Named Reactions And Rearrangements Of Aldehydes And Ketones

Q: How does the Wittig reaction work?

A phosphorus ylide, Ph₃P=CR₂, adds to an aldehyde or ketone to form a betaine that closes to a four-membered oxaphosphetane. This ring breaks into an alkene and triphenylphosphine oxide. The carbonyl oxygen is replaced by the CR₂ group, so the new double bond forms exactly where the C=O group was.

ChemistryAldehydes And KetonesFor NEET aspirants

Named reactions of aldehydes and ketones are the carbonyl reactions that carry a chemist's name: Claisen, Perkin, Wittig, Baeyer-Villiger, pinacol, Beckmann and more. This page groups these named reactions by what they do: build C-C bonds through enolates, turn C=O into C=C, add nitrogen, change the oxidation level, or move a group in a 1,2-shift. Each one comes with its mechanism, the single selectivity rule that examiners test, and worked problems. They are core to JEE Advanced, and NEET students meet several of them in NCERT.

On this page1Claisen, Dieckmann2Perkin, Knoevenagel3Reformatsky, Darzens4Wittig5Benzoin6Nitrogen reagents7Oxidation, reduction8C-to-C shifts9C-to-N shifts10At a glance11Solved examples

Key Formulas - Quick Reference

★ Must learnClaisen (ester with two $α$ -H, one full equivalent of base):
$2C H_{3} COO C_{2} H_{5} (i) NaO C_{2} H_{5} (ii) H_{3} O^{+} C H_{3} COC H_{2} COO C_{2} H_{5} + C_{2} H_{5} OH$
Dieckmann (intramolecular Claisen): a 1,6-diester gives a 5-membered and a 1,7-diester a 6-membered cyclic $β$ -keto ester.
Perkin:
$C_{6} H_{5} CHO + (C H_{3} CO)_{2} O C H_{3} COONa, 443-453 K C_{6} H_{5} CH = CHCOOH + C H_{3} COOH$
Knoevenagel:
$RCHO + C H_{2} (COO C_{2} H_{5})_{2} piperidine RCH = C (COO C_{2} H_{5})_{2} + H_{2} O$
Reformatsky (Zn, not Mg) gives a $β$ -hydroxy ester:
$R_{2} C = O + BrC H_{2} COO C_{2} H_{5} (i) Zn (ii) H_{3} O^{+} R_{2} C (OH) C H_{2} COO C_{2} H_{5}$
★ Must learnWittig:
$R_{2} C = O + P h_{3} P = C H_{2} R_{2} C = C H_{2} + P h_{3} P = O$
★ Must learnBaeyer-Villiger:
$C_{6} H_{5} COC H_{3} + RC O_{3} H C H_{3} COO C_{6} H_{5} + RCOOH$
Migratory aptitude:
$H > 3^{\circ} > 2^{\circ} \approx Ph > 1^{\circ} > C H_{3}$
★ Must learnPinacol:
$(C H_{3})_{2} C (OH) C (OH) (C H_{3})_{2} H^{+} (C H_{3})_{3} CCOC H_{3} + H_{2} O$
Benzilic acid:
$C_{6} H_{5} COCO C_{6} H_{5} + NaOH (C_{6} H_{5})_{2} C (OH) COONa$
★ Must learnBeckmann (the group anti to OH migrates to N):
$R R^{'} C = NOH H_{2} S O_{4} or PC l_{5} RCONH R^{'}$

1. Claisen and Dieckmann Condensations

1.1 Claisen condensation

When two molecules of an ester condense in the presence of a strong base such as sodium ethoxide, the reaction is the Claisen condensation. The product is a $β$ -keto ester.

Ethyl propanoate gives ethyl 2-methyl-3-oxopentanoate:

2C H_{3} C H_{2} COO C_{2} H_{5} (i) NaO C_{2} H_{5} (ii) H_{3} O^{+} C H_{3} C H_{2} COCH (C H_{3}) COO C_{2} H_{5} + C_{2} H_{5} OH

The first steps resemble aldol addition: an enolate attacks a carbonyl carbon. After that, the two reactions differ. In the Claisen condensation the negatively charged oxygen re-forms the C=O $π$ bond and expels the $O C_{2} H_{5}$ group, because alkoxide is a leaving group.

That elimination is reversible, since alkoxide can attack the keto group of the $β$ -keto ester. The reaction is driven to completion by removing a proton from the $β$ -keto ester. Its central $C H_{2}$ , flanked by two carbonyl groups, is much more acidic than the $α$ -hydrogens of the starting ester. A successful Claisen condensation therefore needs an ester with two $α$ -hydrogens and a full equivalent of base, not a catalytic amount. Acid added at the end re-protonates the anion.

Figure 1: Claisen condensation of ethyl ethanoate. Steps 1 and 2 are reversible; step 3 is not, because the

β

-keto ester (pK

_{a} \approx 11

) is far more acidic than ethanol (pK

_{a} \approx 16

). That is why one full equivalent of

C_{2} H_{5} O^{-}

is needed.

Why a full equivalent of base? Compare the acids in the flask: ester $α$ -H (p $K_{a} \approx 25$ ), ethanol (p $K_{a} \approx 16$ ) and the $β$ -keto ester (p $K_{a} \approx 11$ ). Ethoxide can remove only a tiny amount of ester enolate, but it removes the $β$ -keto ester's proton almost completely. That last, one-way step uses up one mole of base per mole of product.

1.2 Mixed Claisen and ketone-ester condensations

A mixed Claisen condensation is useful when one ester has no $α$ -hydrogen and can only act as the electrophile. Ethyl benzoate and ethyl butanoate give ethyl 2-benzoylbutanoate:

C_{6} H_{5} COO C_{2} H_{5} + C H_{3} C H_{2} C H_{2} COO C_{2} H_{5} (i) NaO C_{2} H_{5} (ii) H_{3} O^{+} C_{6} H_{5} COCH (C_{2} H_{5}) COO C_{2} H_{5} + C_{2} H_{5} OH

A ketone can also condense with an ester. The $α$ -hydrogens of a ketone are more acidic than those of an ester, so mainly one product forms if the ketone is added slowly to a mixture of the base and excess ester.

Ketone + ester: a $β$ -diketone, as in cyclohexanone + ethyl ethanoate giving 2-acetylcyclohexan-1-one.
Ketone + formate ester: a $β$ -keto aldehyde, as in cyclohexanone + ethyl methanoate giving 2-oxocyclohexane-1-carbaldehyde.
Ketone + diethyl carbonate: a $β$ -keto ester, ethyl 2-oxocyclohexane-1-carboxylate.

1.3 Dieckmann condensation (intramolecular Claisen)

Adding base to a 1,6-diester causes an intramolecular Claisen condensation that forms a five-membered cyclic $β$ -keto ester. A 1,7-diester gives a six-membered ring. An intramolecular Claisen condensation is called a Dieckmann condensation.

Figure 2: Dieckmann condensation. The

α

-carbon (C-2) attacks the other ester carbon, so the ring contains one carbon fewer than the chain: a 1,6-diester closes a 5-membered ring and a 1,7-diester a 6-membered ring.

