Worked examples — Aldehydes and ketones — preparation; nucleophilic addition; aldol, Cannizzaro, Wittig, Claisen-Schmidt, Mannich, Reforma

Before anything, two pieces of shorthand we lean on constantly:

Read the figure below like this: the black dot in the centre is the carbonyl carbon (labelled "delta +" in orange — it is electron-poor). The teal dot up top is oxygen (labelled "delta -" — it hogs electrons). The plum dot to the left is the α-carbon, and the small orange dot hanging off it is the α-H. The plum arrow shows the base swooping in to grab exactly that H. If a molecule has no orange α-H dot, none of the enolate reactions on this page can even start — that single visual absence is the fork every example below hinges on.

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The scenario matrix

Every question in this topic is really asking "which lever gets pulled?". The levers are: does the molecule have an α-H, what nucleophile shows up, and how crowded/polarised is the carbonyl carbon. Here is the full grid of cases we must cover.

Cell	Case class	Decides the outcome	Example
A	Symmetric self-aldol (both partners identical, have α-H)	α-H present + base	Ex 1
B	Crossed aldol, only ONE partner has α-H	one enolate source only	Ex 2
C	Zero α-H on both aldehydes + conc. base	disproportionation	Ex 3
D	Zero α-H + CN⁻ catalyst (umpolung)	nucleophile flip	Ex 4
E	Reactivity ordering (sign of $\delta +$, steric size)	limiting/comparative	Ex 5
F	Wittig — regiochemically exact C=C	pick the fragments	Ex 6
G	Reformatsky vs base (degenerate "which reagent")	mild Zn enolate	Ex 7
H	Crossed Cannizzaro with HCHO (extreme donor)	limiting hydride-donor	Ex 8
I	Mannich — enol + iminium electrophile	3-component C–C bond	Ex 9
J	Real-world word problem (synthesis planning)	multi-step chaining	Ex 10
K	Exam twist — trap that looks like aldol	must not misfire	Ex 11

Cross off each cell as we hit it.

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Ex 1 — Cell A: symmetric self-aldol

Forecast: two identical aldehydes each could be enolate or electrophile — guess the carbon count of the product before reading on.

First, one piece of jargon in step 3:

1. Base removes an α-H from one ${\text{CH}}_3\text{CHO}$, giving the enolate $[{\text{CH}}_2\text{-CHO}{]}^{-}$ (the negative charge sits on the α-carbon, delocalised onto the carbonyl oxygen). Why this step? Acetaldehyde HAS α-H (the ${\text{CH}}_3$), so it can form an enolate — this is Cell A's defining trait.
2. Enolate carbon attacks the carbonyl carbon of a second acetaldehyde $\Rightarrow$ β-hydroxy aldehyde ${\text{CH}}_3{\text{CH(OH)CH}}_2\text{CHO}$ (the aldol, 4 carbons). Why this step? The enolate is nucleophilic C; the untouched aldehyde's C is $\delta +$ electrophile.
3. Warm $\Rightarrow$ E1cb dehydration to ${\text{CH}}_3\text{CH=CH-CHO}$ (but-2-enal). Why this step? Losing water gives a conjugated C=C–C=O, whose overlapping π systems make it extra-stable, so heat drives the equilibrium toward it.

Verify: carbons in = $2\times 2=4$; carbons out $=4$. ✓ No atoms lost except the water in step 3. Product is conjugated as claimed.

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Ex 2 — Cell B: crossed aldol, only one α-H source

Forecast: four possible pairings exist — will we get a mess?

1. List who can be the enolate. $\text{PhCHO}$ has no α-H (the carbon next to C=O is the aromatic ring, no removable H). Acetophenone's ${\text{CH}}_3$ has α-H. Why this step? Only a molecule with α-H can be nucleophile; this immediately kills 3 of the 4 pairings.
2. Acetophenone becomes the enolate, benzaldehyde is forced to be the electrophile. Why this step? Since only acetophenone can lose an α-H, the roles are fixed — enolate = acetophenone, electrophile = benzaldehyde — which is exactly why no product mixture forms.
3. Enolate attacks $\text{PhCHO}$, then dehydrates $\Rightarrow$ chalcone $\text{PhCH=CH-COPh}$ (this is Claisen–Schmidt). Why this step? The nucleophilic enolate C bonds to the $\delta +$ carbonyl C of benzaldehyde; heat then does the same E1cb dehydration as Ex 1 to give a conjugated enone.

