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

Definition Symbols & shorthand used on this page (read once, then you never get stuck)

$Δ$ — a triangle written over or beside a reaction arrow means simply "heat the mixture" (Δ is the Greek letter delta, long used by chemists as the sign for applied heat).
$≫$ — read as =="much greater than"==. So " $A ≫ B$ " means A is enormously bigger than B, not just a little. In basicity, $N H_{3} ≫ aniline$ means aniline is far weaker, by orders of magnitude — not a close call.
$>$ — plain "greater than": bigger, but not by a landslide.
+I — an alkyl group pushing electron density toward N (donating). −I — a group pulling density away.
+R / −R — the same push/pull but through resonance (delocalisation into a $π$ system), not through the $σ$ -bonds.
β-hydrogen — a hydrogen on the carbon next door to the carbon bearing the leaving group.

Level 1 — Recognition

Problem 1.1

Which nitrogen below is least basic in water, and in one phrase say why? $(a) C H_{3} N H_{2} (b) (C H_{3})_{2} N H (c) N H_{3} (d) C_{6} H_{5} N H_{2}$

Recall Solution 1.1

Answer: (d) aniline ( $C_{6} H_{5} N H_{2}$ ). Why: in aniline the nitrogen lone pair is delocalised into the benzene ring by resonance, so it is not sitting freely on N ready to grab $H^{+}$ . A lone pair "busy" in resonance is a poor base. Also, protonating aniline destroys that resonance stabilisation, so the conjugate acid is high-energy → the base is weak. Reading the master order: $2° > 1° > 3° > N H_{3} ≫ aniline$ . The three " $>$ " signs are ordinary gaps; the " $≫$ " (recall from the symbols box: "much greater than") marks a huge drop — aniline is weaker by many orders of magnitude, not by a hair. See Resonance & Inductive Effects.

Problem 1.2

Name the test that gives a foul-smelling isocyanide and say which class of amine gives a positive result.

Recall Solution 1.2

Test: the carbylamine (isocyanide) test — heat the amine with $C H C l_{3}$ + alcoholic KOH. Positive for: primary amines only (1° aliphatic or aromatic). Reason (one line): KOH + $C H C l_{3}$ makes dichlorocarbene $: C C l_{2}$ ; a 1° amine has exactly two N–H bonds needed to build the $R - N \equiv C$ linkage. 2° and 3° amines lack the right number of N–H bonds → no smell.

Level 2 — Application

Problem 2.1

Rank in order of decreasing basicity in water: $N H_{3}, C H_{3} N H_{2}, (C H_{3})_{2} N H, (C H_{3})_{3} N$

Recall Solution 2.1

Answer: $(C H_{3})_{2} N H > C H_{3} N H_{2} > (C H_{3})_{3} N > N H_{3}$ . Why, step by step (this page carries the reasoning — no need to leave for the parent note):

The +I effect (alkyl groups donating electron density into N) wants: more methyls → richer lone pair → $(C H_{3})_{3} N$ strongest. Alone, this would give $3° > 2° > 1° > N H_{3}$ .
But protonation makes a cation, and that cation needs stabilising by water. Water stabilises the cation by making hydrogen bonds to each remaining N–H bond. Count them: $C H_{3} N H_{3}^{+}$ has 3 N–H bonds; $(C H_{3})_{2} N H_{2}^{+}$ has 2; $(C H_{3})_{3} N H^{+}$ has only 1. Fewer N–H bonds = fewer water H-bonds = less solvation stabilisation.
So a base's real strength in water = (+I gain) − (solvation loss). The two effects pull in opposite directions as you add methyls: +I keeps rising, solvation keeps falling. Their sum peaks at the secondary amine.
$N H_{3}$ has no +I donation at all → weakest. The competition of these two energy terms is drawn out explicitly in the figure below — look at how the blue (+I) line rises while the pink (solvation) line falls, and how their sum bends over at 2°.

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

Problem 2.2

An amine gives $p K_{b} = 3.4$ . Compute (a) its $K_{b}$ , and (b) the $p K_{a}$ of its conjugate acid at 25 °C (use $p K_{a} + p K_{b} = 14$ ).

