Which one of the following compounds will liberate $\mathrm{CO}_{2}$, when treated with $\mathrm{NaHCO}_{3}$ ?

(C) $\left(\mathrm{CH}_{3}\right)_{3} \stackrel{\oplus}{\mathrm{~N}} \stackrel{\ominus}{\mathrm{HCl}}$

(A) $\left(\mathrm{CH}_{3}\right) 4 \stackrel{\oplus}{\mathrm{~N}} \stackrel{\ominus}{\mathrm{O}} \mathrm{H}$

(B) $\mathrm{CH}_{3} \mathrm{NH}_{2}$

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Aldehydes Ketones and Carboxylic Acids

Which one of the following compounds will liberate

CO_{2}

, when treated with

NaHCO_{3}

(A) $(CH_{3}) 4 N \oplus O ⊖ H$
(B) $CH_{3} NH_{2}$
(C) $(CH_{3})_{3} N \oplus HCl ⊖$ ✓ Correct
(D)(see structure)

Answer: (C)

Solution

Sodium hydrogencarbonate gives off carbon dioxide only to a substance that can hand it a proton, the carbonic acid so formed breaking up into water and the gas. Tetramethylammonium hydroxide is a strong base and methylamine is a base, so neither has a proton to give; in ethanamide the nitrogen lone pair is delocalised into the carbonyl group and its

N - H

bonds are not acidic. Trimethylammonium chloride is the salt of a weak base with a strong acid, so it is the one acidic species in the set and its

N + - H

proton passes to the hydrogencarbonate ion. The compound that liberates carbon dioxide with sodium hydrogencarbonate is trimethylammonium chloride,

(C H_{3})_{3} N H^{+} C l^{-}

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Single correct option

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JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Aldehydes Ketones and Carboxylic Acids

Solution

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