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

Before we start, two small vocabularies we will lean on the whole page:

Definition The four "electron-push/pull" labels (+I, –I, +M, –M)

Chemists shorthand how a group moves electron density with two letters. Read them as plain English:

+I ("inductive donate"): a group pushes electron density toward the atom of interest through the σ-bonds (like a gentle nudge along the chain). Alkyl groups do this — they make the N lone pair richer.
–I ("inductive withdraw"): the opposite — pulls electron density away through the σ-bonds (e.g. $- N O_{2}$ ).
+M ("mesomeric/resonance donate"): a group pushes a lone pair into the π-system (into the ring), e.g. $- O C H_{3}$ .
–M ("mesomeric/resonance withdraw"): a group drains π-electrons out of the ring into itself, e.g. $- N O_{2}$ . Rule of thumb for basicity: anything that adds electron density near N (+I, +M) makes it a stronger base; anything that removes it (–I, –M) makes it weaker. (More in Resonance & Inductive Effects.)

The scenario matrix

Every question this topic can ask is one of these cells:

#	Cell class	What makes it tricky	Hit by example
A	Basicity ranking in water	2° beats 3° (solvation twist)	Ex 1
B	Basicity ranking in gas phase	order flips to pure +I	Ex 2
C	Aryl vs alkyl basicity	resonance kills the lone pair	Ex 3
D	Substituent on aniline (limiting/edge)	–NO₂ vs –OCH₃ push the number further	Ex 4
E	Hofmann elimination — which alkene	least-substituted (anti-Zaitsev)	Ex 5
F	Degenerate elimination case	no β-H ⇒ no alkene at all	Ex 6
G	Carbylamine — positive vs negative	only 1° gives foul smell	Ex 7
H	Hinsberg — all three classes at once	soluble / insoluble / no-reaction	Ex 8
I	Real-world word problem	pick the right amine for a job	Ex 9
J	Exam twist (combined trap)	phase-swap + resonance in one	Ex 10

Ex 1 — Cell A: ranking four aliphatic amines in water

Forecast: you might guess "more methyls = more basic, so $(C H_{3})_{3} N$ wins." Hold that thought — water disagrees.

Count the two competing effects for each. Why this step? In water, basicity $=$ (+I electron donation) $-$ (solvation loss of the cation, i.e. lost N–H·····water H-bonds — see the "solvation loss" definition above). We must weigh both, not just +I.

Amine # of $C H_{3}$ (+I) N–H bonds in cation (solvation)

$N H_{3}$ 0 4

$C H_{3} N H_{2}$ 1 3

$(C H_{3})_{2} N H$ 2 2

$(C H_{3})_{3} N$ 3 1
Find the sweet spot. Going $N H_{3} \to$ 1° $\to$ 2°, +I keeps rising and solvation only drops a little — basicity climbs. Going 2° $\to$ 3°, the cation now has just one N–H to H-bond, so solvation collapses and outweighs the extra +I. Why this step? This identifies the maximum: the balance tips at 2°.
Write the order. $(C H_{3})_{2} N H > C H_{3} N H_{2} > (C H_{3})_{3} N > N H_{3}$

Amine	# of $C H_{3}$ (+I)	N–H bonds in cation (solvation)
$N H_{3}$	0	4
$C H_{3} N H_{2}$	1	3
$(C H_{3})_{2} N H$	2	2
$(C H_{3})_{3} N$	3	1

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

Figure s01 — read the four bars one by one. Each bar is a $p K_{b}$ value, and since smaller $p K_{b}$ = stronger base, a shorter bar means a stronger base. Trace them left to right: the green 2° bar is the shortest — this is the champion, exactly as step 2 predicted. The blue 1° bar is a hair taller (slightly weaker). Then comes the orange 3° bar, which has jumped up — that rise is the solvation penalty biting: its cation has only one N–H to anchor water. Finally the gray $N H_{3}$ bar is tallest of the four (weakest), sitting just above 3°; that tiny 3°→NH₃ gap is exactly how little three methyls end up buying you in water once solvation is spent. Read together, the bar heights are the ranking of step 3.

Verify: with real data $p K_{b}$ = $3.27, 3.36, 4.20, 4.75$ . Smaller = stronger, so the ascending- $p K_{b}$ list is $(C H_{3})_{2} N H, C H_{3} N H_{2}, (C H_{3})_{3} N, N H_{3}$ — exactly our order. ✓ (see Solvation and Hydrogen Bonding)

Ex 2 — Cell B: same four amines, but in the gas phase

Forecast: does removing water change anything? Guess before reading.

