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JEE Main levelSingle correct optionDifficulty 2 of 3o3128

Match the reaction in List I with the stereochemical outcome in List II. List I: P (S)-2-bromooctane + in DMF; Q (R)-3-bromo-3-methylhexane in aqueous acetone (neutral); R (S)-1-phenylethanol + /pyridine; S (S)-1-phenylethanol + with no base (SNi). List II: 1 inversion of configuration; 2 racemisation; 3 retention of configuration.

  1. AP-1, Q-2, R-1, S-3
  2. BP-3, Q-2, R-1, S-1
  3. CP-2, Q-1, R-1, S-3
  4. DP-1, Q-2, R-3, S-1

Answer: A

P: secondary bromide with azide in a polar aprotic solvent is SN2, so inversion. Q: tertiary bromide in neutral aqueous acetone ionises to a planar cation, so racemisation. R: pyridine gives free chloride, so SN2 inversion. S: without base the SNi ion-pair path delivers Cl from the same face, giving retention.
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