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JEE Advanced 2023 · Paper 1, Chemistry question 16

Match the reactions in List-I with the features of their products in List-II and choose the correct option.

List-I

  • P
    (−)-1-Bromo-2-ethylpentane (single enantiomer) →SN2 reactionaq. NaOH\xrightarrow[\mathrm{S_N2\ reaction}]{\text{aq. NaOH}}
  • Q
    (−)-2-Bromopentane (single enantiomer) →SN2 reactionaq. NaOH\xrightarrow[\mathrm{S_N2\ reaction}]{\text{aq. NaOH}}
  • R
    (−)-3-Bromo-3-methylhexane (single enantiomer) →SN1 reactionaq. NaOH\xrightarrow[\mathrm{S_N1\ reaction}]{\text{aq. NaOH}}
  • S
    3-Bromo-3,5-dimethylheptane, drawn with both stereocentres defined (single enantiomer) →SN1 reactionaq. NaOH\xrightarrow[\mathrm{S_N1\ reaction}]{\text{aq. NaOH}}
    3-Bromo-3,5-dimethylheptane drawn as a single enantiomer: a seven-carbon zigzag chain with C5 bearing a wedged Me and a hashed H, and C3 bearing a wedged Me and a hashed Br (two stereocentres).

List-II

  • 1
    Inversion of configuration
  • 2
    Retention of configuration
  • 3
    Mixture of enantiomers
  • 4
    Mixture of structural isomers
  • 5
    Mixture of diastereomers
Choose one option
Answer and worked solution

Official answer: B

(P) In 1-bromo-2-ethylpentane, the stereocentre is C2 but substitution happens at C1, so no bond to the stereocentre is broken. The configuration is retained → 2.

(Q) SN2\mathrm{S_N2} at the stereocentre of 2-bromopentane goes by backside attack, so the configuration is inverted → 1.

(R) SN1\mathrm{S_N1} at the stereocentre of 3-bromo-3-methylhexane goes through a planar carbocation. Attack from either face gives a mixture of enantiomers → 3.

(S) The molecule (3-bromo-3,5-dimethylheptane) has two stereocentres. The SN1\mathrm{S_N1} reaction scrambles C3 while C5 is unchanged, so the two products are diastereomers → 5.

P→2, Q→1, R→3, S→5.

Topics:SN1 and SN2Optical isomerism and chirality
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