Visual walkthrough — Amines — basicity (alkyl - NH₃ - aryl in water; reverse in gas phase), Hofmann elimination, carbylamine, Hinsberg te
This page is the picture-by-picture derivation behind the parent topic. Read it slowly; each step has a figure that carries the argument.
Step 1 — What "base" even means, drawn as a handshake
WHAT. An amine is ammonia () with some of its hydrogen atoms swapped for carbon groups. The one feature that matters here is the lone pair — two electrons on the nitrogen that are not used in any bond. Draw them as a little pair of dots sitting on top of the N.
WHY we start here. A base, in the sense we need, is something that grabs a proton — a bare hydrogen nucleus, written . To grab it, the base offers its two spare electrons and forms a new bond. So "how good a base is this?" becomes "how eagerly does the lone pair reach out and grab ?"
PICTURE. In the figure, the lone pair (magenta dots) reaches across and clamps onto the orange . The result is a positively charged particle called the conjugate acid.
Step 2 — The real question: is the product comfortable?
WHAT. Whether the balance sits to the right depends on how happy (low energy, stable) the conjugate acid is. A stable cation has no wish to spit the proton back out, so the balance stays right — strong base.
WHY this reframing. It is easier to reason about a finished charged particle than about a fuzzy "eagerness." So we replace the question "how good a base?" with two concrete sub-questions:
- Before grabbing — is the lone pair rich (easy to donate)?
- After grabbing — is the resulting + charge well looked-after?
PICTURE. The figure splits into two panels: LEFT shows the neutral amine with its lone pair; RIGHT shows the charged conjugate acid, now needing help to feel stable. Two different physical effects act on these two panels — that is the crux of everything that follows.
Step 3 — Effect A: alkyl groups push electrons inward (+I)
WHAT. A carbon group like is slightly electron-pushing. It nudges electron density along the bond toward the nitrogen. This gentle push through bonds is the inductive effect, written (the + means "donates toward").
WHY it strengthens a base. More electron density piled onto N means a richer, fatter lone pair, which reaches out for more eagerly. So each extra alkyl group should, by this effect alone, make the amine a stronger base.
PICTURE. Green arrows point inward from each toward the central N. Count them: ammonia has zero, methylamine one, dimethylamine two, trimethylamine three. By Effect A alone the ranking is strictly .
Wanting to go deeper on why bonds transmit this push, see Resonance & Inductive Effects.
Step 4 — Effect B: water hugs the cation through its N–H bonds
WHAT. Now look only at the RIGHT panel — the charged . Water molecules surround any + charge and stabilise it. The specific glue here is the hydrogen bond: a water oxygen (electron-rich) reaches toward each N–H hydrogen and holds it. This wrapping-in-water is called solvation.
WHY it matters — and why it counts N–H bonds. Each N–H bond is a handle water can grab. So the number of N–H bonds left on the cation sets how many hugs it receives:
| Cation | Formula | N–H bonds (handles) | Water hugs |
|---|---|---|---|
| from | 4 | most | |
| from 1° | 3 | many | |
| from 2° | 2 | some | |
| from 3° | 1 | few |
Notice: more alkyl groups mean FEWER N–H handles, so Effect B runs in the opposite direction to Effect A.
PICTURE. The figure shows the tertiary cation with just one lonely N–H being hugged by water, next to the secondary cation with two N–H bonds getting hugged. Two hugs beat one.
Step 5 — In water: solvation wins the 2° vs 3° fight
WHAT. In liquid water both effects are live. Going 1° → 2°, Effect A's gain is large enough to still win, so basicity rises. But going 2° → 3°, the loss of that last water handle is bigger than the small extra inductive push — so basicity drops.
WHY 2° comes out on top. It is the sweet spot: enough alkyl groups for a strong +I boost, still enough N–H handles for good solvation. The tertiary amine over-committed to alkyl groups and paid for it in lost hugs.
PICTURE. A bar chart of basicity in water. The bars climb , then fall back at 3°. The peak is at 2°. Aniline (lone pair busy in a ring) sits far below everyone — we treat that separately in the parent note.
Step 6 — Take the water away: only Effect A survives
WHAT. In the gas phase there is no water. No water means no hydrogen bonds, which means Effect B (solvation) is switched off entirely. Only the through-bond push, Effect A, remains.
WHY the order snaps back to clean. With Effect B gone, the tug-of-war has only one contestant. Basicity now tracks the number of green inward arrows exactly — the monotonic ladder from Step 3 returns, unopposed.
PICTURE. The same four amines, but now floating in empty space (no water molecules drawn). The bar chart is a clean staircase: 3° highest, then 2°, 1°, lowest. Compare it side-by-side with the water chart from Step 5 — the 2° and 3° bars have swapped order.
Step 7 — The degenerate check: what if R does nothing?
WHAT. Consider the limiting case where the alkyl group's push is turned off — i.e. plain ammonia, , no carbon at all. Then Effect A contributes nothing and leans entirely on solvation (it has the most N–H handles: four).
WHY this matters. It confirms the two effects are truly independent. Ammonia is the most solvated cation yet the weakest aliphatic base, because with zero inductive help its lone pair was poor to begin with. Great solvation cannot rescue a lone pair that was never enriched.
PICTURE. shown wrapped in four water hugs — the most of anyone — yet its basicity bar (from Step 5) is the shortest of the aliphatic set. Maximum hugs, minimum push: it sits at the bottom.
The one-picture summary
WHAT. One figure holds the entire argument: two number lines. The top line (water) puts 2° at the peak with 3° knocked down below 1°. The bottom line (gas) is a clean staircase with 3° on top. An arrow between them is labelled "remove solvent → Effect B off."
Recall Feynman retelling — say it out loud in plain words
A nitrogen has a spare pair of electrons; a base uses that pair to grab a stray proton. Two things decide how good it is. One: the carbon groups attached push electrons inward, fattening the pair — more groups, more push, better base. Two: once the amine has grabbed a proton and gone positive, water hugs it through its N–H bonds — but every carbon group you add removes one N–H handle, so more groups means fewer hugs. In water both forces are alive and they pull opposite ways, so the winner is the secondary amine — enough push, still enough hugs — while the tertiary amine, having traded away its handles, slips below even the primary. Now pour the water out. No water, no hugs — the second force vanishes. Only the inward push is left, so the order becomes a plain staircase and the tertiary amine, with the most push, climbs to the top. The famous "flip" is nothing mysterious: it is just what happens when you switch off one of two competing forces.
Active Recall
Recall Cover and test yourself
Which effect is switched OFF in the gas phase? ::: Solvation (Effect B — the water hugs / hydrogen bonds). Which amine wins in water and why? ::: 2° — best balance of inductive push and enough N–H handles for solvation. Which amine wins in the gas phase and why? ::: 3° — only the inductive push counts, and it has the most alkyl groups. Why is the most solvated yet the weakest aliphatic base? ::: Most N–H handles (four) but zero inductive push, so its lone pair was never enriched. In one line, what causes the flip? ::: Removing the solvent removes Effect B, leaving only the monotonic inductive order.
Related vault topics: Acid–Base Theory & pKa/pKb, Resonance & Inductive Effects, Solvation and Hydrogen Bonding.