Consider the following reactions:[P] xrightar | vedclass

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(B) The reaction of $[P]$ with $NaNO_2/HCl$ at $0-5^oC$ followed by coupling with $\beta$-naphthol to form a colored solid indicates that $[P]$ is a p

Vedclass · Accepted Answer

$3-$methylaniline

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(B) The reaction of $[P]$ with $NaNO_2/HCl$ at $0-5^oC$ followed by coupling with $\beta$-naphthol to form a colored solid indicates that $[P]$ is a primary aromatic amine.The reaction of $[P]$ with $Br_2/H_2O$ to form $C_7H_6NBr_3$ (a tribromo derivative) confirms that the amino group is activating the ring towards electrophilic substitution.Comparing the options:$m$-toluidine ($3$-methylaniline) reacts with $NaNO_2/HCl$ to form a diazonium salt,which then couples with $\beta$-naphthol to form an azo dye (colored solid).It also reacts with $Br_2/H_2O$ to form $2,4,6$-tribromo-$3$-methylaniline $(C_7H_6NBr_3)$.Therefore,$[P]$ is $3$-methylaniline.

Consider the following reactions:
$[P] \xrightarrow[(i) \ NaNO_2/HCl, \ 0 - 5^oC]{(ii) \ \beta-naphthol/NaOH} \text{Colored Solid}$
$[P] \xrightarrow{Br_2/H_2O} C_7H_6NBr_3$
The compound $[P]$ is:

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Consider the following reactions:$[P] \xrightarrow[(i) \ NaNO_2/HCl, \ 0 - 5^oC]{(ii) \ \beta-naphthol/NaOH} \text{Colored Solid}$$[P] \xrightarrow{Br_2/H_2O} C_7H_6NBr_3$The compound $[P]$ is: