Question

$EtN{H}_2+MeMgI\underset{\text{in the presence of pyridine}}{\overset{\text{Heated at high temp.}}{\to }}$ GAS (A) The volume of gas (A) obtained at S.T.P. when $0.45gm$ of $EtN{H}_2$ reacts with $MeMgI$ is

• A. $224ml$
• B. $22.4ml$
• C. $448ml$
• D. $44.8ml$

Answer 1

Answer: (C)

Only one hydrogen atom in $1^{\circ }RN{H}_2$ reactes at room temperature. In the presence of pyridine, both $H$ atoms in $1^{\circ }RN{H}_2$ react

One mole of $EtN{H}_2$ reacts with $2mol$ of $MeMgI$ to give $2mol$ of $C{H}_4$ (g). Molecular mass of $EtNH45gm$
Therefore, $45gm$ of $EtN{H}_2$ gives $2\times 22.4$ litres of $C{H}_4$ at S. T. P.
$0.45$ gm of EtNH ${}_2$ gives $\frac{2\times 22.4\times 0.45}{45}=0.448$ litres $=448ml$