Question

JEE ADVANCED CHEMISTRY QUESTION SEPTEMBER 2020

Answer 1

Which of the following plots is/are correct for the given reaction ?

CH3 - (CH3)C(CH3)-Br + NaOH → CH3-(CH3)C(CH3)-OH + NaBr

solution : given reaction is an example of SN¹ reaction. we know, The number of molecules involved in Rate determining step decides the order of reaction. in SN¹ reaction, only one molecules’ properties affect the rate determining step. so, SN¹ is first order reaction.

as we know, half life of first order , T½ = ln2/k, where k is rate constant, always remains same with respect to time.

so , option (A) is correct choice.

option (B) is incorrect because initial rate depends on [P]₀. but shows initial rate doesn't depend on initial concentration of P.

CH3 - (CH3)C(CH3)-Br + NaOH → CH3-(CH3)C(CH3)-OH + NaBr

at t = 0, P₀ 0

at t = eq , P P₀ - P

now rate = k [reactant ]

⇒-kt = ln[P/P₀]

⇒P/P₀ = e^-kt ......(1)

concentration of product at equilibrium, [Q] = P₀ - P

initial concentration of reactant , [P]₀ = P₀

now [Q]/[P] = (P₀ - P)/(P₀) = 1 - P/P₀

from equation (1),

1 - e^-kt , graph (C) and (D) are incorrect.

Therefore only option (A) is correct.