Q.Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanal?

Concept: Grignard reagents react with methanal (formaldehyde) to give primary alcohols after hydrolysis. The Grignard reagent supplies the alkyl group that attaches to the carbonyl carbon, and the —CH₂OH group comes from the formaldehyde.

Reasoning:

1. Methanal has the structure H–CHO. A Grignard reagent R–MgX adds to the carbonyl, forming an alkoxide intermediate.
2. Acidic hydrolysis (H₃O⁺) converts the alkoxide to the primary alcohol R–CH₂OH.
3. To get a specific alcohol, choose the Grignard reagent with the same R group as the alkyl part of the alcohol.

For (i) $CH_3-CH(CH_3)-CH_2OH$ (2-methylpropan-1-ol): removing the $-CH_2OH$ unit leaves $CH_3-CH(CH_3)-$, the isopropyl group. Use isopropylmagnesium bromide with methanal.

For (ii) $C_6H_{11}-CH_2OH$ (cyclohexylmethanol): The alkyl group is cyclohexyl ($C_6H_{11}-$). Use cyclohexylmagnesium bromide with methanal.

This is a classic application of the Grignard reaction with formaldehyde. The key idea: methanal (HCHO) has no alkyl groups attached to the carbonyl carbon. When a Grignard reagent (RMgX) attacks it, the product after hydrolysis is always a primary alcohol with the structure R–CH₂OH — that is, the R group from the Grignard ends up attached to the –CH₂OH unit.

So to prepare a given primary alcohol of the form R–CH₂OH, you simply need the Grignard reagent R–MgX. The problem gives you the alcohol and asks you to work backwards to find the suitable Grignard.

Let’s do each one.

1. For alcohol (i): $CH_3-CH(CH_3)-CH_2OH$

This is 2-methylpropan-1-ol. Write it as R–CH₂OH. Here R is the group attached to the –CH₂OH carbon. Remove the –CH₂OH part: the remaining group is $CH_3-CH(CH_3)-$ (isopropyl group). So R = isopropyl.

Therefore the Grignard reagent needed is isopropylmagnesium halide: $(CH_3)_2CH-MgX$ (where X = Cl, Br, or I). The reaction:

2. For alcohol (ii): $C_6H_{11}-CH_2OH$ (cyclohexylmethanol)

Here the –CH₂OH is attached to a cyclohexyl ring. So R = cyclohexyl ($C_6H_{11}-$). The Grignard reagent is cyclohexylmagnesium halide: $C_6H_{11}-MgX$.

The reaction:

Method: Grignard Reaction with Methanal (Formaldehyde)

This is a nucleophilic addition reaction. Methanal ($HCHO$) is the simplest aldehyde — it has no alkyl groups attached to the carbonyl carbon. When a Grignard reagent ($RMgX$) attacks methanal, the product after hydrolysis is always a primary alcohol with one extra carbon.

General Steps

1. Identify the target alcohol's carbon skeleton — the alcohol carbon ($-CH_2OH$) comes from methanal. The rest of the molecule ($R-$) comes from the Grignard reagent.
2. Remove the $-CH_2OH$ group and replace it with a $MgX$ group to get the required Grignard reagent.
3. React the Grignard reagent with methanal, then hydrolyse.

(i) $CH_3-CH(CH_3)-CH_2OH$ (Isobutyl alcohol)

Step 1: Identify the $R$ group attached to $-CH_2OH$

$R = CH_3-CH(CH_3)-$ (isopropyl group)

Step 2: Required Grignard reagent

$CH_3-CH(CH_3)-MgX$ (isopropylmagnesium halide)

Step 3: Reaction with methanal

Result: Isobutyl alcohol is obtained.

(ii) $C_6H_{11}-CH_2OH$ (Cyclohexylmethanol)

Step 1: Identify the $R$ group

$R = C_6H_{11}-$ (cyclohexyl group)

Step 2: Required Grignard reagent

$C_6H_{11}-MgX$ (cyclohexylmagnesium halide)

Step 3: Reaction with methanal

Result: Cyclohexylmethanol is obtained.

Key Exam Point

Methanal always gives a primary alcohol with one extra carbon — the Grignard reagent's $R$ group attaches directly to the $-CH_2OH$ formed from $HCHO$.

