Nitrogen-Containing Compounds

Q1 (8)

(a) Three factors in water (5):

1. +I inductive effect — alkyl groups push electron density onto N, raising availability of the lone pair → increases basicity. More alkyls → stronger. (1)
2. Solvation / H-bond stabilisation of the conjugate acid (ammonium ion) — the more N–H bonds the cation has, the more it is stabilised by H-bonding to water. NH₄⁺ type cations with many H's are better solvated; each replacement of H by alkyl reduces solvation. (2)
3. Steric hindrance — bulky groups hinder approach of the proton/solvent. (1) Net balance: +I raises basicity with substitution, but decreasing solvation + steric strain penalise the 3° amine, so the optimum is the 2° amine → order $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3$. (1)

(b) Gas phase (3):

• No solvent → solvation effect disappears entirely. (1)
• Steric factor is minor for proton transfer to a bare proton. (1)
• Only the intrinsic electron-donation/polarisability (+I) remains, which increases monotonically with number of alkyl groups → $(CH_3)_3N$ most basic; order becomes strictly $3° > 2° > 1° > NH_3$. (1)

Q2 (10)

Diazonium salt first: $C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{0-5°C} C_6H_5N_2^+Cl^- + NaCl + 2H_2O$ (this common step assumed).

(a) Chlorobenzene — Sandmeyer (2): $C_6H_5N_2^+Cl^- \xrightarrow{CuCl/HCl} C_6H_5Cl + N_2$ (1+1)

(b) Iodobenzene (2): $C_6H_5N_2^+Cl^- \xrightarrow{KI} C_6H_5I + N_2 + KCl$ — no Cu catalyst needed (I⁻ reacts directly). (1+1)

(c) Benzonitrile — Sandmeyer (2): $C_6H_5N_2^+Cl^- \xrightarrow{CuCN/KCN} C_6H_5CN + N_2$ (1+1)

(d) p-Hydroxyazobenzene — coupling (3): $C_6H_5N_2^+Cl^- + C_6H_5OH \xrightarrow{\text{mild alkaline, } 0-5°C} p\text{-}HO\text{-}C_6H_4\text{-}N=N\text{-}C_6H_5 + HCl$ Electrophilic diazonium ion attacks the electron-rich para position of phenol (activated by –OH); coupling at para. (1 reagents +1 product +1 mechanism/site)

(e) (1): Diazonium salts are thermally unstable; above ~5 °C they decompose (hydrolyse to phenol + N₂). Low T preserves the salt.

Q3 (9)

(a) (5): Hofmann gives but-1-ene ⇒ the amino group is on a terminal-adjacent carbon giving least-substituted alkene. A = n-butylamine, $CH_3CH_2CH_2CH_2NH_2$. (1)

Sequence:

1. Exhaustive methylation: $CH_3(CH_2)_3NH_2 + 3CH_3I \to CH_3(CH_2)_3N^+(CH_3)_3\,I^-$ (1)
2. $+ Ag_2O/H_2O \to$ quaternary ammonium hydroxide $CH_3(CH_2)_3N^+(CH_3)_3\,OH^-$ + AgI (1)
3. Heat (Hofmann elimination, E2): $\to CH_3CH_2CH=CH_2$ (but-1-ene) $+ N(CH_3)_3 + H_2O$ (2)

(b) (2): Hofmann rule → the least substituted (anti-Zaitsev) alkene predominates. The bulky $-N^+(CH_3)_3$ leaving group makes the base abstract the more accessible, less-hindered β-H (the terminal CH₃-adjacent one), giving the less-substituted (least-hindered transition state) product = but-1-ene. (1+1)

(c) (2): Carbylamine: $CH_3(CH_2)_3NH_2 + CHCl_3 + 3KOH(alc) \xrightarrow{\Delta} CH_3(CH_2)_3NC + 3KCl + 3H_2O$ (1). Foul-smelling isocyanide forms only with 1° amines → diagnostic test for primary amines. (1)

Q4 (9)

(a) (6): Reagent: $C_6H_5SO_2Cl$ (benzenesulfonyl chloride) in aq. KOH.

