JEE Mains Chemistry · Aldehydes, Ketones and Carboxylic Acids

Nucleophilic Addition and Carbonyl Derivatives

A nucleophile attacks the carbonyl carbon; aldehydes react faster than ketones, and the products include cyanohydrins, bisulphite adducts, acetals, oximes, hydrazones, semicarbazones, imines and enamines.

All Aldehydes, Ketones and Carboxylic Acids notes JEE Mains Chemistry strategy

Why this matters

Twenty-one PYQs, two numerical, two from 2026. Five rank carbonyl compounds by reactivity or ask about hydrates and bisulphite adducts; seven follow a cyanohydrin through hydrolysis or reduction; nine form an acetal, an oxime, a semicarbazone or an enamine.

Concept 1 of 3: Reactivity towards nucleophilic addition

A nucleophile attacks the carbonyl carbon because it carries a partial positive charge. Anything that makes that charge larger speeds up addition. Anything that feeds electrons in, or crowds the carbon, slows it down.

Definition

Aldehydes are more reactive than ketones. One alkyl group instead of two means less electron donation (+I) and less crowding at the carbon.
Methanal, with no alkyl group, is the most reactive. Among ketones, bigger groups are slower: propanone reacts faster than a ketone carrying a tert-butyl group.
Aromatic aldehydes and ketones are less reactive than aliphatic ones. The ring conjugates with the C=O and lowers the positive charge on the carbon.
On the ring, an electron-withdrawing group ( $\mathrm{{-}NO_2}$ , $\mathrm{{-}CN}$ , $\mathrm{{-}Cl}$ ) raises reactivity; a donor ( $\mathrm{{-}CH_3}$ , $\mathrm{{-}OCH_3}$ ) lowers it.
Hydration shows the same order. Methanal is mostly hydrated in water; propanone hardly at all. Strong −I groups give a stable hydrate: chloral forms chloral hydrate, $\mathrm{Cl_3CCH(OH)_2}$ .
$\mathrm{NaHSO_3}$ adds to aldehydes and unhindered ketones to give a crystalline bisulphite adduct. Dilute acid or alkali gives the carbonyl compound back, so the adduct is used to purify it.

Order of reactivity towards nucleophiles

\mathrm{HCHO > RCHO > RCOR'} \qquad \mathrm{RCHO > ArCHO} \qquad \text{ring: EWG} > \text{H} > \text{EDG}

Worked example

Arrange in order of reactivity towards HCN: propanone, methanal, ethanal, 3,3-dimethylbutan-2-one.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 24 Jan 2025 · Q40Moderate

Example 1 · Aldehydes, Ketones and Carboxylic Acids · Nucleophilic Addition and Carbonyl Derivatives

Which of the following arrangements with respect to their reactivity in nucleophilic addition reaction is correct?

A small donor group still slows addition

A para-methyl group donates electrons by hyperconjugation, so 4-methylbenzaldehyde is less reactive than benzaldehyde. Do not rank by size of the molecule; rank by the charge on the carbonyl carbon.

Aryl ketones are the slowest

An aryl ketone has both handicaps: two groups on the carbon and a ring in conjugation. It sits below every aldehyde and below simple dialkyl ketones.

Drill 4 more on reactivity towards nucleophilic addition

Concept 2 of 3: Cyanohydrins and what they turn into

Cyanide adds to the carbonyl carbon and the oxygen picks up a proton. The CN group is then a handle: hydrolysis turns it into COOH, and reduction turns it into

\mathrm{CH_2NH_2}

Definition

HCN is a weak acid and adds slowly on its own. A trace of base makes $\mathrm{CN^-}$ , the real nucleophile.
The product $\mathrm{R_2C(OH)CN}$ is a cyanohydrin. Hydrolysis (acid, or base then acid) gives the 2-hydroxy acid $\mathrm{R_2C(OH)COOH}$ .
$\mathrm{LiAlH_4}$ reduces the CN group and gives the 2-amino alcohol $\mathrm{R_2C(OH)CH_2NH_2}$ .
Hot concentrated $\mathrm{H_2SO_4}$ hydrolyses and also dehydrates: acetone cyanohydrin gives methacrylic acid, $\mathrm{CH_2{=}C(CH_3)COOH}$ .
The bisulphite adduct $\mathrm{RCH(OH)SO_3Na}$ reacts with NaCN to give the same cyanohydrin, without handling HCN.
When the two groups on the C=O differ, the new carbon is a stereocentre. Cyanide attacks both faces of the flat carbonyl equally, so the product is racemic.

