Question

Given below are two statements:

Statement I: One mole of propyne reacts with excess of sodium to liberate half a mole of \(H_2\) gas.

Statement II: Four g of propyne reacts with \(NaNH_2\) to liberate \(NH_3\) gas which occupies 224 mL at STP.

In the light of the above statements, choose the most appropriate answer from the options given below:

(1) Statement I is correct but Statement II is incorrect.

(2) Both Statement I and Statement II are incorrect

(3) Statement I is incorrect but Statement II is correct

(4) Both Statement I and Statement II are correct.

Answer 1

Correct option is: (1) Statement I is correct but Statement II is incorrect.

\(\underset{\underset{\text{1 mole}}{}}{CH_3} - C \equiv CH + \underset{\underset{\text(excess)}{}}{Na} \rightarrow CH_3 - c \equiv \overset{-}{C} \overset{+}{Na} + \underset{\frac{1}{2}mole \ H_2}{\frac{1}{2}H_2} \uparrow\)

\(\underset{\underset{\underset{\frac{4}{40} = 0.1 mol}{\downarrow}}{\text{4 gm}}}{CH_3} - C \equiv CH + NaNH_2 \rightarrow CH_3C \equiv \overset{-}{C} \overset{+}{Na} + NH_3\)

Mol of \(NH_3 = 0.1 \ mol\)

\(V_{NH_3} = 0.1 \times 22400 = 2240\ ml\)