JEE Advanced 2020 · Paper 2 · Chemistry Q8 · Dronacharya Exams

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ChemistryMultiple correct+4 / -2Question 26 of 54

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

( $[\mathbf P]_0$ is the initial concentration of P)

P, tert-butyl bromide ((CH₃)₃C-Br), reacts with NaOH to give Q, tert-butyl alcohol ((CH₃)₃C-OH), and NaBr.

A

t₁/₂ against [P]₀: a horizontal line (half-life independent of initial concentration).

B

Initial rate against [P]₀: a horizontal line (initial rate independent of initial concentration).

C

[Q]/[P]₀ against time: a curve starting at the origin and rising ever more steeply (concave up).

D

ln([P]/[P]₀) against time: a straight line through the origin with positive slope.

Answer and worked solution

Official answer: A

tert-Butyl bromide is hydrolysed by $\ce{S_N1}$ , so the rate is first order in P only: rate $=k[\mathrm P]$ .

(A) For a first-order reaction $t_{1/2}=\ln2/k$ , independent of $[\mathrm P]_0$ , so the plot is a horizontal line. True.
(B) The initial rate $k[\mathrm P]_0$ grows linearly with $[\mathrm P]_0$ , so it is not constant. False.
(C) $[\mathrm Q]/[\mathrm P]_0=1-e^{-kt}$ rises and levels off (concave down), not concave up. False.
(D) $\ln([\mathrm P]/[\mathrm P]_0)=-kt$ falls with time; the plot shows it rising. False.

(A)

Topics:Integrated rate equations SN1 and SN2 Half-life

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