Hydrocarbons 2 Question 1
1. Consider the following reactions,
$$ A \text { is } $$
(2019 Main, 12 April II)
(a) $\mathrm{CH} \equiv \mathrm{CH}$
(b) $\mathrm{CH}{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}{3}$
(c) $\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CH}$
(d) $\mathrm{CH}{2}=\mathrm{CH}{2}$
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Answer:
Correct Answer: 1. (c)
Solution:
- According to the given conditions, the compound should be alkyne with triple bond present at the terminal. The chemical reactions involved are as follows:
Step 1
$$ \underset{\substack{\text { Prop-1-yne }}}{\mathrm{CH}{3}-\mathrm{C}} \equiv \mathrm{CH} \xrightarrow{\mathrm{Ag}{2} \mathrm{O}} \mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{Ag} \downarrow $$
$$ \text { (A) } $$
Step 2
$$
\mathrm{CH}{3}-\mathrm{C} \equiv \mathrm{CH} \frac{\mathrm{Hg}^{2+}}{\text { dil. } \mathrm{H}{2} \mathrm{SO}_{4}}
$$
$\mathrm{OH}$
Turbidity within 5 minutes
(Insoluble in Lucas reagent)
In step-1, prop-1-yne reacts with $\mathrm{Ag}{2} \mathrm{O}$ to form $\mathrm{CH}{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{Ag}$, that forms white precipitates.
In step 2, prop-1-yne in presence of mercuric sulphate and dil $\cdot \mathrm{H}{2} \mathrm{SO}{4}$ produces carbonyl compound $\left(\mathrm{CH}{3}\right){2} \mathrm{C}=\mathrm{O}$ which produces $\left(\mathrm{CH}{3}\right){2} \mathrm{CH}-\mathrm{OH}$ in presence of $\mathrm{NaBH}_{4} \cdot 2^{\circ}$ alcohol on reaction with Lucas reagent produces turbidity in about 5 min.
