The correct order for acid strength of compounds: $\mathrm{CH} \equiv \mathrm{CH}, \mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CH}, \mathrm{CH}_{2}=\mathrm{CH}_{2}$ is as follows:

(D) $\mathrm{HC} \equiv \mathrm{CH}>\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CH}>\mathrm{CH}_{2}=\mathrm{CH}_{2}$

(A) $\mathrm{CH} \equiv \mathrm{CH}>\mathrm{CH}_{2}=\mathrm{CH}_{2}>\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CH}$

(B) $\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CH}>\mathrm{CH} \equiv \mathrm{CH}>\mathrm{CH}_{2}=\mathrm{CH}_{2}$

(C) $\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CH}>\mathrm{CH}_{2}=\mathrm{CH}_{2}>\mathrm{HC} \equiv \mathrm{CH}$

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Hydrocarbons

The correct order for acid strength of compounds:

CH \equiv CH, CH_{3} - C \equiv CH, CH_{2} = CH_{2}

is as follows:

(A) $CH \equiv CH > CH_{2} = CH_{2} > CH_{3} - C \equiv CH$
(B) $CH_{3} - C \equiv CH > CH \equiv CH > CH_{2} = CH_{2}$
(C) $CH_{3} - C \equiv CH > CH_{2} = CH_{2} > HC \equiv CH$
(D) $HC \equiv CH > CH_{3} - C \equiv CH > CH_{2} = CH_{2}$ ✓ Correct

Answer: (D)

Solution

Acidity of these C-H bonds is governed by the stability of the carbanion left behind, and that stability rises with the

s

-character of the carbon's hybrid orbital, since more

s

-character holds the lone pair closer to the nucleus. The

s p

carbons of the two alkynes make them far more acidic than the

s p^{2}

carbon of ethene. Between the two alkynes, the electron-donating methyl group on propyne slightly destabilises its carbanion, so ethyne is a little more acidic than propyne. The order is therefore

HC \equiv CH > CH X_{3} - C \equiv CH > CH X_{2} = CH X_{2}

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Hydrocarbons

Solution

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