Grignard Reagent: Preparation, Reactions and Mechanism

ChemistryClass 12, Haloalkanes and Haloarenes; AlcoholsFor JEE and NEET aspirants

A Grignard reagent, RMgX, is an alkyl or aryl magnesium halide made by stirring the halide with magnesium in dry ether. Its C–Mg bond is polar with the carbon end negative, so RMgX behaves like a source of the carbanion $R^{-}$ : it adds to the C=O of aldehydes, ketones, esters and carbon dioxide, and to nitriles, building new C–C bonds. Any acidic hydrogen destroys it, giving the hydrocarbon RH. Grignard reagents appear in the Class 12 NCERT chapters on haloalkanes, alcohols and carbonyl compounds, and are tested in JEE Main, JEE Advanced and NEET.

At a glance

Preparation:
$C H_{3} C H_{2} Br + Mg dry ether C H_{3} C H_{2} MgBr$
Solvent: dry diethyl ether or THF. The ether oxygen bonds to Mg and keeps the reagent in solution; any water or alcohol would destroy it.
Halides: reactivity RI $>$ RBr $>$ RCl. Aryl bromides work too ( $C_{6} H_{5} MgBr$ ).
With C=O, then $H_{3} O^{+}$ : methanal gives a 1° alcohol, other aldehydes a 2° alcohol, ketones a 3° alcohol; $C O_{2}$ gives a carboxylic acid.
With acidic H:
$C H_{3} MgBr + H_{2} O C H_{4} + Mg (OH) Br$
Type: the R group is a strong nucleophile and a very strong base.

1. Mechanism

Carbon is more electronegative than magnesium, so in R–MgX the carbon carries a partial negative charge. That carbon attacks the positive carbonyl carbon, exactly as a hydride or cyanide ion would.

Figure 1:

C H_{3} MgBr

with propanone. The R group of RMgX ends up on the old carbonyl carbon, and the oxygen becomes the OH of the alcohol after work-up.

Addition. The C–Mg electrons form a bond to the carbonyl carbon and the C=O $π$ electrons move to oxygen, which ends up bonded to magnesium (a magnesium alkoxide). Magnesium, a Lewis acid, also holds the carbonyl oxygen as the bond forms.
Work-up. Dilute acid (or aqueous ammonium chloride) protonates the alkoxide to the alcohol; magnesium goes into solution as its salts.

The same basic carbon explains why water is forbidden: water (p $K_{a}$ 15.7) is an enormously stronger acid than an alkane (p $K_{a}$ about 50), so $R^{-}$ simply takes a proton and becomes RH.

Exam Trick

Count the carbons and the classes. Methanal and $C O_{2}$ add one carbon, ethylene oxide adds two. Methanal gives a primary alcohol, other aldehydes a secondary one, ketones a tertiary one, and an ester (two moles of RMgX) a tertiary alcohol with two identical R groups (a methanoate ester gives a secondary alcohol).

2. What it gives

Reacts with	Product (after acid work-up)	Example, with methylmagnesium bromide
Methanal, $HCHO$	1° alcohol, one carbon more	Ethanol, $C H_{3} C H_{2} OH$
Other aldehyde, RCHO	2° alcohol	Ethanal gives propan-2-ol, $C H_{3} CH (OH) C H_{3}$
Ketone, $R_{2} CO$	3° alcohol	Propanone gives 2-methylpropan-2-ol, $(C H_{3})_{3} COH$
Carbon dioxide (dry ice)	Carboxylic acid, one carbon more	Ethanoic acid, $C H_{3} COOH$
Nitrile, R'CN	Ketone (through the imine salt)	Ethanenitrile gives propanone, $C H_{3} COC H_{3}$
Ester, R'COOR'' (2 mol RMgX)	3° alcohol with two identical R groups	Ethyl ethanoate gives 2-methylpropan-2-ol
Methanoate ester, HCOOR'' (2 mol RMgX)	2° alcohol, $R_{2} CHOH$	Ethyl methanoate gives propan-2-ol
Ethylene oxide (oxirane)	1° alcohol, two carbons more	Propan-1-ol, $C H_{3} C H_{2} C H_{2} OH$
Water, alcohols, acids, $N H_{3}$ , 1° and 2° amines, terminal alkynes	Hydrocarbon RH (no new C–C bond)	Methane, $C H_{4}$

With a nitrile the reaction stops after one addition. The imine salt carries a negative charge on nitrogen, so its C=N carbon does not accept a second R group, and the ketone forms only on hydrolysis, after all the RMgX has been destroyed. Hydrolysis (dilute acid in practice) gives the ketone:

C H_{3} MgBr + C H_{3} CN dry ether (C H_{3})_{2} C = NMgBr

(C H_{3})_{2} C = NMgBr + 2 H_{2} O C H_{3} COC H_{3} + N H_{3} + Mg (OH) Br

With an ester the first addition gives a ketone, which is more reactive than the ester and takes a second R group straight away, so a tertiary alcohol is the product.

