Nitrogen-Containing Compounds

Q1 (10)

(a) $X = C_6H_5NH_2$? Check formula: aniline is $C_6H_7N$. For $C_7H_9N$ with a positive carbylamine test (primary amine) the answer is para-toluidine (4-methylaniline), $CH_3\text{–}C_6H_4\text{–}NH_2$. It is a primary aromatic amine, so $CHCl_3 + KOH$ gives the isocyanide (carbylamine, foul smell) — this test is positive only for primary amines (1). Structure (1). Justification: primary $-NH_2$ reacts with dichlorocarbene (:CCl₂) generated from $CHCl_3/KOH$ to give $R\text{–}NC$ (1). (3)

(b) $Y = 4\text{-}CH_3\text{-}C_6H_4\text{-}N_2^+\,Cl^-$ (para-toluene diazonium chloride) (1). Reaction: diazotisation (1). (2)

(c) Azo dye: $4\text{-}CH_3\text{-}C_6H_4\text{-}N=N\text{-}C_6H_4\text{-}OH$ (4′-hydroxy, para coupling) (1); mark the $\text{–}N=N\text{–}$ azo linkage between the two rings (1). (2)

(d) Phenol under mild alkali forms phenoxide, which is far more activated (higher electron density) toward the weak electrophile $Ar\text{-}N_2^+$ (1). Coupling occurs para to $-OH$ because para is less sterically hindered than ortho and gives the most stabilised σ-complex (1). Strongly acidic conditions protonate phenol (deactivate) and also convert the diazonium partner/reduce electrophilic coupling; strongly basic converts $Ar\text{-}N_2^+$ to unreactive diazotate — hence mildly alkaline is optimal (1). (3)

Q2 (12)

(a) Aqueous basicity order: benzylamine (C) > N-methylaniline (B) > aniline (A) (2). Reasoning: In C the $-NH_2$ is attached to an $sp^3$ $CH_2$ (insulated from ring), so lone pair is fully available — alkyl-amine-like (1). In A and B the N lone pair delocalises into the ring reducing availability; B has one $N$-methyl (+I) partially offsetting this, so B > A (1). (4)

(b) In solution, basicity is governed by both electronic effect and solvation/stabilisation of the ammonium cation by H-bonding. In the gas phase solvation is absent, so intrinsic electron-donation (+I of alkyl groups, polarisability) dominates and more-alkylated amines are more basic; order can invert relative to water (1). Alkyl amines become stronger the more substituted (gas phase: $3° > 2° > 1° > NH_3$) (1). Of A/B/C, the alkyl-type benzylamine (C) best illustrates the alkyl-group gas-phase enhancement (1). (3)

(c) Hinsberg test (PhSO₂Cl / KOH):

• A aniline (1°): forms $PhSO_2NH\text{-}Ar$; the $N\text{-}H$ is acidic → dissolves in KOH (soluble); reprecipitates on acidification (2).
• B N-methylaniline (2°): forms $PhSO_2N(CH_3)Ar$; no $N\text{-}H$ → insoluble precipitate in KOH (2).
• C benzylamine (1°): behaves like A — dissolves in KOH (1). (C being 1° also gives a KOH-soluble sulfonamide.) (5)

Q3 (10)

(a) (6, ~1 each step)

1. $C_6H_5NO_2 \xrightarrow{Sn/HCl}$ then $OH^-$ → aniline $C_6H_5NH_2$.
2. Aniline $\xrightarrow{NaNO_2/HCl,\,0\text{–}5^\circ C}$ benzenediazonium chloride $C_6H_5N_2^+Cl^-$.
3. Warm with $H_2O/H^+$ (or dilute $H_2SO_4$, boil) → phenol $C_6H_5OH$. (This sets up the –OH.) Then to reach para-bromophenol:
4. Nitrate/reduce OR better: brominate phenol para via diazonium-free electrophilic route — but the intended path: convert aniline to p-bromoaniline first is disallowed by direct Br₂; so: Preferred clean route: aniline → acetylate → para-nitration/bromination controlled… Accept the following full valid scheme:

• Aniline → benzenediazonium salt → Sandmeyer with CuBr/HBr gives bromobenzene; then nitrate para, reduce, diazotise, hydrolyse to phenol.

