Nitrogen-Containing Compounds

Question 1

(a) [4 marks] $pK_a(\text{conj. acid}) = 14 - pK_b$ at 25 °C.

• Methylamine: $14 - 3.36 = 10.64$
• Ammonia: $14 - 4.75 = 9.25$
• Aniline: $14 - 9.38 = 4.62$

[1 each for three values] Basicity order in water (higher $pK_a$ of conjugate acid ⇒ stronger base): methylamine > ammonia > aniline [1].

(b) [6 marks]

• Intrinsic (gas-phase) basicity depends only on the electronic stability of the free base vs its protonated cation — alkyl groups are electron-donating (+I) and stabilise the ammonium cation, so gas-phase basicity increases with alkylation; aniline’s N is still more basic than $NH_3$ intrinsically because polarizability of the ring stabilises the cation. [2]
• In water the measured basicity also includes solvation of the protonated cation by hydrogen bonding. The ammonium ion $R_nNH_{4-n}^+$ is stabilised by H-bond donation from N–H bonds; more alkyl substitution removes N–H bonds and increases steric shielding, reducing solvation stabilisation. [2]
• The net aqueous order (alkyl > NH₃ > aryl) is a balance between electronic (+I) stabilisation and hydration (solvation) of the cation; the reversal on going to the gas phase occurs because the solvation term vanishes. Dominant factor: differential solvation/hydration free energy of the ammonium cation. [2]

(c) [6 marks] Loss of resonance on protonating aniline makes aniline effectively harder to protonate by $\approx 25\ \text{kJ mol}^{-1}$, i.e. $\Delta(\Delta G) = 25\ \text{kJ mol}^{-1}$ favouring cyclohexylamine. [2]

$\ln\frac{K_b(\text{cyclo})}{K_b(\text{aniline})} = \frac{\Delta(\Delta G)}{RT} = \frac{25000}{8.314 \times 298} = 10.09$ [2]

$\frac{K_b(\text{cyclo})}{K_b(\text{aniline})} = e^{10.09} \approx 2.4 \times 10^{4}$ [2]

(Aniline is ~$10^4$ times weaker — consistent with the ~4.7 unit $pK_b$ gap.)

(d) [4 marks]

def pKb_to_pKa(pKb):
    # at 25 C, pKa(conjugate acid) + pKb = 14
    return 14 - pKb

[3] Output for aniline: pKb_to_pKa(9.38) → 4.62 [1].

Question 2

(a) [5 marks] $C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{0-5\,°C} C_6H_5N_2^+Cl^- + NaCl + 2H_2O$ [3] Reagents: $NaNO_2 + HCl$ (generates $HNO_2$ in situ), 0–5 °C. [1] Instability above 5 °C: the $C\text{–}N_2^+$ bond is weak; warming drives loss of $N_2$ (a very stable leaving group) giving a phenyl cation / phenol side products. Low temperature kinetically preserves the salt. [1]

(b) [10 marks — 2 each]

(Full marks require reagent + correct classification. Gattermann = Cu metal/HX for Cl, Br.)

(c) [7 marks] Product: 4-[(4-nitrophenyl)azo]phenol — $O_2N\text{–}C_6H_4\text{–}N=N\text{–}C_6H_4\text{–}OH\ (\text{para,para'})$ [3] (i) Phenol is activated by $-OH$ (strong +M/electron-releasing) directing electrophilic diazonium attack to ortho/para; para is preferred sterically. [2] (ii) The diazonium ion is a weak electrophile, so an activated ring (phenoxide/amine) is needed:

• Strongly alkaline: converts $ArN_2^+$ to unreactive diazotate $Ar\text{–}N=N\text{–}O^-$ / diazohydroxide → no electrophile.
• Strongly acidic: protonates phenol’s $-O^-$ back to $-OH$ (less activating) and (for amine substrates) protonates the amine deactivating the ring. Hence a mildly alkaline medium optimises both electrophile and nucleophile. [2]

Question 3

(a) [8 marks]

• $A = C_7H_7NO_2$ with N and two O ⇒ aromatic nitro compound: A = p-nitrotoluene, $CH_3\text{–}C_6H_4\text{–}NO_2$. Reduction by $Sn/HCl$ converts $-NO_2 \to -NH_2$. [3]
• $B = C_7H_9N$: B = p-toluidine (4-methylaniline), $CH_3\text{–}C_6H_4\text{–}NH_2$. (Formula matches: $C_7H_7NO_2 + 6[H] \to C_7H_9N + 2H_2O$.) [2]
• $B + CHCl_3/KOH(alc)$ = carbylamine reaction → C = p-tolyl isocyanide, $CH_3\text{–}C_6H_4\text{–}NC$ (foul odour confirms isocyanide). [2]
• Positive Hinsberg (primary) and azo coupling with $\beta$-naphthol (aromatic primary amine → diazonium → dye) both confirm primary aromatic amine. [1]

(b) [5 marks] $CH_3C_6H_4NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} CH_3C_6H_4NC + 3KCl + 3H_2O$ [3] Diagnostic only for primary amines because the mechanism requires the amine to attack dichlorocarbene ($:CCl_2$, generated from $CHCl_3 + KOH$) and then eliminate two HCl using two N–H hydrogens. Secondary/tertiary amines lack the two N–H bonds needed to form the terminal isocyanide, so they give no carbylamine. [2]

(c) [5 marks] Index of hydrogen deficiency $= \dfrac{2C + 2 + N - H}{2}$ (O ignored).

• For A ($C_7H_7NO_2$): $\text{IHD} = \frac{2(7)+2+1-7}{2} = \frac{10}{2} = 5$. [2] A benzene ring = 4 (3 π + 1 ring); the $N=O$ of nitro contributes 1 → total 5. ✓ [1]
• For B ($C_7H_9N$): $\text{IHD} = \frac{2(7)+2+1-9}{2} = \frac{8}{2} = 4$ = one aromatic ring exactly. ✓ [2]

[
  {"claim": "pKa conjugate acids: methylamine 10.64, ammonia 9.25, aniline 4.62",
   "code": "vals=[round(14-x,2) for x in [3.36,4.75,9.38]]; result = (vals==[10.64,9.25,4.62])"},
  {"claim": "Kb ratio cyclohexylamine/aniline from 25 kJ/mol is ~2.4e4",
   "code": "import math; r=math.exp(25000/(8.314*298)); result = abs(r-2.4e4)/2.4e4 < 0.05"},
  {"claim": "IHD of C7H7NO2 (A) equals 5",
   "code": "C,H,N=7,7,1; ihd=(2*C+2+N-H)//2; result = (ihd==5)"},
  {"claim": "IHD of C7H9N (B) equals 4 (one aromatic ring)",
   "code": "C,H,N=7,9,1; ihd=(2*C+2+N-H)//2; result = (ihd==4)"}
]