Q2(c). Reduction of a carboxylic acid by LiAlH₄; chemoselectivity of NaBH₄
Outline the mechanism for the following reaction:
RCOOH → (i) LiAlH₄, ii) H₂O) RCH₂OH
In what way is NaBH₄ different from LiAlH₄ in `chemoselectivity'?
Answer in brief
LiAlH₄ first deprotonates the acid (H₂ evolved), then adds hydride to the carboxylate; the tetrahedral intermediate collapses to an aldehyde, which is reduced at once; hydrolysis gives RCH₂OH. NaBH₄ is milder and chemoselective: it reduces aldehydes and ketones but leaves acids, esters, amides and nitriles untouched.
Key points
- the acidic O–H destroys one hydride: RCOOH + LiAlH₄ → RCOO⁻ Li⁺ + AlH₃ + H₂ ↑. (Hydrogen gas is evolved; this is why excess reagent is used.)
- aluminium (a Lewis acid) binds the carboxylate oxygens, making the carbon more electrophilic.
- First hydride transfer (slow, rate-determining): H⁻ adds to the carboxylate carbon: a tetrahedral intermediate RCH(O-Al)₂ (an aluminium acetal).
- one O–Al leaves as O-Al (aluminate), giving an aldehyde RCHO (still complexed).
The full model answer
Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.
Introduction. LiAlH₄ is a source of nucleophilic hydride (H⁻). A carboxylic acid is reduced to the primary alcohol in four stages: acid–base reaction, hydride addition to the carboxylate, collapse to an aldehyde and a fast second reduction; water then liberates the alcohol.
Mechanism
- Acid–base step: the acidic O–H destroys one hydride: RCOOH + LiAlH₄ → RCOO⁻ Li⁺ + AlH₃ + H₂ ↑. (Hydrogen gas is evolved; this is why excess reagent is used.)
- Activation: aluminium (a Lewis acid) binds the carboxylate oxygens, making the carbon more electrophilic.
- First hydride transfer (slow, rate-determining): H⁻ adds to the carboxylate carbon: a tetrahedral intermediate RCH(O-Al)₂ (an aluminium acetal).
- Collapse: one O–Al leaves as O-Al (aluminate), giving an aldehyde RCHO (still complexed).
- Second hydride transfer: the aldehyde is far more reactive than the carboxylate, so it is reduced at once to the alkoxide RCH₂O-Al: the aldehyde cannot be isolated.
- Work-up: H₂O (or dilute acid) hydrolyses the Al–O bonds: RCH₂OH.
Scheme: Overall three hydrides are used per acid: one in the acid–base step and two in the reductions.
Net stoichiometry: 4 RCOOH + 3 LiAlH₄ → 4 RCH₂OH after work-up (with 4 H₂ evolved). Dry ether or THF is required: LiAlH₄ reacts violently with water.
Chemoselectivity: NaBH₄ versus LiAlH₄
4pt4ptxltabular@>X cc@ Functional group & NaBH₄ & LiAlH₄ Functional group (continued) & NaBH₄ & LiAlH₄ aldehyde, ketone → alcohol & reduced & reduced* acid chloride & reduced & reduced ester, lactone → alcohol(s) & not (or very slowly) & reduced carboxylic acid → RCH₂OH & not & reduced amide, nitrile → amine & not & reduced epoxide, alkyl halide, nitro (aliphatic) & usually not & reduced* isolated C=C, C≡C & not & not xltabular
- The Al–H bond is more polar than B–H (Al is more electropositive), so AlH₄⁻ is a much stronger, less selective hydride donor than BH₄⁻.
- Li⁺ is a stronger Lewis acid than Na⁺ and activates C=O further; NaBH₄ can even be used in water or ethanol.
Example: methyl 4-oxopentanoate, CH₃COCH₂CH₂CO₂CH₃. NaBH₄ in methanol reduces only the ketone (methyl 4-hydroxypentanoate, which cyclises to γ-valerolactone); LiAlH₄ reduces both groups (pentane-1,4-diol). So NaBH₄ is chemoselective: it distinguishes aldehydes and ketones from esters, acids, amides and nitriles; LiAlH₄ reduces almost every polar multiple bond.
Conclusion. LiAlH₄ first deprotonates the acid (H₂ evolved), then adds hydride to the carboxylate; the tetrahedral intermediate collapses to an aldehyde, which is reduced at once; hydrolysis gives RCH₂OH. NaBH₄ is milder and chemoselective: it reduces aldehydes and ketones but leaves acids, esters, amides and nitriles untouched.
Application. NaBH₄ is used in the manufacture of drugs (reduction of ketone intermediates in the presence of esters) and in bleaching wood pulp; LiAlH₄ is the laboratory reagent for making primary alcohols and amines from acids, esters and amides.
Practising this answer? Write it in a Mains test and get it checked by VKC. Civil Services Mains test series · Telegram

All parts of Question 2
- 2(a) Hydroboration–oxidation of 1-methylcyclopentene10 marks
- 2(b) Secondary structure of proteins: α-helix and β-pleated sheet10 marks
- 2(c) Reduction of a carboxylic acid by LiAlH₄; chemoselectivity of NaBH₄15 marks
- 2(d) Syntheses of nylon 66, Teflon and Terylene15 marks