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  • Compound ‘A’ with molecular formula C4H9Br is treated with aq. KOH solution. The rate of this reaction depends upon the concentration of the compound ‘A’ only

Compound ‘A’ with molecular formula C_{4}H_{9}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 the concentration of compound and 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 an inverted configuration.

Answers (1)

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

Whereas, ‘B’ is optically active and is an isomer of ‘A’. Therefore, ‘B’ must be 2-Bromobutane. As the concentration of ‘B’ is responsible for the rate of reaction of ‘B’ with aqueous KOH, therefore, the reaction mechanism is S_N2

(ii) Because of the S_N2 reaction, compound ‘B’ will have an inversion of configuration and turn out to be an inverted product.

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