Question

An organic compound contains 69.77% carbon, 11.63% hydrogen and rest, oxygen. The molecular mass of the compound is 86. It does not reduce Tollens' reagent but forms an addition compound with sodium hydrogensulphite and gives positive iodoform test. On vigorous oxidation it gives ethanoic and propanoic acid. The organic compound is

• A. $CH_3 -\overset{\underset{||}{O}}{C}-CH_2 - CH_2 - CH_3$
• B. $CH_3 -\overset{\underset{||}{O}}{C}-CH_2-CH_3$
• C. $CH_3 - CH_2 -\overset{\underset{||}{O}}{C}-CH_2 - CH_3$
• D. $CH_3-CH_2-CH_2CH_2-COOH$

Answer 1

Answer: (A)

$ \% C = 69.77\%; \% H = 11.63\%$

$\therefore \% O = 100 - (69.77 + 11.63) = 18.6\%$

$C : H : O = \frac{69.77}{12} : \frac{11.63}{1} = \frac{18.6}{16}$

$= 5.8 : 11.63 : 1.16 $

$\Rightarrow 5 : 10 : 1$

$\therefore $ The E.F. of the given compound is $C_5H_{10}O$

E.F. mass $= 5 \times 12 + 10 \times 1 + 1 \times 16 = 86$

Since, mol. mass = 86 (given)

$\therefore M.F. = C_5H_{10}O$

Since the given compound on vigorous oxidation gives a mixture of ethanoic acid and propanoic acid, therefore, the methyl ketone is pentan-$2$-one, i.e.