Hydrocarbons - Result Question 5
6. The correct order for acid strength of compounds $\mathrm{CH} \equiv \mathrm{CH}, \mathrm{CH}_3-\mathrm{C} \equiv \mathrm{CH}$ and $\mathrm{CH}_2=\mathrm{CH}_2$ is as follows :
(2019 Main, 12 Jan I)
(a) $\mathrm{CH}_3-\mathrm{C} \equiv \mathrm{CH}>\mathrm{CH}_2=\mathrm{CH}_2>\mathrm{HC} \equiv \mathrm{CH}$
(b) $\mathrm{CH}_3-\mathrm{C} \equiv \mathrm{CH}>\mathrm{CH} \equiv \mathrm{CH}>\mathrm{CH}_2=\mathrm{CH}_2$
(c) $\mathrm{HC} \equiv \mathrm{CH}>\mathrm{CH}_3-\mathrm{C} \equiv \mathrm{CH}>\mathrm{CH}_2=\mathrm{CH}_2$
(d) $\mathrm{CH} \equiv \mathrm{C} \mathrm{H}>\mathrm{CH}_2=\mathrm{CH}_2>\mathrm{CH}_3-\mathrm{C} \equiv \mathrm{CH}$
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Answer:
Correct Answer: 6. (c)
Solution:
- Ethene $\left(\mathrm{H}_2 \mathrm{C}=\mathrm{CH}_2\right)$ is $s p^2$-hybridised and ethyne $(\mathrm{HC} \equiv \mathrm{CH})$ is sp-hybridised. In ethyne, the sp-hybridised carbon atom possesses maximum s-character and hence, maximum electronegativity. Due to which, it attracts the shared electron pair of $\mathrm{C}-\mathrm{H}$ bond to a greater extent and makes the removal of proton easier. Hence, alkyne is much more acidic than alkene.
Presence of electron donating group in alkyne $\left(\mathrm{H}_3 \mathrm{C}-\mathrm{C} \equiv \mathrm{CH}\right)$ decreases the acidic strength of compound. Hence, the correct order of acidic strength is:
$\mathrm{HC} \equiv \mathrm{CH}>\mathrm{H}_3 \mathrm{C}-\mathrm{C} \equiv \mathrm{CH}>\mathrm{CH}_2=\mathrm{CH}_2$
BETA
