Compound 'A' with molecular formula `C_4H_9Br` 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.
Write down the structural formula of both compounds 'A' and 'B'.

Explanation

(i) Alkyl halides on treatment with aq. KOH give alcohols as the major product.
`C_4H_9Br + KOH (aq) to C_4H_9OH + KBr`
In case of compound .A. rate of reaction depends upon the concentration of .A. only, the reaction proceeds via SN mechanism. Since the SN mechanism is favoured by 3° alkyl halides, .A. should be

In case of compound .B. which is an optically active isomer of .A., rate of reaction depends upon the concentration of .B. as well as KOH. Therefore, the reaction occurs by SNP mechanism which is favoured by `1^@ and 2^@` alkyl halides.
Thus .B. can be

However, out of these two, only 2-bromobutene is optically active.
The compound .B. is
`underset("2-Bromobutane(B)")(CH_3CH_2 - overset(Br)overset(|)(CH) - CH_3)`