Cyanohydrin, then hydrolysis or reduction

\mathrm{R_2C{=}O \xrightarrow{HCN,\ OH^-} R_2C(OH)CN} \qquad \xrightarrow{H_3O^+} \mathrm{R_2C(OH)COOH} \qquad \xrightarrow{LiAlH_4} \mathrm{R_2C(OH)CH_2NH_2}

Worked example

Propanone is treated with HCN and a little NaOH, and the product is heated with dilute acid. Give the final product and say whether it is optically active.

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The same idea in a real exam question:

JEE Mains · 2024 · 6 April 2024 · Q137Moderate

Example 2 · Aldehydes, Ketones and Carboxylic Acids · Nucleophilic Addition and Carbonyl Derivatives

{CH}_{3}COOH

is treated in turn with (i)

{LiAlH}_{4}

, (ii) PCC, (iii)

HCN/{OH}^{-}

and (iv)

H_{2}O/{OH}^{-},\ \Delta

. Identify the major product

P

HCN addition does not give an amine

The product of HCN with a carbonyl compound is the cyanohydrin. An amine appears only after a separate reduction of the CN group, for example with

\mathrm{LiAlH_4}

Racemic, not optically active

A cyanohydrin with a new stereocentre is formed as a 50:50 mixture of enantiomers. Unless a chiral reagent is used, the product shows no optical rotation.

Drill 6 more on cyanohydrins and what they turn into

Concept 3 of 3: Acetals, oximes, hydrazones, semicarbazones and enamines

An alcohol or an ammonia derivative adds to the C=O, and then water is lost. With alcohols the product keeps two C–O bonds (an acetal); with nitrogen nucleophiles the oxygen is replaced by a C=N bond.

Definition

One alcohol gives a hemiacetal (OH and OR on one carbon), which usually reverts. A second alcohol, with dry HCl, gives the acetal $\mathrm{R_2C(OR')_2}$ .
An acetal is stable to base, because the group that would have to leave is an alkoxide, a poor leaving group. Dilute aqueous acid hydrolyses it back to the carbonyl compound, so acetals are used to protect a C=O.
Ammonia derivatives $\mathrm{H_2N{-}Z}$ add and then lose water to give $\mathrm{R_2C{=}N{-}Z}$ . The reaction works best in weak acid (pH about 4 to 5): enough acid to activate the C=O, not so much that the amine is protonated.
In semicarbazide, $\mathrm{H_2N{-}NH{-}CO{-}NH_2}$ , only the NH₂ on the NH attacks. The other NH₂ is an amide nitrogen, and its lone pair is delocalised onto the C=O.
A secondary amine cannot form C=N. If the carbonyl compound has an α-hydrogen, it loses that H instead and gives an enamine.

Reagent	Product with a carbonyl compound	What to remember
One $\mathrm{R'OH}$ , dry HCl	Hemiacetal $\mathrm{R_2C(OH)OR'}$	Usually reverts; cyclic hemiacetals (sugars, lactols) are stable
Two $\mathrm{R'OH}$ , dry HCl	Acetal (from a ketone, a ketal) $\mathrm{R_2C(OR')_2}$	Stable to base; dilute acid gives the carbonyl back
Ethane-1,2-diol, dry HCl	Cyclic acetal (ethylene ketal)	Protects a C=O while another group reacts
Hydroxylamine $\mathrm{NH_2OH}$	Oxime $\mathrm{R_2C{=}NOH}$	An aldoxime loses water with $\mathrm{P_2O_5}$ to give a nitrile
Hydrazine $\mathrm{NH_2NH_2}$	Hydrazone $\mathrm{R_2C{=}NNH_2}$	First step of the Wolff–Kishner reduction
Phenylhydrazine $\mathrm{C_6H_5NHNH_2}$	Phenylhydrazone $\mathrm{R_2C{=}NNHC_6H_5}$	Crystalline; used to identify the carbonyl compound
2,4-Dinitrophenylhydrazine (2,4-DNP)	2,4-Dinitrophenylhydrazone	Yellow to orange precipitate: the test for any aldehyde or ketone
Semicarbazide $\mathrm{NH_2NHCONH_2}$	Semicarbazone $\mathrm{R_2C{=}NNHCONH_2}$	Bonds through the NH₂ of the NH–NH₂ end; the product keeps all three N
Primary amine $\mathrm{R'NH_2}$	Imine (Schiff base) $\mathrm{R_2C{=}NR'}$	The C=N carries the amine's R group
Secondary amine $\mathrm{R'_2NH}$	Enamine, C=C–NR′₂	Needs an α-hydrogen on the carbonyl compound

Every entry is addition to C=O followed by loss of water.

