Question

An unknown compound A $C_{8}H_{10}O_{3}$ on acetylation with $CH_{3}COCl/Py$ forms acetyl derivative of A whose MW is 280. A on treat with $CH_{2}N_{2}$ gives methyl ether of B having MW 182. If the number of phenolic hydroxyls and alcoholic hydroxyls in the compound A are X and Y respectively. Find the sum of X + Y here?

Answer 1

Number of hydroxyl group by acetylation method $\text{=} \, \frac{\text{280} - \text{MW} \, \text{of} \, \text{C}_{\text{8}} \text{H}_{\text{10}} \text{O}_{\text{3}}}{\text{42}}$

$=\frac{280 - 154}{42}=\frac{126}{42}=3$

Number of phenolic hydroxyl $\text{=} \, \frac{\text{MW} \, \text{of} \, \text{methyl} \, \text{ether} - \text{MW}}{\text{14}}$

$=\frac{182 - 154}{14}=\frac{28}{14}=2$

Hence, X = 2

Number of alcoholic $-OH$ groups $=3-2=1$

Hence, Y = 1

So X + Y = 2 + 1 = 3.