Q5(b). Chemoselective reduction: NaBH₄ then LiAlH₄

Introduction. The substrate, methyl 2-(4-oxocyclohexyl)acetate, has a ketone and an ester. Sodium borohydride is a mild hydride donor that reduces aldehydes and ketones but not esters; lithium aluminium hydride is strong enough to reduce esters to primary alcohols.

Scheme: Step 1 reduces only the ketone (chemoselective); step 2 reduces the ester to CH₂OH.

1. Step 1. BH₄⁻ transfers hydride to the ketone carbonyl carbon; methanol protonates the alkoxide. The ester carbonyl is less electrophilic (its OMe lone pair donates into C=O), so NaBH₄ leaves it untouched: chemoselectivity.
2. A = methyl 2-(4-hydroxycyclohexyl)acetate, a mixture of cis and trans isomers. The small BH₄⁻ attacks mainly from the axial direction, so the trans isomer (equatorial OH) predominates.
3. Step 2. LiAlH₄ adds hydride to the ester C=O; the tetrahedral intermediate expels methoxide to give an aldehyde, which is reduced at once to the primary alkoxide. The secondary OH is first deprotonated (H₂ evolved) and survives as an alkoxide.
4. Acidic work-up gives B = 2-(4-hydroxycyclohexyl)ethan-1-ol, a diol, with methanol as by-product.

Conclusion. A = methyl 2-(4-hydroxycyclohexyl)acetate (ketone reduced, ester intact); B = 2-(4-hydroxycyclohexyl)ethan-1-ol (ester also reduced). The order of reagents shows the chemoselectivity of NaBH₄ and the power of LiAlH₄.

Application. Chemoselective NaBH₄ reduction of a ketone in the presence of an ester is routine in drug synthesis; selective reductions of this kind are used in making prostaglandin and steroid intermediates.