed to "C–"? only 3, right?

To sum up:
neutral carbon can only have 4 bonds drawn to it (not 5, not 6, not 3, but only 4).

If we put a charge on carbon, we have to adjust the number of bonds.
Based on the above description, C with "+" charge should have 3 bonds drawn to it, and C with "–" charge should have 3 bonds drawn to it.
The opposite is also true: if you see/draw a carbon with 3 bonds to it, a charge is needed.

Hope this clarifies the issues; but if anything, please let me know.

By NotTheBees! on Thursday, September 01, 2022 - 01:24 pm:  Edit

If we are answering what the bond angle is or asked to label bond angles, would it be better to use 109.5 degrees or just 109 degrees for tetrahedral angles? Do you have a preference since those two are so close?

By Sergei on Thursday, September 01, 2022 - 04:50 pm:  Edit

Hi NotTheBees!

either 109.5 or 109 will be taken as the correct angle / correct answer

As I mentioned in the lectures, anything reasonably close to 109 (for sp3; or 120 for sp2; and reasonable = a few degrees) is good enough.

Also, my personal preferences are irrelevant; it's about proposing reasonable and correct things in this course.

By anonymous on Saturday, September 03, 2022 - 07:56 am:  Edit

In Friday's lecture you mentioned that the trends we talked about were for acids, but for bases the trends are the same, they just reverse. Can you elaborate on what this means and how we use it to determine relative base strength?

By Sergei on Saturday, September 03, 2022 - 05:28 pm:  Edit

Hi anonymous,

Recall how we determine the strength of an acid: remove a proton, and evaluate the stability of the conjugate base (delocalization of the negative charge).

For the base, we have to do the opposite, since base is something that should pick up a proton: so lets add a proton (protonate the base), and evaluate the stability of the conjugate acid (i.e., delocalization of the positive charge).

Lets use one of the examples from Friday:
we looked at acidity of R3C–H vs R2N–H vs RO–H vs F–H, and upon removal of the proton, F– was more stable (charge was better held by electronegative F) than R3C–.
So if we look at this situation in reverse:
R3C– should be a stronger base than F–, right?
C is less EN than F, hence it would like to get rid of the charge more than F (in other words, C– cannot hold on to the charge, and thus C– will be less stable/more reactive), thus R3C– is more basic than F–.
Further: R3C–H is the conjugate acid of R3C–, whereas F–H is the conjugate acid of F–.
R3C–H is a weaker acid than F-H

In addition, you might recall from Gen Chem that acidity and basicity are related to each other "numerically"/equation-wise, such that if one goes up the other must go down:
pH + pOH = 14
Ka * Kb = Kw
bc pKa = –logKa, and pKb = –logKb
pKa + pKb = 14 (ish)

By NotTheBees on Wednesday, September 14, 2022 - 01:06 pm:  Edit

For the common names section of the textbook, do we need to know all the common names for the common compounds or will it be acceptable to use the IUPAC names? I suppose they could be used interchangeably, but wanted to ask how we should answer a question like that on a quiz/exam.

By Sergei on Wednesday, September 14, 2022 - 01:11 pm:  Edit

Hi NotTheBees,

As long as name only correlates to one structure, either one should be fine.
Besides, some trivial names are "IUPAC names".

By anon on Sunday, September 18, 2022 - 02:15 am:  Edit

Is there any resource that provides the answers to Mcmurry's book problems?

By anon on Sunday, September 18, 2022 - 11:53 am:  Edit

The schedule says that the exam will cover chapters 1-4 and 6. Should we be reviewing concepts that you have not covered in class yet in preparation for the test? Or will we only be tested on concepts covered in lecture?
Thank you!

By Sergei on Sunday, September 18, 2022 - 02:01 pm:  Edit

Hi,

anon, Sunday, 2.15am:
there is a Study Guide and Solution Manual that goes together with he textbook.
The books store used to get it together with the textbook.

Also, please feel free to post a question here (preferred) or e-mail me or stop by and we could address all the issues with the questions.



Anon, Sunday, 11.53am:
Whatever material will be covered by the end of Wednesday lecture will go onto the exam1.
Whatever is not covered, but if it was planned, would be moved to exam2.

By anonymous on Wednesday, September 21, 2022 - 10:58 am:  Edit

are there any office hours tomorrow?

By sharingisbeneficial on Wednesday, September 21, 2022 - 11:11 am:  Edit

Can you please specify sections in the book that we have already covered?

By anonymous on Wednesday, September 21, 2022 - 01:05 pm:  Edit

What is the most important factor on determining acidity?

By Sergei on Wednesday, September 21, 2022 - 02:59 pm:  Edit

HI,

anonymous, 10.58am:
there are no scheduled office hours tomorrow.
We have a review in LH1 from 6pm to 8pm


sharingisbeneficial:
chapter 1 - all sections
chapter 2 - all sections
chapter 3 - all sections (in 3.5, we only looked at addition, elimination, rearrangements)
chapter 4 - all sections; EXCEPT: 4.6/oxidation of alkenes; 4.7/radicals; in 4.11 - all, except formation of acetylide (although we did cover the acidity of the terminal CH in alkynes, and C– is a Nu, but the reactivity towards specific electrophiles was not covered; we will talk about the reactivity of the spC– later in the course).


anonymous, 1.05pm:
stability/delocalization of the charge in the conjugate base.
Charge delocalization is fundamental.

By anonymous on Wednesday, September 21, 2022 - 06:48 pm:  Edit

Can you please explain again about allylic structure? I tried look for the information in the textbook, but I don't see it talk about it specifically

By anonymous on Wednesday, September 21, 2022 - 08:10 pm:  Edit

Are all LA/LB Bronsted Acid /Base as well?

By Sergei on Wednesday, September 21, 2022 - 08:35 pm:  Edit

Hi,

anonymous, 6.48pm:
textbook: look at the section which talks about HBr and Br2 addition to conjugated dienes, e.g., butadiene.

Essentially, allylic carbocation (that's the one we looked at thus far; if instead of C+ there is an OH group - it's an allylic alcohol, if Br - allyl bromide, etc), is a carbocation, where C+ is a sigma-bond away from the C=C bond, which allows for delocalization of the C+.
Therefore, in the lecture, we labeled such carbocations not as primary, not as secondary, but allylic - a special kind, and in stability it is being very close to a tertiary carbocation.

Similarly, allylic carboanion, where C- is going to be a sigma bond away from C=C, and the resonance delocalization of the negative charge, will have an extra stability.


anonymous, 8.10pm:
No.
The only Bronsted acid is H+. This Bronsted acid is also a Lewis acid.
However, Zn2+ is a Lewis acid, but Zn2+ is not a Bronsted acid.

By Anonymous on Wednesday, September 21, 2022 - 08:49 pm:  Edit

Hi Dr. Dzyuba,

Can Newman projections be used for inorganic molecules? E.g., Br2, SiO2.

By anonymous on Wednesday, September 21, 2022 - 08:49 pm:  Edit

I got it, donate proton means donate H+, accept proton is accept H+. Does it also means LB can also be Bronsted Base?

By Sergei on Wednesday, September 21, 2022 - 09:05 pm:  Edit

Hi,

Anonymous, 8.49pm:
in principle - yes.
But, what information would we expect to get from drawing a Newman projection of Br2, for example.
In fact, we would not be even sure, whether we are looking at Br2, if the Newman projection is drawn looking through Br-Br bond, since BrH, or BrCl, etc. would have the same Newman projection.

Recall that we want to use the Newman projection to get some information about stability of the molecule, potential reactivity, etc.
With that said, we should be able to draw a Newman projection of any molecule.


anonymous, 8.49pm:
yes, proton is H+ , and it's the only thing that is getting exchanged in Bronsted theory of acids and bases.

