Consider the reactions (i) \(\left(\mathrm{CH}_3ight)_2 \mathrm{CH}-\mathrm{CH}_2 \mathrm{Br}…

Consider the reactions (i) \(\left(\mathrm{CH}_3ight)_2 \mathrm{CH}-\mathrm{CH}_2 \mathrm{Br} \xrightarrow{\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}}\left(\mathrm{CH}_3ight)_2 \mathrm{CH}-\mathrm{CH}_2 \mathrm{OC}_2 \mathrm{H}_5+\mathrm{HBr}\) (ii) \(\left(\mathrm{CH}_3ight)_2 \mathrm{CH}-\mathrm{CH}_2 \mathrm{Br} \xrightarrow{\mathrm{C}_2 \mathrm{H}_5 \mathrm{O}^{-}}\left(\mathrm{CH}_3ight)_2 \mathrm{CH} C H_2 \mathrm{OC}_2 \mathrm{H}_5+\mathrm{Br}^{-}\) The mechanism of reactions (i) & (ii) are respectively
  1. \(\mathrm{S}_{\mathrm{N}} 1\) and \(\mathrm{S}_{\mathrm{N}} 2\)
  2. \(\mathrm{S}_{\mathrm{N}} 1\) and \(\mathrm{S}_{\mathrm{N}} 1\)
  3. \(\mathrm{S}_{\mathrm{N}} 2\) and \(\mathrm{S}_{\mathrm{N}} 2\)
  4. \(\mathrm{S}_{\mathrm{N}} 2\) and \(\mathrm{S}_{\mathrm{N}} 1\)

Solution

\(\mathrm{S}_{\mathrm{N}} 2\) and \(\mathrm{S}_{\mathrm{N}} 2\) Since rearrangement do not occur in the given nucleophilic substitution reactions, therefore carbocations are not the intermediate in these reactions. Thus both the reactions occur by \(\mathrm{S}_{\mathrm{N}} 2\) mechanism.

Asked in: JEE-TOPICTESTS-CHEMISTRY

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