An organic compound (A) with molecular formula $C_3H_5N$ on reaction with $C_6H_5MgBr$ followed by hydrolysis, gives a compound (B). Compound (B) forms an orange-red precipitate with 2,4-DNP reagent and does not give iodoform test. It neither reduces Tollens' or Fehling's reagent nor does it decolourise bromine water. On drastic oxidation with chromic acid it gives a carboxylic acid (C) having molecular formula $C_7H_6O_2$. Identify the compounds (A), (B) and (C). Write the reactions of compound (A) with $C_6H_5MgBr$ followed by hydrolysis to give compound (B).

An organic compound (A) with molecular formula C₃H₅N on reaction with C₆H₅MgBr followed by hydrolysis, gives a

Question

C_3H_5N

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Concept understanding — Nucleophilic Addition

Nucleophilic Addition – From Intuition to Precision

Imagine you have a molecule with a carbon–oxygen double bond — a carbonyl group ( $\ce{C=O}$ ). That oxygen is greedy for electrons; it pulls them away from carbon, leaving the carbon slightly positive ( $\delta^+$ ) and the oxygen slightly negative ( $\delta^-$ ). Now, if you bring a species that is rich in electrons (a nucleophile, meaning "nucleus-loving"), it will naturally be attracted to that electron-deficient carbon. The nucleophile attacks the carbon, the $\pi$ bond breaks, and the oxygen picks up a proton (or some other electrophile) to become stable. That, in a nutshell, is nucleophilic addition.

Note

The key idea: a nucleophile adds across a polar multiple bond (usually $\ce{C=O}$ or $\ce{C≡N}$ ), breaking the $\pi$ bond and forming two new sigma bonds.

The Precise Statement

Nucleophilic addition is a reaction in which a nucleophile (an electron-rich species) forms a sigma bond with an electrophilic carbon atom of a polar multiple bond (typically a carbonyl group, $\ce{C=O}$ , or a nitrile, $\ce{C≡N}$ ), while the $\pi$ bond breaks. The resulting intermediate then captures a proton (or another electrophile) to give a neutral product.

In general form:

\ce{R2C=O + Nu^- ->[H^+] R2C(OH)Nu}

The nucleophile ( $\ce{Nu^-}$ ) attacks the carbonyl carbon; the oxygen becomes negatively charged; then a proton ( $\ce{H^+}$ ) from the medium attaches to the oxygen, yielding an alcohol.

Why It Happens – The Driving Force

The carbonyl carbon is electrophilic because:

Oxygen is more electronegative than carbon, so the $\ce{C=O}$ bond is polarised: $\ce{C^{\delta+}=O^{\delta-}}$ .
The $\pi$ bond is weaker than a $\sigma$ bond, so it can break relatively easily.

A nucleophile (like $\ce{OH^-}$ , $\ce{CN^-}$ , or $\ce{NH3}$ ) has a lone pair or a negative charge. It seeks positive centres. The attack forms a new $\sigma$ bond, and the $\pi$ electrons move entirely to oxygen, creating an alkoxide ion ( $\ce{R2C-O^-}$ ). This intermediate is then quenched by a proton.

Watch out

A common mistake: thinking the nucleophile attacks the oxygen. No — oxygen is already electron-rich; the nucleophile goes to the carbon because it is electron-deficient.

A Concrete Example – Addition of HCN to a Ketone

Take acetone ( $\ce{CH3COCH3}$ ) and hydrogen cyanide ( $\ce{HCN}$ ). In the presence of a base, $\ce{CN^-}$ (the nucleophile) attacks the carbonyl carbon:

\ce{CH3COCH3 + CN^- -> CH3C(O^-)(CN)CH3}

The alkoxide intermediate then picks up a proton from $\ce{HCN}$ (or from water) to give a cyanohydrin:

\ce{CH3C(O^-)(CN)CH3 + H^+ -> CH3C(OH)(CN)CH3}

The product is acetone cyanohydrin. Notice: two new sigma bonds formed ( $\ce{C-CN}$ and $\ce{O-H}$ ), and the $\pi$ bond is gone.

What Makes a Good Nucleophile?

Strong nucleophiles are usually negatively charged or have lone pairs:

$\ce{OH^-}$ , $\ce{CN^-}$ , $\ce{NH2^-}$ , $\ce{CH3O^-}$ , $\ce{H^-}$ (from hydride reagents like $\ce{NaBH4}$ or $\ce{LiAlH4}$ )
Neutral but polarisable: $\ce{NH3}$ , $\ce{H2O}$ (weaker, but can add under acidic conditions) …

Why this formula?