LB is an electron pair, and in principle any electron pair could be protonated (i.e., H+ could be added).

By anonymous on Wednesday, September 21, 2022 - 09:30 pm:  Edit

Can I ask why non conjugate diene don't have resonance structure?

By Sergei on Wednesday, September 21, 2022 - 10:17 pm:  Edit

Hi anonymous,

conjugation is another word for resonance, delocalization of the charge. For the delocalization of the electron pair to take place, the C=C bonds must be within a sigma bond from each other. This assures, that the pi-system of one C=C can overlap with another C=C (in other words: one should be able to draw a resonance structure to show that there a conjugation).

In the non-conjugated dienes, two C=C groups are not communicating, i.e., they are more than 1 sigma bond away. If the pi-systems cannot overlap, there is no sharing of the electrons, and if there is no sharing, then there is no resonance.

By Anonymous on Thursday, September 22, 2022 - 08:31 am:  Edit

Why does the chair conformation for cis-1.2-dichlorohexane have to have one axial and one equatorial Cl group? Why can't they both be axial?

By Anonymous on Thursday, September 22, 2022 - 08:41 am:  Edit

When we are drawing intermediates as a Newman projection, how should we indicate partial charge on a carbon?

By Anonymous on Thursday, September 22, 2022 - 09:50 am:  Edit

For the last question of the last quiz (#2), whenever you tell us if the compound could act as LA/LB or BA/BB, do we always have to consider the compound's resonance structures?

By Sergei on Thursday, September 22, 2022 - 11:21 am:  Edit

Hi,

Anonymous, 8.31am:
for 1,2-positions, one axial must go up, the other must go down, i.e., axial substituents will be pointing in different directions. Thus, they would be trans, not cis.

Whether the substituent is Cl or Me (like the one we looked in the lecture) is irrelevant.


Anonymous, 8.41am:
same as the chemical structure, whichever carbon gets a charge/partial charge, put that charge next to the atom in the Newman projection.
If it's a bromonium ion, plus-charge goes in the middle of the 3-membered ring, hence put the plus-charge in the middle of the 3-membered ring in the Newman projection.



Anonymous, 9.50am:
Yes, the resonance structure, in that particular case, tells you that C of the CN can have a positive charge, hence it's a LA/electrophilie.
Since the question states that under certain/specific conditions, we can assume that this conditions could include a negative charge. Thus, a negative charge can attack a positive charge on C of the CN.

By Anonymous on Thursday, September 22, 2022 - 04:44 pm:  Edit

Do we need explanations for the questions, even though most of them are not required in the last couple quizzes. And if we do, will it make us be able to get partial credit even the result is wrong?

By Sergei on Thursday, September 22, 2022 - 08:08 pm:  Edit

Hi Anonymous,

You only need to answer what the question is asking.
If you are providing an additional explanation, only reasonable and relevant information would be considered for partial credit.

For example, if the question is asking to provide the product based on given reagents/reactants, and you know what the product is - you can simply draw the product, and you don't need to provide the mechanism. If you write the mechanism and have the correct product - you would not get additional points, just because you wrote the mechanism.

But, if you don't know what the product should be, then you are better of writing/proposing a mechanism, rather than guessing, and in that case, I could be considering the steps of the mechanism for the partial credit.

By Anonyomus on Monday, September 26, 2022 - 08:17 pm:  Edit

Why does the acidity increase if the conjugate base is aromatic?

By anonymous on Monday, September 26, 2022 - 08:41 pm:  Edit

In today's lecture you said that benzyl is unique like allylic structures. What did you mean by this?

By Sergei on Monday, September 26, 2022 - 11:52 pm:  Edit

Hi,

Anonymous, 8.17pm
Aromaticity assures that all electrons are in conjugation and all electrons are moving in a circle; thus, aromatic conjugate base has a very delocalized negative charge.
Delocalization of the charge leads to increased stability. If the stability of the conjugate base increases, the acidity of the corresponding acid increases (and pKa goes down).


anonymous, 8.41pm:
In the case of benzylic group: CH2 is one sigma bond away from the pi-system, I.e., C=C bond (in allylic the situation is the same).

Recall that in allylic carbocation there is a conjugation, and allylic carbocation is not primary/secondary, but rather allylic (ie unique, in a class of its own)
Benzylic carbocation would behave in the same way (in regard to charge delocalization).

In general, benzylic and allylic compounds would exhibit similar reactivities.

By annon on Tuesday, September 27, 2022 - 11:04 am:  Edit

Hello Dr.Dzuyba when will our tests be ready for pickup ?

By pie on Tuesday, September 27, 2022 - 11:05 am:  Edit

Will we not have a quiz tomorrow ?

By Sergei on Tuesday, September 27, 2022 - 01:40 pm:  Edit

Hi,

annon, 11.04am:
Tentatively: Wed/Thu/Fri.
In reality - as soon as the last person takes it, I'll put the note on D2L and the exams will be ready for pickup.

pie, 11.05am:
Quizzes are unannounced; confirming that we would not be having a quiz was only good for week 1 of the semester.

By Idek on Tuesday, September 27, 2022 - 01:53 pm:  Edit

Bromobenzene upon reaction with Br2/Fe gives 1,4dibromobenzene as the major product. Why is not 1,2 dibromobenzene? Can you please explain this in more details in class as well?

By Sergei on Tuesday, September 27, 2022 - 02:12 pm:  Edit

Hi Idek,

there is a repulsion between two Br s in the case of 1,2-dibromide, which is not present in 1,4-dibromobenzene.

When the positively charged intermediate (C+ of the PhBr) is reacting with Br+, C with "negative charge" in the ortho position is more statically hindered than C in the para-position.
Nucleophile-electrophile interactions are quite sensitive to sterics (i.e., sizes of + and –), thus, nucleophile would prefer to react with electrophile with the least sterics interference.

By BB on Wednesday, September 28, 2022 - 08:14 am:  Edit

Hi Dr. Dzuyba,

When PhBr react with Br2/FeBr3, is it possible to form 1,3-bromobenzene?

By BB on Wednesday, September 28, 2022 - 08:17 am:  Edit

And why PhBr is less Nu than C6H6?

By Sergei on Wednesday, September 28, 2022 - 08:58 am:  Edit

Hi BB,

8.14am:
No, consider the resonance structures (we looked at several ways of presenting where the electron rich, i.e., Nu-centers will be).
Those electron rich centers are on carbons which are ortho and para to C-Br. Thus, only o- and p- dibromobenzenes are possible.


8.17am:
Br is an EWG: EN(Br) > EN (C).
Although Br also has +R effect (hence the electron density is concentrated in o/p positions), the +R effect is not very strong, because Br is in the 4th period, and C is in the second. Hence, the orbital overlap is not great.

By BB on Wednesday, September 28, 2022 - 09:45 pm:  Edit

So the +R effect of Br is not very strong because +R(C) is stronger and more ability to withdraw to electrons toward itself?

By Sergei on Wednesday, September 28, 2022 - 09:56 pm:  Edit

Hi BB,

No. +R effect of Br is not strong bc it’s not a very good orbital overlap between C (2nd period) and Br (4th period).

If C were to have a +R effect then Br should be getting a negative charge, and will have two bonds drawn to it. - this will be violating fundamental principles that we have been advocating for from the start of the semester.

By BB on Thursday, September 29, 2022 - 10:00 am:  Edit

That is more make a lot of sense! Thank you Dr. Dzyuba!

By anonymous on Thursday, September 29, 2022 - 02:10 pm:  Edit

For Toluene, is it because methyl group is EDG so it moves the electron density toward the ring so that's why we have (-) charges in o/p positions?