Delete the solvation term. Why this step? No water means there is no H-bond stabilisation of the cation. The subtraction in "(+I) $-$ (solvation loss)" loses its second term entirely.
Rank by +I alone. More methyls = more electron donation into the N lone pair, full stop. $(C H_{3})_{3} N > (C H_{3})_{2} N H > C H_{3} N H_{2} > N H_{3}$ Why this step? With only one effect left, the order becomes monotonic — every extra methyl helps.

Verify: compare with Ex 1. The 2°/3° pair swapped: in water $(C H_{3})_{2} N H > (C H_{3})_{3} N$ ; in gas $(C H_{3})_{3} N > (C H_{3})_{2} N H$ . Exactly one inversion, and it is the pair whose difference was the solvation term. ✓

Ex 3 — Cell C: alkyl amine vs aniline

Forecast: aniline "looks bigger" — does bigger mean more basic? No.

Ask "is the lone pair free?" Why this step? Basicity needs an available lone pair. In aniline the lone pair is delocalised into the benzene ring (a –M / resonance drain), so it is partly spent (see Resonance & Inductive Effects).
Compare $p K_{b}$ . $3.4 < 9.4$ , so $C H_{3} N H_{2}$ is far stronger.
Turn the $p K_{b}$ gap into a ratio. A difference of $6.0$ in $p K_{b}$ means a factor of $1 0^{6.0}$ in $K_{b}$ . $\frac{K _{b} ( C H _{3} N H _{2} )}{K _{b} ( aniline )} = 1 0^{9.4 - 3.4} = 1 0^{6.0} = 1, 000, 000$ Why this step? Because $p K_{b}$ is a log, each unit is a factor of ten; six units is a million-fold difference.

Verify: $K_{b} = 1 0^{- p K_{b}}$ gives $1 0^{- 3.4}$ vs $1 0^{- 9.4}$ ; methylamine's $K_{b}$ is the larger number, so it is the stronger base, and the ratio is $1 0^{6}$ . ✓

Ex 4 — Cell D: substituted anilines (edge/limiting cases)

Forecast: does adding a group to the ring always weaken the base? Not always — it depends on the group's sign.

Classify each substituent's effect on the N lone pair. Why this step? A group that withdraws electrons ( $- N O_{2}$ : both –I through σ-bonds and –M through the ring) drains the lone pair further → weaker base. A group that donates ( $- O C H_{3}$ : +M, pushing its lone pair into the ring) partly refills the density → stronger base than plain aniline.
Order them. $most basic p - C H_{3} O - C_{6} H_{4} - N H_{2} > C_{6} H_{5} N H_{2} > least basic p - O_{2} N - C_{6} H_{4} - N H_{2}$
Sanity-check the extreme. p-Nitroaniline is one of the weakest common anilines ( $p K_{b} \approx 13$ ): the $- N O_{2}$ is the limiting electron-sink case. Why this step? Checking the extreme confirms the trend direction, not just the middle.

Verify: literature $p K_{b}$ : p-anisidine $\approx 8.7$ , aniline $\approx 9.4$ , p-nitroaniline $\approx 13.0$ . Ascending $p K_{b}$ = descending basicity = our order. ✓

Ex 5 — Cell E: Hofmann elimination product

Forecast: E2 usually gives the more-substituted (Zaitsev) alkene. Does Hofmann agree? No.

Locate the β-hydrogens. The α-carbon (the one bearing $N^{+} (C H_{3})_{3}$ ) has two neighbours — its β-carbons: a $-CH_{2} -$ (leading to but-2-ene) and a $-CH_{3}$ (leading to but-1-ene). Why this step? Elimination removes a β-H; we must see which β-carbons exist first.
Apply Hofmann's rule — bulky, charged leaving group. $N^{+} (C H_{3})_{3}$ is huge. The base $O H^{-}$ reaches the least hindered β-H, which is on the terminal $C H_{3}$ . Why this step? The transition state is sterically controlled (see E2 Elimination & Zaitsev vs Hofmann Orientation); crowding, not alkene stability, decides.
Name the product: removing the terminal $C H_{3}$ β-H gives the less substituted alkene, but-1-ene ( $C H_{2} = C H - C H_{2} - C H_{3}$ ).

