4.3.7 · D5Halides and Oxygenated Derivatives
Question bank — Aldehydes and ketones — preparation; nucleophilic addition; aldol, Cannizzaro, Wittig, Claisen-Schmidt, Mannich, Reforma
Recall the one fact that runs through everything: a is polarized — carbon is (electrophile) and oxygen — and any on the ==-carbon== (the carbon next to C=O) is acidic because the leftover charge is soaked up by the C=O as an enolate.
Before the traps, we anchor the four words that appear over and over. Look at every figure first — the arrows are the reasoning; the text just narrates them.
True or false — justify
True or false: A ketone is more reactive toward a nucleophile than an aldehyde.
False. Ketones carry two alkyl groups that push electrons in (+I effect, lowering the on carbon) and physically block the approaching nucleophile, so the order is .
True or false: and PCC both convert a 1° alcohol to an aldehyde.
False. PCC is anhydrous and stops at the aldehyde; aqueous lets the aldehyde form a hydrate (the gem-diol, two OH on one carbon — see figure s02) which is re-oxidized all the way to the carboxylic acid.
True or false: Benzaldehyde can act as the enolate partner in a crossed aldol.
False. Benzaldehyde has no -H (the carbon next to C=O is part of the ring), so it can never form an enolate — it can only be the electrophile that gets attacked.
True or false: An alkene C=C reacts with nucleophiles just like a carbonyl C=O does.
False. A C=C is non-polar (two similar carbons), so it presents no electron-poor target; nucleophiles are electron-rich and only attack electron-poor carbons, which a polarized C=O provides.
True or false: In the Cannizzaro reaction, the same molecule is both oxidized and reduced.
False. Two different molecules of the same aldehyde are involved: one donates a hydride (and becomes the carboxylate — oxidized), the other accepts it (and becomes the alcohol — reduced). This is disproportionation.
True or false: The Wittig reaction is driven by the strength of the new C=C bond.
False. The driving force is formation of the extremely stable bond in ; the new alkene is a bonus, and its position is guaranteed to be at the former carbonyl carbon.
True or false: The Wittig always gives a single stereoisomer of the alkene.
False. The position is fixed, but the E/Z geometry depends on the ylide: a stabilized ylide (e.g. carbonyl-conjugated) gives mainly the E alkene, while a non-stabilized ylide gives mainly the Z alkene — so "no ambiguity" refers to regiochemistry, not stereochemistry.
True or false: The Reformatsky reaction needs a strong base to make the enolate.
False. It uses zinc, which forms a mild zinc enolate. A strong base would deprotonate and self-condense the ester; the mild zinc enolate attacks the aldehyde cleanly without that side reaction.
True or false: In benzoin condensation the carbonyl carbon behaves as an electrophile as usual.
False. After CN⁻ adds, that former carbonyl carbon becomes nucleophilic — this polarity reversal is called umpolung, and it lets one aldehyde attack another.
Spot the error
"Aldol dehydration is unfavourable because you break a stable C–OH bond." — what's wrong?
The product is an -unsaturated carbonyl with a conjugated system (figure s01); that conjugation stabilizes the product, so dehydration is actually favoured, especially on heating.
"Ketones give acetals easily with one equivalent of alcohol." — fix it.
One equivalent gives only the hemiacetal; you need two equivalents of alcohol and acid catalysis to reach the acetal (the acid protonates OH, water leaves as an oxocarbenium ion, then the second alcohol adds).
"DIBAL-H reduces an ester all the way to a 1° alcohol." — correct this.
At low temperature DIBAL-H delivers only one hydride, stopping at a tetrahedral intermediate that gives the aldehyde on workup — that controlled single delivery is the whole point of using DIBAL.
"Rosenmund reduction uses ordinary Pd, giving an alcohol from the acid chloride." — spot the flaw.
The catalyst is poisoned (Pd on , often with a sulfur/quinoline poison) precisely to stop at the aldehyde; unpoisoned Pd would over-reduce past the aldehyde.
"Gattermann–Koch gives an aryl ketone, just like Friedel–Crafts acylation." — error?
Gattermann–Koch uses and installs a group, giving an aryl aldehyde; Friedel–Crafts acylation uses and gives an aryl ketone.
"A 1° amine plus a ketone gives an enamine." — correct the product.
A 1° amine gives an imine (Schiff base, ); you need a 2° amine (no N–H left after C=N forms) to force the double bond into the enamine position instead (figure s03).