Nucleophilic Addition: Why the Mechanism Works the Way It Does

Let's build this from first principles — understanding why nucleophilic addition happens, not just memorising the steps.

1. The Core Problem: Why Does Addition Happen at All?

A carbonyl group ( $\ce{C=O}$ ) has a polarised double bond:

Oxygen is more electronegative than carbon → it pulls electron density toward itself.
This creates a partial positive charge on carbon ( $\delta^+$ ) and a partial negative charge on oxygen ( $\delta^-$ ).

\ce{C^{\delta+}=O^{\delta-}}

Key insight: The carbon is electron-deficient — it wants electrons. A nucleophile (Nu⁻) is electron-rich — it wants to give electrons. This is a natural match.

2. The Two-Step Mechanism (Why Two Steps?)

Step 1: Nucleophilic Attack (Slow, Rate-Determining)

The nucleophile donates its lone pair to the electrophilic carbonyl carbon.

\ce{Nu^- + C=O -> [Nu-C-O]^-}

Why this happens:

The $\pi$ bond between C and O breaks — the electrons move entirely to oxygen.
Oxygen now has a full negative charge (alkoxide ion).
The carbon changes from $sp^2$ (trigonal planar) to $sp^3$ (tetrahedral).

This step is slow because the $\pi$ bond must break — it requires energy.

Step 2: Protonation (Fast)

The negatively charged oxygen picks up a proton ( $\ce{H^+}$ ) from the solvent or acid.

\ce{[Nu-C-O]^- + H^+ -> Nu-C-OH}

Why this happens:

The alkoxide ion is a strong base — it wants to neutralise its charge.
Protonation gives a stable neutral alcohol product.

3. The Key Formula: Rate Law Derivation

For a general nucleophilic addition:

\ce{Nu^- + R2C=O ->[k] products}

The rate law comes from the slow step (Step 1):

\text{Rate} = k[\text{Nu}^-][\ce{R2C=O}]

Why this form?

The reaction is bimolecular — two species must collide with correct orientation.
Doubling either concentration doubles the rate (first order in each).
This is second order overall.

Exam tip: This is why nucleophilic addition is often called addition-elimination when followed by loss of a leaving group (like in acyl substitution), but here it's just addition.

4. Why the Tetrahedral Intermediate Forms (And Why It's Unstable)

The intermediate is tetrahedral ( $sp^3$ hybridised carbon):

Bond angles: ~109.5°
Four groups around carbon: Nu, R, R', O⁻

Why it's unstable:

The negative charge on oxygen is high-energy.
The tetrahedral geometry is sterically crowded (especially with bulky R groups).
The intermediate collapses quickly — either back to starting materials or forward to product. …

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The key is that compound (A) is propionitrile ( $CH_3CH_2CN$ ), which undergoes nucleophilic addition with $C_6H_5MgBr$ to give, after hydrolysis, a ketone (B) — propiophenone ( $C_6H_5COCH_2CH_3$ ). This ketone gives a positive 2,4-DNP test, negative iodoform test, and does not reduce Tollens'/Fehling's reagent. Drastic oxidation of (B) yields benzoic acid (C), $C_7H_6O_2$ .

The problem is a classic detective puzzle in organic chemistry: you are given a molecular formula, a set of chemical tests, and a reaction sequence. The goal is to identify three compounds. Let's walk through the clues one by one.

Why nucleophilic addition?

The reaction of an organic compound (A) with $C_6H_5MgBr$ (a Grignard reagent) followed by hydrolysis is a hallmark of nucleophilic addition to a polar multiple bond. Grignard reagents attack electrophilic carbon atoms in compounds like carbonyls, nitriles, and epoxides. Here, (A) has formula $C_3H_5N$ — that's three carbons, five hydrogens, one nitrogen. The only common functional group with nitrogen in such a small molecule is a nitrile ( $-CN$ ) or an amine. But amines don't react with Grignard reagents in this way. So (A) is almost certainly a nitrile: $CH_3CH_2CN$ (propionitrile) or $(CH_3)_2CHCN$ (isobutyronitrile). Let's see which fits.

