The major product obtained in the following r | vedclass

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(D) The reactant is $2$-amino-$1$-phenylpropan-$1$-ol. It contains both an amino group $(-NH_2)$ and a hydroxyl group $(-OH)$.When treated with acetic

Vedclass · Accepted Answer

$A$ structure showing an acetamide group at the amino position with the hydroxyl group intact.

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(D) The reactant is $2$-amino-$1$-phenylpropan-$1$-ol. It contains both an amino group $(-NH_2)$ and a hydroxyl group $(-OH)$.When treated with acetic anhydride $((CH_3CO)_2O)$ in the presence of pyridine at room temperature,acetylation occurs.Since the nucleophilicity of the amino group $(-NH_2)$ is significantly higher than that of the hydroxyl group $(-OH)$,the amino group undergoes acetylation preferentially to form an amide.Therefore,the major product is the $N$-acetylated derivative,where the $-NH_2$ group is converted to $-NHCOCH_3$ while the $-OH$ group remains unchanged.

The major product obtained in the following reaction is:

Similar Questions

$Y \xleftarrow[(ii)\,KOD, Br_2, \Delta]{(i)\,NH_3, \Delta} \text{Benzoic acid}$ $\xrightarrow[(ii)\,KOH, Br_2, \Delta]{(i)\,ND_3, \Delta} X$
What are $X$ and $Y$?

The product of the above reaction is:

Assertion : Nitration of aniline can be conveniently done by protecting the amino group by acetylation.
Reason : Acetylation increases the electron density in the benzene ring.

The final product $C$,obtained in the following reaction sequence,would be:
$p$-Toluidine $\xrightarrow{Ac_2O} A$ $\xrightarrow{Br_2/CH_3COOH} B$ $\xrightarrow{H_2O/H^+} C$

What are $X$ and $Y$ in the following reaction?

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