Aldehydes and ketones — preparation; nucleophilic addition; aldol, Cannizzaro, Wittig, Claisen-Schmidt, Mannich, Reforma: Visual walkthrough

Step 0 — The cast of characters (nothing assumed)

Before any arrow moves, let us name every part with a picture, because the whole reaction is a story about these three regions of one small molecule.

Our molecule is acetaldehyde, written $CH_{3} CHO$ . Read that left to right:

$CH_{3}$ — a carbon with three hydrogens. Call this the methyl carbon.
$C$ with a double line to $O$ and a single $H$ — this is the carbonyl carbon. "Carbonyl" is just the name for a carbon double-bonded to oxygen, $C=O$ .

In the figure the $C=O$ is coloured to show the charge split: oxygen pulls, so it wears $δ -$ (coral) and the carbonyl carbon wears $δ +$ (lavender). The three $α$ -hydrogens on the methyl are circled in mint — remember them, they are the trigger for everything.

Step 1 — WHY one α-hydrogen is unusually easy to remove

WHAT we do: a base (hydroxide, $OH^{-}$ ) removes one $α$ -hydrogen as $H^{+}$ , leaving its two electrons behind on the $α$ -carbon.

WHY it works here and not on an ordinary $CH_{3}$ : when the electron pair is left on the $α$ -carbon, it does not have to sit there lonely. It can slide over into the neighbouring $C=O$ , pushing the $π$ electrons up onto oxygen. The negative charge ends up parked on oxygen, which — being greedy — is delighted to hold it. A charge that can spread out over two atoms is far more stable than one stuck on one atom. That spreading is called delocalisation, and the resulting anion is the enolate.

PICTURE:

Step 2 — WHY the enolate is a nucleophile with a target

WHAT we do: we now have two kinds of molecule floating in the flask — some enolate anions (electron-rich at the $α$ -carbon) and plenty of un-reacted acetaldehyde (electron-poor at its carbonyl carbon).

WHY they will react: opposites attract. The enolate's $α$ -carbon is loaded with a lone pair — it is a nucleophile. A fresh acetaldehyde's carbonyl carbon is $δ +$ — an electrophile. This is the same nucleophilic-addition logic from the parent note, only here the nucleophile is itself a carbon, which is how a new C–C bond gets built.

PICTURE:

Recall Why does carbon attack carbon here, but not for two ordinary alkanes?

Because one carbon has an excess pair (enolate) and the other is electron-starved ( $δ +$ from the C=O). Ordinary carbons are neither, so nothing pulls them together. ::: The polarity created by the carbonyl is the entire reason a C–C bond can form.

Step 3 — The attack: forming the new C–C bond

WHAT we do: the enolate's $α$ -carbon reaches over and forms a bond to the carbonyl carbon of the second aldehyde. To make room, the $C=O$ $π$ electrons flop up onto oxygen — that oxygen becomes a negative alkoxide ( $O^{-}$ ).

WHY the $π$ bond breaks and not something else: the $π$ bond is the weakest bond available and its electrons have a wonderful place to go (onto electronegative oxygen). Nature always sends the electrons where they are happiest.

WHAT IT LOOKS LIKE: two three-carbon-and-two-carbon fragments zip together into one four-carbon chain, with a fresh $O^{-}$ dangling off the middle.

Step 4 — Grab a proton: the aldol appears

WHAT we do: the alkoxide $O^{-}$ is a strong base; it plucks a proton from a water molecule, becoming a neutral $-OH$ . This regenerates a hydroxide $OH^{-}$ , so the base is a catalyst — used and handed back.

WHY this happens instantly: a bare $O^{-}$ in water is unstable; protonation gives the calmer, neutral alcohol.

WHAT IT LOOKS LIKE: the molecule now has an $-OH$ on one carbon and a $-CHO$ two carbons away — an "alcohol" plus an "aldol". That is where the name aldol comes from: aldehyde + alcohol.

Step 5 — Heat it: dehydration to the conjugated product

WHAT we do: on warming, the molecule loses one water ( $-OH$ from $β$ -carbon + one $H$ from the $α$ -carbon) and forms a C=C double bond between the $α$ - and $β$ -carbons.

