Do the following conversions in not more than two steps : (a) CH₃CN (acetonitrile) to CH₃-CO-CH₃ (propan-2-one

Question

Q. Do the following conversions in not more than two steps :

(a)

CH_3CN

(acetonitrile) to

CH_3-CO-CH_3

(propan-2-one)

(b)

C_6H_5COOH

(benzoic acid, drawn as a benzene ring bearing

-COOH

) to

C_6H_6

(benzene, drawn as a plain benzene ring)

CBSECBSE Class XII Board 2023Subjective· 2mImportance★★★★★

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Concept understanding — Clemmensen Reduction Cannizzaro Reaction

Clemmensen Reduction & Cannizzaro Reaction: Two Completely Different Reactions

These two reactions are often grouped together in textbooks because they both involve carbonyl compounds (C=O), but they do entirely different things. Let's take them one at a time.

Clemmensen Reduction

Intuition first. Imagine you have a ketone or aldehyde — a molecule with a C=O group. You want to remove that oxygen entirely and replace the C=O with two hydrogen atoms, turning it into a simple hydrocarbon chain. That's a reduction (adding hydrogen, removing oxygen). The Clemmensen reduction is a brute-force way to do this using a strongly acidic, reducing environment.

The precise reaction:

A ketone or aldehyde is heated with zinc amalgam (Zn-Hg) and concentrated hydrochloric acid (HCl). The C=O group is reduced to a CH₂ group.

\ce{R-C(=O)-R' + 4[H] ->[Zn(Hg), HCl, heat] R-CH2-R' + H2O}

\text{Aldehyde or Ketone} \xrightarrow{\text{Zn(Hg), conc. HCl, } \Delta} \text{Hydrocarbon}

Key points for exams:

Works only for ketones and aldehydes that are stable in strong acid.
Does not work for compounds that get destroyed by conc. HCl (e.g., acid-sensitive groups like esters, nitriles).
The mechanism is complex and not usually tested in detail — just know it's a reductive removal of C=O.
The product is always a saturated hydrocarbon (alkane).

Watch out

Clemmensen reduction cannot reduce carboxylic acids, esters, or amides. Only aldehydes and ketones.

Example:

Acetophenone ( $\ce{C6H5-CO-CH3}$ ) → Ethylbenzene ( $\ce{C6H5-CH2-CH3}$ )

Cannizzaro Reaction

Intuition first. This is a disproportionation reaction — one molecule of aldehyde gets oxidised (to a carboxylic acid) while another gets reduced (to an alcohol). It happens only with aldehydes that have no alpha-hydrogen atoms (i.e., the carbon next to the C=O has no H). Why? Because if there were alpha-hydrogens, the aldehyde would undergo aldol condensation instead.

The precise reaction:

An aldehyde without $\alpha$ -hydrogen is treated with concentrated aqueous or alcoholic base (NaOH/KOH). Two molecules of aldehyde react: one becomes a carboxylate salt, the other becomes a primary alcohol.

\ce{2 R-CHO + OH- -> R-COO- + R-CH2OH}

After acidification, the carboxylate salt gives the carboxylic acid.

2\,\text{HCHO} \xrightarrow{\text{conc. NaOH}} \text{HCOONa} + \text{CH}_3\text{OH}

Key points for exams:

Only works for aldehydes with no $\alpha$ -hydrogen: formaldehyde, benzaldehyde, trimethylacetaldehyde, etc.
The base must be concentrated (dilute base won't work).
Formaldehyde is the most common example — it gives formic acid (as formate) and methanol.
Crossed Cannizzaro: When formaldehyde is mixed with another aldehyde (like benzaldehyde), formaldehyde is always the one that gets oxidised (to formate), and the other aldehyde gets reduced (to alcohol). This is because formaldehyde is the strongest reducing agent among aldehydes.

Tip

In a crossed Cannizzaro, formaldehyde always becomes the carboxylate. The other aldehyde becomes the alcohol. This is a common exam question.

Example: …

Why this formula?

Okay, let's break down these two very different reactions. They are often studied together because they both involve carbonyl compounds ( $\ce{C=O}$ ), but their mechanisms and purposes are completely opposite.

