Hydrocarbons 2 Question 6
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) $\mathrm{CH} \equiv \mathrm{CH}>\mathrm{CH}{2}=\mathrm{CH}{2}>\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CH}$
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Answer:
Correct Answer: 6. (a)
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:
$$ \mathrm{HC} \equiv \mathrm{CH}>\mathrm{H}{3} \mathrm{C}-\mathrm{C} \equiv \mathrm{CH}>\mathrm{CH}{2}=\mathrm{CH}_{2} $$
