Question bank — Amines — basicity (alkyl - NH₃ - aryl in water; reverse in gas phase), Hofmann elimination, carbylamine, Hinsberg te

False. In water this is only partly true — the tertiary cation loses H-bond solvation, so (CH₃)₂NH ($p{K}_{b}3.29$) beats (CH₃)₃N ($p{K}_{b}4.28$). Only in the gas phase (no solvation) is it strictly true.

True. With no solvent there is no solvation term — only the +I inductive push remains, and three methyls donate most, so 3° > 2° > 1° > NH₃.

False. Aniline's lone pair is delocalised into the ring, so it's far less available to grab H⁺; aniline ($p{K}_b\approx 9.4$, i.e. very weak) sits well above ammonia's $p{K}_b\approx 4.75$ — remember larger $p{K}_b$ = weaker base.

True. A stable R–NH₃⁺ is reluctant to hand H⁺ back, so the equilibrium sits toward the protonated form — that is exactly what "strong base" (large $K_b$, small $p{K}_b$) means.

False. The bulky, positively charged N⁺(CH₃)₃ leaving group forces the base to grab the least hindered β-H, giving the least substituted (Hofmann) alkene. (Full Zaitsev-vs-Hofmann contrast: E2 Elimination & Zaitsev vs Hofmann Orientation.)

False. It needs two N–H bonds on the same nitrogen to build the R–N≡C linkage; a 2° amine has only one, so no foul-smelling isocyanide forms.

False. A 3° amine has no N–H, so even if N attacks sulfur there is no proton to lose; it reverts and stays as a free (oily) amine — no reaction.

True. The 1° sulfonamide keeps one N–H made acidic by the electron-withdrawing (−I) SO₂ group, so NaOH removes it to give a soluble salt.

False. Once protonated its nitrogen has no free lone pair, so it cannot accept another H⁺ — it can only act as an acid (give H⁺ back). The point is that aniline is hard to protonate in the first place: protonation destroys the ring resonance, which costs energy, and that cost is exactly why aniline is a weak base.

The +I of 3° is larger and helps basicity — the error is the cause. 3° falls because its cation has only one N–H, so it is poorly solvated by water; solvation loss, not induction, sinks it.

Induction is minor here; the real reason is resonance — the lone pair is delocalised into the ring and locked away from H⁺.

Hofmann is sterically (kinetically) controlled by the bulky leaving group, not thermodynamically — it deliberately gives the less stable, less substituted alkene.

The active species is dichlorocarbene :CCl₂, generated by KOH removing HCl from CHCl₃; the carbene, not CHCl₃, reacts with the amine.

The 2° sulfonamide R₂N–SO₂C₆H₅ has no N–H, so it cannot form a salt — it stays an insoluble precipitate.

Reversed. Since $p{K}_b=-\log K_b$, a smaller $p{K}_b$ means a larger $K_b$ = stronger base.

The leaving group is deliberately huge (N⁺(CH₃)₃); its bulk is the whole point — it steers the base to the least hindered β-H.

No — Ag₂O is an ion-exchange reagent: it swaps the salt's I⁻ for OH⁻ (Ag⁺ traps I⁻ as AgI). The base that later abstracts the β-H, on heating, is the OH⁻ counter-ion.

Water stabilises the cation by H-bonding to its N–H bonds; 3° has fewest N–H so loses most solvation. Remove the water (gas phase) and that penalty vanishes, leaving pure +I → 3° on top.

The strongly electron-withdrawing (−I) SO₂ group stabilises the conjugate base R–N⁻–SO₂C₆H₅ by pulling charge off nitrogen, making that N–H acidic enough for NaOH to remove.

Only a 1° amine has the two N–H bonds needed to condense with dichlorocarbene into the isocyanide R–N≡C, whose stench is the diagnostic sign; 2°/3° cannot make it.

I⁻ is a weak, non-basic counter-ion and cannot abstract the β-H. Moist Ag₂O exchanges I⁻ for OH⁻ (a strong base), so on heating the OH⁻ counter-ion removes the β-H and elimination proceeds.

Grabbing H⁺ ties up the lone pair as an N–H bond; with no free lone pair, the nitrogen can no longer attack electrophiles. This is the same lone pair doing both jobs.

It caps nitrogen with three methyls to build the bulky N⁺(CH₃)₃ leaving group; without a positive, bulky leaving group there is nothing for OH⁻ to eliminate against.

NH₃ ($p{K}_{b}4.75$) is weaker than all three simple alkyl amines but still far stronger than aniline: 2° > 1° > 3° > NH₃ ≫ aniline.

Nothing diagnostic — with no two-N–H pattern it forms no isocyanide, so there is no foul smell (same negative result as a 2° amine).

Positive. The test works for any primary amine, aliphatic or aromatic, since only two N–H bonds are required. (Aniline's other reactions: Diazonium Salts & Aniline Reactions.)

E2-style Hofmann elimination is impossible without a β-H; instead it undergoes substitution/other pathways rather than forming an alkene.

No. Nitrogen has no lone pair left (all four bonds to carbon), so it cannot accept H⁺; it is a non-basic cation, not an amine base.

Add NaOH first: the 1° sulfonamide dissolves (acidic N–H → salt) giving a clear solution, while the 2° sulfonamide stays as a precipitate — the solubility difference distinguishes them immediately.

No. Aprotic solvents cannot H-bond to the cation's N–H bonds, so the solvation term that penalised 3° largely disappears; the order shifts back toward the gas-phase +I order (3° highest). The 2°>3° quirk is a protic-water artefact.

Weaker rescue. Less polar / less protic → less stabilisation of the charged conjugate acid, so the pure inductive (+I) order is progressively unmasked — this is the smooth bridge between the water order and the gas-phase order.