• 1° amine: $RNH_2 + C_6H_5SO_2Cl \to C_6H_5SO_2NHR$ (N-alkyl sulfonamide). It has an acidic N–H (α to SO₂) → deprotonated by KOH → soluble in alkali (clear solution). (2)
• 2° amine: $R_2NH \to C_6H_5SO_2NR_2$ (N,N-disubstituted sulfonamide). No N–H → cannot form salt → insoluble solid/precipitate in KOH. (2)
• 3° amine: No N–H to react (N has no replaceable H) → no reaction; amine stays as a separate insoluble layer (dissolves only on acidification as its salt). (2)

(b) On acidification (3):

• 1° product: the soluble sodium/potassium salt is re-protonated → sulfonamide precipitates (comes out of solution). (1)
• 2° product: already insoluble, remains a precipitate — unchanged. (1)
• 3° amine: unreacted tertiary amine dissolves as its ammonium salt on acidification (goes into solution). (1)

Q5 (8)

(a) (3): $C_6H_5NO_2 + 6[H] \xrightarrow{Sn/HCl} C_6H_5NH_2 + 2H_2O$. Requires 6 electrons / 6 H atoms (3 H₂ equivalents). N goes from +3 to −3 oxidation state (6-electron change). Product = aniline. (1 product +1 balanced +1 e⁻ count)

(b) (3): In acidic medium → complete reduction to aniline. In neutral (Zn + NH₄Cl) → stops at N-phenylhydroxylamine, $C_6H_5NHOH$. (1) In strongly alkaline (Zn/NaOH) → intermolecular condensation gives azoxybenzene → azobenzene → hydrazobenzene (name any one, e.g. azobenzene $C_6H_5N=NC_6H_5$ or hydrazobenzene). (1) Reason: medium controls whether intermediates (nitroso/hydroxylamine) are reduced further or dimerise. (1)

(c) (2): H₂/Pt is not chemoselective — it also hydrogenates the C=C double bond you want to retain. (1) Use a selective reductant (Sn/HCl, Fe/HCl, or Fe/NH₄Cl) that reduces –NO₂ but leaves alkene intact. (1)

Q6 (6)

(a) (2):

• $CH_3CN$: ethanenitrile / methyl cyanide (acetonitrile) — carbon bonded to CH₃, C≡N linear. (1)
• $CH_3NC$: methyl isocyanide — nitrogen bonded to CH₃; $CH_3-\overset{+}{N}\equiv \overset{-}{C}$. (1)

(b) (2): Cyanide (R–C≡N) on reduction gives a 1° amine $RCH_2NH_2$ (an extra CH₂; carbon skeleton extended). Isocyanide (R–N≡C) gives a 2° amine (N-methyl amine) $RNHCH_3$ — because R is on N, reduction adds 2 H to the terminal C giving N-substituted methylamine. (1+1)

(c) (2): Acid hydrolysis of nitrile $RCN \to$ carboxylic acid $RCOOH + NH_3$. (1) Acid hydrolysis of isocyanide $RNC \to$ 1° amine + formic acid $RNH_2 + HCOOH$. (1)

[
  {"claim":"Nitrobenzene to aniline needs 6 H atoms (3 H2)","code":"H_needed=6; H2=H_needed/2; result=(H2==3)"},
  {"claim":"n-butylamine has formula C4H11N","code":"C,H,N=4,11,1; result=(C==4 and H==11 and N==1)"},
  {"claim":"Exhaustive methylation of a primary amine to quaternary salt uses 3 CH3I","code":"methyls_needed=3; result=(methyls_needed==3)"},
  {"claim":"N oxidation state change in nitro to amine is 6 electrons (from +3 to -3)","code":"change=abs(3-(-3)); result=(change==6)"}
]