How can you tell where the nucleophile will attack in these reactions? Either th

Question

How can you tell where the nucleophile will attack in these reactions? Either the H off the alpha carbon or the carbonyl? Here's an two situations where OH- attacks different positions. I only notice that the Leaving group or lack of leaving group changes. However, I would think the OH- would be less inclined to attack the carbonyl with the actual leaving group attached because of steric hinderance and would attack the alpha H on the opposite side for (b) and (c) respectively. Appreciate the help.

Explanation / Answer

In (a) and (c), the base attacks the alpha hydrogen, because we use only a catalytic amount of the base so that only a part of the carbonyl compound and ester is converted to enolate ion. Now this enolate ion reacts with rest of the carbonyl compound or ester to form the desired adduct.

Also in presence of a catalytic amount of base/acid the reaction becomes reversible so that only a part of the compound is converted to enolate and the rest of the compoud react with the enolate to form the adduct in the second step.

In (a) and (c) if we use a stoichiometric amount of the base/acid or a concentrated base/acid, then the reaction will not be reversible and no adduct will be formed. Also in this case the base will directly attack to the carbonyl carbon as it is occuring in (b)

In case of (b) we use a strong and catalytic amount of acid/base so that it attacks the carbonyl carbon resulting a hydrolysis reaction to form acid and alcohol.

Hence in presence of catelytic amount of acid/base, it reacts with the alpha-hydrogen and the reaction is reversible. In presence of strong and stoichiometric amount of acid/base, it reacts with the carbonyl carbon.

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