Explain the stereochemical outcome of the reduction of 2‑butanone (CH3COCH2CH3) with NaBH4 in ethanol. Identify the configuration of the newly formed chiral centre and draw the major product.

hard 5 marks Class 12 · Chemistry · Aldehydes Ketones and Carboxylic Acids

Answer

racemic mixture of (R)- and (S)-2‑butanol

The correct answer is racemic mixture of (R)- and (S)-2‑butanol — Step 5: Because both enantiomers are formed in equal amounts, the major product is a racemic mixture of (R)- and (S)-2‑butanol.

Step-by-step solution

Step 1: 2‑butanone has a planar carbonyl carbon with two different alkyl substituents (CH3 and CH2CH3).
Step 2: NaBH4 donates a hydride ion to the electrophilic carbonyl carbon from either the Re or Si face, producing a tetrahedral alkoxide intermediate.
Step 3: No chiral influence (e.g., a chiral catalyst or substrate) biases the attack, so both faces are attacked equally.
Step 4: Subsequent protonation of the alkoxide gives 2‑butanol with a new stereogenic centre at C‑2.
Step 5: Because both enantiomers are formed in equal amounts, the major product is a racemic mixture of (R)- and (S)-2‑butanol.

Common mistake: Assuming a single enantiomer is produced or neglecting the formation of a racemic mixture

Concept tested: Stereochemistry of hydride reduction of ketones

Published 9 September 2026 · How this solution was prepared and checked

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