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JEE Main levelSingle correct optionDifficulty 2 of 3o3124

Optically pure (R)-3-bromo-3-methylhexane is left in neutral methanol (solvolysis). Assume that the reaction goes entirely through a free planar carbocation and ignore elimination. The substitution product, 3-methoxy-3-methylhexane, is

  1. Aoptically pure, with retained configuration
  2. Bachiral, since the carbon is tertiary
  3. Ca racemic mixture, optically inactive
  4. Doptically pure, with inverted configuration

Answer: C

The tertiary carbocation is planar and achiral at the cationic carbon, so methanol attacks both faces equally. The carbon bears four different groups (methyl, ethyl, propyl, OCH3) in the product, so it is a stereocentre, and the product is a 50:50 mixture of the two enantiomers, which is optically inactive.
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