3. Conditions and limits

Dry everything. Water, alcohols, carboxylic acids, ammonia, 1° and 2° amines and terminal alkynes all carry an acidic H and turn RMgX into RH. Even the $C O_{2}$ and moisture of air matter, so the reagent is made and used under dry conditions.
Nothing reactive in the same molecule. RMgX cannot be made from a halide that also carries OH, NH, COOH, a terminal C≡C–H, C=O, CN, $N O_{2}$ or an epoxide: the reagent would destroy itself.
An OH elsewhere uses up one mole. If the carbonyl compound also has an acidic H, the first mole of RMgX only removes that H (giving RH); a second mole is needed for the C=O.
Putting in deuterium. With $D_{2} O$ instead of water, RMgX gives RD: a way to place deuterium on a chosen carbon.

JEE Advanced

Counting active hydrogens. $C H_{3} MgI$ gives one mole of $C H_{4}$ for every mole of acidic H in a compound (O–H, N–H, S–H, terminal C≡C–H). Measuring the methane volume counts them: ethane-1,2-diol gives two moles, ethanoic acid one.

Unsaturated carbonyl compounds. Grignard reagents usually add directly to the C=O (1,2-addition) of $α, β$ -unsaturated aldehydes and ketones, giving allylic alcohols. Hindered enones such as benzalacetophenone, or a copper(I) catalyst, switch them to 1,4-addition.

4. Worked examples

Planning a synthesis. To make an alcohol, cut one of the C–C bonds to the carbon that carries the OH. One piece becomes RMgX; the other becomes the carbonyl compound, with the OH carbon as its C=O carbon. A tertiary alcohol with three different groups can be made three ways; an ester route needs two identical groups.

Solved Example 1

Give the Grignard routes to 2-methylbutan-2-ol,

(C H_{3})_{2} C (OH) C H_{2} C H_{3}

Solution:

The OH carbon carries two $C H_{3}$ groups and one $C_{2} H_{5}$ group.

Cut	Grignard reagent	Carbonyl compound
One $C H_{3}$	$C H_{3} MgBr$	Butan-2-one, $C H_{3} COC H_{2} C H_{3}$
The $C_{2} H_{5}$	$C_{2} H_{5} MgBr$	Propanone, $C H_{3} COC H_{3}$
Both $C H_{3}$ (ester route)	$C H_{3} MgBr$ , 2 mol	Ethyl propanoate, $C H_{3} C H_{2} COO C_{2} H_{5}$

Each is followed by work-up with dilute acid.

Solved Example 2

0.46 g of ethanol is treated with excess

C H_{3} MgI

. What volume of methane forms at STP (22.4 L per mole)?

Solution:

Ethanol has one acidic H (the O–H). Moles of ethanol $= 0.46/46 = 0.010$ mol, so 0.010 mol of $C H_{4}$ forms: $0.010 \times 22.4 = 0.224$ L, that is, 224 mL.

Solved Example 3

4-Hydroxybutan-2-one,

HOC H_{2} C H_{2} COC H_{3}

, is treated with 2 moles of

C H_{3} MgBr

and then with dilute acid. What are the products?

Solution:

The first mole of $C H_{3} MgBr$ reacts with the acidic O–H and gives one mole of methane. The second mole adds to the C=O. After work-up both oxygens are OH: 3-methylbutane-1,3-diol, $HOC H_{2} C H_{2} C (C H_{3})_{2} OH$ , plus $C H_{4}$ .

5. Exam traps

Watch out

Water, alcohols, acids, ammonia, 1° and 2° amines and terminal alkynes destroy RMgX and give RH. A 3° amine has no N–H and does not. Ether is the solvent, never ethanol.
Methanal gives a 1° alcohol, other aldehydes a 2° alcohol, ketones a 3° alcohol.
$C O_{2}$ gives a carboxylic acid with one more carbon than R, not an alcohol.
Esters use two moles of RMgX and give a 3° alcohol (methanoate esters a 2° alcohol), not a ketone.
Nitriles give ketones after hydrolysis, not alcohols and not amines.
A halide that also carries OH, NH, COOH, C≡C–H or C=O cannot be turned into a Grignard reagent.
Count carbons: the product has the carbons of R plus those of the carbonyl compound. For an ester, add two R groups to the acyl part R'CO (the OR'' part leaves as R''OH).
With $D_{2} O$ , RMgX gives RD, not ROD.

Past-paper questions on Grignard reagent

Our past papers: JEE Main 2025-2026, JEE Advanced 2022-2026, NEET 2018-2019, 2022-2026. We found 16 questions in them that test this:

JEE Main 8 (2025-2026)
JEE Advanced 5 (2023-2025)
NEET 3 (2022-2023, 2025)

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