Question

Compound ‘A’ with molecular formula \[\ce{C4H9Br}\] is treated with aq. \[\ce{KOH}\] solution. The rate of this reaction depends upon the concentration of the compound ‘A’ only. When another optically active isomer ‘B’ of this compound was treated with aq. \[\ce{KOH}\] solution, the rate of reaction was found to be dependent on concentration of compound and \[\ce{KOH}\] both.

(i) Write down the structural formula of both compounds ‘A’ and ‘B’.

(ii) Out of these two compounds, which one will be converted to the product with inverted configuration.

Answer 1

(i) The molecular formulae of isomers of \[\ce{C4H9Br}\] are \[\ce{CH3}\]

\[\begin{array}{cc}
\ce{CH3}\phantom{......}\\
|\phantom{.........}\\
\ce{CH3 - C - Br}\phantom{...........}\\
|\phantom{.........}\\
\ce{\underset{2 - Bromo-2-methyl propane (A)}{CH3}}\phantom{......}
\end{array}\]

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

Since the rate of reaction of compound ‘A’ \[\ce{(C4H9Br)}\] with aqueous \[\ce{KOH}\] depends upon the concentration of compound ‘A’ only, therefore, the reaction occurs by S_N1 mechanism and compound ‘A’ is tertiary bromide i.e., 2-Bromo-2-methylpropane.

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

Rate = \[\ce{A:[(CH3)3CBr]}\]

(ii) Since compound ‘B’ is optically active and is an isomer of compound ‘A’ \[\ce{(C4H9Br)}\], therefore, compound ‘B’ must be 2-Bromobutane. Since the rate of reaction of compound ‘B’ with aqueous \[\ce{KOH}\] depends upon the concentration of compound ‘B’ and \[\ce{KOH}\], therefore, the reaction occurs by S_N2 mechanism and product of hydrolysis will have inverted configuration.

\[\begin{array}{cc}
\ce{CH3CH2CHCH3 + KOH -> CH3CH2CHCH3 + KBr}\\
\phantom{..}|\phantom{...............................}|\phantom{...}\\
\phantom{..}\ce{Br}\phantom{.............................}\ce{OH}\phantom{.}
\end{array}\]

Rate =