Question

An unknown compound A $C_8{H}_{10}{O}_3$ on acetylation with $C{H}_{3}COCl/Py$ forms acetyl derivative of A whose MW is 280. A on treat with $C{H}_2{N}_2$ 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

Answer: 3

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.