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The following is a reduction reaction. In the reduction of 2 to 3, keto group undergoes reduction by nucleophillic addition.

Ester could undergo nucleophillic acyl substitution to give an aldo group, further reduced via nucleophillic addition again.

My question is, during such reductions, is keto group reduced first or ester group gets reduced first?

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I believe you understand $\ce{LiAlH4}$ is stronger base than $\ce{NaBH4}$. In the reduction of 2 to 1, $\ce{NaBH4}$ failed to reduce ester function at all, but reduce 3-keto function to 3-hydroxy function. That result tells us that isolated carbonyl group is easy to reduce than that in ester (not even to relevant aldehyde). Thus, you can conclude that isolated carbonyl function reduced first before hard to reduce ester function in the presence of stronger reducing agent such as $\ce{LiAlH4}$, which can able to reduce both.

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  • $\begingroup$ I agree with your instincts but the only sure way to know is to reduce the ketoester with LAH as the limiting reagent (5-10 mol%) and then analyze the products. In which case there will be recovered ketoester, hydroxyester and perhaps some diol in that an intermediate aldehyde would reduce like a ketone. $\endgroup$ – user55119 May 27 '19 at 19:24
  • $\begingroup$ Great point and I agreed. But, I answered according to the data given. :-) $\endgroup$ – Mathew Mahindaratne May 27 '19 at 20:07

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