Nitrogen Compounds (Amines) Previous Year Questions (JEE Main 2026)
Solved JEE Main 2026 previous year questions on nitrogen compounds and amines with step-by-step solutions covering diazonium chemistry, basicity, boiling point order and the Hofmann bromamide reaction.
A curated set of JEE Main 2026 previous-year questions on Nitrogen Compounds (amines, diazonium salts and coupling reactions), each solved step by step so you can check both the final answer and the reasoning.
Solutions are AI-generated and pending review.
Solution
Step 1 — Bromination (A): Aniline is strongly activating, and with $\text{Br}_2$ in water (no catalyst) all three positions ortho/para to $-\text{NH}_2$ are brominated. The major product is 2,4,6-tribromoaniline.
Step 2 — Diazotization (B): $\text{NaNO}_2/\text{HCl}$ at $273\text{-}278\,\text{K}$ converts $-\text{NH}_2$ into the diazonium group, giving 2,4,6-tribromobenzenediazonium chloride ($\text{Ar–N}_2^{+}$).
Step 3 — Replacement of $-\text{N}_2^{+}$ by $-\text{NO}_2$ (C): Treatment with $\text{HBF}_4$ forms the diazonium fluoroborate; subsequent $\text{NaNO}_2/\text{Cu},\Delta$ replaces the diazonium group by a nitro group (nitro-de-diazoniation). The $-\text{NO}_2$ takes the position formerly held by $-\text{NH}_2$ (position 1), with the three bromines at 2, 4 and 6.
Numbering from the nitro group, the bromines sit at 3, 4-adjacent positions giving 1,3,5-tribromo-2-nitrobenzene.
Answer: B
Solution
Boiling point is governed by the strength of intermolecular forces (mainly hydrogen bonding here), for molecules of comparable molar mass:
- C — $n\text{-}C_4H_{10}$ (butane): only weak London dispersion forces, no H-bonding $\Rightarrow$ lowest bp ($\approx 0\,^\circ\text{C}$).
- D — $C_2H_5NHC_2H_5$ (diethylamine): a secondary amine with one $\text{N–H}$, so weak $\text{N}\cdots\text{H–N}$ hydrogen bonding ($\approx 56\,^\circ\text{C}$).
- B — $n\text{-}C_4H_9NH_2$ (butan-1-amine): a primary amine with two $\text{N–H}$ bonds, giving stronger H-bonding than a secondary amine ($\approx 78\,^\circ\text{C}$).
- A — $n\text{-}C_4H_9OH$ (butan-1-ol): $\text{O–H}$ is more polar than $\text{N–H}$, so alcohols H-bond most strongly $\Rightarrow$ highest bp ($\approx 118\,^\circ\text{C}$).
Increasing order: $C < D < B < A$.
Answer: C
Solution
For a base $B$ and its conjugate acid $BH^{+}$:
$$K_a(BH^{+}) \times K_b(B) = K_w \quad\Rightarrow\quad \text{weaker base} \Rightarrow \text{stronger conjugate acid}.$$So we need the weakest base among the substituted anilines:
- $-\text{OCH}_3$ (para): electron-donating (by resonance) $\Rightarrow$ stronger base.
- $-\text{CH}_3$ (para): weakly electron-donating $\Rightarrow$ slightly stronger base than aniline.
- $-\text{H}$ (aniline): reference.
- $-\text{NO}_2$ (para): strongly electron-withdrawing ($-M,\,-I$), it delocalises the lone pair on $\text{N}$ and destabilises $BH^{+}$ $\Rightarrow$ weakest base.
Since 4-nitroaniline is the weakest base, its conjugate acid ($4\text{-}O_2N\text{-}C_6H_4\text{-}NH_3^{+}$) is the strongest.
Answer: C
Solution
Statement I — FALSE. Benzamide undergoing the Hofmann bromamide degradation loses one carbon (the carbonyl carbon leaves as carbonate):
$$C_6H_5CONH_2 \xrightarrow{Br_2,\,NaOH} C_6H_5NH_2\ (\text{aniline}).$$The product is aniline, not benzylamine. (Benzylamine, $C_6H_5CH_2NH_2$, would require the same number of carbons, which the degradation does not give.)
Statement II — TRUE. In the strongly acidic nitrating mixture, aniline is protonated to the anilinium ion $C_6H_5NH_3^{+}$. The $-\overset{+}{N}H_3$ group is deactivating and meta-directing, so a substantial fraction of meta product forms. The observed distribution is roughly $\approx 47\%$ meta, $\approx 51\%$ para and only $\approx 2\%$ ortho. Hence m-nitroaniline is formed in higher amount than o-nitroaniline.
Statement I is false and Statement II is true.
Answer: D
Solution
The balanced Hofmann bromamide equation is:
$$R\text{-}CONH_2 + Br_2 + 4\,NaOH \longrightarrow R\text{-}NH_2 + Na_2CO_3 + 2\,NaBr + 2\,H_2O$$Evaluating each statement:
- A — FALSE. Per mole of amide, $4\ \text{mol NaOH}$ but only $1\ \text{mol Br}_2$ is consumed (not 2).
- B — FALSE. Alkyl amides do undergo the reaction (e.g. $CH_3CONH_2 \to CH_3NH_2$). It is not restricted to aryl amides.
- C — TRUE. The reaction converts a primary amide $R\text{-}CONH_2$ into a primary amine $R\text{-}NH_2$ (with one fewer carbon).
- D — FALSE. The rearrangement needs a primary amide ($-CONH_2$). N-substituted (secondary) amides lack the required $\text{N–H}$/migration pathway to give the analogous amine.
- E — TRUE. The by-products are exactly $Na_2CO_3$, $NaBr$ and $H_2O$, as shown in the balanced equation.
Only C and E are true.
Answer: C