Exam Trick

Dieckmann drops one. Number the diester chain from one C=O carbon to the other: the ring holds one carbon fewer than the chain, because C-1 stays outside as the ester group. 1,6 gives 5, 1,7 gives 6. Five- and six-membered rings form easily; a 1,5-diester (4-ring) does not cyclise.

Aldol addition

Electrophile: an aldehyde or ketone.
The alkoxide is protonated.
Product: $β$ -hydroxy carbonyl compound.
Base can be catalytic.

Claisen condensation

Electrophile: an ester.
The alkoxide expels $O R^{-}$ .
Product: $β$ -keto ester.
One full equivalent of base.

Key idea

An ester enolate attacks a second ester; the tetrahedral intermediate throws out alkoxide; the acidic

β

-keto ester is deprotonated, and that last step drives the Claisen and Dieckmann condensations.

2. Perkin and Knoevenagel Reactions

2.1 Perkin reaction

In the Perkin reaction, an aromatic aldehyde condenses with an aliphatic acid anhydride in the presence of the sodium or potassium salt of the same acid, giving an $α$ , $β$ -unsaturated aromatic acid. The anhydride needs at least two $α$ -hydrogens. With only one, dehydration is impossible and an aldol-type product is isolated.

C_{6} H_{5} CHO + (C H_{3} CO)_{2} O C H_{3} COONa 443-453 K, 5 h C_{6} H_{5} CH = CHCOOH + C H_{3} COOH

Furfural gives 3-(2-furyl)acrylic acid in the same way, and phthalic anhydride with ethanoic anhydride and sodium ethanoate gives phthalylacetic acid.

Acetate ion removes a proton from the $α$ -carbon of the anhydride.
The carbanion attacks the aldehyde carbonyl carbon; the alkoxide takes a proton, giving an aldol-type compound.
This dehydrates in the hot anhydride, and hydrolysis of the mixed anhydride gives the unsaturated acid.

Prolonged heating (about 5 hours) at high temperature is needed because a weak base (acetate) must react with a weak acid (the anhydride). Electron-withdrawing groups on the ring make the carbonyl carbon more positive and speed the reaction up.

2.2 Knoevenagel reaction

The Knoevenagel reaction is the condensation of an aldehyde or ketone with a compound containing an active methylene group (such as diethyl malonate or ethyl cyanoacetate) in the presence of a weak base, giving an $α$ , $β$ -unsaturated compound. The base may be ammonia or an amine: primary, secondary or tertiary amines, pyridine or piperidine. The methylene group is so reactive that the aldehyde does not self-condense.

C_{6} H_{5} CHO + C H_{2} (COO C_{2} H_{5})_{2} piperidine C_{6} H_{5} CH = C (COO C_{2} H_{5})_{2} + H_{2} O

C_{6} H_{5} CH = C (COO C_{2} H_{5})_{2} + 2 H_{2} O H_{3} O^{+}, heat C_{6} H_{5} CH = CHCOOH + C O_{2} + 2 C_{2} H_{5} OH

Mechanism. The amine removes a proton from the active methylene group. The stabilised carbanion adds to the carbonyl carbon, the alkoxide is protonated by $R_{3} N H^{+}$ , and dehydration gives the alkylidene malonate. Hydrolysis and decarboxylation give the unsaturated acid.

Figure 3: Two roads to cinnamic acid. Both are aldol-type condensations followed by dehydration; the Knoevenagel runs under milder conditions because its

C H_{2}

is flanked by two ester groups (p

K_{a} \approx 13

3. Reformatsky and Darzens Reactions

3.1 Reformatsky reaction

The Reformatsky reaction is an organometallic addition that uses zinc. An aldehyde or ketone is treated with an $α$ -bromo ester and zinc metal, usually in benzene or ether. The intermediate is an organozinc reagent, $BrZnC H_{2} COOR$ , which adds to the carbonyl group like a Grignard reagent. The zinc alkoxide formed first is hydrolysed to a $β$ -hydroxy ester, so the carbon skeleton is extended.

R_{2} C = O + BrC H_{2} COO C_{2} H_{5} + Zn ether R_{2} C (OZnBr) C H_{2} COO C_{2} H_{5}

R_{2} C (OZnBr) C H_{2} COO C_{2} H_{5} + H_{2} O H_{3} O^{+} R_{2} C (OH) C H_{2} COO C_{2} H_{5} + Zn (OH) Br

Organozinc reagents are less reactive than Grignard reagents, so the ester group of the reagent is not attacked. A Grignard reagent made from the same bromo ester would destroy itself by attacking its own ester group.

3.2 Darzens glycidic ester condensation

A ketone or aldehyde reacts with an $α$ -chloro ester and a strong base (such as potassium tert-butoxide) to give an $α$ , $β$ -epoxy ester, called a glycidic ester. The ester enolate adds to the carbonyl group, and the new alkoxide displaces chloride intramolecularly (an $S_{N} 2$ step) to close the three-membered epoxide ring.

Figure 4: One

α

-halo ester, two products. Zinc gives an organozinc reagent and a

β

-hydroxy ester (Reformatsky); a strong base keeps the halogen on the carbon, so the new alkoxide can displace it and close an epoxide (Darzens).

Reformatsky

Reagent: $α$ -bromo ester + Zn.
Nucleophile: organozinc $BrZnC H_{2} COOR$ .
After addition: the O-Zn is hydrolysed.
Product: $β$ -hydroxy ester.

Darzens

Reagent: $α$ -chloro ester + strong base.
Nucleophile: ester enolate, Cl still attached.
After addition: $O^{-}$ displaces $C l^{-}$ .
Product: glycidic (epoxy) ester.

Quick Recall: tap to check

Why does a Claisen condensation need one full equivalent of base?

The

β

-keto ester (p

K_{a} \approx 11

) is deprotonated completely; this last step pulls every earlier equilibrium forward and consumes the base.

Which ring does diethyl heptanedioate give in a Dieckmann condensation?

A six-membered ring: ethyl 2-oxocyclohexane-1-carboxylate (a 1,7-diester gives a 6-ring).

Why is zinc, not magnesium, used in the Reformatsky reaction?

The organozinc reagent is too weak to attack an ester, so it survives with its own ester group and adds only to the aldehyde or ketone.

4. Wittig Reaction

The Wittig reaction converts an aldehyde or ketone into an alkene with a phosphorus ylide,

P h_{3} P = C R_{2}

(an alkylidenetriphenylphosphorane). The C=O oxygen is replaced by the

= C R_{2}

group, and the new double bond forms exactly where the carbonyl group was.