Recall Solution 2.2

(a) $p K_{b} = - lo g K_{b} \Rightarrow K_{b} = 1 0^{- p K_{b}} = 1 0^{- 3.4} \approx 3.98 \times 1 0^{- 4}$ . (b) $p K_{a} = 14 - p K_{b} = 14 - 3.4 = 10.6$ . Read it as: a small $p K_{b}$ (3.4) means a fairly strong base; its conjugate acid holds the proton tightly, so a high $p K_{a}$ (10.6) — the two are mirror images through 14.

Level 3 — Analysis

Problem 3.1

In the gas phase the order of methylamine basicity is not the same as in water. Give the gas-phase order and explain what changes.

Recall Solution 3.1

Gas-phase order: $(C H_{3})_{3} N > (C H_{3})_{2} N H > C H_{3} N H_{2} > N H_{3}$ . What changed — grounded, not deferred: Recall from Problem 2.1 that a base's strength in water is a tug-of-war between two energy contributions: the +I inductive term (rises with more methyls) and the solvation term (the stabilisation water gives the cation by H-bonding to its N–H bonds, which falls with more methyls). In the gas phase there is no water at all, so the solvation term is simply zero — there is nothing to H-bond to the cation. Removing that falling term leaves only the +I term, which rises monotonically. So the order becomes strictly $3° > 2° > 1° > N H_{3}$ . Read it off the figure of Problem 2.1: delete the pink (solvation) line entirely and you are left with just the blue (+I) line, which climbs straight up — that is the gas-phase order. The water anomaly ( $2° > 3°$ ) only existed because the pink line was there pulling 3° down.

Problem 3.2

2,4-dinitroaniline has $p K_{b}$ much larger (weaker base) than aniline itself. Explain using resonance/induction.

Recall Solution 3.2

Answer: the two $N O_{2}$ groups make it a much weaker base than aniline. Why:

$N O_{2}$ is strongly electron-withdrawing by both −I (induction, pulling density through the $σ$ -bonds) and −R (resonance, pulling density through the ring's $π$ system).
Placed ortho/para, the nitro groups pull the nitrogen lone pair deeper into the ring (extra resonance structures place positive character on N). The lone pair is even less available than in plain aniline.
Therefore it grabs $H^{+}$ even more reluctantly → higher $p K_{b}$ (weaker base). This is the same logic as aniline, amplified. Compare with the diazonium/aniline chemistry in Diazonium Salts & Aniline Reactions.

Level 4 — Synthesis

Problem 4.1

Starting from butan-2-amine $C H_{3} C H_{2} C H (N H_{2}) C H_{3}$ , carry out a Hofmann elimination. Draw the three steps and predict the major alkene.

Recall Solution 4.1

Step 1 — Exhaustive methylation (excess $C H_{3} I$ ): $C H_{3} C H_{2} C H (N H_{2}) C H_{3} excess C H_{3} I C H_{3} C H_{2} C H (N + (C H_{3})_{3}) C H_{3} I^{-}$ Step 2 — Convert to hydroxide (moist $A g_{2} O$ ): swap $I^{-}$ for $O H^{-}$ . Step 3 — Heat (the " $Δ$ " over a reaction arrow just means apply heat — see the symbols box at the top): $O H^{-}$ removes a β-hydrogen (a hydrogen on the carbon next to the one carrying the leaving group) and neutral $N (C H_{3})_{3}$ leaves.

The nitrogen is on C2. There are two β-carbons: C1 (a $C H_{3}$ ) and C3 (a $C H_{2}$ ).

Remove a β-H from C1 → but-1-ene ( $C H_{2} = C H - C H_{2} C H_{3}$ ), the less substituted alkene.
Remove a β-H from C3 → but-2-ene, the more substituted alkene.

Hofmann's rule: the huge $N^{+} (C H_{3})_{3}$ leaving group makes the base seek the least hindered β-H (the C1 methyl H's) → but-1-ene is the major product. The figure below draws the carbon chain, marks where the bulky leaving group sits, and shows the two β-H choices — follow the blue arrow (easy-to-reach C1 H, major) versus the pink arrow (crowded C3 H, minor). Contrast with ordinary E2 on an alkyl bromide, which would favour Zaitsev (but-2-ene). See E2 Elimination & Zaitsev vs Hofmann Orientation.