By anonymous on Thursday, September 29, 2022 - 04:13 pm:  Edit

Please ignore the question above Dr. Dzyuba. I just figure out because one is secondary carbocation and the other is tertiary carbocation

By Sergei on Thursday, September 29, 2022 - 04:49 pm:  Edit

Hi,

anonymous, 2.10pm:
Me group has +I-effect, hence it's an EDG. Thus, Me-group is an activating group, i.e., it makes the aromatic ring more nucleophilic (more than benzene) in EAS.

When Me-group pushes the electron density onto the aromatic ring, not every carbon of the aromatic ring is getting the same amount of electron density. In order to figure out which carbon of the aromatic ring will be more nucleophilic, we have been using resonance forms.

The split resonance structures showed that carbons that will have "excess" of electron density (as a result +I-effect of the methyl group) are those in ortho/para positions to the Me-group.

That's why bromination of toluene gives o-bromo- and p-bromotoluenes.


anonymous, 4.13pm:
after reading this posts - it cannot be ignored.

There are NO carbocations in toluene's resonance structures!
Please recall that double sided arrows are only for resonance structures; these are imaginary structures, that are used to identify the reactive centers; essentially locate +/– charges so that it would be possible to predict the outcome of chemical reactions, which are taking place by – reacting with +.

Carbocations are real compounds/intermediates. They are formed as a result of specific chemical reactions (the bonds actually have to be broken and formed, and the atoms must move), under certain set of conditions. In general, reaction arrow will be one sided, and equilibrium processes (forward/reverse reactions) will be indicated by two one-sided arrow pointing into the opposite directions.

By jack on Monday, October 03, 2022 - 09:05 pm:  Edit

Hello Dr. Dzyuba,

I have a few questions:

1) Nitration, sulfonation, alkylation, acylation, bromination, and etc. are all forms of EAS, right?

2) When determining substituent placement on aromatics, is the rule (just to clarify) meta substituted for EWG, and para/ortho (depending on if there's other substituent(s)) substituted for EDG? Is para always more favorable than ortho, if steric hindrance isn't an issue?

3) Is resonance what we use to determine substituent placement, like the justification?

4) When you were showing functional group interconversion with an aromatic compound with benzylic hydrogens gaining a CO2H with use of KMnO4 as a reagent, what is that process called?

5) Could you explain again what happens when you add an acid like sulfuric or phosphoric (one where the CB acts as a base, not a nu) to an alkene?

By Sergei on Tuesday, October 04, 2022 - 05:28 am:  Edit

Hi jack,

1) yes; these are examples of EAS. The only difference between those reactions is the nature of E+, subsequent mechanism of the reaction with the aromatic ring is the same.

2) yes; EWG will direct electron density on the aromatic to meta-C, and EDG will direct the electron density to ortho/para-C.
However, we are trying to understand why certain bonds form, why reaction happen, rather than just follow the rules. That's the reason for drawing the resonance structures.

Sterics: in general, carbon with smaller substituents is a better Nu, than carbon with bigger substituents. In other words, less statically hindered position is preferred, and it does not matter whether it's ortho or para.
In the examples that we've looked there was more steric interference in the ortho, and less in the para, that's why the major product was para.

3) yes; but the resonance structures are not just justification, and we use them to predict the outcome of the chemical reactions.

4) oxidation

5) H2SO4 and H3PO4 are acids (H+ is a Bronsted, and also a Lewis acid). Alkene is a Lewis base.
Reaction between H+ and Alkene leads to formation of a carbocation.
Carbocations will do 3 things: react with Nu (of one is available), react with a base (i.e., eliminate), and rearrange (but only to a more stable carbocation).
In the case of H2SO4 (and H3PO4), the conjugate base (HSO4– and H2PO4–) is only a base (i.e., these are not nucleophilic; to be more precise - they prefer to react as bases much more than as Nu), thus they will deprotonate H (pick up H+) from the carbon that is one sigma bond away from the carbocation, and the resulting electron pair on C will form an alkene (formation of the most stable alkene should be assumed). Important: deprotonation and the C=C bond formation is a concerted process - both steps happen at the same time, i.e., as HSO4– picks up a proton, C=C forms (there is no formation of C– at any time, since carbanions don't exist in acidic media!!).
Also, keep in mind that this is a reversible process, and since we are concerned with most thermodynamically stable product (unless instructed otherwise), most stable alkene will be the major product.

By anonymous on Wednesday, October 05, 2022 - 07:29 am:  Edit

Will we have the quiz on Wed Dr. Dzyuba?

By anonymous on Wednesday, October 05, 2022 - 07:31 am:  Edit

Can you explain gain why F.C. Acylation have no rearrangement but Alkylation does?

By Sergei on Wednesday, October 05, 2022 - 07:54 am:  Edit

Hi,

anonymous, 7.29am:
no changes have been made to the syllabus, so quizzes are still going to be announced.


anonymous, 7.31am:
F.-C. alkylation – the reaction proceeds via a carbocation (this is the electrophile that reacts with nucleophilic aromatic compound). Carbocations react, eliminate and rearrange (always, if the rearrangement leads to a more stable carbocation).

F.-C. acylation – the reaction proceeds via asylum ion RC(+)=O.
This type of C+/carbocation does not rearrange, bc it is already resonance stabilized (recall the resonance structures we drew in the lecture; even though + charge is on O in one of them, it's still a resonance driven delocalization). This is probably sufficient to rule out rearrangement.
Another way to look at it, is to say: if a rearrangement does happen, what would be the stability of the resulting/rearranged carbocation (keep it mind that rearranged carbocation must be more stable/charge must be more delocalized for the rearrangement to take place).
If you draw a rearranged carbocation from an acylum ion (lets say RCH2C(+)=O), you will see that C+ is now located next to C=O (RCH(+)CHO), which is an electron withdrawing group, and there is no additional stabilization from resonance; this means that rearranged carbocation is not as stable as the original one. Thus, rearrangement should not take place.
Hope the description makes sense, if anything, let me know.

By anonymous on Wednesday, October 05, 2022 - 08:07 am:  Edit

Thank you Dr. Dzyuba,
And how we know O+ doesn't have vacant orbital but S+ does?

By Sergei on Wednesday, October 05, 2022 - 05:33 pm:  Edit

Hi anonymous,

O is in the 2nd period, S is in the 3rd.
No only s and p orbital in 2nd, but s, p, and d in the 3rd, and d-orbitals are not full for S, hence more than 4 bonds is ok.

In fact in SO3 - 3 S=O bonds, i.e, 6 bonds, once it's protonated, and we move one e-pair onto O, S is 5 bonded, hence + charge

By anonymous on Monday, October 10, 2022 - 07:42 am:  Edit

Are the quizzes and the exam available for pick up Dr. Dzyuba?

By Sergei on Monday, October 10, 2022 - 07:49 am:  Edit

Hi anonymous,

Yes.
In general, once the copy of the quiz/exam, the key, and the stats are posted on D2L, quizzes/exams are available for pick-up.

By anonymous on Monday, October 10, 2022 - 08:10 am:  Edit

do stats currently include the quiz 4 or not Dr. Dzyuba?

By Sergei on Monday, October 10, 2022 - 08:19 am:  Edit

Hi anonymous,

Stats are posted for each quiz, exam.

As stated on D2L, current tentative grade distribution includes Q1-3 and E1 (and as such Q4 is not included in those stats).

Tentative grade distributions, as we discussed during the 1st lecture, will be done after each exam (not after each quiz).

By anonymous on Thursday, October 13, 2022 - 06:15 pm:  Edit

Can you explain again why N3- is nu only, but not base?

