How can propan-2-one be converted into tert-butyl alcohol?

NCERT Exemplar · Q34

Q. How can propan-2-one be converted into tert-butyl alcohol?

Mizoram MbseShort· 2mImportance★★★★★

64% · 86/135 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Electrophilic Addition Reactions

Electrophilic Aromatic Substitution – The First Meeting

Imagine you have a benzene ring — that perfect, flat hexagon of six carbons with alternating double bonds. It's stable, almost stubbornly so. You want to attach something new to it, say a bromine atom or a nitro group. But benzene doesn't react like an alkene. It doesn't just add across a double bond. Instead, it does something more elegant: it kicks out a hydrogen and keeps its aromatic ring intact.

That's the heart of Electrophilic Aromatic Substitution (EAS).

The Intuition: Why "Substitution" and Not "Addition"?

Benzene's stability comes from its delocalised $\pi$ electrons — a cloud above and below the ring. This cloud is electron-rich, so it attracts electrophiles (electron-loving species). But if an electrophile simply added to a double bond, the ring would break its aromaticity, losing that huge stabilisation. That would be energetically costly.

So benzene does something smarter: it lets the electrophile attack, temporarily breaks aromaticity to form a high-energy intermediate (the arenium ion), and then loses a proton to restore the aromatic ring. The net result? A hydrogen is replaced by the electrophile. The ring is back to its stable, aromatic self.

Note

The key trade-off: temporary loss of aromaticity is acceptable because the final product regains it. Addition reactions would permanently destroy aromaticity — benzene avoids that.

The Precise Statement

Electrophilic Aromatic Substitution is a reaction in which an electrophile ( $E^+$ ) replaces a hydrogen atom on an aromatic ring, proceeding through a sigma complex (arenium ion) intermediate, and restoring aromaticity after deprotonation.

The general equation:

\ce{Ar-H + E+ -> Ar-E + H+}

where Ar represents an aromatic ring.

The Mechanism in Three Steps

Generation of the electrophile – Many EAS reactions need a catalyst to create a strong enough $E^+$ . For example, bromination uses $\ce{FeBr3}$ to polarise $\ce{Br2}$ into $\ce{Br^+}$ .
Attack by the aromatic ring – The $\pi$ electrons of benzene attack the electrophile, forming a sigma complex (also called the arenium ion or Wheland intermediate). This intermediate is non-aromatic — it has four $\pi$ electrons delocalised over five carbons, and one $sp^3$ carbon bearing the electrophile and a hydrogen.

\text{Benzene} + E^+ \longrightarrow \text{Sigma complex (non-aromatic)}

Deprotonation – A base (often the counterion of the catalyst, like $\ce{FeBr4-}$ ) removes the proton from the $sp^3$ carbon. The pair of electrons from the C–H bond flows back into the ring, restoring the aromatic sextet.

\text{Sigma complex} + \text{Base} \longrightarrow \text{Product} + \ce{H+}

Watch out

The sigma complex is not aromatic. It's a high-energy intermediate. Students often mistakenly think it's still aromatic — it isn't. That's why the step is fast and the complex is short-lived.

Why This Matters for Exams

EAS is the gateway to understanding how to put groups onto benzene rings. The rate-determining step is usually the formation of the sigma complex (step 2). The regiochemistry (where the electrophile goes) depends on whether the ring already has a substituent — that's the topic of activating/deactivating groups and ortho/para vs. meta directors. …

Why this formula?

Electrophilic Addition Reactions: Why the Mechanism Works

Electrophilic addition is a cornerstone of alkene and alkyne chemistry. Instead of memorising the "arrow pushing," let's understand why the reaction proceeds the way it does — driven by electron density, stability, and charge.

1. The Core Idea: Why Alkenes React This Way

Alkenes have a π-bond — a cloud of electrons above and below the plane of the σ-bond. This π-electron cloud is:

Electron-rich (nucleophilic)
Exposed (not shielded by σ-bonds like in alkanes)

An electrophile (electron-lover) is attracted to this high electron density. The reaction is electrophilic addition because the electrophile attacks first.

Key principle: The π-bond acts as a Lewis base (electron donor). The electrophile is a Lewis acid (electron acceptor).

2. The General Mechanism (Two-Step)

Step 1: Formation of a Carbocation (or Bridged Intermediate)

The electrophile (E⁺) attacks the π-bond. The π-electrons form a new σ-bond to E⁺, leaving the other carbon with a positive charge — a carbocation.

