Which of the following compounds is most reactive towards nucleophilic addition reactions? (i) CH₃CHO (ii) CH₃

NCERT Exemplar · Q2

Q. Which of the following compounds is most reactive towards nucleophilic addition reactions?

(i)

CH_3CHO

(ii)

CH_3COCH_3

(iii)

C_6H_5CHO

(iv)

C_6H_5COCH_3

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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)

Tip

In exams, remember: hydride ion ( $\ce{H^-}$ ) is a nucleophile in reductions (e.g., $\ce{NaBH4}$ reduces aldehydes/ketones to alcohols via nucleophilic addition).

The Big Picture – Why This Matters

Nucleophilic addition is the fundamental reaction of carbonyl compounds. It is how:

Aldehydes and ketones form alcohols (with $\ce{NaBH4}$ or $\ce{LiAlH4}$ )
Cyanohydrins are made (important in organic synthesis)
Grignard reagents ( $\ce{RMgX}$ ) add to carbonyls to form new carbon–carbon bonds
Hemiacetals and acetals form (key in carbohydrate chemistry)

Every time you see a $\ce{C=O}$ group, think: this carbon is a target for nucleophiles.

Final Answer

Nucleophilic addition is a reaction where an electron-rich nucleophile attacks the electrophilic carbon of a polar multiple bond (typically $\ce{C=O}$ or $\ce{C≡N}$ ), breaking the $\pi$ bond and forming two new sigma bonds — one to the nucleophile and one to a proton (or other electrophile). The driving force is the polarity of the $\ce{C=O}$ bond and the stability gained by forming stronger sigma bonds.

Nucleophilic addition is a foundational mechanism in the NCERT Class 12 Chemistry chapter on Aldehydes, Ketones and Carboxylic Acids, and ‘nucleophilic addition reaction mechanism’ or ‘nucleophilic addition class 12 chemistry’ are common searches among students preparing for CBSE boards, JEE Main and NEET. Understanding why carbonyl carbons are electrophilic is the key idea tested across most important questions on this chapter.

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.

Reversibility: If the nucleophile is a poor leaving group (like $\ce{OH^-}$ ), the addition is reversible. If it's a good leaving group (like $\ce{CN^-}$ in cyanohydrin formation), the equilibrium favours product.

5. Why Different Nucleophiles Give Different Products

Nucleophile	Product Type	Why?
$\ce{H^-}$ (from NaBH₄)	Alcohol	Hydride adds, then protonation
$\ce{CN^-}$	Cyanohydrin	CN⁻ adds, stable C-CN bond
$\ce{RO^-}$	Hemiacetal	Alkoxide adds, then protonation
$\ce{NH_3}$	Imine (after water loss)	N adds, then elimination of H₂O

The pattern: The nucleophile always attacks the same carbon — the product differs only in what group is attached.

6. The "Why" in One Sentence

Nucleophilic addition happens because the carbonyl carbon is electron-deficient ( $\delta^+$ ) and the nucleophile is electron-rich — they attract, the $\pi$ bond breaks, and the resulting negative charge on oxygen is neutralised by protonation.

Quick Exam Checklist

✓ Rate depends on both [Nu⁻] and [carbonyl] — second order
✓ Carbon changes hybridisation: $sp^2 \rightarrow sp^3$
✓ Tetrahedral intermediate is key — unstable, short-lived
✓ Protonation is fast — always the second step
✓ Reversibility depends on nucleophile — poor leaving groups make it reversible

The reactivity in nucleophilic addition depends on the electrophilicity of the carbonyl carbon. Acetaldehyde ( $CH_3CHO$ ) is the most reactive because it has the least steric hindrance and the strongest electron-withdrawing effect from the alkyl group, making option (i) correct.

