Q.How are the following conversions carried out?

(i) Propene → Propan-2-ol

Key idea: Electrophilic addition of water (acid-catalysed hydration) follows Markovnikov's rule.

Steps:

1. Propene ($\text{CH}_3\text{CH}=\text{CH}_2$) reacts with $\text{H}_2\text{O}$ in the presence of a dilute acid ($\text{H}_2\text{SO}_4$).
2. The proton adds to the less substituted carbon (terminal), forming a more stable secondary carbocation. …

The Core Idea: Electrophilic Carbon, Nucleophilic Attack

Every conversion here is about creating a new carbon–oxygen bond (an alcohol) by making a carbon atom electrophilic (electron-deficient) and then attacking it with a nucleophile that carries an oxygen atom. The specific reagent and conditions depend on how we generate that electrophilic carbon.

Let’s go through each one.

1. Propene → Propan-2-ol

Propene is an alkene — a flat molecule with a $\pi$ bond. The $\pi$ electrons are loosely held and act as a nucleophile. To add water across the double bond, we first need to make water a better nucleophile (it isn’t, on its own) and also make the alkene more electrophilic.

We use dilute sulphuric acid ($H_2SO_4$). The acid protonates the alkene, forming a carbocation. The key point: the proton adds to the less substituted carbon (Markovnikov’s rule), because that gives the more stable carbocation (secondary, in this case). The carbocation is then attacked by water (the nucleophile), and a final deprotonation gives the alcohol.

The overall reaction:

2. Benzyl chloride → Benzyl alcohol

Benzyl chloride has a chlorine atom attached to a carbon that is directly bonded to a benzene ring. That carbon is benzylic — it can form a very stable carbocation (resonance-stabilised by the ring). This makes $S_N1$ substitution extremely favourable.

We simply treat benzyl chloride with aqueous NaOH (or any source of $OH^-$). The hydroxide ion attacks the benzylic carbon, displacing chloride. Because the carbocation is so stable, the reaction is fast even without a strong nucleophile.

The reaction:

3. Ethyl magnesium chloride → Propan-1-ol

Ethyl magnesium chloride is a Grignard reagent ($CH_3CH_2MgCl$). The carbon–magnesium bond is highly polarised, making the carbon strongly nucleophilic (and basic). To get a primary alcohol with one more carbon than the Grignard reagent, we need to react it with formaldehyde ($HCHO$).

Formaldehyde has a carbonyl carbon that is very electrophilic (no alkyl groups to donate electron density). The Grignard carbon attacks this carbonyl, forming an alkoxide intermediate. Acidic workup ($H_3O^+$) then protonates the alkoxide to give the alcohol. …

Here is a clear, concept-first solution for each conversion, using the named reaction method and stepwise reasoning.

(i) Propene → Propan-2-ol

Method: Acid-catalysed hydration (Electrophilic Addition of water, following Markovnikov’s rule)

Why this works:

Propene is an unsymmetrical alkene. In electrophilic addition, the proton ($H^+$) adds to the less substituted carbon (to form the more stable carbocation). Water then attacks the carbocation.

Steps:

1. Protonation: Propene reacts with dilute $H_2SO_4$ (or $H_3O^+$).

(Secondary carbocation is formed — more stable than primary.)

2. Nucleophilic attack: Water attacks the carbocation.

3. Deprotonation: Loss of $H^+$ gives the alcohol.

Final product: Propan-2-ol (isopropyl alcohol).

(ii) Benzyl chloride → Benzyl alcohol

Method: Nucleophilic substitution ($S_N2$ or $S_N1$, depending on conditions — here, simple hydrolysis)

Why this works:

Benzyl chloride has a benzylic carbon that is highly reactive towards nucleophiles (due to resonance stabilisation of the carbocation in $S_N1$ or low steric hindrance in $S_N2$). Hydroxide ion ($OH^-$) replaces chlorine.

Steps:

1. Reaction with aqueous NaOH or KOH:

2. Mechanism (typically $S_N1$ due to benzylic stabilisation):
   • Step 1: Loss of $Cl^-$ forms a resonance-stabilised benzylic carbocation.
   • Step 2: Water (or $OH^-$) attacks the carbocation.
   • Step 3: Deprotonation yields benzyl alcohol.