\ce{C=C} + \ce{E+} \longrightarrow \ce{C+-C-E}

Why does this happen?

The π-bond is weaker than a σ-bond (~260 kJ/mol vs ~350 kJ/mol). Breaking the π-bond to form a σ-bond is energetically favourable because the new σ-bond is stronger. The carbocation is a high-energy intermediate, but it's stabilised by:

Hyperconjugation (alkyl groups donate electron density)
Inductive effect (alkyl groups push electrons toward the positive carbon)

Step 2: Nucleophilic Attack

A nucleophile (Nu⁻) attacks the carbocation, forming a second σ-bond.

\ce{C+-C-E} + \ce{Nu-} \longrightarrow \ce{C-Nu-C-E}

Why does this happen?

The carbocation is electron-deficient (positive charge). The nucleophile is electron-rich. Opposite charges attract — this is electrostatic and orbital overlap driven.

3. The Key "Formula" — Markovnikov's Rule

Statement: In the addition of HX to an unsymmetrical alkene, the hydrogen attaches to the carbon with more hydrogens already, and the X attaches to the carbon with fewer hydrogens.

Why does this rule hold? (The reasoning)

Consider propene: $\ce{CH3-CH=CH2}$ + HBr.

Possible carbocations:
1. Primary carbocation: $\ce{CH3-CH+-CH2Br}$ (less stable)
2. Secondary carbocation: $\ce{CH3-CHBr-CH2+}$ (more stable)

The more substituted carbocation (secondary > primary) is more stable due to:

Hyperconjugation: More alkyl groups = more C–H σ-bonds that can donate electron density into the empty p-orbital of the carbocation.
Inductive effect: Alkyl groups are electron-donating, stabilising the positive charge.

Result: The reaction proceeds via the more stable carbocation, leading to Markovnikov addition.

Markovnikov's rule is not a law — it's a consequence of carbocation stability.

4. The "Anti-Markovnikov" Exception (Why It Happens)

With HBr in the presence of peroxides (ROOR), the addition is anti-Markovnikov — Br goes to the less substituted carbon.

Why? The mechanism changes from ionic to free-radical.

Peroxide decomposes to radicals: $\ce{ROOR -> 2 RO•}$
RO• abstracts H from HBr: $\ce{RO• + HBr -> ROH + Br•}$
Br• adds to the alkene — at the less substituted carbon (because the radical formed is more stable — tertiary > secondary > primary).
The new radical abstracts H from another HBr, regenerating Br•. …

Start your 14-day free trial to unlock the full solution →

The conversion of propan-2-one (acetone) to tert-butyl alcohol requires a two-step strategy: first, a Grignard reaction with methylmagnesium halide adds a methyl group to the carbonyl carbon, forming a tertiary alcohol after hydrolysis. The final product is 2-methylpropan-2-ol, commonly called tert-butyl alcohol.

The Concept: Building Carbon Skeletons with Grignard Reagents

Propan-2-one (acetone) is a simple ketone with the structure $\ce{CH3-CO-CH3}$ . Tert-butyl alcohol is 2-methylpropan-2-ol, $\ce{(CH3)3C-OH}$ . The key difference? Acetone has two methyl groups attached to the carbonyl carbon; tert-butyl alcohol has three. So we need to add one more methyl group to that central carbon.

This is a classic carbon–carbon bond-forming problem. The most reliable tool for adding an alkyl group to a carbonyl compound is the Grignard reaction. A Grignard reagent ( $\ce{RMgX}$ ) acts as a carbanion equivalent — it's strongly nucleophilic and attacks the electrophilic carbonyl carbon.

Watch out

Common Mistake

Students often try direct reduction (e.g., $\ce{NaBH4}$ ) or oxidation. Reduction of acetone gives isopropyl alcohol (a secondary alcohol), not tert-butyl alcohol. Oxidation would give carboxylic acids or nothing useful here. You must add a carbon, not just change the oxidation state.

Step-by-Step Solution

1. Identify the target carbon skeleton.

Tert-butyl alcohol: $\ce{(CH3)3C-OH}$ . The central carbon is quaternary, bonded to three methyl groups and one OH. Acetone: $\ce{CH3-CO-CH3}$ . The carbonyl carbon has two methyls and one oxygen. To get three methyls, we need to add one $\ce{CH3}$ group to that carbon, and simultaneously convert the $\ce{C=O}$ into $\ce{C-OH}$ .