The Core Idea: Two Paths from a Carbonyl

Think of a carbonyl group ( $\ce{C=O}$ ) as a reactive hub. The carbon is electrophilic (electron-loving) because the oxygen pulls electron density away. The reactions it undergoes depend entirely on the conditions (acidic, basic, reducing) and the structure of the molecule (does it have an $\alpha$ -hydrogen?).

Clemmensen Reduction is about removing the oxygen entirely.
Cannizzaro Reaction is about disproportionating the molecule (one gets reduced, one gets oxidized).

1. Clemmensen Reduction: Why it Removes Oxygen

What it does: Converts a carbonyl group ( $\ce{C=O}$ ) in an aldehyde or ketone into a methylene group ( $\ce{CH2}$ ).

\ce{R2C=O ->[Zn(Hg)/HCl, heat] R2CH2}

Why this formula holds (The Mechanism):

The key is the reducing power of zinc amalgam in a strongly acidic environment.

Protonation: The carbonyl oxygen is basic. In the strong $\ce{HCl}$ , it gets protonated first.

\ce{R2C=O + H+ -> R2C=OH+}

This makes the carbon *even more* electrophilic.

2. Electron Transfer from Zinc: Zinc metal ( $\ce{Zn}$ ) is a good reducing agent. It donates electrons to the electron-deficient carbon. This is a single electron transfer (SET) process, not a simple hydride transfer.

- The zinc inserts itself, forming an organozinc intermediate (a carbenoid species).

- This intermediate is highly reactive.

Protonation and Elimination: The acidic medium provides plenty of $\ce{H+}$ ions. The intermediate gets protonated, and the oxygen (now as $\ce{H2O}$ ) is eliminated. The zinc is oxidized to $\ce{Zn^{2+}}$ .

The "Why" in a nutshell: The strong acid activates the carbonyl, and the zinc metal provides the electrons needed to break the $\ce{C=O}$ bond and replace it with two $\ce{C-H}$ bonds. The reaction does not work under basic conditions because you need the acid to protonate the oxygen first.

2. Cannizzaro Reaction: Why it Disproportionates

What it does: An aldehyde without an $\alpha$ -hydrogen (like formaldehyde $\ce{HCHO}$ or benzaldehyde $\ce{C6H5CHO}$ ) reacts with a strong base to give a carboxylic acid and an alcohol.

\ce{2 HCHO ->[conc. NaOH] HCOONa + CH3OH}

Why this formula holds (The Mechanism):

The key is the absence of $\alpha$ -hydrogens. If there were an $\alpha$ -hydrogen, the base would deprotonate that instead, leading to an aldol reaction. Here, the base has no choice but to attack the carbonyl itself.

Nucleophilic Attack: The strong base ( $\ce{OH-}$ ) attacks the electrophilic carbonyl carbon.

\ce{RCHO + OH- -> R-CH(OH)O-}

This forms a **tetrahedral intermediate** (an alkoxide).

2. The Crucial Hydride Transfer: This is the unique step. The tetrahedral intermediate is unstable. It can't lose $\ce{OH-}$ (that would just give back the aldehyde). Instead, it acts as a hydride donor ( $\ce{H-}$ ).

- The carbon bearing the negative charge (from the $\ce{OH-}$ attack) is very electron-rich. It kicks out a hydride ion ( $\ce{H-}$ ) to a second molecule of aldehyde.

- This is a hydride shift.

$\ce{R-CH(OH)O- + RCHO -> RCOOH + RCH2O-}$ …

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Both conversions rely on classic carbonyl chemistry: (a) acetonitrile to acetone via a Grignard reaction followed by hydrolysis, and (b) benzoic acid to benzene via decarboxylation using soda lime. The final products are propan-2-one and benzene, respectively.

The Concept Behind Each Conversion

These two problems test your understanding of how to transform functional groups using reagents that either build up or break down carbon chains. The key is to see the target molecule and work backwards: what functional group change is needed, and what reagent accomplishes it in one or two steps?

For (a), acetonitrile ( $CH_3CN$ ) has a nitrile group, while acetone ( $CH_3COCH_3$ ) has a ketone. The nitrile carbon is electrophilic and can be attacked by a Grignard reagent, adding an alkyl group. Hydrolysis then converts the resulting imine intermediate into the ketone. This is a classic two-step chain extension.

