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

WHAT to do: find the carbon(s) directly attached to $\text{C=O}$ and ask "do they carry any H?"

• (a) $\text{PhCHO}$: the only carbon touching $\text{C=O}$ is the benzene ring carbon — a ring carbon has no free H to give. NO.
• (b) ${\text{CH}}_3\text{CHO}$: the ${\text{CH}}_3$ sits next to $\text{C=O}$ and has 3 H. YES.
• (c) $\text{HCHO}$: both sides of $\text{C=O}$ are just H — there is no carbon next to $\text{C=O}$ at all, so no $\alpha$-carbon and no $\alpha$-H. NO.
• (d) acetone: two ${\text{CH}}_3$ groups flank $\text{C=O}$, six $\alpha$-H total. YES.

WHY it matters: NO $\alpha$-H (a, c) $\Rightarrow$ Cannizzaro / benzoin territory (and Wittig always works). YES (b, d) $\Rightarrow$ aldol family is available.

${\text{CN}}^{-}$ attacks the $\delta +$ carbon; the $\pi$ electrons drop onto oxygen giving an alkoxide ${\text{O}}^{-}$ (that ${\text{C-O}}^{-}$ we just defined); that oxygen grabs ${\text{H}}^{+}$. The product is a cyanohydrin: ${\text{CH}}_3\text{CH(OH)CN}$ One carbon now carries both an $-\text{OH}$ and a $-\text{CN}$ — the signature of a cyanohydrin.

• Stop at butanal ${\text{CH}}_3{\text{CH}}_2{\text{CH}}_2\text{CHO}$: use PCC (anhydrous).
• Overshoot to butanoic acid ${\text{CH}}_3{\text{CH}}_2{\text{CH}}_2\text{COOH}$: use ${\text{KMnO}}_4$ or ${\text{K}}_2{\text{Cr}}_2{\text{O}}_7/{\text{H}}_2\text{O}$.
• Why: in water the aldehyde forms a hydrate ${\text{RCH(OH)}}_2$, which is itself an alcohol-like species and gets re-oxidized. PCC has no water, so no hydrate forms, and oxidation halts at the aldehyde.

Step 1 (WHAT): base removes an $\alpha$-H from one propanal. The $\alpha$-carbon of propanal is the ${\text{CH}}_2$; removing one of its two H leaves a carbanion, written unambiguously as the enolate ${\text{CH}}_3{\text{CH}}^{-}\text{CHO}$ — i.e. that carbon now carries ${\text{CH}}_3$, one remaining H, and a lone-pair/negative charge, still bonded to $\text{CHO}$. Step 2 (WHY): this nucleophilic carbon attacks the $\delta +$ carbonyl carbon of a second propanal, forming a new C–C bond and a $\beta$-hydroxy aldehyde (the aldol): ${\text{CH}}_3{\text{CH}}_2\text{CH(OH)}\underset{\alpha }{{\text{CH(CH}}_3)}\text{CHO}$ Step 3 (heat — E1cb dehydration): using the E1cb route just defined, base removes the $\alpha$-H (now flanked by the C=O), and the resulting carbanion expels the $\beta$-$\text{OH}$ as ${\text{OH}}^{-}$, giving: ${\text{CH}}_3{\text{CH}}_2{\text{CH=C(CH}}_3)\text{CHO}(\text{2-methylpent-2-enal})$ WHY the conjugated enal is favoured (thermodynamics): in the product the new $\text{C=C}$ sits right next to the $\text{C=O}$ ($\text{C=C-C=O}$). These two double bonds share their $\pi$ electrons (conjugation), spreading electron density over four atoms and lowering the energy. That extra stabilization is the reward that pulls the reversible dehydration forward — heat simply pushes the system to this most stable, conjugated sink. Carbon count check: propanal is ${\text{C}}_3$; two of them make a ${\text{C}}_6$ product. ✓

The deciding question every time: is there an $\alpha$-H?