"Cyanide catalyses aldol condensation of acetaldehyde." — what's confused here?
Cyanide catalyses the benzoin condensation of aldehydes without -H (like benzaldehyde); acetaldehyde has -H and undergoes base-catalysed aldol, a completely different pathway.
Why questions
Why does formaldehyde (HCHO) get oxidized, not reduced, in a crossed Cannizzaro?
HCHO has two H's and no electron-donating alkyl group, making it the best hydride donor; it always gives up a hydride and becomes formate, so the other aldehyde is the one reduced to alcohol.
Why is the -hydrogen acidic while a normal alkane C–H is not?
When base removes the -H, the negative charge lands next to the C=O and delocalizes onto the electronegative oxygen (the enolate, figure s03); an alkane carbanion has nowhere to spread charge, so it stays much less stable.
Why does Claisen–Schmidt avoid the messy product mixture a plain crossed aldol would give?
One partner (e.g. benzaldehyde) has no -H so it cannot be the enolate; only the other partner (e.g. acetophenone) makes the enolate, so there is only one combination and one product.
Why does adding HCN to a carbonyl give a cyanohydrin rather than just an alkoxide salt?
CN⁻ adds to the carbon forming an alkoxide, then that O⁻ grabs a proton from HCN (or solvent), leaving a neutral next to the — the cyanohydrin (figure s03).
Why does the Mannich reaction need formaldehyde and an amine together first?
The amine and formaldehyde condense and lose water to form an iminium ion , a strong electrophile; the enol of the ketone then attacks that, not the plain carbonyl.
Why does the Wittig let you place a C=C with no positional ambiguity, unlike an E1/E2 elimination?
The new double bond forms exactly at the old carbonyl carbon because that is where the ylide bonds and where leaves; elimination reactions can lose H from either neighbouring carbon, giving positional isomers.
Why can't a nucleophile simply add to a carboxylic acid the way it does to an aldehyde?
The acidic proton is snatched by the nucleophile first (acid–base wins over addition), and the resulting carboxylate is electron-rich, repelling further nucleophilic attack.
Why would you pick LDA at low temperature instead of NaOH to make an enolate?
LDA is bulky and strong: it grabs the least hindered -H irreversibly, giving the kinetic enolate (figure s04); NaOH is weaker and reversible, letting the system drift to the more stable thermodynamic enolate you may not want.
Edge cases
What happens if you try an aldol with an aldehyde that has no -H under strong base?
With no enolate possible, it cannot self-aldol; under conc. NaOH it instead disproportionates via Cannizzaro (one oxidized to acid, one reduced to alcohol).
What if a ketone has no -H at all (e.g. a fully substituted neighbour) — can it self-condense?
No — an aldol/Claisen–Schmidt needs an -H to make the enolate; with none, the ketone can only be an electrophile and must partner with something that supplies an enolate.
What is the product when both aldehyde partners in a "crossed" reaction have -H and similar reactivity?
You get a statistical mixture of all four aldol products (two self, two crossed), which is exactly why crossed aldols are only clean when one partner lacks -H or is far more electrophilic.
What if you use only one equivalent of a Grignard reagent on an ester?
The ester adds once, expels the alkoxide leaving group to give a ketone, but that ketone is more reactive and grabs a second Grignard — so with excess RMgX you overshoot to a 3° alcohol; stopping at the ketone requires a milder route.
What happens to Cannizzaro if the aldehyde does have an -H?
Cannizzaro is out-competed — the -H lets base make an enolate, so the faster aldol pathway dominates instead of hydride transfer.
Which enolate forms if you deprotonate an unsymmetrical ketone with a bulky base at −78 °C?
The kinetic enolate — the bulky base cannot reach the crowded side, so it removes the less hindered -H fast and, being cold and irreversible, that product is trapped before it can equilibrate to the thermodynamic one.
Zero-water limiting case: why is anhydrous condition essential for PCC to stop at the aldehyde?
With no water present, the aldehyde cannot form a gem-diol (hydrate, figure s02); since only the hydrate is re-oxidizable to acid, the reaction is frozen at the aldehyde stage.
Recall One-line memory hook for the tree
No -H splits three ways by reagent (NaOH→Cannizzaro, CN⁻→Benzoin, ylide→Wittig); with -H the enolate does aldol-type chemistry.