Nucleophilic addition to a carbonyl compound is all about how easily a nucleophile can attack the electrophilic carbon of the C=O group. The key factors are: (1) the electron density on the carbonyl carbon (more positive = more reactive), and (2) the steric hindrance around it (less bulky = easier attack). Let’s see how each compound stacks up.

Compare the substituents on the carbonyl carbon.

In $CH_3CHO$ (acetaldehyde), one side is a hydrogen atom and the other is a methyl group ( $CH_3$ ). Hydrogen is small and doesn’t donate electrons much, so the carbonyl carbon remains fairly electron-deficient.

In $CH_3COCH_3$ (acetone), both sides are methyl groups. Methyl groups are electron-donating via hyperconjugation and inductive effect, which reduces the positive charge on the carbonyl carbon. Plus, two methyl groups create more steric bulk, making it harder for a nucleophile to approach.
Now look at the aromatic compounds.

$C_6H_5CHO$ (benzaldehyde) has a phenyl ring attached. The phenyl ring can delocalize the positive charge on the carbonyl carbon through resonance — the lone pair on oxygen can be pushed into the ring, but more importantly, the ring’s $\pi$ electrons can interact with the carbonyl. This resonance stabilizes the carbonyl group, making it less electrophilic.

$C_6H_5COCH_3$ (acetophenone) has both a phenyl ring and a methyl group. The phenyl ring still provides resonance stabilization, and the methyl group adds electron donation and steric hindrance. So it’s even less reactive.
Rank them by reactivity.

The general order for nucleophilic addition reactivity is:

$HCHO > CH_3CHO > C_6H_5CHO > CH_3COCH_3 > C_6H_5COCH_3$

(formaldehyde is not in the options, but it’s the most reactive).

Among the given, $CH_3CHO$ has the smallest substituent (H) on one side and only one electron-donating methyl group, so it’s the most electrophilic and least hindered.

Watch out

A common mistake is to think that the phenyl ring withdraws electrons (it does inductively), but its resonance donation actually decreases the carbonyl’s electrophilicity. So benzaldehyde is less reactive than acetaldehyde, not more.

Tip

Remember: For nucleophilic addition, less substitution on the carbonyl carbon means higher reactivity. Aldehydes (with at least one H) are generally more reactive than ketones (two alkyl/aryl groups). Among aldehydes, those with smaller alkyl groups are more reactive.

Confirm with a classic example. In the reaction with HCN or NaHSO $_3$ , acetaldehyde reacts readily, acetone reacts slowly, and benzaldehyde reacts even slower. Acetophenone is the least reactive of the lot.

✓Final answer

CH_3CHO

Method: Steric and Electronic Effect Analysis for Nucleophilic Addition Reactivity

Concept First (Why this method works)

Nucleophilic addition to a carbonyl group depends on two factors:

Steric hindrance around the carbonyl carbon — less hindrance = easier attack
Electronic effects (inductive and resonance) — more positive carbonyl carbon = faster attack

Steps

Step 1: Identify the carbonyl compounds

Compound	Type
$CH_3CHO$	Aliphatic aldehyde
$CH_3COCH_3$	Aliphatic ketone
$C_6H_5CHO$	Aromatic aldehyde
$C_6H_5COCH_3$	Aromatic ketone

Step 2: Compare steric hindrance

Aldehydes ( $RCHO$ ) have one alkyl/aryl group → less crowded
Ketones ( $RCOR'$ ) have two groups → more crowded

So: aldehydes > ketones (sterically)

Step 3: Compare electronic effects

In $C_6H_5CHO$ and $C_6H_5COCH_3$ , the phenyl ring donates electrons via resonance → reduces carbonyl carbon's positive charge → decreases reactivity
In $CH_3CHO$ and $CH_3COCH_3$ , alkyl groups donate electrons inductively (+I effect) but less effectively than phenyl's resonance donation

Step 4: Combine both factors

Least hindered + least electron donation = most reactive
$CH_3CHO$ has: smallest steric hindrance + weakest electron donation

Final Answer

Most reactive: $\boxed{\text{(A) } CH_3CHO}$ (acetaldehyde)

Quick Comparison Table

Compound	Steric hindrance	Electronic deactivation	Reactivity rank
$CH_3CHO$	Low	Low	1st
$C_6H_5CHO$	Low	High (resonance)	2nd
$CH_3COCH_3$	High	Low	3rd
$C_6H_5COCH_3$	High	High (resonance)	4th

Here is a breakdown of the common mistakes students make on this question, along with the correct reasoning to avoid them.

