My teacher told - "To reduce carboxylic acid to alcohol, $\ce{LiAlH4}$ is used. But since $\ce{LiAlH4}$ is an expensive reagent, it is commercially reduced by forming an ester and then to alcohol". I asked that we can also use $\ce{NaBH4}$ as it also contains 4 hydrogen bonds. He said that $\ce{LiAlH4}$ is a strong reducing agent that $\ce{NaBH4}$. I was not satisfied by his answer. I want an answer in terms of a reaction mechanism.

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    $\begingroup$ Reaction mechanism won't help you much here imo, but thermodynamics could. $\endgroup$ – Mithoron Oct 9 '15 at 17:49

The answer is polarity.

Consider the two ions $\ce{AlH4-}$ and $\ce{BH4-}$ (as they exist as ions in a solution).

Now the electro-negativity of $\ce{Al}$ is $1.5$, and that of $\ce{B}$ is $2$, and that of $\ce{H}$ is $2.1$.

So the electro-negativity difference ($\Delta\mathrm{EN}$) between $\ce{Al-H}$ bond is $0.6$ and that of $\ce{B-H}$ bond is $0.1$. It is clearly visible that the $\Delta\mathrm{EN}$ of the $\ce{Al-H}$ bond is greater than that of the $\ce{B-H}$ bond.

Now considering the mechanism, due to the small polarity induced by the carbonyl group of the carboxylic acid, the $\ce{H}$ atom in $\ce{Al-H}$ bond is readily broken due to its high polarity, while the $\ce{H}$ in $\ce{B-H}$ can't do the same because it has a low polarity.

This is the reason why $\ce{NaBH4}$ cannot be used instead of $\ce{LiAlH4}$ for reducing carboxylic acids to alcohols.

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    $\begingroup$ Borane $\ce{BH_3}$ can reduce carboxylic acids to alcohols $\endgroup$ – K_P Oct 10 '15 at 12:20

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