(C_{6} H_{5})_{2} C = O + P h_{3} P = C H_{2} (C_{6} H_{5})_{2} C = C H_{2} + P h_{3} P = O

The ylide is made in two steps. Triphenylphosphine displaces bromide from an alkyl halide ( $S_{N} 2$ ), and a strong base such as $C_{6} H_{5} Li$ , BuLi, $NaN H_{2}$ , NaH or $C_{2} H_{5} ONa$ then removes a proton from the carbon next to phosphorus:

P h_{3} P + C H_{3} Br [P h_{3} PC H_{3}]^{+} B r^{-}

[P h_{3} PC H_{3}]^{+} B r^{-} + C_{6} H_{5} Li P h_{3} P = C H_{2} + C_{6} H_{6} + LiBr

The nucleophilic ylide carbon attacks the carbonyl carbon, giving a dipolar betaine.
The oxide oxygen bonds to the positive phosphorus, closing a four-membered oxaphosphetane ring.
The ring breaks into the alkene and triphenylphosphine oxide. Forming the very strong P=O bond drives the reaction.

An optically active phosphonium salt gives a phosphine oxide with retention of configuration at phosphorus, which supports this cyclic pathway.

Figure 5: Wittig reaction. The ylide carbon replaces the carbonyl oxygen; the strong P=O bond formed in

P h_{3} P = O

pays for the whole sequence.

Phosphorus ylides react in the same way with other multiple bonds:

Ketenes ( $R_{2} C = C = O$ ) give allenes, $R_{2} C = C = C R_{2}$ .
Isocyanates ( $R^{'} N = C = O$ ) give ketenimines, $R^{'} N = C = C R_{2}$ .
Nitroso compounds ( $R^{'} N = O$ ) give imines, $R^{'} N = C R_{2}$ .
Imines ( $R_{2} C = NR$ ) give alkenes, $R_{2} C = C R_{2}$ .

Key idea

To write a Wittig product, erase the O of C=O and the

PP h_{3}

of the ylide, then join the two carbons with a double bond.

5. Benzoin Condensation

The benzoin condensation dimerises two aromatic aldehydes under the catalytic influence of cyanide ions (ethanolic KCN), giving benzoin, $C_{6} H_{5} CH (OH) CO C_{6} H_{5}$ .

2 C_{6} H_{5} CHO C N^{-} (ethanolic KCN) C_{6} H_{5} CH (OH) CO C_{6} H_{5}

The aldehyde hydrogen is not acidic enough to remove. Once cyanide adds to the carbonyl carbon, however, that hydrogen sits $α$ to a nitrile and becomes relatively acidic. The carbanion attacks the carbonyl carbon of a second aldehyde molecule in the rate-determining step. The cyanohydrin of benzoin then breaks down into benzoin and HCN, releasing cyanide.

The reaction is not catalysed by hydroxide or by bases in general, only by cyanide, because cyanide (a) is a good nucleophile, (b) stabilises the carbanion, and (c) leaves easily in the last step. The carbonyl carbon, normally an electrophile, has been turned into a nucleophile: this reversal of polarity is called umpolung.

Figure 6: Benzoin condensation. Cyanide reverses the polarity of the aldehyde carbon (umpolung); only

C N^{-}

can add, stabilise the carbanion and then leave, so

O H^{-}

cannot catalyse the reaction.

6. Reactions with Nitrogen Reagents

6.1 Enamines from secondary amines

An aldehyde or ketone reacts with a secondary amine to give an enamine, an $α$ , $β$ -unsaturated amine. The name joins "ene" and "amine". The double bond is in the part of the molecule that came from the carbonyl compound.

The mechanism follows imine formation up to the iminium ion. That ion has no N-H to lose, so a proton is lost from the $α$ -carbon instead. Cyclopentanone with diethylamine gives N,N-diethylcyclopent-1-en-1-amine, and cyclohexanone with pyrrolidine gives 1-(cyclohex-1-en-1-yl)pyrrolidine. In aqueous acid, an enamine is hydrolysed back to the carbonyl compound and the protonated secondary amine.

6.2 Reductive amination

The imine from ammonia is not stable, because nitrogen needs a substituent other than hydrogen to stabilise it. It is still a useful intermediate. If the reaction with $N H_{3}$ is run with $H_{2}$ and a metal catalyst such as Raney nickel, hydrogen adds to the C=N bond as soon as it forms, giving a primary amine. Secondary and tertiary amines are made the same way by reducing imines or enamines. Butan-2-one with ammonia and $H_{2}$ /Ni gives butan-2-amine, and benzaldehyde with ethylamine and $H_{2}$ /Ni gives N-ethylbenzylamine. This is reductive amination.

Figure 7: What decides the product is how many H atoms the nitrogen still carries after losing water:

RN H_{2}

gives an imine,

R_{2} NH

an enamine, and

N H_{3}

with

H_{2}

/Ni an amine.

6.3 Methanal and ammonia: urotropine

Methanal reacts with ammonia to form hexamethylenetetramine, known as urotropine, a cage of four nitrogen atoms joined by six $C H_{2}$ groups:

6HCHO + 4N H_{3} (C H_{2})_{6} N_{4} + 6 H_{2} O

Urotropine is used as a urinary antiseptic. Controlled nitration of urotropine gives RDX, a powerful explosive.

7. Selective Oxidations and Reductions

7.1 Baeyer-Villiger oxidation

Aldehydes and ketones are oxidised by peroxy acids. The reaction is especially useful for ketones, which are converted into esters. Acetophenone gives phenyl ethanoate:

C_{6} H_{5} COC H_{3} + RC O_{3} H C H_{3} COO C_{6} H_{5} + RCOOH

The product shows that phenyl migrates more readily than methyl; otherwise it would be $C_{6} H_{5} COOC H_{3}$ . This tendency is called migratory aptitude:

H > 3^{\circ} alkyl > 2^{\circ} alkyl \approx phenyl > 1^{\circ} alkyl > C H_{3}

Groups migrate with their electron pair and keep their configuration. Aldehydes give carboxylic acids because hydrogen migrates, and cyclic ketones give lactones (the ring grows by one oxygen).

Figure 8: Baeyer-Villiger oxidation. The oxygen atom is inserted on the side of the group with the higher migratory aptitude, so acetophenone gives

C H_{3} COO C_{6} H_{5}

, not

C_{6} H_{5} COOC H_{3}

Exam Trick

Oxygen sits next to the bigger boss. In a Baeyer-Villiger product, the new O lands between the carbonyl carbon and the group of higher migratory aptitude: usually the more substituted carbon, and always H in an aldehyde. Pinacolone gives tert-butyl ethanoate, not methyl 2,2-dimethylpropanoate.

7.2 Selenium dioxide oxidation

Aldehydes and ketones with a $C H_{3}$ or $C H_{2}$ group next to the carbonyl group are oxidised by selenium dioxide to 1,2-dicarbonyl compounds. The reaction is usually carried out in acetic acid, and the actual reagent is selenous acid. It probably proceeds through the enol, which attacks selenium, followed by loss of water and selenium.

C H_{3} CHO + Se O_{2} OHC - CHO + Se + H_{2} O

C H_{3} COC H_{3} + Se O_{2} C H_{3} COCHO + Se + H_{2} O

7.3 Meerwein-Ponndorf-Verley (MPV) reduction

Aldehydes and ketones are reduced to alcohols by aluminium isopropoxide in excess propan-2-ol. The equilibrium is shifted forward by distilling out the acetone formed. The reaction is mild and rapid, side reactions are negligible, and it is specific for C=O: C=C and $N O_{2}$ groups in the substrate are unaffected. If a compound contains two C=O groups, one can be protected as an acetal while the other is reduced. Ketones with a high enol content, such as $β$ -diketones and $β$ -keto esters, do not react.

