The number of primary,secondary,and tertiary | vedclass

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(C) For the molecular formula $C_3H_9N$,the possible isomers are:$1$. Primary amines $(1^{\circ})$: $CH_3CH_2CH_2NH_2$ (propan$-1-$amine) and $CH_3CH(

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$2, 1, 1$

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(C) For the molecular formula $C_3H_9N$,the possible isomers are:$1$. Primary amines $(1^{\circ})$: $CH_3CH_2CH_2NH_2$ (propan$-1-$amine) and $CH_3CH(NH_2)CH_3$ (propan$-2-$amine). Total = $2$.$2$. Secondary amine $(2^{\circ})$: $CH_3CH_2NHCH_3$ ($N$-methylethanamine). Total = $1$.$3$. Tertiary amine $(3^{\circ})$: $(CH_3)_3N$ ($N,N$-dimethylmethanamine). Total = $1$.Thus,the number of primary,secondary,and tertiary amines are $2, 1, 1$ respectively.

List-$I$	List-$II$
$1.$ Aniline	$a.$ Used in the preparation of azo dyes
$2.$ Nitrobenzene	$b.$ Sulpha drug
$3.$ Sulphanilamide	$c.$ Solvent in Friedel-Crafts reaction
$4.$ Trinitrotoluene	$d.$ Used as an explosive

The number of primary,secondary,and tertiary amines possible with the molecular formula $C_3H_9N$ is:

Similar Questions

How will you convert:
$(i)$ Benzene into aniline
$(ii)$ Benzene into $N,N$-dimethylaniline
$(iii)$ $Cl-(CH_2)_4-Cl$ into hexane-$1,6$-diamine?

In the given reaction,$A$ is:
$A$ $\xrightarrow{\text{Reduction}} B$ $\xrightarrow{HNO_2} C_2H_5OH$

Match List-$I$ with List-$II$ and select the correct answer.

List-$I$ List-$II$

$1.$ Aniline $a.$ Used in the preparation of azo dyes

$2.$ Nitrobenzene $b.$ Sulpha drug

$3.$ Sulphanilamide $c.$ Solvent in Friedel-Crafts reaction

$4.$ Trinitrotoluene $d.$ Used as an explosive

Given below are some nitrogen-containing compounds. Each of them is treated with $HCl$ separately. $1.0 \ g$ of the most basic compound will consume $........$ $mg$ of $HCl$. ($Given$ molar mass in $g \ mol^{-1}$: $C:12, H:1, O:16, Cl:35.5$)

This reaction is an example of:

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