The Core Concept: Why Reactivity Varies

Nucleophilic addition to a carbonyl group ( $C=O$ ) is controlled by electrophilicity of the carbonyl carbon. The more positive (electron-deficient) this carbon is, the faster a nucleophile will attack.

The key factors are:

Inductive Effect: Electron-withdrawing groups (EWG) make the carbon more positive (more reactive). Electron-donating groups (EDG) make it less positive (less reactive).
Steric Hindrance: Bulky groups attached to the carbonyl carbon physically block the nucleophile from attacking.

Mistake #1: Ignoring Steric Hindrance (The Most Common Error)

The Mistake: Students often rank reactivity based only on inductive effects, forgetting that a bulky group physically blocks the attack.

Example: They might think $CH_3COCH_3$ (acetone) is more reactive than $CH_3CHO$ (acetaldehyde) because two methyl groups donate more electron density, but they forget the size issue.

The Correct Reasoning:

$CH_3CHO$ (Acetaldehyde): Has one small $H$ and one $CH_3$ group. Very little steric hindrance.
$CH_3COCH_3$ (Acetone): Has two $CH_3$ groups. This creates significant steric hindrance, making it less reactive than acetaldehyde.

How to Avoid: Always draw the structure. If the carbonyl carbon is attached to two large groups (like two alkyl groups or an aromatic ring), expect low reactivity due to steric hindrance, even if the inductive effect is favorable.

Mistake #2: Misjudging the Inductive Effect of the Phenyl Ring ( $C_6H_5$ )

The Mistake: Students assume the phenyl ring is a strong electron-withdrawing group (like a nitro group) and therefore makes the carbonyl carbon very positive.

The Correct Reasoning:

The phenyl ring is electron-withdrawing by induction (due to its $sp^2$ carbons being more electronegative than $sp^3$ ).
However, it is electron-donating by resonance. The $\pi$ electrons of the ring can delocalize into the carbonyl group, partially neutralizing the positive charge on the carbonyl carbon.
Net effect: The resonance donation is stronger than the inductive withdrawal. This makes the carbonyl carbon in $C_6H_5CHO$ (benzaldehyde) less electrophilic than in $CH_3CHO$ (acetaldehyde).

How to Avoid: Remember the "Resonance Rule": If a group can donate electrons via resonance into the carbonyl, it decreases reactivity towards nucleophilic addition. The phenyl ring does this.

Mistake #3: Forgetting the "Ketone vs. Aldehyde" Rule

The Mistake: Students treat all ketones and aldehydes as having similar reactivity.

The Correct Reasoning:

Aldehydes ( $RCHO$ ) are generally more reactive than ketones ( $RCOR'$ $R C O R^{'}$ ) because:
1. Sterics: Aldehydes have one small $H$ atom, ketones have two alkyl/aryl groups.
2. Electronics: The $H$ atom is not electron-donating, while alkyl groups are. This makes the carbonyl carbon in aldehydes more positive.

How to Avoid: Memorize the general trend: Aldehyde > Ketone (for similar alkyl groups). This immediately tells you that $CH_3CHO$ is more reactive than $CH_3COCH_3$ .

