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I was going through some PYQ'S of JEE Advanced where I encountered this question.

enter image description here

I was not able to decide whether I should go for an SN2 reaction to substitute $\ce{-Cl}$ or a nucleophilic attack on the ester.

The solution provided by a reputed coaching center says that we should prefer the nucleophilic attack on the ester but I don't understand why.

What is the reason for this?

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  • $\begingroup$ @Waylander I tried using that but in the end product that I am getting I have 2-hydroxy ethanoic acid instead of carboxylic acid (cooh) grp . It would be of great help if you could elaborate on what you said.Thanks $\endgroup$
    – okk
    Aug 23 at 9:01
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    $\begingroup$ If you run the first step of the reaction with 2 or more eq of KCN both sites will react. The benzyl chloride will give the phenyl acetonitrile and the ester will give the acyl cyanide. The acid hydrolysis which follows convert the phenylacetonitrile to the carboxylic acid and the acyl cyanide will hydrolyse to the carboxylic acid. If you run reaction 1 with 1 eq of KCN then you get attack at the benzyl chloride. $\endgroup$
    – Waylander
    Aug 23 at 9:12
  • $\begingroup$ @Waylander "" the acyl cyanide will hydrolyse to the carboxylic acid. "",,I am not geting this part ,I was confused as how to hydrolyze this but then I found "pubs.rsc.org/en/content/articlelanding/1967/j2/…." in which it says that it would hydrolyze to acetic acid but according to the solution I shoul be getting carboxylic acid. Please look into this. $\endgroup$
    – okk
    Aug 23 at 9:25
  • $\begingroup$ It depends on the conditions of hydrolysis. Acyl cyanides under strong acid conditions will give the alpha keto acid. Under mild basic conditions they lose cyanide to give the carboxylic acid as they are acylating agents. $\endgroup$
    – Waylander
    Aug 23 at 9:46
  • $\begingroup$ @Waylander I now undersand the reaction if KCN is in excess but I fail to understand why would benzyl chloride would be attacked if KCN is 1 eq. is it beacuse of the electronic factors mentioned in the answer below or is there anything else? $\endgroup$
    – okk
    Aug 23 at 9:49

1 Answer 1

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I would prefer to go along with $\mathrm{S_N2}$ because the benzylic carbon is much more electron deficient than the carbon associated with the ester. This is because the phenyl ring is made up of $\mathrm{sp^2}$ carbons, which are more electronegative than the $\mathrm{sp^3}$ benzylic carbon. Thus it exerts a negative inductive effect which decreases the electron density on the carbon. The same effect, albeit much stronger, is also exerted on the same carbon by the chlorine.

On the other hand, the acyl carbon of the compound receives a positive mesomeric effect due to the phenyl ring. This should somewhat stabilise the electron density on the carbon atom. Here’s a diagram; check out the last three canonical forms: enter image description here

Thus, the benzylic carbon, with lower electron density , is much more susceptible to the $\ce{CN^-}$ nucleophile and hence the reaction should favour an $\mathrm{S_N2}$ pathway. Then cyanide and ester hydrolysis give the respective acid groups.

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  • $\begingroup$ Thanks for the answer, I understand the electronic effects that are coming into play into this question,but I have been taught that for a particulr nucleophile we first consider nucleophilic addition reactions rather than sn2 that's why I was confused. $\endgroup$
    – okk
    Aug 23 at 9:06
  • $\begingroup$ This definitely gives an answer to my question but I am waiting so as to receive any other possible explanations to my question. $\endgroup$
    – okk
    Aug 23 at 9:34

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