Question
A student needs to synthesise 2-phenylethanol (PhCH₂CH₂OH) starting from benzyl chloride (PhCH₂Cl) and any other reagents available in the lab. They decide to use a Grignard route. Show the preparation of the Grignard reagent and its reaction with a suitable carbonyl compound to get the target alcohol.
Solution — Step by Step
React benzyl chloride with magnesium turnings in dry diethyl ether under anhydrous conditions:
The ether isn’t just a solvent — it coordinates to the magnesium through its lone pairs, stabilising the highly reactive organomagnesium species. Without this coordination, the reagent won’t form cleanly.
We need 2-phenylethanol: Ph–CH₂–CH₂–OH. The Grignard adds a carbon to whatever carbonyl we use, so we need the simplest option — formaldehyde (HCHO).
Formaldehyde gives a primary alcohol and adds exactly one carbon. If we had used ethanal (CH₃CHO), we’d get a secondary alcohol — wrong product.
The intermediate is a magnesium alkoxide salt, not the free alcohol. Add dilute aqueous acid (H₃O⁺) to hydrolyse it:
The final product is 2-phenylethanol — a primary alcohol with one more carbon than our starting Grignard.
| Starting material | Carbons | Carbonyl used | Carbons | Product | Carbons |
|---|---|---|---|---|---|
| PhCH₂Cl (benzyl chloride) | 7 | HCHO | 1 | PhCH₂CH₂OH | 8 |
This carbon-counting check is the fastest way to verify your route in JEE — count carbons before and after.
Why This Works
The key to Grignard chemistry is understanding what makes RMgX so reactive. Carbon in RMgX is bonded to magnesium, which is more electropositive. This makes the C–Mg bond highly polarised: carbon carries a partial negative charge, making it a powerful nucleophile (almost like a carbanion).
When this nucleophile attacks the electrophilic carbonyl carbon, a new C–C bond forms. This is why Grignard reactions are so valuable — they build carbon skeletons, which few other reactions do so directly.
The choice of carbonyl compound directly controls the degree of the resulting alcohol:
- HCHO → primary alcohol (adds one C)
- RCHO → secondary alcohol
- R₂CO (ketone) → tertiary alcohol
This pattern has appeared in JEE Main 2023 and CBSE board papers repeatedly — examiners love asking students to identify which carbonyl gives which alcohol class.
Alternative Method
We can reach the same product using a different Grignard + different carbonyl combination. Instead of PhCH₂MgCl + HCHO, use:
A cleaner alternative: start from PhMgBr (phenylmagnesium bromide, made from bromobenzene + Mg) and react with ethylene oxide:
Grignard + ethylene oxide is a classic route to add two carbons and always gives a primary alcohol. This specific reaction (ArMgX + ethylene oxide) appeared in JEE Advanced 2019. Recognise the pattern: if the product has –CH₂CH₂OH attached to an aryl group, ethylene oxide was used.
Common Mistake
Forgetting anhydrous conditions. The single most common error in Grignard problems — students write the reaction but forget that even trace moisture destroys the reagent:
The Grignard reacts with water instead of your carbonyl compound. In exam answers, always state “dry ether” and “anhydrous conditions” explicitly — CBSE marking schemes deduct marks if these are missing. In JEE MCQs, options that omit this detail are usually wrong.