Identify (A) from its reaction with Grignard reagent A nitrile reacts with a Grignard reagent in two steps: first, the Grignard adds to the nitrile carbon, forming an imine intermediate. Then, hydrolysis gives a ketone. If (A) is $CH_3CH_2CN$ , then:

CH_3CH_2CN + C_6H_5MgBr \rightarrow \text{intermediate} \xrightarrow{H_3O^+} C_6H_5COCH_2CH_3

That product is propiophenone (ethyl phenyl ketone).

If (A) were $(CH_3)_2CHCN$ , the product would be $C_6H_5COCH(CH_3)_2$ (isopropyl phenyl ketone). Both are structurally possible at this stage, so we need the tests on (B) — and the formula of (A) itself — to decide.

Compound (B) gives an orange-red precipitate with 2,4-DNP

This is a positive test for a carbonyl group (aldehyde or ketone). So (B) is indeed a ketone (or aldehyde). That confirms the Grignard addition worked.
Compound (B) does NOT give the iodoform test

The iodoform test is positive for methyl ketones ( $CH_3CO-$ ) and for ethanol/acetaldehyde. Propiophenone ( $C_6H_5COCH_2CH_3$ ) has an ethyl group next to the carbonyl, not a methyl group — so it gives a negative iodoform test. Isopropyl phenyl ketone ( $C_6H_5COCH(CH_3)_2$ ) also has no $CH_3CO-$ group, so it too would be negative. So this test doesn't distinguish them yet.
It neither reduces Tollens' or Fehling's reagent

These tests are for aldehydes (and some α-hydroxy ketones). Ketones do not reduce them. So (B) is a ketone, not an aldehyde. That's consistent.
It does not decolourise bromine water

Bromine water tests for unsaturation (C=C or C≡C). So (B) has no carbon-carbon double or triple bond. That's fine for a simple ketone.
On drastic oxidation with chromic acid, (B) gives a carboxylic acid (C) with formula $C_7H_6O_2$

Drastic oxidation (hot, concentrated $K_2Cr_2O_7/H_2SO_4$ ) cleaves carbon chains next to the carbonyl. For a ketone, oxidation breaks the bond between the carbonyl carbon and one of the alkyl groups, producing two carboxylic acids (or one acid and $CO_2$ ).

The product (C) has formula $C_7H_6O_2$ . That's benzoic acid ( $C_6H_5COOH$ ). So one fragment of (B) must be a phenyl group ( $C_6H_5-$ ). The other fragment, after oxidation, becomes a smaller acid. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 17 on this concept.

CBSE 2026Set A1 markMCQ
$Q. When chloroform reacts with acetone then which of the following is formed ?$
(a) Ethylene dichloride
(b) Mesitylene
(c) Chloretone
(d) Chloral
›Reveal solutionSolution

Chloroform adds across the carbonyl of acetone to give chloretone, 1,1,1-trichloro-2-methyl-2-propanol.

Chloroform (CHCl3) in the presence of a base loses a proton and its CCl3 carbanion adds to the carbonyl carbon of acetone. The addition product is chloretone (also written chlorbutol), a well-known hypnotic/preservative.

…

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CBSE 2026Set ANNUAL1 mark
$Q. Write True or False: Contrary to electrophilic addition reactions observed in alkenes, the aldehydes and ketones undergo nucleophilic addition reactions.$
›Reveal solutionSolution

True - the polar C=O of aldehydes/ketones is attacked by nucleophiles.

In alkenes the C=C double bond is electron-rich, so it attracts electrophiles (electrophilic addition). In aldehydes and ketones the carbonyl C=O bond is polar: oxygen is electronegative and pulls electrons, leaving the carbonyl carbon partially positive (electron-deficient). Therefore th …

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CBSE 2025Set ANNUAL1 mark
$Q. Passage: Aldehydes are generally more reactive than ketones in nucleophilic addition reactions due to steric and electronic reasons. Sterically, the presence of two relatively large substituents in ketones hinders the approach of nucleophile to carbonyl carbon than in aldehydes having only one such substituent. Electronically, aldehydes are more reactive than ketones because two alkyl groups reduce the electrophilicity of the carbonyl carbon more effectively than in former (i.e. than one alkyl group does). A nucleophile attacks the electrophilic carbon atom of the polar carbonyl group from a direction approximately perpendicular to the plane of sp2 hybridised orbitals of carbonyl carbon. The hybridisation of carbon changes from sp2 to sp3 in this process and a tetrahedral alkoxide intermediate is produced. This intermediate captures a proton from the reaction medium to give the electrically neutral product.$
(b) What product is formed when CH3CHO reacts with NaHSO3? Give chemical equation.
›Reveal solutionSolution

Bisulfite ion adds across the carbonyl of acetaldehyde to give a crystalline addition compound.