R_{2} C = O + (C H_{3})_{2} CHOH Al (OCH (C H_{3})_{2})_{3} R_{2} CHOH + (C H_{3})_{2} C = O

A hydride ion moves from the $α$ -C-H of an isopropoxide group to the carbonyl carbon through a six-membered cyclic transition state, giving a mixed alkoxide. Excess propan-2-ol exchanges with it to release the product alcohol. So one hydrogen of the product comes from the isopropoxide group and the other from the solvent.

7.4 Oppenauer oxidation

The Oppenauer oxidation is the reverse of MPV reduction. A secondary alcohol is refluxed with a ketone (acetone, butanone or cyclohexanone) as hydrogen acceptor and a base such as aluminium tert-butoxide, aluminium isopropoxide or potassium tert-butoxide, in benzene or toluene. The alcohol is dehydrogenated to a ketone; an excess of acetone favours oxidation.

Primary alcohols can be oxidised to aldehydes if a better hydrogen acceptor such as p-benzoquinone is used. The first step involves the alcoholic OH group, so hindered alcohols react less readily; in cyclohexanols, axial OH groups are attacked more slowly.

Figure 9: MPV reduction and Oppenauer oxidation are one equilibrium run in opposite directions. The hydride travels inside a six-membered ring, which is why only the C=O group is touched.

MPV reduction

Ketone + propan-2-ol, $Al (OP r^{i})_{3}$ .
Acetone is distilled off.
C=O becomes CH-OH; C=C, $N O_{2}$ untouched.

Oppenauer oxidation

2° alcohol + excess acetone, $Al (OB u^{t})_{3}$ .
Same equilibrium, run backwards.
CH-OH becomes C=O; C=C untouched.

Key idea

Baeyer-Villiger adds an oxygen next to the best migrating group;

Se O_{2}

turns the

α

C H_{2}

into C=O; MPV and Oppenauer move a hydride through one six-membered ring in either direction.

Quick Recall: tap to check

What does acetophenone give with a peroxy acid?

Phenyl ethanoate,

C H_{3} COO C_{6} H_{5}

: phenyl migrates before methyl.

What does selenium dioxide do to propanone?

It oxidises the

C H_{3}

next to C=O, giving methylglyoxal (2-oxopropanal),

C H_{3} COCHO

Why does MPV reduction leave a C=C bond untouched?

The hydride is delivered only inside the six-membered ring formed with Al coordinated to the C=O oxygen, so only the carbonyl group can accept it.

8. Rearrangements I: Carbon-to-Carbon Shifts

8.1 Pinacol-pinacolone rearrangement

The acid-catalysed rearrangement of 1,2-diols into ketones or aldehydes, with loss of water, is the pinacol-pinacolone rearrangement. It is named after the classic conversion of pinacol into pinacolone:

(C H_{3})_{2} C (OH) C (OH) (C H_{3})_{2} H^{+} (dil. H_{2} S O_{4}) heat (C H_{3})_{3} CCOC H_{3} + H_{2} O

(C_{6} H_{5})_{2} C (OH) C H_{2} OH H^{+} (C_{6} H_{5})_{2} CHCHO + H_{2} O

Elimination of water without rearrangement, the normal reaction of alcohols, needs drastic conditions: $A l_{2} O_{3}$ at 450 °C gives 2,3-dimethylbuta-1,3-diene and only a little pinacolone.

Mechanism. One OH is protonated and lost as water, giving a tertiary carbocation (I). A methyl group then shifts from the neighbouring carbon, producing cation (II). Although (I) is already tertiary, (II) is preferred because the lone pair on OH shares the positive charge, so every atom has an octet. Loss of a proton gives the ketone.

Figure 10: Pinacol-pinacolone rearrangement. Cation I is already tertiary, but the 1,2-methyl shift still happens because cation II is an oxocarbenium ion with a full octet on every atom.

Any carbocation with the positive charge on the carbon next to a C-OH undergoes the same rearrangement, which supports this mechanism. The cation can be generated by treating an amino alcohol with $NaN O_{2} /HCl$ at 5 °C, or a chlorohydrin with $A g^{+}$ .

The migrating group probably never becomes completely free; it is partly bonded to both carbons in a bridged structure. The evidence: a chiral migrating group keeps its configuration, and no cross-over products form when a mixture of two similar 1,2-diols is treated with acid.

Migratory aptitude: in general H > aryl > alkyl.
Among aryl groups: the more electron-rich group migrates better: p-anisyl > p-tolyl > phenyl > p-chlorophenyl. Electron-withdrawing groups retard migration.
Among alkyl groups: $(C H_{3})_{3} C$ > $(C H_{3})_{2} CH$ > $C H_{3}$ .
Carbocation stability comes first. The OH that leaves is the one giving the more stable cation, and this can override migratory aptitude. In 2-methyl-1,1-diphenylpropane-1,2-diol the resonance-stabilised $(C_{6} H_{5})_{2} C^{+}$ forms, so a methyl group migrates, giving 3,3-diphenylbutan-2-one.
Steric hindrance matters too: p-anisyl migrates about 1000 times faster than o-anisyl.

Exam Trick

Cation first, migrant second. In a pinacol problem, never pick the migrating group first. Step 1: remove the OH that leaves the more stable carbocation (benzylic, then 3°). Step 2: from the other carbon, move the best group (H > aryl > alkyl). Step 3: that carbon becomes C=O.

8.2 Benzilic acid rearrangement

A strong base adds to a carbonyl group to give an anion, and reversal of the charge can expel an attached group. In a 1,2-diketone, the group can instead migrate to the neighbouring electron-deficient carbonyl carbon, forming an $α$ -hydroxy acid. Benzil with strong base gives the salt of benzilic acid, which gives the rearrangement its name.

C_{6} H_{5} COCO C_{6} H_{5} + NaOH (C_{6} H_{5})_{2} C (OH) COONa

C_{6} H_{5} COCO C_{6} H_{5} + C H_{3} ONa then H^{+} gives methyl benzilate (C_{6} H_{5})_{2} C (ONa) COOC H_{3}

Bases: barium and thallium hydroxides work better than NaOH or KOH. Alkoxides (methoxide, tert-butoxide) give the corresponding esters, while phenoxide is too weak a nucleophile to attack.
Substrates: aromatic 1,2-diketones, aliphatic and heterocyclic diketones, and o-quinones all rearrange. Cyclic 1,2-diketones undergo ring contraction.
Kinetics: rate = k[benzil][ $O H^{-}$ ], and in $H_{2}^{18} O$ benzil exchanges $^{18}$ O faster than it rearranges. So $O H^{-}$ adds in a fast reversible first step, migration is the slow step, and a rapid proton transfer completes the process. The reaction is an intramolecular analogue of the Cannizzaro reaction.
Unsymmetrical benzils: the carbonyl group attached to the less electron-releasing aryl group is more positive and is attacked by $O H^{-}$ , so the less electron-donating aryl group is the one that migrates.