• (a) $\text{PhCHO}$ — no $\alpha$-H $\Rightarrow$ Cannizzaro: $2\text{PhCHO}\to {\text{PhCOO}}^{-}+{\text{PhCH}}_2\text{OH}$.
• (b) ${\text{CH}}_3\text{CHO}$ — has $\alpha$-H $\Rightarrow$ aldol, then dehydration to ${\text{CH}}_3\text{CH=CH-CHO}$ (but-2-enal).
• (c) pivaldehyde — the neighbouring carbon is a quaternary ${\text{C(CH}}_3{)}_3$ carbon: it has no H. So no $\alpha$-H $\Rightarrow$ Cannizzaro: $2({\text{CH}}_3{)}_3\text{CCHO}\to ({\text{CH}}_3{)}_3{\text{CCOO}}^{-}+({\text{CH}}_3{)}_3{\text{CCH}}_2\text{OH}$.

WHAT this teaches: "$\alpha$-H present" is a structural question about the neighbour carbon, not about whether the molecule "looks reactive."

Mechanism recap: ${\text{OH}}^{-}$ adds to one aldehyde to make an alkoxide; that alkoxide donates a hydride (${\text{H}}^{-}$) to a second aldehyde. The hydride donor is oxidized (becomes carboxylate); the acceptor is reduced (becomes alcohol).

• $\text{HCHO}$ is the best hydride donor (smallest, least hindered, most electron-poor alkoxide pushes ${\text{H}}^{-}$ out easily), so it is oxidized to formate ${\text{HCOO}}^{-}$.
• Therefore $\text{PhCHO}$ is reduced to benzyl alcohol ${\text{PhCH}}_2\text{OH}$.

Products: ${\text{HCOO}}^{-}{\text{Na}}^{+}+{\text{PhCH}}_2\text{OH}$.

Use benzaldehyde $\text{PhCHO}$ + acetophenone ${\text{PhCOCH}}_3$ with dilute base (Claisen–Schmidt).

• $\text{PhCHO}$ has no $\alpha$-H $\Rightarrow$ it can only be the electrophile.
• ${\text{PhCOCH}}_3$ has $\alpha$-H $\Rightarrow$ it only forms the enolate (the nucleophile).
• Because each partner has exactly one role, only the crossed product forms; no self-aldol scramble. $\text{PhCHO}+{\text{PhCOCH}}_3\stackrel{\text{dil. NaOH}}{\to }\text{PhCH=CH-COPh}+{\text{H}}_2\text{O}$ The final C=C is conjugated with both the ring and the C=O — that stability drives dehydration.

The C=C forms at the former carbonyl carbon, so split the target there:

• Carbonyl = benzophenone ${\text{Ph}}_2\text{C=O}$ (supplies the ${\text{Ph}}_2\text{C}$ end).
• Ylide = methylenetriphenylphosphorane ${\text{Ph}}_3{\text{P=CH}}_2$ (supplies the ${\text{CH}}_2$ end). ${\text{Ph}}_2\text{C=O}+{\text{Ph}}_3{\text{P=CH}}_2⟶{\text{Ph}}_2{\text{C=CH}}_2+{\text{Ph}}_3\text{P=O}$ By-product: triphenylphosphine oxide ${\text{Ph}}_3\text{P=O}$ — the very stable $\text{P=O}$ bond is the driving force. See Grignard reagents and organometallics for the carbanion-character parallel of the ylide carbon.

E/Z stereochemistry (the edge case): E/Z labels describe which side the two ends of a double bond sit on — they only exist when each alkene carbon carries two different groups. Here the new double bond is ${\text{Ph}}_2{\text{C=CH}}_2$: one carbon has two identical $\text{Ph}$ groups, the other has two identical $\text{H}$. With a matched pair on at least one carbon, there is no E vs Z distinction — this specific target is stereochemically unambiguous. General rule to remember: when E/Z does arise (both carbons bear two different groups), an unstabilized ylide like ${\text{Ph}}_3\text{P=CHR}$ tends to give the Z (cis) alkene, while a stabilized ylide (R = ester/carbonyl) gives the E (trans) alkene. We dodge that here only because of the symmetry.