WHY heat drives it and why it is favoured: removing water is entropically helped by heat, but the deeper reason is the product. The new $C=C$ sits right next to the old $C=O$ : pattern $C=C-C=O$ . When a double bond neighbours another double bond, their $π$ systems overlap and share electrons — this is conjugation, and conjugated molecules are noticeably more stable (lower energy). The system rolls downhill into that stability.

WHY the α-H leaves (not some other H): only the $α$ -hydrogen can leave and still leave behind an anion stabilised by the neighbouring carbonyl — the same enolate trick as Step 1. The mechanism is called E1cb ("elimination, conjugate base") precisely because it goes through an enolate before kicking out the $OH$ as water.

WHAT IT LOOKS LIKE: the $-OH$ and an adjacent $-H$ pinch off together as $H_{2} O$ , and a flat, conjugated four-carbon molecule remains.

Step 6 — Edge case A: a partner with NO α-hydrogen (why the reaction can fail)

WHAT changes: suppose one of our aldehydes has no $α$ -hydrogen — for example benzaldehyde, $PhCHO$ , whose carbonyl carbon is attached to a benzene ring and an $H$ , with nothing removable next door.

WHY it matters: no $α$ -H means no enolate can ever form on that molecule. It cannot be the nucleophile. It can only sit and wait as the electrophile.

Consequence — the crossed (Claisen–Schmidt) fix: pair the α-H-less aldehyde (benzaldehyde) with a partner that does have α-H (e.g. acetaldehyde or acetophenone). Now only one enolate is possible, so only one product forms — clean, no messy mixture. This is exactly Claisen–Schmidt in the parent note.

Second consequence — total dead end: if both partners lack an α-H (e.g. two benzaldehydes with just base), no enolate exists at all, so aldol is impossible; the molecules take an entirely different escape route (Cannizzaro disproportionation). See Step 7.

Recall Why can't two benzaldehydes do an aldol?

Neither has an α-hydrogen, so neither can become an enolate; with no nucleophilic carbon, the C–C bond of Step 3 can never form. ::: No α-H ⇒ no enolate ⇒ no aldol.

Step 7 — Edge case B: no α-H on either partner ⇒ Cannizzaro instead

WHAT happens: with two α-H-less aldehydes and concentrated $NaOH$ , hydroxide finally does what it could not in Step 1 — it acts as a nucleophile and adds directly to a carbonyl carbon (there was no α-H to grab, so this is the only move left).

WHY this is a different story: the alkoxide that forms now hands a hydride ( $H^{-}$ , a hydrogen with its electrons) to a second aldehyde. The donor is oxidised to a carboxylate; the acceptor is reduced to an alcohol. One reactant, two opposite fates — a disproportionation.

PICTURE:

The one-picture summary

The single diagram above compresses the whole journey: one branch (α-H present) marches through enolate → C–C attack → aldol → conjugated product; the other branch (no α-H) forks off to Cannizzaro. The fork is decided by one yes/no question asked at the very start.

Recall Feynman retelling — say it to a friend with no notation

Imagine each acetaldehyde as a little person. On one shoulder it has a magnet (the C=O oxygen pulling electrons, leaving the carbon slightly positive), and in one hand it holds three loose hydrogens (the α-hydrogens next door).

A base walks by and gently takes one loose hydrogen. Now that hand is empty and electron-rich — the person has become a grabber (an enolate).

The grabber reaches out and grabs the positive shoulder of a second, still-normal person. That handshake is the new carbon–carbon bond. The oxygen on that shoulder, pushed out of the way, briefly becomes grumpy and negative, then snatches a hydrogen from water and calms down. Two people are now joined into one taller molecule with an $-OH$ and a $-CHO$ — the aldol.

Warm the room, and the molecule sheds a drop of water, tightening into a sleek, flat, extra-stable shape where two double bonds sit side by side (conjugation) — the final product.

But some aldehydes (like benzaldehyde) have no loose hydrogen in hand — nothing for the base to take. They can't become grabbers. If their partner also has none, they play a different game entirely: one hands a hydrogen to another, one becomes an acid, one becomes an alcohol — that's Cannizzaro.

The single question that decides everything: does this aldehyde have a hydrogen next to its C=O? ::: α-H → enolate → aldol → conjugated enal. No α-H → Cannizzaro. That fork is the whole chapter in one sentence.