Question

An organic compound was used with sodium metal and extracted with distilled water. On adding freshly prepared $FeS{O}_4$ solution followed by the addition of $FeC{l}_3$ and dil. $HCl$ produced a greenish blue solution. 0.30g of the organic compound after Kjeldahlisation evolved a gas $\left(X\right)$ which was passed on 100 ml of 0.1 M $H_{2}S{O}_4$. The excess of acid required 20 ml of 0.5 M $NaOH$ for neutralization.
The organic compound which suitably fits into the above data is:

• A. $C{H}_3-\stackrel{\stackrel{O}{||}}{C}-N{H}_2$
  
• B. $C_6{H}_5-\stackrel{\stackrel{O}{||}}{C}-N{H}_2$
  
• C. $H_{2}N-\stackrel{\stackrel{S}{||}}{C}-N{H}_2$
• D. $H_{2}N-\stackrel{\stackrel{O}{||}}{C}-N{H}_2$

Answer 1

The correct option is D $H_{2}N-\stackrel{\stackrel{O}{||}}{C}-N{H}_2$

$H_{2}S{O}_4$= is dibasic
meq of acid$=2\times 0\cdot 1\times 100=20$ meq
meq of Base $=1\times 0\cdot 5\times 20=10$ meq

Excess of acid used to nuetralize $NaOH=20-10=10$ meq

$\therefore$ Acid used for nutralization of evolved gas $X$ = 10 meq

% $N=\frac{1.4\times \text{meq of excess acid}}{\text{weight of compound}}=\frac{1.4\times 10}{0.3}$

% $N=46.67$ %

% $N$ in $N{H}_2-CO-N{H}_2=\frac{28}{60}\times 100$

% $N=46.67$%
Hence, the organic compound is Urea. Option D is correct.