Verify: count C: $\text{PhCHO}=7$, ${\text{PhCOCH}}_3=8$; chalcone $\text{PhCH=CHCOPh}=15$; $7+8=15$ minus water (only H, O lost). ✓ Single clean product because only one enolate was possible.

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Ex 3 — Cell C: zero α-H, disproportionation

Forecast: no α-H means no aldol — so what fills the gap?

1. Check α-H: none. So the enolate route is closed. Why this step? This is the fork — no α-H ⇒ Cannizzaro, not aldol.
2. ${\text{OH}}^{-}$ adds to one $\text{PhCHO}$ giving a tetrahedral alkoxide ${\text{PhCH(O}}^{-}\text{)OH}$. Why this step? With no enolate available, the only nucleophile around is ${\text{OH}}^{-}$ itself, so it does a plain nucleophilic addition onto the carbonyl carbon.
3. Hydride transfer: that alkoxide shoves its C–H (as ${\text{H}}^{-}$) onto a second $\text{PhCHO}$. Why this step? The alkoxide's lone pair pushes down to reform a C=O, expelling the C–H as a hydride; this is the disproportionation — one molecule oxidised, one reduced.
4. Donor → ${\text{PhCOO}}^{-}$ (oxidised); acceptor → ${\text{PhCH}}_2\text{OH}$ (reduced). Why this step? The hydride donor loses an H and gains overall bonds to O (oxidised to carboxylate); the acceptor gains an H (reduced to alcohol).

Verify: oxidation state of the carbonyl C: starts at $+1$ (aldehyde). In acid ${\text{PhCOO}}^{-}$ it is $+3$; in ${\text{PhCH}}_2\text{OH}$ it is $-1$. Average $=(+3-1)/2=+1$ = starting value. ✓ Disproportionation conserves total oxidation state.

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Ex 4 — Cell D: zero α-H + CN⁻ (umpolung)

Forecast: same reactant (no α-H) but different reagent — guess whether we get a C–C bond.

1. ${\text{CN}}^{-}$ adds to $\text{PhCHO}$, giving $\text{PhCH(OH)CN}$ region. Why this step? CN⁻ is a good nucleophile AND, crucially, can leave later — so it can act as a temporary "helper" that we get back at the end.
2. The former carbonyl H becomes acidic — CN stabilises the resulting carbanion. The carbonyl carbon, once $\delta +$, is now nucleophilic. This polarity flip is umpolung. Why this step? Only by flipping the polarity can the originally electrophilic carbonyl carbon become the nucleophile that forms the new C–C bond — that flip is the whole point of using CN⁻.
3. This carbanion attacks a second $\text{PhCHO}$, forming a C–C bond. Why this step? Now that carbon is nucleophilic, it targets the still-electrophilic carbonyl C of a fresh benzaldehyde, joining the two molecules.
4. ${\text{CN}}^{-}$ leaves (regenerated catalyst), unmasking a ketone $\Rightarrow$ benzoin $\text{PhCH(OH)-CO-Ph}$. Why this step? CN⁻ was only a temporary handle; when it departs, the C=O it was masking is revealed, and the catalyst is free to start again.

Verify: C count: $2\times 7=14$; benzoin $\text{PhCH(OH)COPh}=14$. ✓ CN⁻ appears and disappears (catalyst), net atoms = two benzaldehydes joined.

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Ex 5 — Cell E: reactivity ordering (limiting/comparative)

Forecast: two effects fight — write which two before reading.

First the electronic idea we use in step 1:

1. Effect 1 — how much $\delta +$? Each ${\text{CH}}_3$ donates electrons (+I effect), reducing the positive charge on carbonyl C. Why this step? Less $\delta +$ ⇒ weaker target for Nu⁻.
2. Effect 2 — crowding. More/larger R groups physically block the incoming nucleophile. Why this step? A blocked carbonyl carbon is physically harder to reach, so addition slows even if the charge were the same.
3. Count R (alkyl) groups: HCHO has 0 alkyl, acetaldehyde 1, acetone 2. Why this step? Both effects above scale with the number of alkyl groups, so simply counting them tells us the reactivity order at a glance — more alkyls means both less $\delta +$ AND more crowding, so slower.

Read the bar chart below: each bar's height is relative reactivity (taller = faster addition). Notice the height drops as the "number of alkyl groups" label under each bar rises from 0 → 1 → 2. The take-home: the two effects (less $\delta +$ AND more crowding) both point downward together, which is why the trend is so reliable.

Verify (sanity): the order matches the general rule from the parent, $\text{HCHO}>\text{RCHO}>{\text{R}}_2\text{CO}$. Fewer alkyls ⇒ more reactive. ✓

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Ex 6 — Cell F: Wittig, exact double-bond placement

Forecast: the C=C can be split two ways — pick the disconnection.

