Question

The increasing order of reactivity in the $ S_{N}{2} $ hydrolysis of the following alkylbromides is
(1) $tert$-Butyl bromide
(2) Methyl bromide
(3) $sec$-Butyl bromide
(4) $ n $ - Propyl bromide

• A. $ (1) <\, (3) <\, (4) < \,(2) $
• B. $ (2) <\, (4) <\, (3) <\, (1) $
• C. $ (1) <\, (4) <\, (3) <\, (2) $
• D. $ (2) <\, (3) <\, (4) < \,(1) $

Answer 1

Answer: (A)

Reactivity order in $S_{N}2$ reaction will be
$1^{\circ}>\, 2^{\circ}>\,3^{\circ}$
More bulky substrates are least reactive in $S_{N}2$ reactions


Therefore, order of reactivity is
$(1) <\, (3) <\, (4) <\, (2)$