Step 1: ${\text{CN}}^{-}$ adds to $\text{PhCHO}$ $\Rightarrow$ the alkoxide ${\text{Ph-CH(O}}^{-}\text{)-CN}$, then protonation to $\text{Ph-CH(OH)-CN}$. The old carbonyl carbon now carries an H, an OH, and a CN. Step 2 (the trick): that C–H is now acidic, because the resulting carbanion is stabilized by the electron-withdrawing $\text{CN}$ group. Removing it gives ${\text{Ph-C}}^{-}\text{(OH)-CN}$ — a carbanion sitting on the former carbonyl carbon. This charge reversal ($\delta +\to$ nucleophile) is umpolung. Step 3: this carbanion attacks the $\delta +$ carbon of a second $\text{PhCHO}$, forming the new C–C bond. Step 4: ${\text{CN}}^{-}$ leaves (regenerated as catalyst) and a C=O re-forms, giving benzoin $\text{PhCH(OH)-CO-Ph}$. Why only CN⁻: it must both add reversibly and stabilize the carbanion by resonance — cyanide does both; hydroxide cannot stabilize the carbanion, so it fails.

Read the three clues:

1. Silver mirror with Tollens $\Rightarrow$ X is an aldehyde (has a $-\text{CHO}$).
2. Undergoes aldol $\Rightarrow$ X has an $\alpha$-H (can form an enolate).
3. ${\text{C}}_2$ aldehyde from alkyne hydration $\Rightarrow$ the only two-carbon aldehyde is acetaldehyde.

Combine: the only ${\text{C}}_2$ aldehyde with an $\alpha$-H is acetaldehyde ${\text{CH}}_3\text{CHO}$. (The other ${\text{C}}_2$/${\text{C}}_1$ candidate, formaldehyde $\text{HCHO}$, is ${\text{C}}_1$ and has no $\alpha$-H, so it fails clues 2 and 3.)

• Alkyne source: hydration of ethyne (acetylene) $\text{HC}\equiv \text{CH}$ with ${\text{HgSO}}_4/{\text{H}}_2{\text{SO}}_4$. It adds water Markovnikov to give the enol ${\text{CH}}_2=\text{CHOH}$, which tautomerizes to ${\text{CH}}_3\text{CHO}$ — see Enols and enolates — keto-enol tautomerism.
• Aldol then dehydration: $2{\text{CH}}_3\text{CHO}\stackrel{\text{dil. NaOH}}{\to }{\text{CH}}_3{\text{CH(OH)CH}}_2\text{CHO}\stackrel{\Delta }{\to }{\text{CH}}_3\text{CH=CH-CHO}$ The dehydrated product is but-2-enal ("crotonaldehyde"). Consistency check: two ${\text{C}}_2$ units $\to$ a ${\text{C}}_4$ product. ✓

Order: $\text{HCHO}>{\text{CH}}_3\text{CHO}>({\text{CH}}_3{)}_2\text{CO}$. Two factors, both pointing the same way:

1. Electronic (+I): each ${\text{CH}}_3$ pushes electron density onto the carbonyl carbon, shrinking its $\delta +$ and making it a weaker target. $\text{HCHO}$ has zero ${\text{CH}}_3$; acetone has two.
2. Steric: each ${\text{CH}}_3$ blocks the incoming nucleophile's path. Fewer alkyls $=$ more open carbon. So more R groups $\Rightarrow$ slower addition. $\text{HCHO}$ (no R) wins; acetone (two R) loses.

No α-H + conc. NaOH ::: Cannizzaro (disproportionation) No α-H + CN⁻ ::: Benzoin (umpolung) Carbonyl + ylide ::: Wittig, C=C at old carbonyl C (works with or without α-H) One partner no α-H + base ::: Claisen–Schmidt (clean crossed aldol) Reactivity order to Nu ::: HCHO > RCHO > R₂CO Crossed Cannizzaro oxidized species ::: HCHO (best hydride donor)