Final product: Benzyl alcohol.

(iii) Ethyl magnesium chloride → Propan-1-ol

Method: Grignard reaction with formaldehyde followed by hydrolysis

Why this works:

Grignard reagents ($RMgX$) act as strong nucleophiles/carbanions. Formaldehyde ($HCHO$) is the simplest aldehyde — it gives a primary alcohol after reaction with a Grignard reagent.

Steps:

1. Formation of Grignard reagent (if not given):

2. Reaction with formaldehyde:

3. Hydrolysis (dilute acid):

Final product: Propan-1-ol (primary alcohol).

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Here are the common mistakes students make for each conversion, along with the correct conceptual approach to avoid them.

(i) Propene → Propan-2-ol

Common Mistake: Using hydration with a strong acid (like conc. $\text{H}_2\text{SO}_4$) without controlling the temperature, or incorrectly predicting the product as Propan-1-ol.

• Why it’s wrong: Direct hydration of propene with $\text{H}_2\text{SO}_4$ follows Markovnikov’s rule. The $\text{H}^+$ adds to the less substituted carbon (the $\text{CH}_2$ group), and the $\text{OH}$ (from water) adds to the more substituted carbon (the $\text{CH}$ group). This gives Propan-2-ol, not Propan-1-ol. If you try to force the anti-Markovnikov product, you need a different reagent (like $\text{B}_2\text{H}_6$ followed by $\text{H}_2\text{O}_2/\text{OH}^-$), which is hydroboration-oxidation, not simple hydration.

How to avoid:

• Remember the rule: For electrophilic addition to an unsymmetrical alkene, the hydrogen goes to the carbon with more hydrogens already.
• Use the correct reagent: For Propan-2-ol, use dilute $\text{H}_2\text{SO}_4$ (or $\text{H}_3\text{PO}_4$) at a controlled temperature (around 80–100°C). Alternatively, use oxymercuration-demercuration ($\text{Hg(OAc)}_2/\text{H}_2\text{O}$, then $\text{NaBH}_4$) which also gives Markovnikov addition without rearrangement.

Correct Method:

(ii) Benzyl chloride → Benzyl alcohol

Common Mistake: Trying to use aqueous $\text{NaOH}$ (simple nucleophilic substitution) and forgetting that benzyl chloride undergoes $\text{S}_\text{N}1$ or $\text{S}_\text{N}2$ easily, but the product may be contaminated with ethers or side products if conditions are wrong.

• Why it’s wrong: Benzyl chloride is highly reactive. If you use strong base (like $\text{NaOH}$) in water, the reaction works, but the yield can be low due to competing elimination or ether formation. Also, students often forget that the reaction is nucleophilic substitution, not electrophilic addition.

How to avoid:

• Use mild conditions: The best method is aqueous $\text{KOH}$ or $\text{NaOH}$ under reflux (or simply warm). The $\text{OH}^-$ attacks the benzylic carbon, displacing $\text{Cl}^-$.
• Alternative: Use moist $\text{Ag}_2\text{O}$ (silver oxide) which gives a clean $\text{S}_\text{N}1$ reaction without side products.

Correct Method:

(iii) Ethyl magnesium chloride → Propan-1-ol

Common Mistake: Using water directly on the Grignard reagent, or forgetting that the product is an alcohol with one more carbon than the Grignard reagent.

• Why it’s wrong: Ethyl magnesium chloride ($\text{CH}_3\text{CH}_2\text{MgCl}$) has 2 carbons. To get Propan-1-ol (3 carbons), you need to add one carbon via a carbonyl compound. If you just add water, you get ethane ($\text{CH}_3\text{CH}_3$), not an alcohol. Students often mistakenly think the Grignard reagent itself becomes the alcohol.

How to avoid:

• Remember the carbon count: Grignard reagents add to formaldehyde ($\text{HCHO}$) to give a primary alcohol with one extra carbon.
• Reaction sequence: $\text{RMgX} + \text{HCHO} \rightarrow \text{RCH}_2\text{OH}$ after hydrolysis.

Correct Method:

(iv) Methyl magnesium bromide → 2-Methylpropan-2-ol

Common Mistake: Using formaldehyde (which gives a primary alcohol) or acetaldehyde (which gives a secondary alcohol) instead of the correct carbonyl compound. …