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An aromatic compound ' \( \mathrm{A} \) ' \( \left(\mathrm{C}_{7} \mathrm{H}_{9} \mathrm{~N}\right) \) on reacting with \( \mathrm{NaNO}_{2} / \mathrm{HCl} \) at \( 0^{\circ} \mathrm{C} \) forms benzyl
alcohol and nitrogen gas. The number of isomers possible for the compound ' \( A \) ' is
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alcohol and nitrogen gas. The number of isomers possible for the compound ' \( A \) ' is
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The correct answer is:
\( 55 \)
The reaction involved is
No. of isomeric forms of \( C_{7} H_{9} N \)
o-Toluidine
\( m \)-Toluidine
\( \mathrm{CH}_{3} \)
\( p \)-Toluidine
\( \mathrm{CH}_{2} \mathrm{NH}_{2} \)
No. of isomeric forms of \( C_{7} H_{9} N \)
o-Toluidine
\( m \)-Toluidine
\( \mathrm{CH}_{3} \)
\( p \)-Toluidine
\( \mathrm{CH}_{2} \mathrm{NH}_{2} \)
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