Problem 4.2

You are handed three unlabelled bottles: 1° amine, 2° amine, 3° amine. Design a single reagent test to tell all three apart, and state exactly what you would observe.

Recall Solution 4.2

Reagent: the Hinsberg reagent = benzenesulfonyl chloride $C_{6} H_{5} S O_{2} C l$ , then treat the mixture with aqueous NaOH/KOH. Observations (the key is: how many N–H bonds survive?):

1° amine → $R - N H - S O_{2} C_{6} H_{5}$ : one N–H remains, made acidic by the electron-withdrawing $S O_{2}$ . It dissolves in NaOH (forms a soluble sodium salt) → clear solution. Acidify → precipitate returns.
2° amine → $R_{2} N - S O_{2} C_{6} H_{5}$ : no N–H left → cannot form a salt → insoluble precipitate persists in NaOH.
3° amine → no reaction (no N–H to react); the amine stays as an oily, insoluble free base and does dissolve on adding dilute acid (because it's still a basic amine). Summary: dissolves in base = 1°; stays solid in base = 2°; unreacted oil = 3°.

Level 5 — Mastery

Problem 5.1

An unknown liquid amine:

gives no foul smell with $C H C l_{3}$ /alc. KOH,
with Hinsberg reagent + NaOH gives an insoluble precipitate. Identify the amine class and justify using both results together.

Recall Solution 5.1

Class: secondary (2°) amine. Reading the two clues together:

No carbylamine smell → it is not a primary amine (needs two N–H bonds to make an isocyanide). So it is 2° or 3°.
Hinsberg gives an insoluble precipitate that does NOT dissolve in NaOH → the sulfonamide formed has no N–H to lose ( $R_{2} N - S O_{2} C_{6} H_{5}$ ). A 3° amine would have given no reaction at all (an oily free base that dissolves in acid, not a persistent solid in base).
Only a 2° amine matches both: fails carbylamine and forms a base-insoluble sulfonamide.

Problem 5.2

For $C H_{3} N H_{2}$ , given $K_{b} = 4.4 \times 1 0^{- 4}$ , and for aniline $K_{b} = 4.0 \times 1 0^{- 10}$ : (a) compute both $p K_{b}$ values, (b) how many powers of ten stronger a base is methylamine, (c) in one line, what physical difference causes this gap?

Recall Solution 5.2

(a) $p K_{b} (C H_{3} N H_{2}) = - lo g (4.4 \times 1 0^{- 4}) \approx 3.36$ $p K_{b} (aniline) = - lo g (4.0 \times 1 0^{- 10}) \approx 9.40$ (b) Ratio of $K_{b}$ 's: $\frac{4.4 \times 1 0 ^{- 4}}{4.0 \times 1 0 ^{- 10}} = 1.1 \times 1 0^{6}$ → methylamine is about a million times (six orders of magnitude) the stronger base. Equivalently, the $p K_{b}$ gap $9.40 - 3.36 = 6.04 \approx 6$ . (c) In methylamine the lone pair sits freely on N (helped by +I of $C H_{3}$ ); in aniline it is delocalised into the ring by resonance and hence far less available to bind $H^{+}$ .

Active Recall

Recall One-line answers (cover them)

Order of methylamine basicity in water? ::: $(C H_{3})_{2} N H > C H_{3} N H_{2} > (C H_{3})_{3} N > N H_{3}$ Same order in gas phase? ::: $(C H_{3})_{3} N > (C H_{3})_{2} N H > C H_{3} N H_{2} > N H_{3}$ (pure +I, no solvation) What does $Δ$ over a reaction arrow mean? ::: apply heat What does $≫$ mean in a basicity order? ::: "much greater than" — a huge gap, orders of magnitude Hofmann elimination gives which alkene? ::: the least substituted (Hofmann) alkene Carbylamine test is positive for? ::: primary amines only In Hinsberg, which amine dissolves in NaOH? ::: primary (its sulfonamide N–H is acidic) How to tell 2° from 3° after Hinsberg? ::: add dilute HCl — 3° free amine dissolves, 2° sulfonamide does not