Acetaldehyde undergoes nucleophilic addition with saturated sodium bisulphite solution: the bisulphite ion ( $HSO_3^-$ ) acts as the nucleophile, attacking the carbonyl carbon and forming a tetrahedral addition compound, which is a white crystalline solid (the 'bisulphite addition product'):

$CH_3CHO + NaHSO_3 \to CH_3CH(OH)SO_3Na$

…

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CBSE 2025Set A1 mark
$Q. Write True or False: Ketones containing carbonyl group.$
›Reveal solutionSolution

By definition, a ketone is a carbonyl compound in which the C=O group is bonded to two carbon (alkyl/aryl) groups.

The carbonyl group (a carbon doubly bonded to oxygen, >C=O) is the functional group common to aldehydes, ketones, and carboxylic acids. In a ketone, this carbonyl carbon is attached to two other carbon atoms (R–CO–R′), unlike an aldehyde, where the carbonyl carbon is attached to at least one hydr …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following is not a characteristic of carbonyl compounds?$
(a) They have a polarized C=O bond.
(b) They undergo nucleophilic addition reactions.
(c) They show geometric isomerism.
(d) They can be reduced to alcohol.
›Reveal solutionSolution

Geometric (cis-trans) isomerism about a C=O needs two distinguishable groups on BOTH ends of the double bond, but the oxygen end carries only a lone pair on a single atom, so plain aldehydes/ketones cannot show it.

Carbonyl compounds genuinely have a polarized C=O bond (a), readily undergo nucleophilic addition at the electrophilic carbonyl carbon (b), and can be reduced to alcohols (d) — all true. But geometric (cis–trans) isomerism requires restricted rotation about a double bond WITH two different substituents on each doubly-bonded atom; in a simple aldehyde/ketone (>C=O), the oxygen end carries only a lone pair (not tw …

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CBSE 2024Set 56/1/11 markMCQ
$Q. The formation of cyanohydrin from an aldehyde is an example of: (A) nucleophilic addition (B) electrophilic addition (C) nucleophilic substitution (D) electrophilic substitution$
›Reveal solutionSolution
Cyanohydrin formation involves the cyanide ion ( $\ce{CN^-}$ ) attacking the electrophilic carbonyl carbon of an aldehyde — a textbook case of nucleophilic addition. The answer is (A).

Why this is nucleophilic addition

The carbonyl group ( $\ce{C=O}$ ) in aldehydes is polarized: oxygen is more electronegative than carbon, so the carbon carries a partial positive charge ( $\delta^+$ ) and becomes electron-deficient. This makes it a prime target for nucleophiles — species that are electron-rich and "love" positive centers.

When we treat an aldehyde with a source of cyanide ion (typically $\ce{HCN}$ or $\ce{NaCN}$ ), the $\ce{CN^-}$ acts as a nucleophile. It donates its electron pair to the carbonyl carbon, and the $\pi$ -bond of the carbonyl breaks, with both electrons moving onto the oxygen. The result? A new $\ce{C-CN}$ bond forms, and we add two groups across the original double bond — the hallmark of an addition reaction.

The key distinction: nothing leaves the molecule. In substitution reactions, one group replaces another; here, we're simply adding to the existing structure.

Step-by-step mechanism

Generation of the nucleophile In aqueous or alcoholic medium, $\ce{HCN}$ dissociates (or $\ce{NaCN}$ provides) the cyanide ion:

$\ce{HCN <=> H^+ + CN^-}$
The $\ce{CN^-}$ is a strong nucleophile with a lone pair on carbon.

Nucleophilic attack on the carbonyl carbon The cyanide ion attacks the electrophilic carbonyl carbon of the aldehyde:

$\ce{R-CHO + CN^- -> R-CH(O^-)-CN}$
The $\pi$ -electrons of the $\ce{C=O}$ bond shift entirely onto oxygen, forming an alkoxide intermediate ( $\ce{O^-}$ ).

