Question

Paragraph 4
A substance (X) contains $41.37 \%$ C, $6.89 \%$ H. $0.116\, gm$ of $(X)$ gave $NH _{3}$, which was absorbed in $50\, ml$ of $N / 10 H _{2} SO _{4}$. The excess of acid required $30 ml$ of $N / 10\, NaOH$ for neutralisation. (X) on treatment with $HNO _{2}$ gave succinic acid. (X) on heating lost $NH _{3}$ to give (A). (A) reacts with $Br _{2}$ and $NaOH$ to give (B) containing $41.02 \% C , 5.88 \% H$, and $11.96 \% N$. (B) on further treatment with $Br _{2}$ and $NaOH$ gives (C) (3-amino propanoic acid). (C) reacts with $HNO _{2}$ to give $\beta$-hydroxy-propanoic acid.
Compound (C) is:

• A. (V)
• B. (VI)
• C. (VII)
• D. (VIII)

Answer 1

Answer: (C)

Percentage of $N=$ ?
Total acid $=50 \times \frac{1}{10}=5\, mEq$.
Excess acid $=30 \times \frac{1}{10}=3\, mEq$.
Acid used $=5-3=2\, mEq$.
Percentage of $N =\frac{1.4 \times mEq \text {. of acid used }}{\text { Weight of compound }}$
$=\frac{1.4 \times 2}{0.116}=24.38 \%$
$C =41.37 \% H =6.89 \% N =24.38 \%$
$O=100-(41.37+6.89+24.38)=27.36$
Formula $= C _{4} H _{8} O _{2} N _{2}$; degree of unsaturation $=2^{\circ}$.