Question

$0.15$ g of an organic compound (A) gave $0.33$ g of $C{O}_2$ and $0.18$ g of $H_{2}O$. The molecular weight of (A) is 60. The compound (A) on dehydration gave ahydrocarbon (B) containing $85.7$% of C. (B), on successive treatment with hydriodic acid and silver hydroxide gave a product (C), isomeric with (A). Find structural formulae of (A), (B) and (C).

Answer 1

Organic compound $\to C{O}_2+H_{2}O$

A(0.15 g) 0.33 g 0.18 g

$\Rightarrow$ % of C $=\frac{\frac{12}{44}\times 0.33}{0.15}\times 100$ = 60%

$\Rightarrow$ % of H $=\frac{\frac{2}{18}\times 0.18}{0.15}\times 100=\frac{2}{0.15}$ = 13.33%

% of O = 100 - (60 + 13.33) 100 - 73.33 = 26.67%

Since compound A undergoes dehydration, it must have -OH group, so have oxygen atom.