By Sergei on Thursday, October 13, 2022 - 06:56 pm:  Edit

Hi anonymous,

We will cover the specifics of how to assess Nu vs B tomorrow.

But based on what we know at this point (since we've been covering O-Chem from the fundamental standpoint, we should be able to provide a reasonable explanation to almost any problem):
N3– is small (linear, or almost linear anion), hence it should not have a any difficulty getting to the electrophilic center, hence it's a Nu.

If it were a base, then the conjugate acid would be HN3.
As a base, N3– would be deprotonating betaCH, i.e., C-H (sp3 CH!). When you compare HN3 and that CH2, HN3 must be a stronger acid, and you cannot make a stronger acid from a weaker one (and you can also look at it from the perspective of the N3– vs C–, in the extreme case, when H is removed; and N– is a weaker based than C–)

using numbers as last resort: if you look for pKa of HN3, it's probably around 4.5, i.e., close to acetic acid; obviously acetic acid is more acidic than beta-CH of an alkyl halide (pKas of those sp3 CH are likely to be in the order of 30-50 units, the exact number is not important, bc it's going to be significantly higher than 4-5).

Thus, N3– is not a great base for E2, yet it's OK as a Nu for SN2, hence - N3– is a Nu only.

By anonymous on Thursday, October 13, 2022 - 07:44 pm:  Edit

So is it something with MeS- Dr. Dzyuba? because MeSH is strong acid so MeS- is more stable and is an weak base?

And MeO- is the B/Nu while t-buO- is Base only is that because t-bu is much larger than Me so it push more e density toward O and lead to larger EN on Oxygen so decrease nucleophilic?

By Sergei on Friday, October 14, 2022 - 07:44 am:  Edit

Hi anonymous,

Yes.
Except, MeSH is a stronger acid (not just strong; it's all relative).
This is a bit nuanced situation, which we will address today.

tBuO– vs MeO– will be covered today, but briefly:
tBu has a bigger +I-effect than Me, hence negative charge (not EN!) is more localized on O of the tBuO– than MeO– (however, if you recall previous evaluations of I-effects, e.g., in acids, the pKa difference between Pr, Et, and Me containing acids wasn't huge).
Thus, the electronics alone cannot be used to explain the drastic difference between tBuO– and MeO–; which means that we need to use sterics.
tBu is much much bigger than Me, hence it's much harder for tBuO– to approach the electrophilic C than H. That's why tBuO– is a base.

By Anonymous on Sunday, October 16, 2022 - 01:17 pm:  Edit

In class on Friday, you said that in NaBH3, the hydride would be soft because boron is soft. In NaH, hydride would be hard because Na+ is hard. Why is B in NaBH3 not hard, even though it is also connected to sodium? In other words, when is the hardness/softness of an atom determined by the atom it is connected to and when is the hardness/softness determined by its location on the periodic table/elemental properties?

By Sergei on Sunday, October 16, 2022 - 02:18 pm:  Edit

Hi Anonymous,

NaBH4 (Na+ and BF4–).

B is not at the edge of the periodic table, it's closer to the middle/carbon. hence B is softer than Na+.

In NaH, H is "connected" to Na, i.e., there is nothing else, only Na and H, hence Na+ is the neighbor of H–; that's why H– in NaH is hard, and hence NaH is the base only.

In NaBH4, H– is not "connected" directly to Na+, H– is a neighbor of B. Bc B is softer than Na, H– in NaBH4 is softer, and therefore, H– in NaBH4 is a Nu (but it's also a base, if there is an acidic H, e.g., OH, NH, SH; bc the conjugate acid will be H2, and the conjugate base will be O–, N–, and S–, all are EN, and can handle negative charge).

By Anonymous on Sunday, October 16, 2022 - 06:54 pm:  Edit

Can you explain again why C+ more softer than H+? is the softness increase as we move from right to left?

By Sergei on Sunday, October 16, 2022 - 07:11 pm:  Edit

Hi Anonymous,

Recall that softness is related to charge delocalization (and also softness is more related to electrophlicity/nucleophlicity).

In H+ charge is localized (there is nothing that could delocalize that positive charge).
In C+ charge is more delocalized, because there are other groups on C, and they could donate something to delocalize the positive charge.

Right/left of the periodic table are typically hard(er), middle is soft(er).
Softness is increasing by going down in the periodic table.

By Anonymous on Monday, October 17, 2022 - 07:41 am:  Edit

Is there any relationship between nucleophilic and good/bad leaving group? or they are distinct concept?

By Sergei on Monday, October 17, 2022 - 07:56 am:  Edit

Hi Anonymous,

It depends...

On one hand, these are different species, bc one is replacing the other: Nu should have a more concentrated/localized charge, and the leaving group should have a more delocalized charge, such that when it leaves, it's stabilized.

On the other hand, Nu/LG are related, bc everything relates to charge delocalization. A LG could also be a Nu (e.g., I–: it's delocalized charge, hence soft, and therefore a good Nu; but it's also a good leaving group, bc it's a delocalized charge). If this is the case, then one should expect a reversible reaction.

Also keep in mind that it's relative, bad/good leaving group is a relative concept. What is a good LG in one reaction could be a bad LG in the other.

By BB on Thursday, October 20, 2022 - 12:49 pm:  Edit

For the Finkelstein rxn, is it possible for reaction between alkyl fluoride (or alkyl bromide) and NAI happen?

By Sergei on Thursday, October 20, 2022 - 02:53 pm:  Edit

Hi BB,

Regretfully, no.
Finkelstein is still an SN2 reaction, and as such it should follow the rules of the SN2, which require the presence of a good leaving group.
F– is localized charge, compared to Cl–, Br–, I–; hence it's not viable LG in SN2.

By anonymous on Monday, October 24, 2022 - 11:33 am:  Edit

Can a carbocation rearrange if the mechanism is concerted like in SN2/E2?

By Sergei on Monday, October 24, 2022 - 12:50 pm:  Edit

Hi anonymous,

SN2/E2 do not proceed via a carbocation; hence no rearrangement should be happening.

By Anonymous on Monday, October 24, 2022 - 09:05 pm:  Edit

Can hard/soft acid/base theory be applied to distinguish between SN1/E1 reactions or is it only to distinguish between SN2/E2 reactions?

By Sergei on Monday, October 24, 2022 - 11:01 pm:  Edit

Hi Anonymous,

In principle, yes. But it’s not needed.
Carbocation is the most reactive species (both as an electrophile and an acid). Thus, the softness/hardness of the Nu and B is irrelevant.
Recall that when reactivity goes up, selectivity goes down. C- is soft and prefers to react with a soft electrophile, but has no problem deprotonating an acidic H (which is hard), and prefers to do that rather than attack E.

Same for C+ : C is trying to get rid of the charge by all means.

By jack on Tuesday, October 25, 2022 - 08:36 am:  Edit

In EAS, halogen groups are ortho and para-directing because, even though they are EWG by induction (-I), they are +R by resonance, right? So therefore can we call them ortho and para-directing deactivators?

By Tina on Tuesday, October 25, 2022 - 08:46 am:  Edit

Do we need to know the mechanism of the benzylic alkyl to benzylic carboxylic acid through KMnO4-, H+ and heat acts as a reagents for exam 2? Thank you

By Tina on Tuesday, October 25, 2022 - 08:53 am:  Edit

Do we also need to know the mechanism of reduction of aromatic nitro groups to amino group also?

By Sergei on Tuesday, October 25, 2022 - 09:15 am:  Edit

Hi,

Jack, 8.36am:
Yes.
Halogens are deactivating groups (in EAS!) because of the –I-effect.
Halogens direct the electron density to o/p-position due to +R-effect.


Tina, 8.46am:
no.

