Question

Which of the following compound will give $S_{N}2$ mechanism?

• A. $C{H}_3-Br$
• B. $C{H}_3-C{H}_2-C{H}_2-Br$
• C. $C_6{H}_5-C{H}_2-Br$
• D. All of these

Answer 1

Answer: (A), (B)

$C{H}_3-C{H}_2-C{H}_2-Br$
$C{H}_3-Br$

$S{N}^1$ mechanism proceeds through carbocation formation, as stability of carbocation increases, reaction favors $S{N}^1$ mechanism.

Benzylic carbocation is most stable due to more resonating structures and allylic carbocations are more stable than aliphatic carbocation due to resonance so order of stability of carbocations is benzyll > allyl > $3^0>2^0>1^0$.

($3^0$ carbocation are more stable than others or stability order of carbocations is $3^0>2^0>1^0$ so order of alkyl halides towards $S{N}^1$ reaction is $3^0>2^0>1^0$.)

So order of reactivity for $S{N}^1$ mechanism is
benzyll > allyl > $3^0>2^0>1^0$.

As $1^0$ are least reactive for $S{N}^1$ mechanism so they are most reactive for $S{N}^2$ mechanism.
So the correct answer is only A & B.
C is benzylic halide so they will perform $S{N}^1$ mechanism.