Ncert Exercises & Solutions

Compound ‘X with molecular formula C,H, Br is treated with aq. 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. KOH solution, the rate of reaction was

found to be dependent on concentration of compound and KOH both.

1. Write down the structural formula of both compounds ‘A’ and ’B’,

2. Out of these two compounds, which one will be converted to the

product with inverted configuration.

(i) mechanism and the given compound will be

tertiary alky! halide i.e., 2-bromo-2-methylpropane and the structure is as follow

$C H_{3}$
$|$
$H_{3} C — C — B r (A)$
$|$
$C H_{3}$

Optically active isomer of (A) is 2-bromobutane (8) and its structural formula is

$C H_{2} — C H_{2} — C H C H_{3}$
$(R) |$
$B r$

(ii) As compound (B) is opically active therefore, compound (14) must be 2-bromobutane.

Since. the rate of reaction of compound (B) depends both upon the concentration of

compound (B) and KOH, hence, the reaction follow S $_{N}$ 2 mechanism. In S $_{N}$ 2 reaction,

nucleophile attack from, the back side, therefore, the product of hydrolysis will have

‘opposite configuration