Two words you need first:

1. Cut the target at the double bond. The two fragments are $\text{PhCH=}$ and ${\text{=CHCH}}_3$. Why this step? In Wittig the new C=C sits exactly between the old carbonyl C and the ylide C — so we reverse-engineer from the bond.
2. Choose: carbonyl = $\text{PhCHO}$ (benzaldehyde), ylide = ${\text{Ph}}_3{\text{P=CHCH}}_3$ (from ethyl-triphenylphosphonium + base). Why this step? One fragment must come from a carbonyl (the C=O carbon) and the other from an ylide; benzaldehyde has no α-H so it is the clean, unambiguous carbonyl partner.
3. Combine: ylide C attacks carbonyl C → oxaphosphetane → collapses, ejecting ${\text{Ph}}_3\text{P=O}$. Why this step? The very stable P=O bond is the thermodynamic reward that drives the ring collapse and locks the C=C exactly where the carbonyl used to be.

Verify: C in target $=6(\text{Ph})+3=9$. Ph in ${\text{Ph}}_3\text{PO}$ is a by-product, not in target. Benzaldehyde gives 7 C, the ylide's ${\text{=CHCH}}_3$ gives 2 C → $7+2=9$. ✓ Double bond lands precisely where carbonyl was — no isomer ambiguity.

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Ex 7 — Cell G: Reformatsky vs base (degenerate "which reagent")

Forecast: both could make an enolate — what goes wrong with base?

1. Base danger: $\text{NaOEt}$ would deprotonate the ester and trigger the ester self-condensing (Claisen ester condensation) — messy. Why this step? Strong base makes a reactive ester enolate that attacks another ester molecule instead of the aldehyde — an unwanted side reaction that ruins yield.
2. Zn-enolate formation: $\text{Zn}$ metal inserts into the C–Br bond of ${\text{BrCH}}_2\text{COOEt}$ to give an organozinc ${\text{BrZnCH}}_2\text{COOEt}$, i.e. a mild zinc enolate $[{\text{CH}}_2\text{-COOEt}{]}^{-}\cdot {\text{ZnBr}}^{+}$. (See Grignard reagents and organometallics for the organometallic idea.) Why this step? The zinc enolate is nucleophilic enough for the more reactive aldehyde carbonyl but too weak to touch the less reactive ester — so we get the C–C bond we want and nothing else.
3. Add to $\text{PhCHO}$: the zinc enolate carbon bonds to the aldehyde's $\delta +$ carbon, giving a zinc alkoxide. Why this step? The nucleophilic zinc-enolate carbon attacks the aldehyde's electrophilic carbonyl carbon — the C–C bond-forming step we planned.
4. Aqueous workup (${\text{H}}_3{\text{O}}^{+}$) protonates the alkoxide → ${\text{PhCH(OH)CH}}_2\text{COOEt}$. Why this step? The negative oxygen picks up ${\text{H}}^{+}$ from water to become the neutral β-hydroxy ester alcohol.

Verify: C count: $\text{PhCHO}=7$, the ${\text{CH}}_2\text{COOEt}$ fragment $=4$; product ${\text{PhCH(OH)CH}}_2\text{COOEt}=7+4=11$ C. ✓ Br leaves (as ZnBr(OH) salts on workup), Zn is removed in workup; exactly one controlled C–C bond is formed and the ester survives intact.

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Ex 8 — Cell H: crossed Cannizzaro with HCHO (extreme donor)

Forecast: two aldehydes both lack α-H — who wins the hydride tug-of-war?

1. Both lack α-H ⇒ Cannizzaro, not aldol. Why this step? Neither partner can make an enolate, so the enolate route is closed and the hydride-transfer (Cannizzaro) route takes over.
2. HCHO is the best hydride donor (its carbon is least hindered, most willing to give up ${\text{H}}^{-}$). Why this step? The donor is the one that gets oxidised; HCHO's tiny, unhindered carbon gives up hydride most readily, so it always volunteers as donor.
3. HCHO → formate ${\text{HCOO}}^{-}$ (oxidised); PhCHO → ${\text{PhCH}}_2\text{OH}$ (reduced, our target). Why this step? Having donated its hydride, HCHO ends up oxidised to formate; benzaldehyde, having received a hydride, is reduced to benzyl alcohol.

