Q.40 The reaction of (R)-2-bromobutane with CN proceeds by ________. (A) retention of configuration (B) inversion of configuration (C) formation of CH2 =CH(CH2 CH3 ) (D) formation of (S)-2-methylbutanenitrile

Q.40 The reaction of (R)-2-bromobutane with CN proceeds by ________.
(A) retention of configuration
(B) inversion of configuration
(C) formation of CH2 =CH(CH2 CH3 )
(D) formation of (S)-2-methylbutanenitrile

The reaction of (R)-2-bromobutane with CN⁻ proceeds by inversion of configuration via an SN2 mechanism. This substitutes Br with CN while flipping the stereochemistry at the chiral center. The correct answer is (B).

Reaction Mechanism

(R)-2-bromobutane is a secondary alkyl halide with CN⁻ as a strong nucleophile, favoring SN2 over SN1. In SN2, backside attack by CN⁻ forms a pentacoordinate transition state, displacing Br⁻ and inverting configuration from R to S. The product is (S)-2-cyanobutane (also called butane-2-carbonitrile), not 2-methylbutanenitrile.

Option Analysis

  • (A) Retention of configuration: Incorrect, as SN2 causes inversion, not retention (seen in SN1 racemization or double inversion).

  • (B) Inversion of configuration: Correct, hallmark of SN2 for chiral secondary halides like this.

  • (C) Formation of CH₂=CH(CH₂CH₃): Incorrect, that’s 1-butene from E2 elimination, but CN⁻ prefers substitution over elimination here.

  • (D) Formation of (S)-2-methylbutanenitrile: Incorrect, as 2-methylbutanenitrile (CH₃CH₂CH(CH₃)CN) requires a branched substrate; product is linear CH₃CH(CN)CH₂CH₃.

(R)-2-bromobutane with CN⁻ reaction is a classic SN2 example in organic chemistry, key for CSIR NET Life Sciences preparation. This substitution showcases inversion of configuration at the chiral center.

SN2 Mechanism Breakdown

CN⁻ attacks the secondary carbon from the back, inverting R to S configuration in one concerted step. No carbocation forms, ensuring stereospecificity. Energy diagram shows a single transition state.

Stereochemistry Details

Starting with (R)-2-bromobutane (CH₃CHBrCH₂CH₃), the product is (S)-2-cyanobutane (CH₃CH(CN)CH₂CH₃). Priority shifts (CN > CH₂CH₃ > CH₃ > H) confirm S after inversion.

Why Not Other Pathways?

SN1 would racemize; E2 yields alkenes like CH₂=CHCH₂CH₃. Conditions (polar aprotic solvent implied) favor clean SN2.

Exam Relevance

For CSIR NET, recognize secondary halides + strong nucleophiles = SN2 inversion. Practice drawing wedges/dashes for stereochemistry.

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