Presence of which reagent will affect the reversibility of the following reaction, and change it to a irreversible reaction: $\mathrm{CH}_{4}+\mathrm{I}_{2} \underset{\text{Reversible}}{\stackrel{h\nu}{\rightleftharpoons}} \mathrm{CH}_{3}\mathrm{I}+\mathrm{HI}$

(D) Concentrated $\mathrm{HIO}_{3}$

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Hydrocarbons

Presence of which reagent will affect the reversibility of the following reaction, and change it to a irreversible reaction:

CH_{4} + I_{2} Reversible ⇌ h ν CH_{3} I + HI

(A)HOCl
(B)Liquid $NH_{3}$
(C)dilute $HNO_{2}$
(D)Concentrated $HIO_{3}$ ✓ Correct

Answer: (D)

Solution

Iodination of an alkane is reversible because the hydrogen iodide formed alongside the iodoalkane is a strong reducing agent and converts

C H_{3} I

back to

C H_{4}

. Removing HI as it is produced leaves nothing to drive the reverse step, and the way to remove it is to oxidise its iodide back to iodine, which calls for a strong oxidising agent. Hypochlorous acid, liquid ammonia and dilute nitrous acid do not serve: the first two are not used as oxidants here and dilute nitrous acid is too weak and is itself readily reduced. Concentrated

HI O_{3}

is a strong oxidising agent and destroys HI as fast as it forms, so its presence makes the iodination irreversible.

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Single correct option

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JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Hydrocarbons

Solution

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