Figure 11: Benzilic acid rearrangement. The 1,2-diketone becomes an

α

-hydroxy acid: one carbon ends up carrying both phenyl groups and the OH, the other becomes COOH.

8.3 Acid-catalysed cyclisation of unsaturated aldehydes

A protonated aldehyde is a strong electrophile. If the same molecule contains a C=C double bond at a suitable distance, the double bond can act as the nucleophile and close a ring. The resulting carbocation is captured by water. Citral-type aldehydes cyclise this way to cyclohexene diols (Solved Example 13).

9. Rearrangements II: Carbon-to-Nitrogen Shifts

9.1 Beckmann rearrangement

The acid-catalysed conversion of a ketoxime into an N-substituted amide is the Beckmann rearrangement. It is catalysed by acidic reagents such as

H_{2} S O_{4}

, polyphosphoric acid,

SOC l_{2}

and

PC l_{5}

, and it is highly stereospecific: the group anti (trans) to the OH migrates to nitrogen.

R R^{'} C = NOH H_{2} S O_{4} or PC l_{5} RCONH R^{'}

The configuration of 2-chloro-5-nitrobenzophenone oxime was fixed independently, by its easy conversion into a nitro-substituted phenylbenzisoxazole, which showed that the nitrated ring lies on the same side as the OH. In the Beckmann rearrangement of this oxime, the phenyl group (anti to OH) migrates, not the nitrated ring. Anti migration is so reliable that the product identifies the configuration of an oxime: the two geometrical isomers of 4-methoxybenzophenone oxime give different anilides.

Aldoximes: both syn- and anti-benzaldoxime give benzamide with polyphosphoric acid. The syn form partly converts to the anti form, which then rearranges with hydrogen migration.
Ring enlargement: cyclohexanone oxime gives $ε$ -caprolactam, the monomer of nylon-6.

Figure 12: Beckmann rearrangement. Only the group anti to the leaving group can reach the back of the N-O bond, so the oxime geometry decides the amide. A cyclic oxime grows by one atom: cyclohexanone oxime gives the 7-membered

ε

-caprolactam.

JEE Advanced

How we know the mechanism. Three classic experiments pin it down:

Oxime esters rearrange too. The benzenesulphonate of benzophenone oxime rearranges with no acid at all, and rates follow the strength of the esterifying acid ( $C_{6} H_{5} S O_{3} H$ > $C_{6} H_{5} COOH$ > $C H_{3} COOH$ ). So ionisation of the N-O bond is rate-controlling.
Configuration is retained. Kenyon and Young converted optically active 2-ethylhexanoic acid two ways: its amide by Hofmann rearrangement, and its methyl ketone (via $SOC l_{2}$ and $(C H_{3})_{2} Cd$ ) by oxime formation and Beckmann rearrangement. Both gave the same enantiomer of heptan-3-amine, so C-C bond breaking and C-N bond making are synchronous.
The oxygen is lost. With $PC l_{5}$ in the presence of $H_{2}^{18} O$ , benzophenone oxime gives benzanilide containing $^{18}$ O: the nitrilium ion takes its oxygen from the solvent, so the overall process is not intramolecular.

9.2 Schmidt reaction

A carbonyl compound reacts with hydrazoic acid, $H N_{3}$ , in the presence of concentrated sulphuric acid. Ketones give amides. Aldehydes give nitriles together with N-formyl derivatives of primary amines:

RCO R^{'} + H N_{3} conc. H_{2} S O_{4} RCONH R^{'} + N_{2}

RCHO + H N_{3} conc. H_{2} S O_{4} RCN + N_{2} + H_{2} O

RCHO + H N_{3} conc. H_{2} S O_{4} RNHCHO + N_{2}

The mechanism proposed by Smith (1948) is a 1,2-shift from carbon to nitrogen. The protonated ketone adds $H N_{3}$ , water is lost to give an iminodiazonium ion, and the group anti to the $N_{2}^{+}$ migrates as $N_{2}$ leaves. The nitrilium ion adds water to give the amide, exactly as in the Beckmann rearrangement. When the two groups differ, the ratio of isomeric amides depends on the ratio of the two geometrical isomers of the iminodiazonium ion.

Key idea

In every 1,2-shift the migrating group takes its bonding pair with it and keeps its configuration; what changes from reaction to reaction is only the rule that decides which group goes.

Quick Recall: tap to check

Which group migrates when pinacol-type diols have one benzylic OH?

First the benzylic OH leaves (more stable cation); then the best group on the other carbon migrates, even a methyl if that is all there is.

What is the benzilic acid product of benzil with

O H^{-}

The benzilate ion,

(C_{6} H_{5})_{2} C (OH) CO O^{-}

; acid gives benzilic acid.

Which amide does the oxime of acetophenone with OH anti to phenyl give?

Phenyl migrates to N: acetanilide,

C H_{3} CONH C_{6} H_{5}

10. Named Reactions at a Glance

Reaction	Substrates	Reagent	Product
Claisen	ester with 2 $α$ -H	$NaO C_{2} H_{5}$ (1 equiv), then $H^{+}$	$β$ -keto ester
Dieckmann	1,6- or 1,7-diester	$NaO C_{2} H_{5}$ , then $H^{+}$	cyclic $β$ -keto ester (5 or 6 ring)
Perkin	ArCHO + anhydride	RCOONa, heat	$ArCH = CHCOOH$
Knoevenagel	C=O + active $C H_{2}$	weak amine base	$α$ , $β$ -unsaturated compound
Reformatsky	C=O + $BrC H_{2} COOR$	Zn, then $H_{3} O^{+}$	$β$ -hydroxy ester
Darzens	C=O + $ClC H_{2} COOR$	strong base	glycidic (epoxy) ester
Wittig	C=O + $P h_{3} P = C R_{2}$	(the ylide itself)	alkene + $P h_{3} PO$
Benzoin	2 ArCHO	ethanolic KCN	$ArCH (OH) COAr$
Baeyer-Villiger	ketone	$RC O_{3} H$	ester (lactone from a cyclic ketone)
$Se O_{2}$	$C H_{3}$ / $C H_{2}$ next to C=O	$Se O_{2}$	1,2-dicarbonyl
MPV / Oppenauer	C=O / 2° alcohol	$Al (OP r^{i})_{3}$ + propan-2-ol / $Al (OB u^{t})_{3}$ + acetone	alcohol / ketone
Pinacol	1,2-diol	$H^{+}$	ketone or aldehyde
Benzilic acid	1,2-diketone	$O H^{-}$ , then $H^{+}$	$α$ -hydroxy acid
Beckmann	ketoxime	$H_{2} S O_{4}$ , $PC l_{5}$ , $SOC l_{2}$ , PPA	N-substituted amide
Schmidt	ketone	$H N_{3}$ , conc. $H_{2} S O_{4}$	amide + $N_{2}$

The four rearrangements differ only in the rule that picks the migrating group:

Rearrangement	Group moves	Which group moves	Configuration
Pinacol	C to C (cation)	best group on the carbon next to the more stable cation (H > aryl > alkyl)	retained
Benzilic acid	C to C (C=O)	aryl on the carbon attacked by $O H^{-}$ (the less electron-rich aryl)	not applicable
Baeyer-Villiger	C to O	higher aptitude: H > 3° > 2° $\approx$ Ph > 1° > $C H_{3}$	retained
Beckmann / Schmidt	C to N	group anti to the leaving OH (or $N_{2}$ )	retained

Figure 13: Rearrangement flowchart. Identify the reaction from the substrate and reagent first; each reaction has its own rule for which group moves.