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The same idea in a real exam question:

JEE Mains · 2023 · 25 Jan 2023 · Q36Moderate

Example 3 · Aldehydes, Ketones and Carboxylic Acids · Nucleophilic Addition and Carbonyl Derivatives

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason

R

: Assertion A: Acetal / Ketal is stable in basic medium. Reason

R

: The high leaving tendency of alkoxide ion gives the stability to acetal/ ketal in basic medium. In the light of the above statements, choose the correct answer from the options given below:

Acetals survive base because alkoxide leaves badly

An assertion–reason item may say acetals are stable in base because alkoxide leaves easily. The reason is reversed: they are stable because alkoxide is a poor leaving group.

Which end of semicarbazide bonds

The product is

\mathrm{R_2C{=}N{-}NH{-}CO{-}NH_2}

. A structure written as

\mathrm{R_2C{=}N{-}CO{-}NH{-}NH_2}

has bonded through the amide nitrogen, which is not nucleophilic.

Drill 8 more on acetals, oximes, hydrazones, semicarbazones and enamines

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Reactivity towards nucleophilic addition
Order of reactivity towards nucleophiles
$\mathrm{HCHO > RCHO > RCOR'} \qquad \mathrm{RCHO > ArCHO} \qquad \text{ring: EWG} > \text{H} > \text{EDG}$
Cyanohydrins and what they turn into
Cyanohydrin, then hydrolysis or reduction
$\mathrm{R_2C{=}O \xrightarrow{HCN,\ OH^-} R_2C(OH)CN} \qquad \xrightarrow{H_3O^+} \mathrm{R_2C(OH)COOH} \qquad \xrightarrow{LiAlH_4} \mathrm{R_2C(OH)CH_2NH_2}$

Reference tables (1)

Acetals, oximes, hydrazones, semicarbazones and enamines10 rows

Reagent	Product with a carbonyl compound	What to remember
One $\mathrm{R'OH}$ , dry HCl	Hemiacetal $\mathrm{R_2C(OH)OR'}$	Usually reverts; cyclic hemiacetals (sugars, lactols) are stable
Two $\mathrm{R'OH}$ , dry HCl	Acetal (from a ketone, a ketal) $\mathrm{R_2C(OR')_2}$	Stable to base; dilute acid gives the carbonyl back
Ethane-1,2-diol, dry HCl	Cyclic acetal (ethylene ketal)	Protects a C=O while another group reacts
Hydroxylamine $\mathrm{NH_2OH}$	Oxime $\mathrm{R_2C{=}NOH}$	An aldoxime loses water with $\mathrm{P_2O_5}$ to give a nitrile
Hydrazine $\mathrm{NH_2NH_2}$	Hydrazone $\mathrm{R_2C{=}NNH_2}$	First step of the Wolff–Kishner reduction
Phenylhydrazine $\mathrm{C_6H_5NHNH_2}$	Phenylhydrazone $\mathrm{R_2C{=}NNHC_6H_5}$	Crystalline; used to identify the carbonyl compound
2,4-Dinitrophenylhydrazine (2,4-DNP)	2,4-Dinitrophenylhydrazone	Yellow to orange precipitate: the test for any aldehyde or ketone
Semicarbazide $\mathrm{NH_2NHCONH_2}$	Semicarbazone $\mathrm{R_2C{=}NNHCONH_2}$	Bonds through the NH₂ of the NH–NH₂ end; the product keeps all three N
Primary amine $\mathrm{R'NH_2}$	Imine (Schiff base) $\mathrm{R_2C{=}NR'}$	The C=N carries the amine's R group
Secondary amine $\mathrm{R'_2NH}$	Enamine, C=C–NR′₂	Needs an α-hydrogen on the carbonyl compound

Every entry is addition to C=O followed by loss of water.

Watch out for (6)

A small donor group still slows addition
→ Reactivity towards nucleophilic addition
Aryl ketones are the slowest
→ Reactivity towards nucleophilic addition
HCN addition does not give an amine
→ Cyanohydrins and what they turn into
Racemic, not optically active
→ Cyanohydrins and what they turn into
Acetals survive base because alkoxide leaves badly
→ Acetals, oximes, hydrazones, semicarbazones and enamines
Which end of semicarbazide bonds
→ Acetals, oximes, hydrazones, semicarbazones and enamines

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