For (b), benzoic acid ( $C_6H_5COOH$ ) has a carboxyl group attached to a benzene ring, and benzene ( $C_6H_6$ ) is just the bare ring. The carboxyl group must be removed entirely. Decarboxylation — loss of $CO_2$ — is the direct route, but it requires heating with a strong base like soda lime ( $NaOH + CaO$ ). This is a one-step conversion.

Watch out

A common mistake in (a) is to try a direct reduction of the nitrile to a ketone. That’s not possible in one step — nitriles reduce to amines or aldehydes, not ketones. The Grignard approach is the correct two-step path.

Step-by-Step Solution

(a) $CH_3CN$ to $CH_3COCH_3$

Step 1: Grignard addition to the nitrile

Acetonitrile has a polar $C \equiv N$ triple bond. The carbon is electrophilic. When we add methylmagnesium iodide ( $CH_3MgI$ , a Grignard reagent), the carbanion ( $CH_3^-$ ) attacks the nitrile carbon. This forms an imine intermediate (a magnesium salt of an imine).

The reaction is:

CH_3CN + CH_3MgI \rightarrow CH_3C(=NMgI)CH_3

Step 2: Acidic hydrolysis

Treating the imine intermediate with dilute acid (e.g., $H_3O^+$ ) hydrolyses the $C=N$ bond to a $C=O$ bond, giving the ketone. Water adds across the imine, and ammonia (as $NH_3$ ) is eliminated.

CH_3C(=NMgI)CH_3 + H_3O^+ \rightarrow CH_3COCH_3 + MgI_2 + NH_3

Tip

The Grignard reagent must be freshly prepared and used in anhydrous conditions. Any water destroys it before it can react with the nitrile. Also, the nitrile must be dry. …

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

CBSE 2024Set 56/1/12 marks
$Q. Write the chemical equation when: (1+1=2)$
(a) Butan-2-one is treated with Zn(Hg) and conc. HCl.
(b) Two molecules of benzaldehyde are treated with conc. NaOH.
›Reveal solutionSolution
The key idea is that Zn(Hg)/conc. HCl reduces a carbonyl group to a methylene group (Clemmensen reduction), while conc. NaOH on benzaldehyde without an α-hydrogen triggers the Cannizzaro reaction, giving benzyl alcohol and sodium benzoate.

Let’s unpack each reaction separately, starting with the concept behind the reagent.

1. Butan-2-one with Zn(Hg) and conc. HCl

Concept: This is the Clemmensen reduction. It’s a classic method to reduce a carbonyl group ( $\ce{C=O}$ ) in a ketone or aldehyde all the way to a methylene group ( $\ce{CH2}$ ). The reagent — zinc amalgam in concentrated hydrochloric acid — provides a strongly acidic, reducing environment. The zinc metal donates electrons, and the acid protonates intermediates, ultimately replacing the oxygen with two hydrogens.

Why does this work? The carbonyl oxygen is first protonated, making the carbon more electrophilic. Zinc then transfers electrons, breaking the $\ce{C=O}$ bond and forming a carbene-like intermediate that gets further reduced. The net result: the ketone becomes an alkane.

Step-by-step:

Identify the substrate: Butan-2-one is $\ce{CH3COCH2CH3}$ . The carbonyl is at the second carbon.

Apply the reduction: The $\ce{C=O}$ group is replaced by $\ce{CH2}$ . So the product has the same carbon skeleton, but the carbonyl carbon becomes a $\ce{CH2}$ group.

Write the product: The carbon chain is $\ce{CH3-CH2-CH2-CH3}$ , which is butane.

Watch out

A common mistake is to think the product is an alcohol. Clemmensen reduction goes all the way to the alkane — it does not stop at the alcohol stage. Also, this reaction works best for ketones and aldehydes that are stable to strong acid; acid-sensitive groups (like esters) would be destroyed.