Figure 14: The whole page on one map. Sort any question first by what the reagent does to the carbonyl group: makes a C-C bond, swaps O for C, adds nitrogen, changes the oxidation level, or moves a group.

11. Solved Examples

Solved Example 1

(a) Give the product of the Claisen condensation of ethyl propanoate. (b) Ethyl 2-methylpropanoate gives almost no Claisen product with sodium ethoxide. Why?

Solution:

(a) The enolate of one ester (at $C H_{2}$ ) attacks the C=O of another and ethoxide is expelled: ethyl 2-methyl-3-oxopentanoate, $C H_{3} C H_{2} COCH (C H_{3}) COO C_{2} H_{5}$ .

(b) $(C H_{3})_{2} CHCOO C_{2} H_{5}$ has only one $α$ -hydrogen. After condensation, the carbon between the two carbonyl groups carries no hydrogen, so the product cannot be converted into the stabilised anion. Without that final deprotonation the reversible steps are not pulled forward, so the equilibrium lies on the side of the starting ester.

Solved Example 2

Cyclohexanone is treated with sodium ethoxide and (a) excess ethyl ethanoate, (b) ethyl methanoate, (c) excess diethyl carbonate, followed by acid. Give the products.

Solution:

The more acidic ketone forms the enolate, which attacks the ester and expels ethoxide.

(a) 2-acetylcyclohexan-1-one, a $β$ -diketone. (b) 2-oxocyclohexane-1-carbaldehyde, a $β$ -keto aldehyde. (c) Ethyl 2-oxocyclohexane-1-carboxylate, a $β$ -keto ester. Ethanol is the by-product each time.

Solved Example 3

2-Acetyl-2-methylcyclohexan-1-one is heated with aqueous acid to give A. A is:
(A) 1-methyl-2-oxocyclohexane-1-carbaldehyde
(B) 2-methylcyclohexan-1-one
(C) 1-methylcyclohexane-1-carbaldehyde
(D) 3-hydroxy-2-acetyl-2-methylcyclohexanal

Solution:

Answer: (B). This $β$ -diketone has no hydrogen on the carbon between its two C=O groups, so it cannot form a stabilised enol. Hot aqueous acid cleaves it (a retro-Claisen reaction): water adds to the acetyl C=O, the C-C bond to the ring breaks with the enol of 2-methylcyclohexanone as the leaving group, and ethanoic acid is released. The ketone left is 2-methylcyclohexan-1-one.

Solved Example 4

C_{7} H_{6} O

[A] with alcoholic KCN gives

C_{14} H_{12} O_{2}

[B]. With sodium ethanoate and ethanoic anhydride, A gives the aromatic acid

C_{9} H_{8} O_{2}

[C], which adds

B r_{2}

to give

C_{9} H_{8} O_{2} B r_{2}

[D]. Identify A to D.

Solution:

A = benzaldehyde, $C_{6} H_{5} CHO$ . Benzoin condensation gives B = benzoin, $C_{6} H_{5} CH (OH) CO C_{6} H_{5}$ . The Perkin reaction gives C = cinnamic acid, $C_{6} H_{5} CH = CHCOOH$ . Bromine adds to the C=C to give D = 2,3-dibromo-3-phenylpropanoic acid, $C_{6} H_{5} CHBrCHBrCOOH$ .

Solved Example 5

Complete: 2-methylcyclohexan-1-one + Zn +

BrC H_{2} COO C_{2} H_{5}

, then

H_{2} O

Solution:

Zinc forms $BrZnC H_{2} COO C_{2} H_{5}$ , which adds to the ketone carbonyl group to give a zinc alkoxide. Hydrolysis gives the $β$ -hydroxy ester ethyl 2-(1-hydroxy-2-methylcyclohexyl)ethanoate.

Solved Example 6

Cyclohexanone +

ClC H_{2} COO C_{2} H_{5}

, with t-BuOK in t-BuOH, gives A. A is:
(A) ethyl cyclohexylethanoate
(B) a chloro ketone
(C) ethyl 1-oxaspiro[2.5]octane-2-carboxylate
(D) cyclohexylethanoic acid

Solution:

Answer: (C). t-BuO⁻ forms the enolate $ClC H^{-} COO C_{2} H_{5}$ , which adds to the ketone. The resulting alkoxide displaces chloride intramolecularly, giving a spiro epoxide: the glycidic ester of a Darzens condensation.

Solved Example 7

Complete: (a) cyclohexanone +

(C_{6} H_{5})_{3} P = C H_{2}

; (b) benzaldehyde +

C H_{2} = CH - CH = PP h_{3}

Solution:

(a) The C=O oxygen is replaced by $= C H_{2}$ : methylenecyclohexane + $(C_{6} H_{5})_{3} PO$ .

(b) The oxygen is replaced by $= CH - CH = C H_{2}$ : $C_{6} H_{5} CH = CH - CH = C H_{2}$ (1-phenylbuta-1,3-diene) + $P h_{3} PO$ .

Solved Example 8

Complete: (a) butan-2-one +

C H_{3} C H_{2} C H_{2} CH = PP h_{3}

; (b) cyclopentanone +

C H_{3} CH = PP h_{3}

; (c) propenal +

P h_{3} P = C (C H_{3}) COOC H_{3}

Solution:

Join the carbonyl carbon and the ylide carbon with a double bond; $P h_{3} PO$ is the other product each time.

(a) $C H_{3} C H_{2} C (C H_{3}) = CHC H_{2} C H_{2} C H_{3}$ , 3-methylhept-3-ene. (b) Ethylidenecyclopentane. (c) $C H_{2} = CH - CH = C (C H_{3}) COOC H_{3}$ , methyl 2-methylpenta-2,4-dienoate.

Solved Example 9

Give the Baeyer-Villiger products with a peroxy acid of (a) cyclohexanone, (b) 3,3-dimethylbutan-2-one, (c) benzaldehyde.

Solution:

(a) A ring $C H_{2}$ migrates to oxygen, enlarging the ring: $ε$ -caprolactone (oxepan-2-one).

(b) tert-Butyl has a higher migratory aptitude than methyl, so it migrates: tert-butyl ethanoate, $C H_{3} COOC (C H_{3})_{3}$ .