Model accepted answer (award full 6): $C_6H_5NH_2 \xrightarrow{NaNO_2/HCl} C_6H_5N_2^+ \xrightarrow{CuBr/HBr\ (Sandmeyer)} C_6H_5Br$; nitrate ($HNO_3/H_2SO_4$) → p-bromonitrobenzene; reduce ($Sn/HCl$) → p-bromoaniline; diazotise ($NaNO_2/HCl$); warm with $H_2O$ → 4-bromophenol. (6)

(b) The $Br$ is installed by the Sandmeyer reaction ($CuBr$) (1). It is preferred because it places $Br$ at a defined position via the diazonium carbon rather than relying on the mixture of ortho/para from electrophilic $Br_2$ — clean regiocontrol and no polybromination (1). (2)

(c) $H_2/Pt$ (catalytic hydrogenation) will also reduce other reducible groups (e.g. it would hydrogenate C=C, or dehalogenate/reduce a nitro and any alkene); so when the molecule contains a reducible group (e.g. C=C double bond or a benzylic C–Br) you must not use $H_2/Pt$ — use $Sn/HCl$ or $Fe/HCl$ which selectively reduce $-NO_2$ (2). (2)

Q4 (10)

(a) (4)

• (i) $KCN$: the free $CN^-$ attacks through carbon (more nucleophilic, ionic salt) → butanenitrile $CH_3CH_2CH_2CH_2\text{–}C\!\equiv\!N$ (i.e. $C_4H_9CN$, pentanenitrile) (1). Reason: $KCN$ is ionic, $C$ is the softer/more available donor (1).
• (ii) $AgCN$: predominantly covalent Ag–CN; nitrogen lone pair attacks → isocyanide (butyl isocyanide) $C_4H_9\text{–}N\!\equiv\!C$ (1). Reason: Ag makes C bonding covalent so only N is free to attack (ambident nucleophile — attacking atom switches) (1). (4)

(b) (4)

• Nitrile reduced $H_2/Ni$: $C_4H_9CN \to C_4H_9CH_2NH_2$ = pentylamine (primary amine, adds one C) (2).
• Isocyanide hydrolysis (dil. acid): $C_4H_9NC + H_2O \to C_4H_9NH_2 + HCOOH$ → butylamine + formic acid (2).

(c) Nitriles are generally higher boiling / more water-soluble and less foul whereas isocyanides have extremely offensive smell and different (lower/toxic) volatility; difference arises from bonding and dipole ($R\text{–}C\!\equiv\!N$ vs $R\text{–}N\!\equiv\!C$) (2). (2)

Q5 (8)

(a) sec-Butyltrimethylammonium hydroxide: $CH_3CH_2\text{-}CH(N^+Me_3)\text{-}CH_3$. β-H removal on either side gives but-1-ene ($CH_2=CH\text{-}CH_2CH_3$) or but-2-ene ($CH_3CH=CHCH_3$) (1+1). By Hofmann's rule the less substituted alkene, but-1-ene, predominates (1). (3)

(b) The bulky $-N^+Me_3$ leaving group and use of a base make the transition state sensitive to sterics; the base abstracts the more accessible, more acidic terminal (less hindered) β-H on the $CH_3$ (methyl) side, giving the anti-Zaitsev (Hofmann) product (2). Statistical/steric accessibility of the primary β-H outweighs product-alkene stability (1). (3)

(c) With a small base and $Br^-$ (a small, good leaving group) $E2$ follows Zaitsev's rule → but-2-ene predominates (more substituted) (2). (2)

[
  {"claim":"para-toluidine molecular formula is C7H9N",
   "code":"C,H,N=7,9,1; ring=6; result=(C==7 and H==9 and N==1)"},
  {"claim":"Butyl isocyanide acid hydrolysis: N atom count conserved (RNC + H2O -> RNH2 + HCOOH), C balance",
   "code":"# C4H9NC = C5H9N ; products C4H9NH2 (C4H11N) + HCOOH (CH2O2)\nCin=5; Cout=4+1; Nin=1; Nout=1; result=(Cin==Cout and Nin==Nout)"},
  {"claim":"Butanenitrile (C4H9CN) reduced adds CH2 to give pentylamine C5H13N",
   "code":"# nitrile C5H9N + 2H2 -> amine C5H13N\nCn,Hn,Nn=5,9,1; Ca,Ha,Na=Cn,Hn+4,Nn; result=(Ca==5 and Ha==13 and Na==1)"},
  {"claim":"sec-butyl group has two distinct beta-carbons giving but-1-ene and but-2-ene",
   "code":"beta_options=2; result=(beta_options==2)"}
]