What happens when (a) Propanone is treated with methylmagnesium iodide and then hydrolysed, and (b) Benzene is

Question

Q. What happens when

(a) Propanone is treated with methylmagnesium iodide and then hydrolysed, and

(b) Benzene is treated with

CH_3COCl

in presence of anhydrous

AlCl_3

CBSECBSE Class XII Board 2020Subjective· 2mImportance★★★★★

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Concept understanding — Grignard Synthesis of Alcohols

The Core Idea: Building a Carbon Skeleton

Imagine you have a molecule and you want to attach a brand new carbon chain to it — specifically, to turn a carbon-oxygen double bond into an alcohol. That's the job of a Grignard reagent. It's one of the most powerful tools in organic chemistry for making carbon-carbon bonds.

The intuition is simple: a Grignard reagent acts like a carbon-based nucleophile (a carbon atom with a strong negative character) that attacks the electrophilic carbon of a carbonyl group ( $\ce{C=O}$ ). After a simple workup with water or acid, you get an alcohol.

What is a Grignard Reagent?

A Grignard reagent has the general formula $\ce{R-Mg-X}$ , where $\ce{R}$ is an alkyl or aryl group and $\ce{X}$ is a halogen (usually $\ce{Cl}$ , $\ce{Br}$ , or $\ce{I}$ ). The carbon-magnesium bond is highly polarised: carbon carries a significant partial negative charge ( $\delta^-$ ) and magnesium carries a partial positive charge ( $\delta^+$ ). This makes the carbon atom strongly nucleophilic and basic.

Watch out

Grignard reagents are extremely sensitive to moisture and air. They react violently with water, so the entire reaction must be carried out in anhydrous (dry) conditions, typically in dry ether as the solvent.

The Reaction: Step by Step

The reaction proceeds in two distinct stages:

Step 1 — Nucleophilic Addition: The Grignard reagent attacks the electrophilic carbonyl carbon. The $\pi$ bond of $\ce{C=O}$ breaks, and the electrons move to oxygen, forming a magnesium alkoxide intermediate.

\ce{R-MgX + R'-C(=O)-R'' -> R'-C(R)(O^-MgX)-R''}

Step 2 — Hydrolysis: The alkoxide intermediate is treated with dilute acid or water. This protonates the negatively charged oxygen, giving the final alcohol.

\ce{R'-C(R)(O^-MgX)-R'' + H2O -> R'-C(R)(OH)-R'' + Mg(OH)X}

\ce{R-MgX + R'-C(=O)-R'' ->[1. dry ether][2. H3O+] R'-C(R)(OH)-R''}

The Three Types of Alcohols You Can Make

The beauty of this method is that by choosing the right carbonyl compound, you can selectively produce primary, secondary, or tertiary alcohols.

1. Primary Alcohols (from Formaldehyde)

Formaldehyde ( $\ce{HCHO}$ ) has two hydrogens on the carbonyl carbon. After Grignard addition and hydrolysis, you get a primary alcohol.

\ce{R-MgX + HCHO ->[1. ether][2. H3O+] R-CH2OH}

Note

The product has one more carbon than the Grignard reagent's $\ce{R}$ group. The new carbon comes from formaldehyde.

2. Secondary Alcohols (from Aldehydes other than Formaldehyde)

Any aldehyde other than formaldehyde ( $\ce{R'-CHO}$ ) gives a secondary alcohol. The carbonyl carbon now has one $\ce{R'}$ group and one $\ce{H}$ .

\ce{R-MgX + R'-CHO ->[1. ether][2. H3O+] R'-CH(OH)-R}

3. Tertiary Alcohols (from Ketones)

Ketones ( $\ce{R'-C(=O)-R''}$ ) have two alkyl/aryl groups on the carbonyl carbon. After addition, you get a tertiary alcohol.

\ce{R-MgX + R'-C(=O)-R'' ->[1. ether][2. H3O+] R'-C(OH)(R)-R''}

A Quick Summary Table

Carbonyl Compound	Type of Alcohol Produced	General Product
Formaldehyde ( $\ce{HCHO}$ )	Primary	$\ce{R-CH2OH}$
Other aldehydes ( $\ce{R'-CHO}$ )	Secondary	$\ce{R'-CH(OH)-R}$
Ketones ( $\ce{R'-C(=O)-R''}$ )	Tertiary	$\ce{R'-C(OH)(R)-R''}$

A Common Mistake to Avoid …

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Both reactions are classic carbon–carbon bond-forming steps: (a) is a Grignard addition to a ketone, giving a tertiary alcohol (2-methylpropan-2-ol);

(b) is a Friedel–Crafts acylation, giving acetophenone.

These two reactions look different — one uses an organometallic reagent, the other uses an acyl chloride with a Lewis acid — but they share a common theme: electrophilic attack on a nucleophilic carbon centre. In (a), the nucleophile is the carbanion-like carbon of the Grignard reagent; in (b), the nucleophile is the electron-rich benzene ring. Let’s walk through each.

(a) Propanone + methylmagnesium iodide, then hydrolysis

What’s happening conceptually:

Methylmagnesium iodide ( $CH_3MgI$ ) is a Grignard reagent. The carbon–magnesium bond is highly polarised, making the methyl carbon strongly nucleophilic (carbanion-like). Propanone (acetone, $CH_3COCH_3$ ) has a carbonyl group — the carbon is electrophilic because of the $\ce{C=O}$ bond’s polarity. The nucleophilic methyl carbon attacks the carbonyl carbon, forming a new $\ce{C-C}$ bond. After hydrolysis (adding water or dilute acid), the alkoxide intermediate picks up a proton to give an alcohol.