Tina, 8.53am:
no.

By BB on Tuesday, October 25, 2022 - 02:37 pm:  Edit

Hi Dr.Dzyuba,
Do you know which parts (pages) of textbook refer the lecture on Monday?
When alkene react with C4H11N, Nu tends to react with sp2 C via Sn2 prime than sp3 C next to Br?

By Sergei on Tuesday, October 25, 2022 - 06:33 pm:  Edit

Hi BB,

SN2' is not in the textbook.
It's not specific to an amine being a Nu, it's with any Nu; but the substrate must be allylic - that's why it's a special case. In other words, this type of reaction is only happening on the allylic type of substrates.

Other examples are not in the book either; but they are also special, i.e., only occurring or not occurring on those exact (or very similar) substrates/compounds.


The last reaction we looked at (HI + PhOCH3, and this being SN2 reaction) - the general topic is covered in chapter 8, p276. And we will revisit it once we start covering ethers.

On Monday it was used as an illustration that a) SN2 can happen under acidic conditions, and it's governed by the substrate, and b) reaction proceeds via the most stable intermediate (forming H3C+ is unfavorable, since this is a high energy intermediate; thus, SN1 is unlikely), and the most stable would be C with partial charge, i.e., SN2.

By Anonymous on Wednesday, October 26, 2022 - 08:02 am:  Edit

When the benzylic alkyl reacts with KMnO4, is it always happen on the nearest C to the benzene ring?

By Anonymous on Wednesday, October 26, 2022 - 08:04 am:  Edit

I am sorry I just find out it happens on Benzylic H, please ignore the question above

By Anonymous on Wednesday, October 26, 2022 - 08:21 am:  Edit

Are we supposed to know how to calculate normalized specific totality tabulated for exam 2?

By anonymous on Wednesday, October 26, 2022 - 11:21 am:  Edit

In the lecture, you said the leaving group should be more delocalized than nucleophile for Sn2, so is it correct when say alkyl iodide will react with F- via Sn2?

By Sergei on Wednesday, October 26, 2022 - 01:03 pm:  Edit

Hi,

Anonymous, 8.02:
When the compound with a benzylic CH (C-H, which us one sigma bond away from the aromatic ring) reacts with KMnO4, it's that benzylic C will be converted to CO2H.


8.21:
no worries


Anonymous 8.21:
Please specify what is the normalized specific totality?
I'm guessing if it's tabulated, there might not be a reason to calculate it; but still without knowing what it is, it's hard to tell.


anonymous, 11.21am:
Yes.
R–I + F– = R–F + I–
this is a legit reaction; when we looked at it in class, we specified that F– is hard, localized, thus the reaction should be done in a softer solvent (DMSO, for example) for it to happen. Still, due to harness, F– is not a terrific Nu, hence we shouldn't expect a smooth, facile process; but it can still happen. In lecture, we don't care about kinetics too much (or as long as we don't have options to choose from).

I– will not removed F–, bc F– is localized charge, and thus not a good LG.

By anonymous on Wednesday, October 26, 2022 - 05:23 pm:  Edit

Will diazonium-salt formation be on exam 2?

By Sergei on Wednesday, October 26, 2022 - 07:16 pm:  Edit

Hi anonymous,

No.

By anonymous on Wednesday, October 26, 2022 - 07:54 pm:  Edit

Can I ask what is the different between SN2' and SN1? how we know which reaction will occur?

By Sergei on Thursday, October 27, 2022 - 07:25 am:  Edit

Hi anonymous,

SN2' is SN2 on sp2C of the allylic halide (and ONLY allylic substrate; in other words, unless you have an allylic halide or tosylate - there is no need to think about SN2'). Otherwise, SN2' follows the exact same rules as those set for SN2.


SN1 must proceed via carbocation, SN2' proceeds through delta plus.

By Anonymous on Thursday, October 27, 2022 - 10:29 am:  Edit

Do we need to know the mechanism for mesylation? Or just that it produces MsO?

By Sergei on Thursday, October 27, 2022 - 10:34 am:  Edit

Hi Anonymous,

Yes.
The mechanism is exactly the same as tosylation.
The difference is that in Ts - there is a tolyl group (C6H4CH3), and in Ms - there is a Methyl/CH3 group. But it does not have an impact on the mechanism/arrow-pushing.

By anonymous on Thursday, October 27, 2022 - 12:36 pm:  Edit

For E1cb, what kinds of bad leaving groups would we commonly have besides OH-, F-, OR-?

Also, is converting a bad leaving group like OH- to Br- with use of PBr3 SN2 and therefore the stereochemistry should invert?

How do we reason through whether an FC alkylation will be a dialkylation or not? How does adding another isopropyl group, for example, stabilize the monoalkylated product if we're adding another EDG that increases the nucleophilicity of the ring?

By Sergei on Thursday, October 27, 2022 - 04:19 pm:  Edit

Hi anonymous,

These are the most common; maybe one more to add would be carboxylate (RCO2–) - but, it's a bit of a special case.
So, HO–/RO–, F– should be sufficient.

Yes, RCH2OH + PBr3 = RCH2Br involved an SN2 step.

Friedel-Crafts alkylation almost always leads to di/tri/etc alkylated products.
This is not related to the stability of the product, but due to the fact, as you correctly point out, alkyl group is an EDG, hence monoalkylated benzene must be more nucleophilic than benzene, for example. Thus, since the product of the 1st EAS is more nucleophilic than the original starting material, then it will continue reacting with the E+ and not the starting material.

Also, both toluene and benzene are stable compounds (both liquids at rt, could be stored for a long time without any changes to their structures), but toluene is more reactive than benzene in EAS.

By anonymous on Tuesday, November 01, 2022 - 08:50 pm:  Edit

Hi Dr. Dzyuba,
Do we need to know the mechanism of the reaction from alkene react with H+, H2O to form alcohol and ether?

By Sergei on Tuesday, November 01, 2022 - 09:26 pm:  Edit

Hi anonymous,

Yes.
These have been discussed in several lectures (and these are based on fundamental acid-base/electrophlie-nucleophile interactions).

By anonymous on Wednesday, November 02, 2022 - 08:25 pm:  Edit

Are esters or carboxylic acids more reactive towards nucleophilic addition?

By Sergei on Wednesday, November 02, 2022 - 09:18 pm:  Edit

Hi anonymous,

In general, esters are more reactive towards Nu attack than acids.

By anonymous on Thursday, November 03, 2022 - 05:33 pm:  Edit

Hi Dr. Dzyuba,
Can you explain again why esters and carboxylic acid will not react with NaBH4?

By anonymous on Thursday, November 03, 2022 - 06:46 pm:  Edit

Hi Dr. Dzyuba,

For the reaction between carboxylic acid and RMgBr, is it because of the greater difference between partial charges of O-H than C=O that make H will more reactive than C -> better acid -> R will react with H to form conjugate base

By Sergei on Thursday, November 03, 2022 - 09:45 pm:  Edit

Hi,

anonymous, 5.33pm:
electrophilicity of the esters is reduced compared to that of the ketones/aldehydes (due to +R-effect of O).
Electrophilicity of the carboxylic acid is reduced because NaBH4 /H– is still a base, and deprotonates the acidic H, forming carboxylate - and in carboxylate, the electron pair on the O is even better for delocalization than in the ester. Thus, carboxylate is even less electrophlic than ester.


anonymous, 6.46pm:
RMgBr is both Nu and B. Deprotonation is energetically easier to do than a Nu addition to an electrophilic center. Thus, deportation takes place faster.
Yes, you can think of the bigger charge on H vs C, as well as ease of accessibility, H is single bonded, hence easier to be approached than C.

