Q.How many moles of methylbromide are required to convert ethanamine to N, N-dimethyl ethanamine ?

Nitrous acid reacts differently with the three classes of aromatic amines; only the secondary amine, N-methylaniline, gives the diagnostic yellow oily liquid (an N-nitrosamine) — this is a classic distinguishing test for 1°, 2°, and 3° amines.

Reaction of amines with $HNO_2$ (generated in situ from $NaNO_2 + HCl$, at $0$–$5°C$):

• Primary aromatic amine (aniline, option d): forms a benzenediazonium chloride, $C_6H_5N_2^+Cl^-$, a colourless/pale-yellow crystalline solid in cold solution — not an oily liquid.
• Secondary aromatic amine (N-methylaniline, option a): the N-H hydrogen is replaced by the nitroso group ($-N=O$), giving N-methyl-N-nitrosoaniline, $C_6H_5N(CH_3)-N=O$ — this N-nitrosamine is a characteristic yellow oily liquid.
• Tertiary aromatic amine (N,N-dimethylaniline, option c): since there is no N-H to substitute, nitrosation occurs on the aromatic ring instead (at the position para to the strongly activating $-N(CH_3)_2$ group), giving p-nitroso-N,N-dimethylaniline, which separates as a green crystalline solid.
• Nitrobenzene (option b) does not react with nitrous acid at all under these conditions (the $-NO_2$ group is already fully oxidised and unreactive here).

So the compound that gives the yellow oily liquid on treatment with nitrous acid is the secondary amine, N-methylaniline.

Aniline undergoes N-acylation with acetic anhydride: the amine nitrogen's lone pair attacks the anhydride's carbonyl carbon, displacing acetate, to give the amide acetanilide.

Aniline, $C_6H_5NH_2$, has a nucleophilic nitrogen lone pair that attacks the electrophilic carbonyl carbon of acetic anhydride, $(CH_3CO)_2O$, in a nucleophilic acyl substitution reaction. One acetyl group is transferred to nitrogen (displacing acetate/acetic acid as the leaving group), converting the -NH$_2$ into an amide -NHCOCH$_3$ group: $C_6H_5NH_2 + (CH_3CO)_2O \rightarrow C_6H_5NHCOCH_3\ (\text{acetanilide}) + CH_3COOH$ The product, N-phenylacetamide, is commonly called acetanilide, and this specific N-acetylation reaction is also used industrially/in the lab to protect (and mildly deactivate) the -NH$_2$ group before performing further electrophilic aromatic substitutions on the ring. The other options (p-/o-/m-aminoacetophenone) describe ring-substituted ketones, which would require a Friedel-Crafts acylation on a different substrate, not the simple N-acylation of aniline's amino group.

Acylation of a primary amine by an acid anhydride produces an amide.

An acid anhydride reacts with a primary amine by nucleophilic acyl substitution: the amine's lone pair attacks the carbonyl carbon of the anhydride, displacing a carboxylate/carboxylic acid and forming an N-substituted amide:

$RNH_2 + (CH_3CO)_2O \rightarrow RNHCOCH_3\ (amide) + CH_3COOH$

The functional group produced is therefore the amide group, –CONH–.

Converting a primary amine to a quaternary ammonium salt by exhaustive methylation consumes 3 moles of CH₃I per mole of amine.

Methylamine, $CH_3NH_2$, has one methyl group and two N–H bonds. Exhaustive (excess) alkylation with methyl iodide replaces both remaining N–H hydrogens and then alkylates the resulting tertiary amine's lone pair to form the quaternary ammonium salt:

$CH_3NH_2 \xrightarrow{CH_3I} (CH_3)_2NH \xrightarrow{CH_3I} (CH_3)_3N \xrightarrow{CH_3I} (CH_3)_4N^+I^-$

Three successive methylations (three moles of $CH_3I$) are required to go from $CH_3NH_2$ to $(CH_3)_4N^+I^-$ (tetramethylammonium iodide).

Vigorous oxidation of aniline with acidified potassium dichromate ($K_2Cr_2O_7/H^+$) oxidises the ring, converting aniline into p-benzoquinone.

Aniline's strongly activating $-NH_2$ group makes the ring very susceptible to oxidative attack. Acidified dichromate (a strong oxidising agent) oxidises aniline not at a side chain (there is none) but at the ring itself, removing the amino nitrogen as part of the process and installing two carbonyl groups at the para positions to give p-benzoquinone ($C_6H_4O_2$). Benzaldehyde (a) and benzoic acid (d) would arise from oxidation of a methyl/aldehyde side chain (as in toluene or benzyl alcohol oxidation), which aniline does not have. Benzyl alcohol (c) is unrelated — it is not a plausible oxidation product of aniline at all.

Primary amine + $CHCl_3$/alcoholic KOH is the carbylamine reaction, giving a foul-smelling isocyanide.

This is the carbylamine (isocyanide) test, characteristic of primary amines only (both aliphatic and aromatic): the amine, chloroform and alcoholic KOH react to form an isocyanide (carbylamine) with a distinctive foul/offensive smell, along with KCl and water:

$RNH_2 + CHCl_3 + 3KOH \rightarrow RNC + 3KCl + 3H_2O$

Secondary and tertiary amines do not give this test, which is why it is used to distinguish primary amines from the other two classes.

The carbylamine reaction (foul-smelling isocyanide) is given only by primary amines.

When a primary amine reacts with chloroform ($CHCl_3$) in the presence of alcoholic KOH, it forms an isocyanide (carbylamine), which has an extremely unpleasant, foul odour — this is the carbylamine test, specific to primary amines only (both aliphatic and aromatic).

$RNH_2 + CHCl_3 + 3KOH \rightarrow RNC + 3KCl + 3H_2O$

Among the given options, ethylamine ($C_2H_5NH_2$) is the only primary amine; diethylamine and ethylmethylamine are secondary amines, and triethylamine is a tertiary amine — none of these give the carbylamine reaction (they lack the required N-H needed for isocyanide formation in this test).