This is a classic exam trap in Grignard reactions. Let's first clarify the concept, then list the common mistakes.

The Core Concept

The question asks: How to prepare these alcohols by reacting a Grignard reagent with methanal (formaldehyde, $\ce{HCHO}$)?

When a Grignard reagent ($\ce{RMgX}$) reacts with methanal, the product after hydrolysis is a primary alcohol with one more carbon than the Grignard reagent:

So the alcohol's carbon skeleton = R (from Grignard) + CH2 (from methanal).

Common Mistakes & How to Avoid Them

1. Mistaking the number of carbons added

• Mistake: Thinking methanal adds 2 or more carbons.
• Why it's wrong: Methanal has only one carbon ($\ce{HCHO}$). It adds exactly one CH2 group.
• How to avoid: Always count: product = R–CH2OH. So if the alcohol has $n$ carbons, the Grignard must have $n-1$ carbons.

2. Choosing the wrong Grignard reagent

• Mistake: For alcohol (i) $\ce{CH3-CH(CH3)-CH2OH}$, students misread the skeleton as a 5-carbon alcohol and pick $\ce{(CH3)2CHCH2-MgX}$ (isobutyl, 4 carbons).
• Why it's wrong: That Grignard would give $\ce{(CH3)2CHCH2-CH2OH}$ — a 5-carbon alcohol (3-methylbutan-1-ol), not the 4-carbon target.

Correct approach:

• Target: $\ce{CH3-CH(CH3)-CH2OH}$ → 4 carbons total.
• Remove the –CH2OH (from methanal) → remaining R = $\ce{CH3-CH(CH3)-}$ (isopropyl, 3 carbons).
• So Grignard = isopropylmagnesium halide ($\ce{(CH3)2CH-MgX}$).

How to avoid: Draw the alcohol, circle the –CH2OH part, and the rest is your Grignard's R group.

3. Forgetting that methanal gives only primary alcohols

• Mistake: Trying to use methanal to make secondary or tertiary alcohols.
• Why it's wrong: Methanal has no alkyl groups on the carbonyl carbon — it always yields a primary alcohol.
• How to avoid: If the target is secondary or tertiary, methanal is not the right carbonyl — use other aldehydes or ketones.

4. Incorrectly handling cyclic alcohols

• Mistake: For (ii) $\ce{C6H11-CH2OH}$ (cyclohexylmethanol), students write the Grignard as $\ce{C6H11-MgX}$ but forget it's cyclohexyl, not phenyl.
• Why it's wrong: $\ce{C6H11}$ is cyclohexyl (saturated), not benzene. The Grignard must be cyclohexylmagnesium halide.
• How to avoid: Draw the ring — if it's saturated (no double bonds), it's cyclohexyl, not phenyl.

5. Writing the wrong product after hydrolysis

• Mistake: Showing the product as $\ce{R-CH2OMgX}$ or $\ce{R-CH2OH}$ without proper hydrolysis step.
• Why it's wrong: The reaction sequence is: Grignard + methanal → alkoxide → acidic hydrolysis gives alcohol.
• How to avoid: Always write the two-step mechanism clearly: (i) dry ether, (ii) $\ce{H3O+}$.

6. Ignoring the "suitable" condition

• Mistake: Using a Grignard that has acidic H (like –OH, –NH, –SH groups).
• Why it's wrong: Grignard reagents are destroyed by acidic protons.
• How to avoid: Ensure the R group has no acidic H (no –OH, –NH2, –COOH, etc.).

Quick Summary Table

Final Correct Answers

(i) $\ce{CH3-CH(CH3)-CH2OH}$

Grignard: Isopropylmagnesium bromide ($\ce{(CH3)2CH-MgBr}$) + methanal → hydrolysis.

(ii) $\ce{C6H11-CH2OH}$ (cyclohexylmethanol)

Grignard: Cyclohexylmagnesium chloride ($\ce{C6H11-MgCl}$) + methanal → hydrolysis.

Key takeaway: Always count carbons and identify the –CH2OH fragment — the rest is your Grignard. Methanal is your friend for making primary alcohols with one extra carbon.