Aniline and methyl amine can be differentiated by: - Vidyadip

MCQ

Aniline and methyl amine can be differentiated by:

A
Reaction with chloroform and aqueous solution of $\text{KOH}.$
✓
Diazotisation followed by coupling with phenol.
C
Reaction with $\ce{HNO_2.}$
D
None of these.

✓

Answer

Correct option: B.

Diazotisation followed by coupling with phenol.

Phenol react with aniline to give diazonium salt by coupling but Methyl amine not react with phenolss.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from Amines→Amines — all question types→Chemistry STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

Which of the following halogens is solid at room temperature

When iodine reacts with $NaF,$ $NaBr$ and $NaCl$

What is an example of camphor in $N_2$ gas?

When acetone is heated with hydroxylamine, the compound formed is

The reagent used for dehydrohalogenation is $........$

Faraday constant

Benzenediazonium chloride on reaction with phenol in weakly basic medium gives

For the first order reaction $A \rightarrow B$ Br the half life is $30\,mm$. The time taken for $75 \%$ completion of the reaction is $.....mm$. (Nearest mteger)

Given : $\log 2=0.3010$

$\log 3=0.4771$

$\log 5=0.6989$

An aqueous solution containing $1\,g$ of urea boils at $100.25\,^oC$. The aqueous solution containing $3\,g$ of glucose in the same volume will boil at ........ $^oC$. (Molecular weight of urea and glucose are $ 60 $ and $180$ respectively)

$\wedge^{{\circ}}_\text{m}\text{H}_2\text{O}$ is equal to $.......$
$a. \wedge^{{\circ}}_{\text{m}(\text{HCl})}+\wedge^{{\circ}}_{\text{m}(\text{NaOH})}-\wedge^{{\circ}}_{\text{m}(\text{NaCl})}$
$b. \wedge^{{\circ}}_{\text{m}(\text{HNO}_3)}+\wedge^{{\circ}}_{\text{m}(\text{NaNO}_3)}-\wedge^{{\circ}}_{\text{m}(\text{NaOH})}$
$c. \wedge^{{\circ}}_{\text{(HNO}_3)}+\wedge^{{\circ}}_{\text{m}(\text{NaOH})}-\wedge^{{\circ}}_{\text{m}(\text{NaNO}_3)}$
$d. \wedge^{{\circ}}_{\text{m}(\text{NH}_4\text{OH})}+\wedge^{{\circ}}_{\text{m}(\text{HCl})}-\wedge^{{\circ}}_{\text{m}(\text{NH}_4\text{Cl})}$