Solved Example 10

Complete: (a)

(C H_{3})_{2} C (OH) C H_{2} OH

with

H^{+}

; (b) 2,3-diphenylbutane-2,3-diol,

C_{6} H_{5} C (C H_{3}) (OH) C (C H_{3}) (OH) C_{6} H_{5}

, with

H^{+}

Solution:

(a) Loss of the tertiary OH gives the tertiary cation $(C H_{3})_{2} C^{+} - C H_{2} OH$ . A hydride shift from $C H_{2} OH$ gives the protonated aldehyde, and loss of $H^{+}$ gives 2-methylpropanal, $(C H_{3})_{2} CHCHO$ .

(b) Either OH leaves to give a benzylic tertiary cation. On the neighbouring carbon, phenyl (higher aptitude) migrates rather than methyl, giving 3,3-diphenylbutan-2-one, $(C_{6} H_{5})_{2} C (C H_{3}) COC H_{3}$ .

Solved Example 11

What is formed when 4-methylbenzil,

C H_{3} C_{6} H_{4} COCO C_{6} H_{5}

, is treated with

O H^{-}

and then acid? What does sodium methoxide give with benzil?

Solution:

The p-tolyl group donates electrons, so the carbonyl group next to the phenyl group is more positive and is attacked by $O H^{-}$ . The phenyl group on that carbon, the less electron-donating aryl group, migrates. The product is 2-hydroxy-2-(4-methylphenyl)-2-phenylethanoic acid, $C H_{3} C_{6} H_{4} C (OH) (C_{6} H_{5}) COOH$ .

With $C H_{3} O^{-}$ , benzil gives the ester methyl benzilate, $(C_{6} H_{5})_{2} C (OH) COOC H_{3}$ .

Solved Example 12

Compound A (

C_{8} H_{8} O

) with

N H_{2} OH

·HCl gives B and C. With acid, B and C rearrange to D and E respectively. B, C, D and E are all

C_{8} H_{9} NO

. D boiled with alcoholic KOH gives an oil F (

C_{6} H_{7} N

), which reacts rapidly with

C H_{3} COCl

to give back D. E boiled with alkali and acidified gives a white solid G (

C_{7} H_{6} O_{2}

). Identify A to G.

Solution:

A = acetophenone, $C_{6} H_{5} COC H_{3}$ . B and C are its two geometrical oximes.

In B, phenyl is anti to OH and migrates, giving D = acetanilide, $C H_{3} CONH C_{6} H_{5}$ . Its hydrolysis gives F = aniline, which is acetylated back to D.

In C, methyl is anti and migrates, giving E = N-methylbenzamide, $C_{6} H_{5} CONHC H_{3}$ . Its hydrolysis gives G = benzoic acid.

Solved Example 13

Suggest a mechanism for the reaction of

(C H_{3})_{2} C = CHC H_{2} C H_{2} C (C H_{3}) = CHCHO

with

H_{3} O^{+}

, which gives a cyclohexene ring carrying OH and

C (C H_{3})_{2} OH

groups.

Solution:

$H^{+}$ protonates the aldehyde oxygen, making the carbonyl carbon strongly electrophilic.
The remote C=C, the $(C H_{3})_{2} C = CH$ group, acts as a nucleophile. Its CH carbon attacks the carbonyl carbon, closing a six-membered ring and leaving a tertiary carbocation on the $(C H_{3})_{2} C$ carbon.
The former carbonyl oxygen is now a ring OH.
Water captures the tertiary carbocation, and loss of $H^{+}$ gives the second OH: 6-(2-hydroxypropan-2-yl)-3-methylcyclohex-2-en-1-ol.

Solved Example 14

2-Methylcyclohexan-1-one is treated with m-chloroperbenzoic acid. The major product is:
(A) 3-methyloxepan-2-one
(B) 7-methyloxepan-2-one
(C) 2-methylcyclohexan-1-ol
(D) 6-oxoheptanoic acid

Solution:

Answer: (B). The two groups on the carbonyl carbon are a secondary ring carbon (CH bearing the methyl) and a primary ring $C H_{2}$ . The secondary carbon has the higher migratory aptitude, so the new oxygen goes in between C=O and CH( $C H_{3}$ ). In the seven-membered lactone (oxepan-2-one: O-1, C=O at C-2) that carbon becomes C-7: 7-methyloxepan-2-one. Option (A) would need the $C H_{2}$ to migrate.

Solved Example 15

The (E)-oxime of butan-2-one is warmed with concentrated

H_{2} S O_{4}

. The product is:
(A) N-ethylethanamide
(B) N-methylpropanamide
(C) butanamide
(D) butan-2-amine

Solution:

Answer: (A). In the (E)-oxime the higher-priority groups, OH on N and ethyl on C, are on opposite sides, so ethyl is anti to OH. Ethyl migrates to nitrogen and methyl stays on the carbonyl carbon: $C H_{3} CONHC H_{2} C H_{3}$ , N-ethylethanamide. The (Z)-oxime would give N-methylpropanamide, (B).

Solved Example 16

Suggest two Wittig routes to 2-methylbut-2-ene,

(C H_{3})_{2} C = CHC H_{3}

, and say how the ylide for each is made.

Solution:

Cut the C=C and put O on one side and $PP h_{3}$ on the other.

Route 1: propanone + $C H_{3} CH = PP h_{3}$ . The ylide comes from bromoethane: $P h_{3} P$ + $C H_{3} C H_{2} Br$ gives the phosphonium salt, and BuLi removes a proton.

Route 2: ethanal + $(C H_{3})_{2} C = PP h_{3}$ , from 2-bromopropane in the same way. Route 1 is preferred because a primary halide undergoes the $S_{N} 2$ step with $P h_{3} P$ far more easily than a secondary one.

Solved Example 17

Cyclohexane-1,2-dione is heated with aqueous NaOH and then acidified. Identify the product and name the reaction.

Solution:

This is a benzilic acid rearrangement. $O H^{-}$ adds to one C=O; a ring carbon then migrates to the other carbonyl carbon, so the ring loses one carbon (ring contraction). The product is 1-hydroxycyclopentane-1-carboxylic acid, $C_{6} H_{10} O_{3}$ : the six ring carbons of the dione become five ring carbons plus the COOH carbon.

Solved Example 18

3-Phenylprop-2-enal (cinnamaldehyde) is treated separately with (a) aluminium isopropoxide in propan-2-ol and (b) excess

H_{2}

over Ni. Give the products.

Solution:

(a) MPV reduction touches only the C=O group: 3-phenylprop-2-en-1-ol (cinnamyl alcohol), $C_{6} H_{5} CH = CHC H_{2} OH$ ; the C=C survives and acetone distils off.

(b) Catalytic hydrogenation reduces both the C=C and the C=O (the benzene ring survives under normal conditions): 3-phenylpropan-1-ol, $C_{6} H_{5} C H_{2} C H_{2} C H_{2} OH$ .