Step-by-step:

Nucleophilic addition: The lone pair on the methyl carbon of $CH_3MgI$ attacks the electrophilic carbonyl carbon of propanone. The $\pi$ bond of $\ce{C=O}$ breaks, and the electrons move to oxygen, forming a magnesium alkoxide intermediate.

\ce{CH3-C(=O)-CH3 + CH3MgI -> CH3-C(OMgI)(CH3)-CH3}

Hydrolysis: Adding water (or dilute acid) protonates the alkoxide oxygen, giving the free alcohol and magnesium salts.

\ce{CH3-C(OMgI)(CH3)-CH3 + H2O -> CH3-C(OH)(CH3)-CH3 + Mg(OH)I}

Product identification: The product is 2-methylpropan-2-ol (tert-butyl alcohol), a tertiary alcohol. Why tertiary? Because the carbon bearing the –OH is attached to three alkyl groups (two methyls from the original ketone, one from the Grignard).

Tip

Grignard additions to ketones always give tertiary alcohols (unless the ketone is formaldehyde, which gives primary alcohols). This is a reliable way to build a carbon skeleton.

Watch out

Grignard reagents are extremely moisture-sensitive. Any water present before the intended hydrolysis will destroy the reagent, giving methane gas instead of the desired addition product. That’s why the reaction is always done in anhydrous conditions.

(b) Benzene + $CH_3COCl$ in presence of anhydrous $AlCl_3$

What’s happening conceptually:

This is a Friedel–Crafts acylation. The acyl chloride ( $CH_3COCl$ ) is activated by the Lewis acid $AlCl_3$ , which coordinates to the chlorine, making the carbonyl carbon even more electrophilic. The benzene ring, rich in $\pi$ electrons, acts as a nucleophile and attacks this activated species. After loss of $HCl$ and regeneration of the catalyst, an acyl group ( $-COCH_3$ ) is attached to the ring.

Step-by-step:

Formation of the acylium ion: $AlCl_3$ pulls a chlorine atom from $CH_3COCl$ , generating a resonance-stabilised acylium ion (the key electrophile). …

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

CBSE 2026Set A1 markMCQ
$Q. Which of the following reagents reacts easily with both Acetaldehyde and Acetone ?$
(a) Fehling solution
(b) Grignard reagent
(c) Schiff's reagent
(d) Tollen's reagent
›Reveal solutionSolution
Grignard reagent adds to any carbonyl group, so it reacts with both aldehydes and ketones.

Acetaldehyde is an aldehyde and acetone is a ketone. A reagent that reacts with both must attack the common C=O group:

Fehling's solution, Tollen's reagent and Schiff's reagent are tests specific to aldehydes (they detect the -CHO group) and do not react with ketones. …
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CBSE 2020Set NC1 markMCQ
$Q. Which of the following reagents will give a primary alcohol on reacting with a Grignard reagent?$
(a) Methanal
(b) Ethanal
(c) Propanone
(d) Benzaldehyde
›Reveal solutionSolution
$RMgX$ adds across $C=O$ ; the class of alcohol formed (1°, 2° or 3°) depends on how many carbon groups were already on the carbonyl carbon. Formaldehyde (methanal) has none, so it alone gives a primary alcohol.

General Grignard addition to a carbonyl compound, followed by aqueous acid work-up:

$R'_2C=O + RMgX \longrightarrow R'_2C(OMgX)R \xrightarrow{H_3O^{+}} R'_2C(OH)R$

The resulting alcohol's class is set by how many carbon groups flank the original carbonyl carbon:

Methanal, $HCH=O$ (no carbon substituent on the carbonyl carbon): $HCHO + RMgX \rightarrow R\text{–}CH_2\text{–}OH$ → primary alcohol.

Ethanal (any other aldehyde), $R'CH=O$ (one carbon substituent): gives $R'CH(OH)R$ → secondary alcohol.

Propanone (a ketone), $R'_2C=O$ (two carbon substituents): gives $R'_2C(OH)R$ → tertiary alcohol. …
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CBSE 2016Set ANNUAL1 mark
$Q. Show how benzaldehyde can be converted into benzophenone?$
›Reveal solutionSolution

Benzaldehyde is first converted to a secondary alcohol via a Grignard reaction, which is then oxidised to the ketone benzophenone.

Benzaldehyde ( $C_6H_5CHO$ ) reacts with the Grignard reagent phenylmagnesium bromide ( $C_6H_5MgBr$ ) at the carbonyl carbon; aqueous acid work-up gives the secondary alcohol diphenylmethanol (benzhydrol):

$C_6H_5CHO + C_6H_5MgBr \rightarrow C_6H_5CH(OMgBr)C_6H_5 \xrightarrow{H_3O^+} (C_6H_5)_2CHOH$ …

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Q.What happens when

(a) Propanone + methylmagnesium iodide, then hydrolysis

(b) Benzene + CH3COClCH_3COClCH3​COCl in presence of anhydrous AlCl3AlCl_3AlCl3​

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(b) Benzene + $CH_3COCl$ in presence of anhydrous $AlCl_3$

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