Protonation of the alkoxide The negatively charged oxygen picks up a proton from the medium (from $\ce{HCN}$ , water, or the solvent):

$\ce{R-CH(O^-)-CN + H^+ -> R-CH(OH)-CN}$
This gives the final cyanohydrin, which contains both a hydroxyl group ( $\ce{-OH}$ ) and a nitrile group ( $\ce{-CN}$ ) on the same carbon. …
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CBSE 2024Set 56/3/11 markMCQ
$\xrightarrow{KCN}$
›Reveal solutionSolution
Cyanide ion is a strong nucleophile and displaces only the reactive benzylic chlorine by $S_N2$ ; the aromatic (aryl) C–Cl is inert under these conditions. The major product is 4-chlorobenzyl cyanide — option (C).

The molecule 4-chlorobenzyl chloride, $Cl\text{–}C_6H_4\text{–}CH_2\text{–}Cl$ , has two very different C–Cl bonds, and the whole question turns on telling them apart.

The benzylic C–Cl is highly reactive. The $-CH_2Cl$ carbon is a primary, benzylic position. Its $S_N2$ transition state is stabilised by the adjacent aromatic ring, so cyanide readily displaces this chlorine:

$Cl\text{–}C_6H_4\text{–}CH_2Cl + CN^- \rightarrow Cl\text{–}C_6H_4\text{–}CH_2CN + Cl^-$

The aryl C–Cl is essentially inert. In an aryl chloride the C–Cl carbon is $sp^2$ and the bond has partial double-bond character from resonance with the ring, so it is short and strong. There is no $S_N2$ at an aromatic carbon, and an aryl cation is far too unstable for $S_N1$ . Nucleophilic aromatic substitution ( $S_NAr$ ) would need strong electron-withdrawing groups (e.g. $-NO_2$ ) ortho/para to the chlorine to stabilise the Meisenheimer intermediate; a weakly withdrawing $-CH_2CN$ group does not provide that, so the ring chlorine survives.

Result. Only the benzylic chlorine is replaced, giving 4-chlorobenzyl cyanide — a benzene ring carrying $-CH_2CN$ with the ring $-Cl$ still para to it. …
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CBSE 2024Set A11 markMCQ
$Q. Nucleophilic attack on carbonyl carbon atom changes its hybridization from :$
(a) $sp$ to $sp^2$
(b) $sp^2$ to $sp^3$
(c) $sp^3$ to $sp^2$
(d) $sp$ to $sp^3$
›Reveal solutionSolution

Nucleophilic addition changes the carbonyl carbon from $sp^2$ to $sp^3$ — option (b).

In a carbonyl group $>\!\text{C}=\text{O}$ the carbon is $sp^2$ hybridised and planar (trigonal, ~120°). When a nucleophile attacks the electrophilic carbonyl carbon, the C=O π bond breaks and a new σ bond forms; the carbon now has four …
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CBSE 2024Set D1 markMCQ
$Q. Chloretone is formed when chloroform reacts with$
(a) Formaldehyde
(b) Acetaldehyde
(c) Acetone
(d) Benzaldehyde
›Reveal solutionSolution

Chloroform + acetone -> chloretone.

Chloroform (CHCl3) adds across the carbonyl group of acetone (in presence of base) to give chloretone, chemically 1,1,1-trichloro-2-methyl-2-propanol, (CH3)2C(OH)CCl3, which is used as a hypnotic/sedative:

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Change occurs in hybridisation state of carbonyl carbon in nucleophilic addition reaction is -$
(a) sp2 to sp
(b) sp to sp2
(c) sp2 to sp3
(d) sp3 to sp2
›Reveal solutionSolution

Nucleophilic addition converts the planar, trigonal carbonyl carbon into a tetrahedral carbon, so its hybridisation changes from sp2 to sp3.

In an aldehyde/ketone, the carbonyl carbon is sp2 hybridised: it forms three sigma bonds (to O and to two other groups) that lie in one plane, plus a pi bond to oxygen, with bond angles close to 120 degrees. …

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CBSE 2023Set 56/2/11 markMCQ
$HCHO$
›Reveal solutionSolution
Nucleophilic addition to carbonyls is controlled by steric hindrance and electronic effects; formaldehyde ( $HCHO$ ) has the least steric hindrance and strongest electrophilic carbon, making it the most reactive. The correct option is (A).

Nucleophilic addition to a carbonyl group ( $\ce{C=O}$ ) is the fundamental reaction of aldehydes and ketones. A nucleophile attacks the electrophilic carbonyl carbon, and the key question is: what makes that carbon more or less attractive to an incoming nucleophile? Two factors dominate: steric hindrance around the carbonyl carbon, and electronic effects (inductive and hyperconjugative) that stabilise or destabilise the carbonyl.