By anonymous on Friday, November 04, 2022 - 07:35 am:  Edit

Can I ask why alcohol can act as weak acid that donate proton to base NaH but NaH can not act as acid to donate proton to alcohol?

By Sergei on Friday, November 04, 2022 - 08:47 am:  Edit

Hi anonymous,

As you correctly state - NaH is a base (bases take protons, not give).
NaH does not have a proton; NaH means Na+ and H– (not Na– and H+).

By Anonymous on Friday, November 04, 2022 - 04:06 pm:  Edit

What happens if PCC is used as an oxidizing agent in acidic media?

By Sergei on Friday, November 04, 2022 - 05:07 pm:  Edit

Hi Anonymous,

Nothing good will happen...
The whole point of using PCC is to convert primary alcohol to an aldehyde, without further oxidation to the carboxylic acid. Thus, there is no need to use the media/solvent that will be working against the reagent.

By unknown on Tuesday, November 08, 2022 - 08:02 am:  Edit

Is the reason there no reaction when tertiary alcohols undergo oxidation because we are separating the 3 substituent groups from each other? Or why do we say there is no reaction?

By Sergei on Tuesday, November 08, 2022 - 08:14 am:  Edit

Hi unknown,

Tertiary alcohols do not undergo oxidation, which means that C=O group is not going to form, because it would mean that C will be pentavalent (since all three groups on the alcohol should be preserved), which obviously is not going to happen.

However, recall that in the lecture we discussed that "no reaction" in this case does not mean that starting material will be recovered. Oxidation reactions, with strong oxidants, takes place under acidic conditions, and tertiary alcohols would get protonated, and H2O is a good leaving, which will lead to the formation of the carbocation, with all relevant consequences.

By anonymous on Tuesday, November 08, 2022 - 05:42 pm:  Edit

For the ethers cleavage under acidic condition, can you explain gain the case R=Me? Why is the carbocation formed first?

By jack on Wednesday, November 09, 2022 - 04:50 pm:  Edit

Is a hydral the equivalent to a hydrate of what a hemiacetal is to an acetal (does a hydral look like a hemiacetal except the OMe group is just an H)?

Could you explain again why the central carbon on, for example, acetal and hydrate is not electrophilic ?

By Sergei on Wednesday, November 09, 2022 - 05:08 pm:  Edit

Hi,

anonymous, 5.42pm:
When R was Me - the carbocation is tertiary (there were two R groups), tertiary carbocations are stable in acidic media, hence the reaction proceeded via SN1 mechanism.


jack, 4.50pm:
Hydrate and hydral are the same; our textbook is using hydrate (and it's most commonly used; but in some sources it's hydral).

yes - hydrate looks the same as acetal; for hydrate H2O was used as the solvent/Nu, for acetal - ROH.

Carbon atom of ketone/aldehyde is an electrophile.
When it's converted to the acetal, that carbon is sp3, and it is hindered (all alkyl groups, Os), this Nu cannot approach that center. Acetals are stable in basic conditions bc there is no acidic hydrogen.
Since there is no accessible electrophilic center, and no acidic H, these compounds could be used to protect C=O of aldehydes/ketones, and allow Nu-addition to other carbonyl containing moieties in the molecule.

By jack on Monday, November 14, 2022 - 01:54 pm:  Edit

I know that Indole is aromatic because the pair of electrons on the N atom is counted towards the number of electrons to determine aromaticity, but why isn't the lone pair of electrons on the N atom in Pyridine counted towards the number of electrons to determine aromaticity? Does it have to do with the fact that in Pyridine, the N is doubled bonded versus the N in Indole is not, so the electrons on the N atom in Indole are in conjugation with the aromatic system?

By Anonymous on Tuesday, November 15, 2022 - 08:52 am:  Edit

When we are deciding how to make a bad leaving group into a good leaving group, how do we know when to use TsCl vs PBr3/ SOCl2?

By Sergei on Tuesday, November 15, 2022 - 09:18 am:  Edit

Hi,

Jack, 1.54pm:
Yes, in pyridine the orbital with non-bonding electrons must be perpendicular to the aromatic pi-system; in other words it cannot be in conjugation. N is sp2, hence it's using the pi-orbital for the aromatic system.

In pyrrole, N is not sp2, but having it's orbital with non-bonding electron in conjugation with the rest of the pi-bonds extends the aromatic system to two rings.



Anonymous, 8.52am:
In principle, either one is fine for primary and secondary alcohols. However, if you look through the notes, mechanistically these reactions are quite different, and this determines which reagent could be used for tertiary alcohols.

PBr3 and SOCl2 proceed via SN2 reaction (Br– or Cl– are displacing the respective HO-PBr2 and HOSOCl) to form the Br or Cl.
Thus, tertiary alcohols could not be converted to Br/Cl using this reaction, since SN2 reaction is not taking place on tertiary C.

Tosylation does not involve the carbon atom, as O is the nucleophile that attacks the electrophilic S. Thus, tertiary alcohols could be tosylated.

By anon on Tuesday, November 15, 2022 - 12:11 pm:  Edit

Hi, I was wondering if you always need a solvent when using an oxidizing agent! I saw in the book that CH2Cl2 should be used as a solvent with periodinane. Should PCC be used with a solvent? What about H2SO4, CrO3, H2Cr2O7, or NO2Cr2O7? Thank you!!

By Sergei on Tuesday, November 15, 2022 - 12:27 pm:  Edit

Hi anon,

In general, reactions are run in solvents. Oxidation reactions are not an exception.
By definition, solvent is something that does not participate in the reaction. Thus, we typically don't specify it, unless it is crucial for the specific transformation (for example, recall Grignard, Finkelstein) or if the solvent also acts as a reagent/reactant (e.g., ring opening of epoxides in MeOH/H+; MeOH is the solvent and reagent).

For the oxidant that you listed: all these reactions are done in solvents. Although DCM is commonly used, since it solubilized a lot of organic compounds, and it's easy to remove (i.e., convenience; and also historically), any solvent that is not prone to oxidation by these reagents is going to be ok to use (e.g., 1,2-dichloroethane, some esters, alkanes, even water).
In the lecture course it will be ok, if you don't specify the solvent for these oxidation reactions.

If anything, let me know.

P.S. Some reactions could be done under solvent-free conditions, but this topic is not a part of this course (or the next one).

By anonymous on Tuesday, November 15, 2022 - 08:46 pm:  Edit

In aldol condensation, if there is only one alpha H, will the beta-hydroxy substituted carbonyl be the final product?

By Sergei on Tuesday, November 15, 2022 - 09:01 pm:  Edit

Hi anonymous,

Yes; no H to deprotonate and form conjugate base for E1cb.

By anonymous on Wednesday, November 16, 2022 - 07:41 am:  Edit

Can you please explain again why extra excess of base will not react with beta-direction compound?

By Sergei on Wednesday, November 16, 2022 - 07:57 am:  Edit

Hi anonymous,

beta-direction compound?
I'm assuming you are referring to beta-diketone case (acetylacetone or 2,4-pentanedione) - but if I'm off, please repost.

For that example there are 3 alpha-carbons/3 alpha CHs (2 CHs are the same, i.e., CH3-groups). CH in between C=O (i.e., CH2) is the most acidic, and it will be deprotonated 1st. One of the CH3 will get deprotonated next. These will be C– (i.e., enolates) which will be delocalized to C=O. But there are only 2 C=O groups, and if there are 2 C– and 2 C=O, then the 3rd C– cannot be delocalized, hence it will be localized. But localized C– is not stable at all. This means that pKa of that corresponding CH (of the CH3) is very high (meaning that the conjugate base is localized). Essentially the last CH is as acidic as CH of an alkane (or close-ish), and thus it's not getting deprotonated by any reasonable amount of base.

