Q. In the mechanism for the reaction of HBr with t-butyl alcohol, pick out the correct statement.

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A

Formation of protonated alcohol in a slow step

B

Formation of $\left(\right.CH_{3}\left(\left.\right)_{3}C^{+}$ is a slow step

C

Formation of $\left(\right.CH_{3}\left(\left.\right)_{3}CBr$ from $\left(\right.CH_{3}\left(\left.\right)_{3}C^{+}$ is a slow step

D

Formation of $\left(\right.CH_{3}\left(\left.\right)_{3}C^{+}$ is a fast step

Solution:

The mechanism of this reaction is as follows
$\left(\left(\text{(CH}\right)_{\text{3}} \text{)}\right)_{\text{3}} \text{COH} \, \text{+} \, \text{HBr} \rightleftharpoons \left(\left(\text{(CH}\right)_{\text{3}} \text{)}\right)_{\text{3}} \text{C} \overset{\text{+}}{\text{O}} \left(\text{H}\right)_{\text{2}} \, \text{+} \, \left(\text{Br}\right)^{-}$ (fast)

Hence, formation on carbocation $\left(\right.CH_{3}\left(\left.\right)_{3}C^{+}$ is slow step (R.D.S.).

Questions from NTA Abhyas 2022

1. An alkene $\left(\right.A\left.\right)$ on ozonolysis gives a mixture of two carbonyl compounds. Mixture on Clemmensen reduction gives just one alkane $\left(\right.B\left.\right)$ . $\left(\right.B\left.\right)$ is the lowest alkane which in pure form can not be prepared by standard Wurtz method. $\left(\right.A\left.\right)$ is

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5. The density of 2 M aqueous solution of NaOH is 1.28 g/cm$^3$. The molality of the solution is
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