Questions on some Ochem Problems

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woox

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Got a few Ochem problems that I found going thorugh ochem roadmaps/destroyer. Would appreciate explanations for these 🙂


Problem 1:
What reaction is this? I don't understand how basically a CH3 gets cleaved off/become a leaving group?
ochemA.jpg


Problem 2:
Since there is a perodixde/ROOR, why doesn't Br attach to CH3, the least sub carbon? Is this a special case for Aromatics where it doesn't matter?
ochemB.jpg


Problem 3:
How can the CH3 just attack the carbon with Cl? Isn't this a aromatic? Is this one of the reactions to just memorize?
Ochemc.jpg


Problem 4: Why wouldn't the chain just attach as shown? Why does it rearrange itself?
ochemD.jpg



Thanks 🙂
 
1) This is an alpha halogenation, therefore you will be introducing halogens on the alpha carbon. Remember, by doing so, you are introducing electronegative atoms that can stabilize a negative charge and will make next alpha hydrogen even more acidic. So its difficult to stop at a monohalogenation product using a base catalyst to form the enolate.The more halogens you introduce, the more easily the next alpha hydrogen can be removed. You will eventually get an alpha carbon that contains three halogens. This -CX3 group acts as a leaving group when the OH- attacks the carbonyl carbon. A carboxylic acid is the result.

2) The hydrogens on the benzylic carbon are resonance stabilized and will reduce the activation energy and rate of the radical formation at that location.

3) This is a Gilman reaction. It is able to perform a substitution reaction through a means that differs from a typical Sn2 reaction. In particular, it can displace a aryl and vinyl leaving group. The last time I remembered, the exact mechanism wasn't fully elucidated. If someone else has any new information, please enlighten us!

4) You're missing the AlCl3 catalyst. Remember, the R-Cl will attack the aluminum of the catalyst (since it has an empty p orbital). This will form a carbocation on the alkane chain. In this case, it is a primary carbocation, which will rearrange via a hydride shift to form a secondary (and more stable) carbocation.
 
Listen man I cant explain it to you with all these concepts like all these other kids on sdn. Cause like I said before, and ill say it again, that's all junk to me. and i can care less about the mechanism or why something actually happens. Ill give it to you the way I understand, which is amazing because if I showed it to me 2 year old cousin he would get it.

For problem number 1, if you see Cl2 excess / OH- or I2 excess/ OH- or other halogens in excess with OH- then you remove a CH3 from a methyl ketone and form the anion. I don't care why it happens, neither do the DAT ppl they just want the answer 🙂 so memorize that. FOREVER. (And for the person who posted above me the only way you get the carboxy acid is if u react it again with H3O+ or any other acid....the steps you posted only give you the ANION)

Problem number 2, whenever you see NBS know that you will remove an alpha hydrogen and add a Br. again, this is all your understanding needs to be for the DAT. Know this, and your golden.

Problem number 3, whenever you see "Cool-ee"" (CuLi) your are going to think ok this is a "cool" reaction. I will replace the halogen with only ONE mole of what is between the parentheses. DONE.

Problem 4, this is alkylation. Whenever you are adding on 3 carbons or more in a linear fashion, we get the reararranged product. (if you were adding an acyl-halide with 3 or more carbons though "Ch3Ch2Ch2Ch2C=OCl", there would be no shift. this is the exception)

Like i said, this may not be the answer you want, but this is the answer that will get you a good score. the DAT will never ask you why this happens, they just want to know what happens. So instead of wasting brain cells on stupid mechanisms use that space for bio 🙂 .....if you understand it the way i showed you, you should have no problem blowing away ANY problem they can give you. Peace.


"If you can't explain it simply, you don't understand it well enough" ---------> Albert Einstein
 
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"(And for the person who posted above me the only way you get the carboxy acid is if u react it again with H3O+ or any other acid....the steps you posted only give you the ANION)"

If by anion you mean the carboxylate anion, then yes. No acid was introduced based on the reaction sequence provided by the OP; therefore, the carboxylic acid will remain deprotonated in a basic environment. If this is not what you mean, then please clarify. Thanks
 
"(And for the person who posted above me the only way you get the carboxy acid is if u react it again with H3O+ or any other acid....the steps you posted only give you the ANION)"

If by anion you mean the carboxylate anion, then yes. No acid was introduced based on the reaction sequence provided by the OP; therefore, the carboxylic acid will remain deprotonated in a basic environment.

👍
 
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