By anonymous on Wednesday, November 16, 2022 - 08:02 am:  Edit

Can I ask why -C=O is harder E than R-Br?

By Sergei on Wednesday, November 16, 2022 - 08:07 am:  Edit

Hi anonymous,

O is harder than Br, hence C connected to O is going to be harder than C connected to Br.

Recall NaH vs NaBH4, where the hardness of the neighbor determined the hardness/softness of H–.

By anonymous on Wednesday, November 16, 2022 - 04:42 pm:  Edit

Hi,
In reference to the 2nd question on the practice set, can PCC oxidize two OH groups at once?

By Sergei on Wednesday, November 16, 2022 - 06:07 pm:  Edit

Hi anonymous,

Yes, there is no problem for PCC to oxidize two alcohols.

In general, if you have two (or more) functional groups that have the same reactivity, both (or whatever) should react.

By anonymous on Wednesday, November 16, 2022 - 08:59 pm:  Edit

Will reduction of -CN will be covered in exam 3 Dr.Dzyuba?

By Sergei on Wednesday, November 16, 2022 - 09:34 pm:  Edit

Hi anonymous,

No.
Neither nitriles nor amides will be on the exam 3.

By anon on Wednesday, November 16, 2022 - 10:18 pm:  Edit

hi, when practicing nomenclature, where should we find the list of common names that we will be responsible for?

By Anonymous on Thursday, November 17, 2022 - 02:03 pm:  Edit

When there is a nucleophilic addition to the carbonyl of the anhydride, what is the leaving group?

By Sergei on Thursday, November 17, 2022 - 07:40 pm:  Edit

Hi,

anon, 10.18pm:
There is no such a list per se.
Those that were given/used in the lectures will be a fair game. Trivial names are also given in the nomenclature chapters of the textbook.

I understand that this is not very definitive, but if you are naming a compound, you can use IUPAC nomenclature.
For example, HCO2H is a formic acid, but methanoic acid is an acceptable/correct name as well. AcOH is typically known as acetic acid, but if you name it as ethanoic acid, you will get a full credit.



Anonymous, 2.03pm:
RCO2–

By anonymous on Tuesday, November 22, 2022 - 07:22 pm:  Edit

Dr. Dzyuba,

Do you by chance know if we will use the same textbook next semester or will use a different one?

By Sergei on Tuesday, November 22, 2022 - 08:48 pm:  Edit

Hi anonymous,

It will be a different one. Please contact Dr. Minter about the title and all the details.

However, you might want to keep your current textbook till May, as it will serve as a compact reference for major chemical reactions.

By anonymous on Sunday, November 27, 2022 - 06:18 pm:  Edit

Will the content on the final exam be equally weighted from everything covered this semester, or will it focus more on content covered since exam 3?

By Sergei on Sunday, November 27, 2022 - 07:01 pm:  Edit

Hi anonymous,

It will be more or less equally distributed.

By Tina on Thursday, December 01, 2022 - 07:57 am:  Edit

Do we need to know the mechanism of amide hydrolysis using aqueous acid? Correct me if Im wrong but is it true that it is harder to hydrolysis under basic conditions but under acidic conditions are easier?

By Sergei on Thursday, December 01, 2022 - 08:26 am:  Edit

Hi Tina,

Yes, mechanisms of amide hydrolysis under acidic and basic conditions is a fair game.
For the purposes of this course (i.e., O-Chem lecture, where chemistry is done on a board, piece of paper, iPad worksheet, etc), either H+/H2O or HO–/H2O should be treated equally.

In reality, hydrolysis of amides under basic conditions is going to be more facile. Protonated amide is still not a terrific electrophile, and water is not a super strong nucleophile.
Under basic conditions, at least HO– is a very strong Nu.

Also, pH of human body, for example, is more on the acidic side, rather than basic. Proteins (where amino acids are connected to each other via amide bonds) do not really degrade//hydrolyze. This indirectly hints that amides should probably be quite stable under acidic conditions.

By Tina on Friday, December 02, 2022 - 08:01 am:  Edit

So for Gabriel synthesis of amine, you were using NaOH as a base to add to the electrophilic center carbonyl group, which resulting in the product contains 2 COO- and a primary amine. Would the next step be a acidic work up? Thank you

By Sergei on Friday, December 02, 2022 - 08:20 am:  Edit

Hi Tina,

Work-up step is a procedure for isolating and separating the products of a reaction.
Typically, organic compounds are neutral, and thus, products of the reaction should be isolated as neutral compounds (i.e., not salts).

If during the reaction there was an excess of protons (reaction was done under acidic conditions, in the presence of an acid, and hence basic centers are protonated), then a work up step should include a basic work-up to get neutral product.

If the reaction was done under basic conditions (i.e., all acidic protons are removed/deprotonated), then an acidic work up should be used to get to neutral product.

In other words, acidic and basic work up are used if addition or removal of a proton gets a neutral form of the product. Thus, there is always a work-up step (addition of water, extraction, filtration, etc), but it does not always have to involve an acid or a base.


In Gabriel synthesis, the product of the reaction (i.e., the compound that we are trying to synthesize) is the amine. Even in the presence of NaOH, amine is in its neutral form. Thus, to isolate the amine a work up step does not require an acid.
In fact, if the acidic work up is used, than NH2 is going to get protonated, and it will be an ammonium salt.

Phthalic acid is typically not considered a product here, and thus, it's not isolated. But if one desires to isolate phthalic acid, then an acidic work up would be required, since the corresponding decarboxylate (2 CO2–) is water soluble, whereas protonation will make it less soluble in water, and more soluble in an organic solvent (this is similar to O-Chem Lab #1, extraction/separation of an acid and a base)

By anonymous on Saturday, December 03, 2022 - 06:20 am:  Edit

Are we responsible for the mechanism for Gabriel synthesis using H2N-NH2? Or just the reactants/products?

By Sergei on Saturday, December 03, 2022 - 09:53 am:  Edit

Hi anonymous,

Yes, the mechanism is a fair game, since it only deals with nucleophilic additions to carbonyl (inter and intra), protonation/deprotonation steps.

By anonymous on Sunday, December 04, 2022 - 01:29 pm:  Edit

Can we use Pd and Pt interchangeably when doing hydrogenation with H2?

By Anonymous on Sunday, December 04, 2022 - 03:21 pm:  Edit

When adding protecting groups to amino acids for peptide synthesis, does it matter which amino acid has the NH2 group protected and which amino acid has the CO2 group protected?

By Sergei on Sunday, December 04, 2022 - 04:13 pm:  Edit

Hi,

anonymous, 1.29pm:
yes, either precious metal will be fine.


Anonymous, 3.21pm:
it depends on which di/tri/polypeptide one tries to make.
Amino acid sequence (i.e., which amino acid is attached to which one, and in what order, determines primary peptide/protein sequence).
Consider that NH2-AA1-AA2-CO2H and H2OC-AA1-AA2-NH2 are two different dipeptides (AA1 and AA2 two different amino acids).

By Tina on Monday, December 05, 2022 - 08:48 am:  Edit

Can hydrolysis of nitriles happened under basic condition?

By Sergei on Monday, December 05, 2022 - 08:50 am:  Edit

Hi Tina,

yes, absolutely.

By anonymous on Monday, December 05, 2022 - 06:28 pm:  Edit

When drawing chair conformations from haworth projections, how do we know if the OH groups are axial or equatorial?

By Sergei on Monday, December 05, 2022 - 07:10 pm:  Edit

Hi anonymous,

recall that we were redrawing substituents on the chair conformation from Haworth projection:
whichever OH was pointing up in the Haworth, it has to be drawn up in the chair conformation (regardless if it's axial or eq, the OH on the chair must point up), and same for HO being down in the Haworth.

