# Q 12.19     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 hydrogen sulphite and give a positive iodoform test. On vigorous oxidation it gives ethanoic and propanoic acid. Write the possible structure of the compound.

M manish

In the question, it is given that,
% of carbon = 69.77
% of Hydrogen = 11.63
% of Oxygen = (100-69.77-11.63) = 18.6

Now, the no. of moles of the components are-
$n_{c}=\frac{69.77}{12}=5.81$
$n_{H}=\frac{11.63}{1}=11.63$
$n_{O}=\frac{18.6}{16}=1.16$
Thus ration of C:H:O = 5.81:11.63:1.16
To get empirical formula we need to divide them by 1.16 .So,
C:H:O= 5:10:1
So, the empirical formula is $C_{5}H_{10}O$
The molecular mass of the compound = $(5\times 12)+(10\times 1)+(1\times 16) = 86$

Since the compound does not give tollens test it means it is not an aldehyde. It is given that, the compound gives positive iodoform test it means the compound must be methyl ketone.
On oxidation, it gives ethanoic acid and propanoic acids. Therefore, the compound should be

Pentan-2-one
the structure of the compound is - $CH_{3}-CO-(CH_{2})_{2}-CH_{3}$

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