Write the structures of the main products of the following reactions : (a) Methyl 3-(2-oxocyclohexyl)propanoate (drawn: a cyclohexanone ring bearing, on the carbon next to the $C{=}O$, the side chain $-CH_2CH_2-\overset{O}{\underset{\|}{C}}-O-CH_3$) $\xrightarrow{NaBH_4}$ (b) $6\

Question

Write the structures of the main products of the following reactions : (a) Methyl 3-(2-oxocyclohexyl)propanoate (drawn: a cyclohexanone ring bearing, on the carbon next to the $C{=}O$, the side chain $-CH_2CH_2-\overset{O}{\underset{\|}{C}}-O-CH_3$) $\xrightarrow{NaBH_4}$ (b) $6\,(CH_3)_3C-OH + 2Al \longrightarrow$

Accepted Answer

> [!TLDR]
> The key idea is to identify the functional group that reacts with the given reagent. NaBH₄ selectively reduces the ketone to an alcohol, leaving the ester untouched. Aluminium reacts with tert-butyl alcohol to form aluminium tert-butoxide and hydrogen gas.

## (a) Methyl 3-(2-oxocyclohexyl)propanoate + NaBH₄

**Concept first:** Sodium borohydride (NaBH₄) is a mild reducing agent. It reduces aldehydes and ketones to alcohols, but it does **not** reduce esters, carboxylic acids, or amides under normal conditions. This selectivity is crucial — you must recognise which carbonyl group will react and which will survive.

The molecule has two carbonyl groups:
- A **ketone** on the cyclohexanone ring (the "2-oxo" part)
- An **ester** in the side chain (the methyl propanoate part)

NaBH₄ will attack only the ketone.

**Step-by-step reasoning:**

1. **Identify the reactive site.** The ketone carbon is electrophilic. NaBH₄ provides a hydride ion (H⁻) that attacks this carbon.

2. **Mechanism in brief:** The hydride adds to the carbonyl carbon, forming an alkoxide intermediate. Aqueous workup (protonation) then gives the alcohol.

3. **Stereochemistry note:** The ketone is on a cyclohexane ring. Hydride attack can occur from either face, producing a mixture of cis and trans alcohols relative to the side chain. In exam contexts, you usually show the racemic product — often drawn with a wedge/dash or simply as the alcohol without specifying stereochemistry unless asked.

4. **The ester remains unchanged.** The ester carbonyl is less electrophilic and is not reduced by NaBH₄.

> [!WARNING]
> A common mistake is to reduce both carbonyls. Remember: NaBH₄ stops at the aldehyde/ketone level. LiAlH₄ would reduce the ester too, but that's a different reagent. …

Q.Write the structures of the main products of the following reactions :

Concept understanding — Preparation of Alcohols by Hydroboration and Carbonyl Reduction

(a) Methyl 3-(2-oxocyclohexyl)propanoate + NaBH₄

Q.Write the structures of the main products of the following reactions :

(a) Methyl 3-(2-oxocyclohexyl)propanoate + NaBH₄

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