Applying the Logic to the Options

Let's rank them from most to least reactive:

$CH_3CHO$ (Acetaldehyde): Aldehyde. One small $H$ , one $CH_3$ . Least steric hindrance, no resonance donation. Most reactive.
$C_6H_5CHO$ (Benzaldehyde): Aldehyde. One small $H$ , one phenyl ring. The phenyl ring causes resonance stabilization of the carbonyl, making it less reactive than acetaldehyde.
$CH_3COCH_3$ (Acetone): Ketone. Two $CH_3$ groups. Steric hindrance and electron donation from two alkyl groups make it less reactive than both aldehydes.
$C_6H_5COCH_3$ (Acetophenone): Ketone. One $CH_3$ , one phenyl ring. Maximum steric hindrance (two bulky groups) and maximum resonance stabilization (from the phenyl ring). Least reactive.

Final Answer: (A) $CH_3CHO$ is the most reactive.

Quick Cheat Sheet to Avoid Mistakes

Compound	Type	Steric Hindrance	Resonance Stabilization	Reactivity Rank
$CH_3CHO$	Aldehyde	Low	None	1 (Highest)
$C_6H_5CHO$	Aldehyde	Low	High (from ring)	2
$CH_3COCH_3$	Ketone	High	None	3
$C_6H_5COCH_3$	Ketone	Very High	High (from ring)	4 (Lowest)

The Golden Rule: When comparing reactivity, Steric Hindrance > Resonance > Inductive Effect in most cases for this specific reaction.

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

GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
$Q. The increasing order of reactivity towards nucleophilic addition is:$
(a) Acetophenone < Benzaldehyde < p-Tolualdehyde < p-Nitrobenzaldehyde
(b) Benzaldehyde < Acetophenone < p-Nitrobenzaldehyde < p-Tolualdehyde
(c) p-Tolualdehyde < Benzaldehyde < p-Nitrobenzaldehyde < Acetophenone
(d) Acetophenone < p-Tolualdehyde < Benzaldehyde < p-Nitrobenzaldehyde
›Reveal solutionSolution
Nucleophilic addition to a carbonyl is favoured by a more electrophilic (less hindered, less electron-rich) carbonyl carbon: aldehydes beat ketones, and electron-withdrawing ring substituents beat electron-donating ones.

Two factors govern reactivity toward nucleophilic addition here:

Aldehyde vs ketone: aldehydes are generally more reactive than ketones — fewer/smaller substituents mean less steric hindrance and less electron donation into the carbonyl carbon, keeping it more electrophilic.

Ring substituent on aromatic aldehydes: an electron-WITHDRAWING group (like –NO2) pulls electron density away from the carbonyl carbon, making it MORE electrophilic (more reactive); an electron-DONATING group (like –CH3) pushes electron density in, making the carbonyl LESS electrophilic (less reactive).

Ranking the four:

Acetophenone (a ketone) — least reactive (steric + extra alkyl/aryl electron donation).

p-Tolualdehyde (aldehyde with EDG –CH3) — less reactive than plain benzaldehyde.

Benzaldehyde (aldehyde, no substituent) — baseline reactivity.

p-Nitrobenzaldehyde (aldehyde with strong EWG –NO2) — most reactive.

Increasing order: Acetophenone < p-Tolualdehyde < Benzaldehyde < p-Nitrobenzaldehyde

✓Final answer
$(d) Acetophenone < p-Tolualdehyde < Benzaldehyde < p-Nitrobenzaldehyde.$
GUJCET 2024Set 131 markMCQ
$'R' + CH_3-CO-CH_3 \xrightarrow{H^+}$
›Reveal solutionSolution

Schiff's base $=$ imine $\;C{=}N$ , formed by a carbonyl reacting with a primary amine.

Concept. A ketone/aldehyde condenses with a primary amine ( $RNH_2$ ) under acid catalysis to give a Schiff's base (imine). Hydrazine and phenylhydrazine give hydrazones, and hydroxylamine gives an oxime — not Schiff bases.

So $R=CH_3{-}NH_2$ (methylamine, a primary amine).