Verify (oxidation states): HCHO carbon $+0\to +2$ in ${\text{HCOO}}^{-}$ (lost H⁻, oxidised ✓). PhCHO carbonyl C $+1\to -1$ in ${\text{PhCH}}_2\text{OH}$ (gained H⁻, reduced ✓). Electron bookkeeping: 2 lost by HCHO, 2 gained by PhCHO — balanced. ✓

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Ex 9 — Cell I: Mannich reaction (enol + iminium)

Forecast: three different molecules go in — guess how many end up joined in the product.

First, the special electrophile:

1. Amine + $\text{HCHO}$ lose water → iminium ion $[{\text{CH}}_2={\text{N(CH}}_3{)}_2{]}^{+}$. Why this step? The amine adds to formaldehyde, then dehydrates; this manufactures the strong, positively charged electrophile the enol needs.
2. Acetone tautomerises to its enol ${\text{CH}}_2={\text{C(OH)CH}}_3$ (nucleophilic α-carbon). Why this step? Acetone HAS α-H, so under mild acid it forms an enol — see Enols and enolates — keto-enol tautomerism — giving a nucleophilic carbon.
3. Enol carbon attacks the iminium carbon → new C–C bond → β-amino ketone ${\text{CH}}_3{\text{COCH}}_2{\text{CH}}_2{\text{N(CH}}_3{)}_2$. Why this step? The nucleophilic enol α-carbon bonds to the electrophilic iminium carbon, stitching all three starting pieces together into one molecule.

Verify: C count: acetone $3$ + HCHO $1$ + dimethylamine $2=6$; product ${\text{CH}}_3{\text{COCH}}_2{\text{CH}}_2{\text{N(CH}}_3{)}_2$ has $3+1+2=6$ C. ✓ Only water is lost; all three fragments end up in one β-amino ketone.

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Ex 10 — Cell J: real-world synthesis word problem

Forecast: two operations are needed — name the intermediate.

1. Oxidise benzyl alcohol → benzaldehyde with PCC (anhydrous, stops at aldehyde; see Alcohols — oxidation and PCC). Why this step? We need a carbonyl electrophile that has no α-H so the crossed aldol is clean.
2. Crossed aldol of $\text{PhCHO}$ + acetaldehyde ${\text{CH}}_3\text{CHO}$ (which supplies the enolate), dilute base, then heat. Why this step? Only acetaldehyde has α-H ⇒ single product, exactly Cell B logic.
3. Dehydration gives conjugated $\text{PhCH=CH-CHO}$. Why this step? E1cb loss of water forms the stabilising conjugated C=C–C=O, delivering the target aroma molecule.

Verify: target C $=9$. $\text{PhCHO}=7$ + ${\text{CH}}_3\text{CHO}=2\Rightarrow 9$ ✓ (water lost, no carbons lost). PCC must be used, not ${\text{KMnO}}_4$, or we'd overshoot to $\text{PhCOOH}$.

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Ex 11 — Cell K: exam twist (trap that looks like aldol)

Forecast: looks like a routine aldol… is it?

1. Look for α-H. The α-carbon here is the quaternary $\text{C}({\text{CH}}_3{)}_3$ — it has no hydrogen (all four bonds used by C's). Why this step? No α-H ⇒ no enolate can form ⇒ no aldol.
2. So the question is a trap. With conc. base this aldehyde instead does Cannizzaro (it is exactly the no-α-H case). Why this step? Once aldol is impossible, the only base-driven fate left for a no-α-H aldehyde is hydride-transfer disproportionation.
3. Correct answer: disproportionation to $({\text{CH}}_3{)}_3{\text{C-COO}}^{-}$ + $({\text{CH}}_3{)}_3{\text{C-CH}}_2\text{OH}$. Why this step? One molecule donates hydride (oxidised to carboxylate), the other accepts it (reduced to alcohol) — the standard Cannizzaro outcome.

Verify (oxidation states): carbonyl C $+1\to +3$ (carboxylate, oxidised) and $+1\to -1$ (alcohol, reduced); average $+1$ conserved. ✓ The "obvious" aldol answer is wrong — this is the classic exam twist.

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Recall One-line decision flow (memorise this)

α-H present? ::: Yes → enolate chemistry (aldol / Claisen–Schmidt / Mannich / Reformatsky). α-H absent, conc. NaOH? ::: Cannizzaro (disproportionation). α-H absent, CN⁻? ::: Benzoin condensation (umpolung). Want an exact C=C? ::: Wittig (ylide + carbonyl). Crossed Cannizzaro donor with excess HCHO? ::: HCHO is oxidised; partner reduced. Amine + HCHO + enolisable ketone? ::: Mannich (β-amino ketone).