Figure s02 — follow the two coloured arrows. The orange blob perched above the middle carbon is the bulky leaving group $N^{+} (C H_{3})_{3}$ ; the carbon under it is the α-carbon (labelled). Now look at its two β-neighbours. The green arrow points to a β-H on the terminal $C H_{3}$ on the right — it sticks out into open space, so the base $O H^{-}$ can swing in and pluck it easily; that path gives but-1-ene, the major product. The red arrow points to a β-H on the internal $C H_{2}$ on the left — it is boxed in by the rest of the chain, so the crowded base struggles to reach it; that path gives but-2-ene, the minor product. The figure's whole message: sterics (reachability), not alkene stability, choose the winner.

Verify: Hofmann on 2-butyl trimethylammonium hydroxide is a textbook case yielding but-1-ene as the major product (~95% vs but-2-ene). Anti-Zaitsev ✓.

Ex 6 — Cell F: degenerate case — no β-hydrogen

Forecast: it's a quaternary ammonium hydroxide, so surely Hofmann elimination → an alkene?

Look for a β-hydrogen. Every carbon attached to N is a $C H_{3}$ ; that carbon is the α-carbon. There is no β-carbon at all — nowhere for a β-H to live. Why this step? Elimination is impossible without a β-H; this is the degenerate input.
Conclude: no alkene. On strong heating it instead undergoes substitution ( $O H^{-}$ attacks a methyl) giving $C H_{3} O H + (C H_{3})_{3} N$ . Why this step? When the intended pathway is blocked, the reagent takes the only other route available.

Verify: the smallest amine that can Hofmann-eliminate needs a $\geq 2$ -carbon alkyl chain (an ethyl or longer). Tetramethylammonium has all one-carbon groups ⇒ β-H count $= 0$ ⇒ elimination forbidden. ✓

Ex 7 — Cell G: carbylamine, positive vs negative

Forecast: guess how many bottles stink.

Recall the requirement. The carbylamine (isocyanide) reaction needs an amine with two N–H bonds to build the $R - N \equiv C$ linkage from dichlorocarbene $: C C l_{2}$ . Why this step? Only a 1° amine ( $R - N H_{2}$ ) has exactly two N–H bonds.
Test each.
- $C H_{3} N H_{2}$ : two N–H → foul-smelling methyl isocyanide $C H_{3} - N \equiv C$ . Positive.
- $(C H_{3})_{2} N H$ : one N–H → no isocyanide. Negative.
- $(C H_{3})_{3} N$ : zero N–H → no reaction. Negative.
Answer: only one bottle (the primary, $C H_{3} N H_{2}$ ) smells foul.

Verify: balance the full equation $C H_{3} N H_{2} + C H C l_{3} + 3 K O H \to C H_{3} N C + 3 K C l + 3 H_{2} O$ atom by atom.

C: left $= 1 (C H_{3} N H_{2}) + 1 (C H C l_{3}) = 2$ ; right $= 2 (C H_{3} N C) = 2$ . ✓
N: left $= 1 (C H_{3} N H_{2})$ ; right $= 1 (C H_{3} N C)$ . ✓
Cl: left $= 3$ (all in $C H C l_{3}$ ); right $= 3$ (all in $3 K C l$ ). ✓
K: left $= 3$ ; right $= 3$ . ✓
H: left $= 5 (C H_{3} N H_{2}) + 1 (C H C l_{3}) + 3 (3 K O H) = 9$ ; right $= 3 (C H_{3} N C) + 6 (3 H_{2} O) = 9$ . ✓ (The two N–H hydrogens of $C H_{3} N H_{2}$ are the ones lost; the three methyl hydrogens ride through unchanged into $C H_{3} N C$ .)

Ex 8 — Cell H: Hinsberg test on all three classes

Forecast: predict "dissolves / precipitate / no reaction" for each before reading.

Count N–H bonds left after the amine attacks sulfur. Why this step? The whole test hinges on whether the sulfonamide product still carries an acidic N–H.
Walk each tube.
- 1° → $R - N H - S O_{2} C_{6} H_{5}$ : one N–H left, made acidic by the electron-withdrawing $S O_{2}$ (a –I pull). KOH removes it → soluble salt, clear solution.
- 2° → $R_{2} N - S O_{2} C_{6} H_{5}$ : no N–H left → cannot form a salt → insoluble precipitate persists.
- 3° → no reaction (no N–H to displace Cl) → stays as an oily free amine, unchanged.
Summarise the diagnostic: dissolves = 1°, precipitate = 2°, oily/no reaction = 3°.