By Anonymous on Tuesday, December 06, 2022 - 09:59 am:  Edit

For which sugars do we need to memorize the structures?

By Sergei on Tuesday, December 06, 2022 - 10:56 am:  Edit

Hi Anonymous,

All that were (will be) shown/named in lecture(s):
D-glucose, D-mannose, D-fructose, and D-ribose.
We might add one or two tomorrow.

By anonymous on Wednesday, December 07, 2022 - 01:03 pm:  Edit

Do we still have review session tomorrow?

By Sergei on Wednesday, December 07, 2022 - 01:08 pm:  Edit

Hi anonymous,

Yes.
LH1, 6pm

By Tina on Wednesday, December 07, 2022 - 10:33 pm:  Edit

Hi Dr. Dzyuba, please correct my statement if I said anything wrong below. I am trying to understand of how to protect amines and carboxylic acid.

For CO2 protection, we using R-OH and H+ to convert the carboxylic acid to ester, which can called fisher esterification process. But, for deprotection, depending on the R group you will use different reagents, such as if R group is Me than you will use NaOH or if R group is Bn(which is abbreviate for benzene ester?) than you use H2 and Pd.

Thank you

By Tina on Wednesday, December 07, 2022 - 10:34 pm:  Edit

correction: benzyl ester not *benzene ester*

By Sergei on Thursday, December 08, 2022 - 06:59 am:  Edit

Hi Tina,

Yes, this is correct.
Formation of the ester on amino acid is no different than formation of the ester on any other acid/CO2H.

Using H2/Pd to remove benzyl ester is also not specific to the ester of an amino acid. Any benzyl ester will be cleaved using H2/Pd. However, this deprotection could be quite advantageous in the case of amino acids/peptides, since these reagents/conditions don't interfere with a number of functional groups.

Same is true for Boc-protection/deprotection of amines. Primary/secondary amines (don't have to be the amine of the amino acid per se) could also be protected by Boc2O, Et3N., and the Boc-group could be removed by TFA.

By Anon on Thursday, December 08, 2022 - 07:25 pm:  Edit

When converting N3 or CN to a NH2, can we used hydrogenation and reduction interchangeably, or are there circumstances where one should be used over the other?

By Sergei on Thursday, December 08, 2022 - 07:30 pm:  Edit

Hi Anon,

For this course, either options could be used.

Experimentally (i.e., if one is actually doing the reaction in a lab, not on the board/ipad/paper), there will be circumstances when one set of conditions could be more advantageous than the other.

By Anonymous on Friday, December 09, 2022 - 09:54 am:  Edit

Do keto-enol tautomerization and imine-enamine tautomerization only occur in acidic conditions?

By Sergei on Friday, December 09, 2022 - 10:46 am:  Edit

Hi Anonymous,

Technically yes. Since enol and enamine are neutral.

However, keep in mind that conversion of ketones/electrophile to enolate/nucleophile takes place under basic conditions.

By Annon on Friday, December 09, 2022 - 11:14 am:  Edit

What is the purpose of keto enol tautomerization and immine formation?

By Sergei on Friday, December 09, 2022 - 11:48 am:  Edit

Hi Annon,

Ketone/aldehyde (C=O) is an electrophile, enol is a nucleophile.
Imine (C=N) is an electrophile, enamine is a nucleophile.

However, ketone, for example, is a more thermodynamically stable form, hence as a starting material (or whatever is sold in a bottle) most ketones/aldehydes are in the keto form/C=O. Thus, keto-enol (imine-enamine) tautomerization allows to convert E to Nu in situ.

By anonymous on Saturday, December 10, 2022 - 12:55 pm:  Edit

Where do carboxylic acids fall (with respect to ketones, esters, amides, etc) in the ranking of reactivity towards nucleophilic attack of the carbonyl?

By Sergei on Saturday, December 10, 2022 - 04:01 pm:  Edit

Hi anonymous,

carboxylic acids will be in between ester and amides/nitriles.
However, there is an acidic H, which might impact the reactivity (Nu is also B).

By Anonymous on Sunday, December 11, 2022 - 12:28 pm:  Edit

Is the final more focused on the content after exam 3, or does it cover all of the topics from the semester somewhat evenly?

By Sergei on Sunday, December 11, 2022 - 02:02 pm:  Edit

Hi Anonymous,

The final is comprehensive; material will be more or less evenly distributed.

By anonymous on Monday, December 12, 2022 - 05:30 pm:  Edit

How do you distinguish between when it is best to use TsCl/Et3N to replace a bad leaving group, and when it is best to use PB3r or SOCl2? I know that PBr3 and SOCl2 are best for SN2 processes, but can TsCl/Et3N be used too?

By Sergei on Monday, December 12, 2022 - 07:03 pm:  Edit

Hi anonymous,

Introduction of Br or Cl is via the SN2 process; with Br- or Cl- acting as a Nu towards a carbon that is connected to a good LG at that point.

With TsCl, it’s the O of the ROH that is acting as a Nu; the carbon that is attached to OH and then to OTs never gets involved. Thus, tosylation works on all alcohols.

By sharingisbeneficial on Monday, December 12, 2022 - 09:09 pm:  Edit

How to know when to use Pt or Pd with H2?

By Sergei on Monday, December 12, 2022 - 09:35 pm:  Edit

Hi sharingisbeneficial,

Either one is ok to use in this course

By IM on Tuesday, December 13, 2022 - 11:09 am:  Edit

Hi Dr. Dzyuba,

I just have a question about what distinguishes E1 from E1cb. I’m just not sure why we don’t call E1cb an E1 reaction, how are they different?

By Tina on Tuesday, December 13, 2022 - 11:33 am:  Edit

Hi Dr. Dzyuba,

On exam 1 2022, for ranking acidity question, I see there are 2 E in your answers. Would the ranking be like this:

C < E < B < D < A

By sharingisbenefical on Tuesday, December 13, 2022 - 02:28 pm:  Edit

what's the difference between amide and imide?

By Sergei on Tuesday, December 13, 2022 - 05:21 pm:  Edit

Hi,

IM, 11.09am:
These are two different eliminations in regard to conditions, intermediates, structural requirements, etc. They never compete (same for E2 and E1, which never compete with each other). "1" indicates that both eliminations are unimolecular (formation of C+ or C– is the rate determining step/slowest step of the reaction).

E1 is taking place under acidic conditions, and it proceeds via a carbocation.

E1cb is taking place under basic conditions, and it proceeds via a carboanion (delocalized, hence the requirement for an EWG to stabilize C–).


Tina, 11.33am:
Yes, A is the most acidic.
My apologies for overlooking it.


sharingisbeneficial, 2.28pm:
Amide: N–C=O
Imide: O=C–N–C=O (recall phthalimide in Gabriel's synthesis of amines).

By sharingisbeneficial on Tuesday, December 13, 2022 - 06:25 pm:  Edit

Are we supposed to know the mechanism for florescein synthesis?

By Sergei on Tuesday, December 13, 2022 - 06:43 pm:  Edit

Hi sharingiscaring,

:-)

No need to memorize mechanisms. These should be rationalized, analyzed to gain fundamental understanding of the elementary steps. Use the mechanism on D2L studying/reviewing relevant steps/processes.

By anonymous on Saturday, December 17, 2022 - 07:09 pm:  Edit

Will the +/- system be used for final grades?

By Sergei on Saturday, December 17, 2022 - 07:44 pm:  Edit

Hi anonymous,

Yes, but in very few cases.
The cut-offs have been e-mailed and will be posted on D2L shortly.


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