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The question is(See image):

Hydrolysis

I know that initially a cyanohydrin will be formed. On reaction with $\ce{H2SO4}$, the cyanohydrin will give a carboxylic acid/an amide.

Here's my doubt:

Since the medium is highly concentrated acid medium(95% $\ce {H2SO4}$), the quantity of water in the medium is less, which means lesser of the nucleophile(water in this case). Hence the cyanohydrin should undergo partial hydrolysis to give an amide(because of lesser water!). Therefore, according to me the answer should be (D).(Had it been dilute acid medium I would have gone with the carboxylic acid product.) However, in this case my answer is not correct...:(

What is incorrect about my thinking?

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So G is the expected cyanohydrim, but the second step looks very vigorous (concentrated sulfuric acid and heat). I would expect two things to happen under these conditions, 1) hydrolysis of the nitrile group to the corresponding carboxylic acid (the reaction conditions are too strong to stop at the intermediate amide, rather the amide is further converted to the corresponding carboxylic acid; keep in mind that an equilibrium will initially exist between the amide and the carboxylic acid, but everytime the acid forms from the amide, ammonia is released and escapes from the system making the reaction irreversible, pushing this reaction further and further to the carboxylic acid side, ultimately only the carboxylic acid will be present.), and 2) acid catalyzed elimination of water from the alcohol portion of the molecule. Taken together I would expect the final product to be the various cis-trans isomers of "A".

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    $\begingroup$ $\ce{H2SO4}$ is typically used in excess since it is the solvent. There should be more than enough water to convert the amide to the acid, especially since the ammonia escapes from the system pushing the equilibrium totally to the acid side. But I agree, the question would have been clearer if the words "excess $\ce{H2SO4}$" had been used. $\endgroup$
    – ron
    May 22 '14 at 15:28
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    $\begingroup$ Would the ammonia stay in the heated solution, or would it be in the vapor phase above the solution? Also an additional equivalent of water is generated during the reaction from the elimination step. $\endgroup$
    – ron
    May 22 '14 at 16:42
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    $\begingroup$ BTW, as I read your link, to me it says that hydrolysis can occur with dilute acid; it doesn't say that dilute acid is required nor that concentrated acid fails. $\endgroup$
    – ron
    May 22 '14 at 16:53
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    $\begingroup$ The ammonia will be protonated by the strongly acidic solution giving ammonium ion. $\endgroup$
    – jerepierre
    May 23 '14 at 0:29
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    $\begingroup$ good point! But either way, protonation or escape, the absence of ammonia in the solution drives the equilibrium to the acid side. $\endgroup$
    – ron
    May 23 '14 at 0:32

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