Practice Questions

Benzil, $C_{6} H_{5} COCO C_{6} H_{5}$ , with $O H^{-}$ gives A. A is: (a) benzoin (b) $(C_{6} H_{5})_{2} C (OH) CO O^{-}$ (c) a diol (d) noneAnswer: (b), the benzilate ion (benzilic acid rearrangement).
Complete: propanone + $ClC H_{2} COO C_{2} H_{5}$ with $NaN H_{2}$ .Answer: Darzens condensation gives the glycidic ester ethyl 3,3-dimethyloxirane-2-carboxylate.
Give the major products: (i) cyclohexanone with $Se O_{2}$ ; (ii) cyclohexanone + $ClC H_{2} COO C_{2} H_{5}$ with base.Answer: (i) The $α$ - $C H_{2}$ becomes C=O: cyclohexane-1,2-dione. (ii) Darzens: ethyl 1-oxaspiro[2.5]octane-2-carboxylate.
Identify C in $C H_{3} CN$ → (Na/ $C_{2} H_{5} OH$ ) A → ( $HN O_{2}$ ) B → ( $KMn O_{4}$ , $H^{+}$ ) C. (a) $C H_{3} COOH$ (b) $C H_{3} C H_{2} NHOH$ (c) $C H_{3} CON H_{2}$ (d) $C H_{3} CHO$ Answer: (a). A = $C H_{3} C H_{2} N H_{2}$ , B = $C H_{3} C H_{2} OH$ (with $N_{2}$ ), C = $C H_{3} COOH$ .
End product of $HC \equiv CH$ → ( $C H_{3} MgBr$ ) → ( $C O_{2}$ / $H_{3} O^{+}$ ) → ( $HgS O_{4}$ / $H_{2} S O_{4}$ ) → ( $A g_{2} O$ , heat): (a) $C H_{3} COCOOH$ (b) $HOOCC H_{2} COOH$ (c) $C H_{3} COCHO$ (d) $OHCC H_{2} COOH$ Answer: (b). The acetylide gives propiolic acid, $HC \equiv CCOOH$ . Water adds to the triple bond conjugated with COOH so that O ends up on the terminal carbon, giving $OHCC H_{2} COOH$ , which $A g_{2} O$ oxidises to malonic acid.
What is the major product when 1-phenyl-2-[2-(bromomethyl)phenyl]ethan-1-one is treated with base?Answer: The enolate oxygen attacks the $C H_{2} Br$ intramolecularly (O-alkylation), closing a six-membered ring: 3-phenyl-1H-isochromene. C-alkylation would need a strained four-membered ring.
An aromatic compound A gives two isomers B and C with $N H_{2} OH$ . C rearranges with $H_{2} S O_{4}$ to D ( $C_{8} H_{9} NO$ ), which hydrolyses to E and F. A with perbenzoic acid gives G, and hydrolysis of G gives H and E. The anhydride of E and its sodium salt condense with PhCHO to give cinnamic acid, and H with phthalic anhydride/ $H_{2} S O_{4}$ gives phenolphthalein. Identify A to H.Answer: A = acetophenone; B, C = its (E) and (Z) oximes; D = acetanilide; E = $C H_{3} COOH$ ; F = aniline; G = phenyl ethanoate (Baeyer-Villiger, phenyl migrates); H = phenol. The anhydride of E with PhCHO is the Perkin reaction.

Common Mistakes to Avoid

Watch out

Using a catalytic amount of base in the Claisen condensation. A full equivalent is needed to deprotonate the $β$ -keto ester and drive the equilibrium.
Expecting a Claisen condensation from an ester with only one $α$ -H, such as ethyl 2-methylpropanoate. Its product cannot be deprotonated, so the reaction does not go to completion.
Using NaOH in the benzoin condensation. Only cyanide catalyses it.
Moving the syn group in a Beckmann rearrangement. The group anti to the OH migrates.
In Baeyer-Villiger reactions, moving the smaller group. The group with higher migratory aptitude moves (tert-alkyl > sec-alkyl $\approx$ phenyl > primary > methyl), and it keeps its configuration.
In pinacol rearrangements, choosing the migrating group before the cation. First decide which OH leaves (the one giving the more stable carbocation), then pick the best migrating group on the neighbouring carbon.
Writing a Grignard reagent instead of zinc in the Reformatsky reaction. The organozinc reagent tolerates the ester group; a Grignard reagent does not.
Using MPV reduction on $β$ -diketones, or expecting it to reduce C=C or $N O_{2}$ . It reduces only the carbonyl group of ordinary aldehydes and ketones.

Frequently Asked Questions

What is the difference between aldol and Claisen condensation?

Both start with an enolate attacking a carbonyl carbon. In the aldol reaction the electrophile is an aldehyde or ketone, and the alkoxide is protonated to a beta-hydroxy carbonyl compound. In the Claisen condensation the electrophile is an ester, so the alkoxide expels an alkoxide group and gives a beta-keto ester; one full equivalent of base drives it.

What is the Perkin reaction?

The Perkin reaction condenses an aromatic aldehyde with an aliphatic acid anhydride in the presence of the sodium salt of the same acid, on long heating at about 443-453 K. Benzaldehyde, ethanoic anhydride and sodium ethanoate give cinnamic acid and ethanoic acid. The anhydride needs at least two alpha hydrogens so that the aldol-type product can lose water.

How does the Wittig reaction work?

A phosphorus ylide, $P h_{3} P = C R_{2}$ , adds to an aldehyde or ketone to form a betaine that closes to a four-membered oxaphosphetane. This ring breaks into an alkene and triphenylphosphine oxide. The carbonyl oxygen is replaced by the $C R_{2}$ group, so the new double bond forms exactly where the C=O group was.

Which group migrates in Baeyer-Villiger oxidation?

The group with the higher migratory aptitude moves from carbon to oxygen: hydrogen first, then tertiary alkyl, then secondary alkyl or phenyl, then primary alkyl, and methyl last. Acetophenone gives phenyl ethanoate, and pinacolone gives tert-butyl ethanoate. The migrating group moves with its electron pair and keeps its configuration.

Why does only cyanide catalyse the benzoin condensation?

Cyanide is a good nucleophile that adds to the aldehyde, it makes the former aldehyde hydrogen acidic so that the carbon becomes a nucleophile, and it is a good leaving group in the final step. Hydroxide and other bases cannot do all three jobs, so they do not catalyse benzoin formation from aromatic aldehydes.

Which group migrates in the Beckmann rearrangement?

The group anti (trans) to the hydroxyl group of the oxime migrates to nitrogen, with retention of configuration, as the N-O bond breaks. The rearrangement is so stereospecific that the product amide reveals the configuration of the oxime. Cyclohexanone oxime gives epsilon-caprolactam, the monomer of nylon-6.

Which named reactions of aldehydes and ketones are most asked in JEE Advanced?

JEE Advanced most often tests the mechanisms and selectivity of Claisen, Dieckmann, Perkin and Knoevenagel condensations, the Wittig reaction, Baeyer-Villiger migratory aptitude, pinacol and benzilic acid rearrangements, and Beckmann anti migration. These usually appear inside multi-step problems that ask you to identify A, B and C.

Are these named reactions needed for NEET?

NEET follows NCERT, which does not require most of these mechanisms. NEET students should still recognise the named products, such as cinnamic acid from the Perkin reaction, benzoin, caprolactam from the Beckmann rearrangement and urotropine, and should know that these reactions build on aldol condensation and nucleophilic addition, which are core NEET topics.

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