Chemical equation:
$\ce{CH3COCH2CH3 + 4[H] ->[Zn(Hg)/conc. HCl] CH3CH2CH2CH3 + H2O}$

2. Two molecules of benzaldehyde with conc. NaOH

Concept: This is the Cannizzaro reaction. It occurs with aldehydes that have no α-hydrogen atoms (i.e., the carbon next to the carbonyl has no hydrogen). Benzaldehyde ( $\ce{C6H5CHO}$ ) is the classic example. In concentrated base, one molecule of aldehyde is oxidized to a carboxylic acid (as its salt), and the other is reduced to a primary alcohol. It’s a disproportionation reaction.

Why does this happen? Without an α-hydrogen, the aldehyde cannot form an enolate (which would lead to an aldol reaction). Instead, the hydroxide ion attacks the carbonyl carbon of one aldehyde molecule, forming a tetrahedral intermediate. This intermediate transfers a hydride ion ( $\ce{H-}$ ) to the carbonyl carbon of a second aldehyde molecule. The result: one aldehyde becomes a carboxylate (after deprotonation), and the other becomes an alcohol. …
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CBSE 2023Set 56/2/12 marks
$Q. Do the following conversions in not more than two steps :$
(a) $CH_3CN$ (acetonitrile) to $CH_3-CO-CH_3$ (propan-2-one)
(b) $C_6H_5COOH$ (benzoic acid, drawn as a benzene ring bearing $-COOH$ ) to $C_6H_6$ (benzene, drawn as a plain benzene ring)
›Reveal solutionSolution

Both conversions rely on classic carbonyl chemistry: (a) acetonitrile to acetone via a Grignard reaction followed by hydrolysis, and (b) benzoic acid to benzene via decarboxylation using soda lime. The final products are propan-2-one and benzene, respectively.

The Concept Behind Each Conversion

These two problems test your understanding of how to transform functional groups using reagents that either build up or break down carbon chains. The key is to see the target molecule and work backwards: what functional group change is needed, and what reagent accomplishes it in one or two steps?

For (a), acetonitrile ( $CH_3CN$ ) has a nitrile group, while acetone ( $CH_3COCH_3$ ) has a ketone. The nitrile carbon is electrophilic and can be attacked by a Grignard reagent, adding an alkyl group. Hydrolysis then converts the resulting imine intermediate into the ketone. This is a classic two-step chain extension.

For (b), benzoic acid ( $C_6H_5COOH$ ) has a carboxyl group attached to a benzene ring, and benzene ( $C_6H_6$ ) is just the bare ring. The carboxyl group must be removed entirely. Decarboxylation — loss of $CO_2$ — is the direct route, but it requires heating with a strong base like soda lime ( $NaOH + CaO$ ). This is a one-step conversion.

Watch out

A common mistake in (a) is to try a direct reduction of the nitrile to a ketone. That’s not possible in one step — nitriles reduce to amines or aldehydes, not ketones. The Grignard approach is the correct two-step path.

Step-by-Step Solution

(a) $CH_3CN$ to $CH_3COCH_3$

Step 1: Grignard addition to the nitrile

Acetonitrile has a polar $C \equiv N$ triple bond. The carbon is electrophilic. When we add methylmagnesium iodide ( $CH_3MgI$ , a Grignard reagent), the carbanion ( $CH_3^-$ ) attacks the nitrile carbon. This forms an imine intermediate (a magnesium salt of an imine).

The reaction is:
$CH_3CN + CH_3MgI \rightarrow CH_3C(=NMgI)CH_3$
Step 2: Acidic hydrolysis

Treating the imine intermediate with dilute acid (e.g., $H_3O^+$ ) hydrolyses the $C=N$ bond to a $C=O$ bond, giving the ketone. Water adds across the imine, and ammonia (as $NH_3$ ) is eliminated.
$CH_3C(=NMgI)CH_3 + H_3O^+ \rightarrow CH_3COCH_3 + MgI_2 + NH_3$
Tip

The Grignard reagent must be freshly prepared and used in anhydrous conditions. Any water destroys it before it can react with the nitrile. Also, the nitrile must be dry. …

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Q.Do the following conversions in not more than two steps :

The Concept Behind Each Conversion

Step-by-Step Solution

(a) CH3CNCH_3CNCH3​CN to CH3COCH3CH_3COCH_3CH3​COCH3​

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(a) $CH_3CN$ to $CH_3COCH_3$

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