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:
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Identify the reactive site. The ketone carbon is electrophilic. NaBH₄ provides a hydride ion (H⁻) that attacks this carbon.
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Mechanism in brief: The hydride adds to the carbonyl carbon, forming an alkoxide intermediate. Aqueous workup (protonation) then gives the alcohol.
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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.
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The ester remains unchanged. The ester carbonyl is less electrophilic and is not reduced by NaBH₄.
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.
Product structure: The cyclohexanone ring becomes a cyclohexanol ring. The side chain remains exactly as it was.
✓Final answer
The product is methyl 3-(2-hydroxycyclohexyl)propanoate — the ketone is reduced to an alcohol, the ester is untouched.
(b) 6 (CH₃)₃C–OH + 2 Al ⟶
Concept first: This is a reaction between a tertiary alcohol and a metal. Aluminium is a reactive metal, but it does not simply dissolve in alcohols the way sodium does. Instead, it reacts with alcohols to form alkoxides, releasing hydrogen gas. The stoichiometry here is key: each aluminium atom can replace three hydroxyl hydrogens.
Step-by-step reasoning:
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Recognise the type of reaction. This is an acid-base / redox reaction. The alcohol acts as a weak acid (the O–H bond breaks), and aluminium acts as a reducing agent (it gets oxidised from Al⁰ to Al³⁺).
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Write the balanced equation conceptually. Each Al atom reacts with three alcohol molecules to give one molecule of aluminium trialkoxide and three atoms of hydrogen (which combine to form H₂ gas).
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Apply to the given numbers. You have 6 molecules of tert-butyl alcohol and 2 atoms of aluminium. That's exactly a 3:1 ratio of alcohol to Al. So all reactants are consumed completely.
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The product: Aluminium tert-butoxide, Al[OC(CH₃)₃]₃, plus hydrogen gas.
2Al+6(CH3)3C–OH⟶2Al[O–C(CH3)3]3+3H2
This is the same type of reaction as sodium with ethanol (forming sodium ethoxide), but aluminium gives a trialkoxide. The tert-butyl group is bulky, but that doesn't hinder the reaction — the O–H bond is still acidic enough.
✓Final answer
The products are aluminium tert-butoxide, Al[O–C(CH3)3]3, and hydrogen gas, H2.