Figure s03 — a decision tree you can run in your head. Start at the single gray box at the top: one reagent, $C_{6} H_{5} S O_{2} C l$ , is added to whichever amine you have. The three branches fan out by the one thing that matters — how many N–H bonds survive the reaction. Follow the left (blue) branch for a 1° amine: its product keeps one acidic N–H, so adding KOH pulls that proton off to make a salt → the bottom box is green: DISSOLVES to a clear solution. Follow the middle (orange) branch for a 2° amine: its product has zero N–H, cannot form a salt → the bottom box is orange: a PRECIPITATE that persists. Follow the right (red) branch for a 3° amine: it has no N–H to react in the first place, so nothing happens → the bottom box is red: an oily free amine, untouched. Pick any tube, drop onto its branch, and the observation at the bottom is forced — that is the whole diagnostic.

Verify: acidity of the 1° sulfonamide N–H is real — its conjugate base $R - N^{-} - S O_{2} C_{6} H_{5}$ is stabilised by $S O_{2}$ , giving $p K_{a} \approx 10$ , acidic enough for KOH ( $p K_{a H} \approx 15.7$ for water) to deprotonate. So 1° dissolves, 2° cannot. ✓ (link: Acid–Base Theory & pKa/pKb)

Ex 9 — Cell I: real-world word problem

Forecast: two conditions must hold at once — narrow it down.

Filter by "strong base in water." From Ex 1's order $(C H_{3})_{2} N H > C H_{3} N H_{2} > (C H_{3})_{3} N > N H_{3} ≫$ aniline, the top two are $(C H_{3})_{2} N H$ and $C H_{3} N H_{2}$ . Aniline is out (weak). Why this step? Condition (i) removes the weakest candidates immediately.
Filter by "positive carbylamine." Only 1° amines pass (Ex 7). Of the survivors, $(C H_{3})_{2} N H$ is 2° (fails), $C H_{3} N H_{2}$ is 1° (passes). Why this step? Condition (ii) is a hard yes/no gate that eliminates the 2° amine.
Answer: $C H_{3} N H_{2}$ (methylamine) — strong aqueous base and primary.

Verify: $C H_{3} N H_{2}$ has $p K_{b} \approx 3.4$ (2nd strongest of the list) and two N–H bonds (carbylamine-positive). Both conditions satisfied; no other candidate satisfies both. ✓

Ex 10 — Cell J: exam twist (phase + resonance combined)

Forecast: the sentence chains two claims. Attack each separately.

Check the gas-phase clause. In the gas phase, solvation is gone; only electronic effects remain. Even so, aniline's ring still delocalises the lone pair (the –M resonance drain), so aniline is not a strong gas-phase base — the resonance drain persists in both phases and is the real weakener. Why this step? The trap smuggles in a false-friendly premise ("gas phase helps aniline").
Check the "therefore." Gas-phase order and water-phase order need not track each other — the whole 2°/3° reversal (Ex 1 vs Ex 2) proves phase can flip rankings. So a gas-phase result cannot be exported to water by logic alone. Why this step? Even if the premise were true, the inference is invalid.
Correct picture: in water, aniline is far weaker than $N H_{3}$ ( $p K_{b} \approx 9.4$ vs $4.75$ ) because its lone pair is tied up in the ring and protonation destroys that resonance.

Verify: measured gas-phase proton affinities (kJ/mol) are aniline $\approx 883$ vs ammonia $\approx 854$ , so aniline is the slightly stronger base in the gas phase — the premise's fact happens to be true. But the water data $p K_{b} (aniline) = 9.4 > p K_{b} (N H_{3}) = 4.75$ makes aniline the weaker base in solution. So the statement's conclusion ("must also beat in water") is false — the phase cannot be transferred. ✓

Active Recall

Recall Cover the answers and self-test

Water vs gas: which class sits on top of each? ::: Water → 2°; Gas → 3°.
A $p K_{b}$ gap of 6 means what factor in $K_{b}$ ? ::: A million-fold ( $1 0^{6}$ ).
Hofmann gives which alkene, and why? ::: Least-substituted; bulky $N^{+} (C H_{3})_{3}$ + base reach the least-hindered β-H.
Which amine gives NO alkene under Hofmann? ::: One with no β-hydrogen, e.g. $(C H_{3})_{4} N^{+} O H^{-}$ .
Hinsberg: dissolves / precipitate / no reaction map to? ::: 1° / 2° / 3°.
Only which amines are carbylamine-positive? ::: Primary (1°) amines — they have two N–H bonds.

Adding an electron-donating group (+M) to aniline does what to

p K_{b}

Lowers it (stronger base), e.g. p-anisidine

p K_{b} \approx 8.7 <

aniline

9.4

Why can't a gas-phase basicity order be copied to water?

Water adds a solvation term that can invert rankings (the 2°/3° reversal proves it).