Let’s examine each compound in the list.

Steric hindrance

The nucleophile must physically approach the carbonyl carbon. If bulky groups are attached to it, they block the path.

In $HCHO$ (formaldehyde), the carbonyl carbon is attached to two hydrogen atoms — the smallest possible substituents. There is almost no steric barrier.

In $CH_3CHO$ (acetaldehyde), one hydrogen is replaced by a methyl group. That methyl is larger than hydrogen, so approach is slightly hindered.

In $CH_3COCH_3$ (acetone), both substituents are methyl groups. The carbonyl carbon is now flanked by two bulky groups, making it significantly more crowded.

In $CH_3COC_2H_5$ (butanone, or ethyl methyl ketone), one methyl is replaced by an ethyl group — even bulkier. This is the most sterically hindered of the four.

So purely on steric grounds, reactivity order is: $HCHO > CH_3CHO > CH_3COCH_3 > CH_3COC_2H_5$ .

Electronic effects

Alkyl groups are electron-donating via the inductive effect (+I) and hyperconjugation. They push electron density toward the carbonyl carbon, making it less electrophilic (less positive). More alkyl groups = more electron donation = lower reactivity toward nucleophiles.

$HCHO$ has zero alkyl groups — the carbonyl carbon is the most electron-deficient.

$CH_3CHO$ has one alkyl group — slightly less electrophilic.

$CH_3COCH_3$ and $CH_3COC_2H_5$ each have two alkyl groups, so they are the least electrophilic. (The ethyl group is slightly more electron-donating than methyl, but the difference is small; both are much less reactive than $HCHO$ .) …
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CBSE 2023Set 56/3/11 markMCQ
$HCHO$
›Reveal solutionSolution
The reactivity of carbonyl compounds toward nucleophilic addition is governed by steric hindrance and electronic effects. Formaldehyde (I) is the most reactive, followed by acetaldehyde (II), and then acetone (III). The correct order is I > II > III, which corresponds to option (B).

Nucleophilic addition to a carbonyl group is one of the most fundamental reactions in organic chemistry. The carbonyl carbon is electrophilic because oxygen is more electronegative and pulls electron density away, leaving the carbon partially positive. A nucleophile attacks this carbon, forming a tetrahedral intermediate.

But why do different carbonyl compounds react at different rates? Two factors matter here: steric hindrance and electronic effects.

Steric hindrance: The nucleophile must physically approach the carbonyl carbon. If the carbon is surrounded by bulky groups, the approach is blocked, and the reaction slows down. Formaldehyde has two small hydrogen atoms attached to the carbonyl carbon — almost no hindrance. Acetaldehyde has one methyl group and one hydrogen — moderate hindrance. Acetone has two methyl groups — maximum hindrance among these three.

Electronic effects: Alkyl groups are electron-donating (through hyperconjugation and inductive effect). More alkyl groups attached to the carbonyl carbon mean more electron density pushed toward that carbon, making it less electrophilic (less positive). This also slows down nucleophilic attack. Formaldehyde has no alkyl groups, so its carbonyl carbon is the most electrophilic. Acetaldehyde has one methyl group, so it's less electrophilic. Acetone has two methyl groups, making it the least electrophilic.

Both factors — steric and electronic — work in the same direction here. So the order is clear.

Let's walk through it step by step.

Identify the carbonyl compounds and their substituents.

(I) HCHO: formaldehyde — two H atoms on the carbonyl carbon.

(II) CH₃CHO: acetaldehyde — one CH₃ and one H.

(III) CH₃COCH₃: acetone — two CH₃ groups.

Consider steric hindrance.

The nucleophile must approach the carbonyl carbon from above or below the plane. In formaldehyde, the two H atoms are tiny — no obstruction. In acetaldehyde, the methyl group is larger than H, so it partially blocks one side. In acetone, two methyl groups crowd the carbon from both sides, making approach difficult.

So steric hindrance increases: I < II < III.

Since more hindrance means slower reaction, reactivity due to sterics: I > II > III.

Consider electronic effects.

Methyl groups donate electrons via hyperconjugation and the inductive effect. More electron donation makes the carbonyl carbon less δ⁺, so less attractive to nucleophiles.

Formaldehyde: no donation → most δ⁺. …
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