Question

When an organic compound [A] having molecular formula C₄H₉Br is treated with aqueous KOH, the rate of reaction depends on concentration of compound [A] only. But when compound [B], with the same molecular formula, reacts with aqueous KOH, the rate of reaction depends on the concentration of compound [B] as well as of KOH. Compound [B] is a structural isomer of compound [A]. 

Identify compounds [A] and [B].

Answer 1

The rate of reaction of compound ‘A’ (C₄H₉Br) with aqueous KOH depends upon the concentration of compound ‘A’ only; therefore, the reaction occurs by the S_N1 mechanism, and compound ‘A’ is a tertiary bromide, i.e., 2-Bromo-2-methylpropane.

\[\ce{(CH3)3CBr + KOH_{(aq)} -> (CH3)3COH + KBr}\]

Rate = A: [(CH₃)₃CBr]

\[\begin{array}{cc}
\phantom{......}\ce{CH3}\\
\phantom{....}|\\
\ce{CH3 - C - Br}\\
\phantom{...}|\\
\phantom{.....}\ce{\underset{Compound A}{\underset{2-Bromo-2-methylpropane}{CH3}}}\\
\end{array}\]

Compound ‘B’ must be 2-Bromobutane since it is optically active and an isomer of compound ‘A’ (C₄H₉Br). Because the rate at which compound ‘B’ reacts with aqueous KOH depends on the concentration of both compound ‘B’ and KOH, the reaction is caused by the S_N2 mechanism, and the hydrolysis product will have an inverted structure.

\[\begin{array}{cc}
\ce{CH3 - CH2 - CH - CH3}\\
\phantom{.........}|\\
\phantom{..........}\ce{\underset{Compound B}{\underset{2-Bromobutane}{Br}}}\\
\end{array}\]