Question

Compound A $({C_7}{H_8}O)$ is insoluble in water, dilute HCL & aqueous $NaHC{O_3}$ , but it dissolves in dilute $NaOH$ . When A is treated with $B{r_2}$ water it is converted into a compound ${C_7}{H_5}OB{r_3}$ rapidly. The structure of A is:
(A)

(B)
(C)
(D)

Answer 1

Hint: Here is this question, Compound A is given as $({C_7}{H_8}O)$ is insoluble in water is given, where as dilute HCL & aqueous $NaHC{O_3}$ as given but it also dissolves in dilute $NaOH$. After which Compound A is reacted with $B{r_2}$ water and as rapidly it is converted into a compound ${C_7}{H_5}OB{r_3}$ . Now we have to justify the structure of Compound A.

Complete Step by Step Solution:
Here A is given as $({C_7}{H_8}O)$,
AS from we can see that, the Oxygen atom will directly got attached to the benzene ring as from which We have reacted HCL and $NaHC{O_3}$ there is no change taking place in the reaction.

As we react the reactant with $NaOH$ it got reacted and got the following reaction as,


As from the above reaction when $({C_7}{H_8}O)$ as in the presence of $B{r_2}$it is directly converted into ${C_7}{H_5}OB{r_3}$.
From which we get the correct structure and positioning of $({C_7}{H_8}O)$ is option (A) as similar to the product of our reaction.
Hence, the correct option is (A).

Note: As we see the importance of nature of any compound in this question when we reacted our main compound A with dilute HCL & aqueous $NaHC{O_3}$ , we got no change as per they are no reactive with each other but when we add dilute $NaOH$ in it got directly soluble in it and got reacted as in the presence of $B{r_2}$ and form a compound as ${C_7}{H_5}OB{r_3}$ . Hence, from this we get that the nature of compounds is very important to understand.