✓Final answer
$CH_3-NH_2$

ANSWER: (C)
GSEB Higher Secondary Certificate (HSC) Examination 2023Set ANNUAL1 markMCQ
$Q. Salicylaldehyde on heating with zinc dust give ___ organic product.$
(a) Benzene
(b) Benzaldehyde
(c) Benzoic acid
(d) Benzyl alcohol
›Reveal solutionSolution

Zn dust distillation removes the -OH of salicylaldehyde, giving benzaldehyde.

Salicylaldehyde is 2-hydroxybenzaldehyde (a benzene ring bearing -CHO and -OH at adjacent positions). Heating a phenol with zinc dust replaces the aromatic -OH group with -H (reduction/removal of OH). Applying this to salicylaldehyde removes only the phenolic -OH, leaving the -CHO group intact:

2-HO-C6H4-CHO + Zn -> C6H5-CHO (benzaldehyde) + ZnO.

✓Final answer
$(b) Benzaldehyde.$
GUJCET 2022Set 171 markMCQ
$C_6H_5-CHO + CH_3CHO \xrightarrow[\Delta]{OH^-}$
›Reveal solutionSolution

Base-catalysed cross-aldol + heat ⇒ dehydration ⇒ cinnamaldehyde $C_6H_5CH=CH-CHO$ .

Concept. Benzaldehyde has no α-H, so acetaldehyde supplies the enolate (nucleophile). The aldol adds, and on heating with base the β-hydroxy aldehyde loses water (condensation) to give the conjugated α,β-unsaturated aldehyde.
$C_6H_5CHO + CH_3CHO \xrightarrow[\Delta]{OH^-} C_6H_5CH=CH-CHO$
✓Final answer
$C_6H_5-CH=CH-CHO$

ANSWER: (B)
GUJCET 2019Set 131 markMCQ
$Q. Which of the major product obtained by hydrolysis of compound formed by reaction between formaldehyde and ethyl magnesium bromide? (A) 2 - Methyl - propan - 2 - ol (B) Propan - 1 - ol (C) Propan - 2 - ol (D) Ethan - 1 - ol$
›Reveal solutionSolution

Formaldehyde with a Grignard reagent gives a primary alcohol; ethyl MgBr adds C₂H₅ to give $C_2H_5CH_2OH$ = propan-1-ol.

Concept — Grignard on formaldehyde. $HCHO$ + $RMgX$ → $RCH_2OMgX$ → (hydrolysis) → $RCH_2OH$ , a primary alcohol with one extra carbon.

Steps. With $R = C_2H_5$ :
$HCHO + C_2H_5MgBr \rightarrow C_2H_5CH_2OMgBr \xrightarrow{H_2O} C_2H_5CH_2OH$
That is $CH_3CH_2CH_2OH$ = propan-1-ol.

✓Final answer
$Option (B) — Propan-1-ol$

ANSWER: (B)
GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQ
$Q. What is the final product 'C' in the following reaction? CH3CHO --HCN--> 'A' --H3O+--> 'B' --soda lime--> 'C'$
(a) Propanol
(b) Ethanol
(c) Propane
(d) Propanoic acid
›Reveal solutionSolution
CH3CHO -> cyanohydrin -> lactic acid -> (soda lime decarboxylation) -> ethanol.

Step by step:

CH3CHO + HCN -> CH3-CH(OH)-CN (A, acetaldehyde cyanohydrin).

A + H3O+ (hydrolysis of nitrile) -> CH3-CH(OH)-COOH (B, lactic acid).

B heated with soda lime (NaOH + CaO) -> decarboxylation, replacing -COOH by -H:

CH3-CH(OH)-COOH -> CH3-CH2-OH + CO2 (as carbonate).

Product C = CH3CH2OH = ethanol.

✓Final answer
$(b) Ethanol.$

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Q.Which of the following compounds is most reactive towards nucleophilic addition reactions?

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