2. Choose the Grignard reagent.

We need to add a methyl group. So the Grignard reagent is methylmagnesium halide, typically $\ce{CH3MgBr}$ (or $\ce{CH3MgI}$ ). This is prepared from methyl bromide (or iodide) and magnesium metal in dry ether.

3. Perform the Grignard reaction with acetone.

The reaction proceeds in two stages:

Nucleophilic addition: The methyl carbanion (from $\ce{CH3MgBr}$ ) attacks the electrophilic carbonyl carbon of acetone. The $\pi$ bond breaks, and the electrons move to oxygen, forming a magnesium alkoxide intermediate.

\ce{CH3-CO-CH3 + CH3MgBr -> (CH3)3C-O-MgBr}

This intermediate is stable in the anhydrous ether solvent.

Hydrolysis (work-up): After the addition is complete, we add dilute acid (e.g., $\ce{H3O+}$ ) or water. This protonates the alkoxide oxygen, giving the free alcohol.

\ce{(CH3)3C-O-MgBr + H3O+ -> (CH3)3C-OH + MgBr(OH)}

The product is 2-methylpropan-2-ol — tert-butyl alcohol.

Tip

Why This Works …

Start your 14-day free trial to unlock the full solution →

Common Mistakes: Propan-2-one → tert-Butyl Alcohol

This is a classic Grignard reaction problem. Let's break down the common errors students make.

✓ The Correct Approach

The conversion requires two methyl groups to be added to the carbonyl carbon of propan-2-one (acetone):

\ce{CH3-CO-CH3 + 2 CH3MgBr ->[H3O+] (CH3)3C-OH}

Key idea: The carbonyl carbon is electrophilic; the Grignard reagent ( $\ce{CH3MgBr}$ ) acts as a nucleophile, adding a methyl group. A second equivalent adds the second methyl, forming the tertiary alcohol.

✗ Mistake #1: Using the Wrong Reagent

Error: Trying $\ce{CH3Cl}$ or $\ce{CH3Br}$ directly (without Mg).

Why it's wrong: Alkyl halides are not nucleophilic — they are electrophilic at carbon. They cannot attack the carbonyl.

How to avoid: Remember: Grignard reagents ( $\ce{RMgX}$ ) are the nucleophilic carbon source. Always use Mg metal in dry ether to generate the reagent.

✗ Mistake #2: Forgetting the Second Equivalent

Error: Adding only one $\ce{CH3MgBr}$ and stopping.

Why it's wrong: Propan-2-one already has two methyl groups. Adding one more gives a secondary alcohol (2-butanol), not the desired tert-butyl alcohol (2-methylpropan-2-ol).

How to avoid: Count the final carbon skeleton:

Propan-2-one: 3 carbons
tert-Butyl alcohol: 4 carbons (three methyls on a central carbon)

You need two methyl additions.

✗ Mistake #3: Skipping the Acidic Workup

Error: Writing only $\ce{CH3MgBr}$ as the final step.

Why it's wrong: The Grignard addition forms an alkoxide intermediate ( $\ce{R3C-O-MgBr}$ ). This must be protonated with dilute acid ( $\ce{H3O+}$ ) to get the free alcohol.

How to avoid: Always include $\ce{H3O+}$ (or $\ce{NH4Cl}$ ) as the second step.

✗ Mistake #4: Using Aqueous Conditions for Grignard

Error: Trying the reaction in water or wet solvent.

Why it's wrong: Grignard reagents react violently with water (forming alkane + Mg(OH)Br). The reaction fails completely.

How to avoid: Use anhydrous ether (dry diethyl ether or THF) and keep all glassware dry.

--- …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ33 — Name the enzymes and write the reactions involved in the preparation o…→Next questionQ35 — Write the structures of the isomers of alcohols with molecular formula…

Q.How can propan-2-one be converted into tert-butyl alcohol?

Concept understanding — Electrophilic Addition Reactions

Electrophilic Aromatic Substitution – The First Meeting

The Intuition: Why "Substitution" and Not "Addition"?

The Precise Statement

The Mechanism in Three Steps

Why This Matters for Exams

Why this formula?

Electrophilic Addition Reactions: Why the Mechanism Works

1. The Core Idea: Why Alkenes React This Way

2. The General Mechanism (Two-Step)

Step 1: Formation of a Carbocation (or Bridged Intermediate)

Step 2: Nucleophilic Attack

3. The Key "Formula" — Markovnikov's Rule

Why does this rule hold? (The reasoning)

4. The "Anti-Markovnikov" Exception (Why It Happens)

The Concept: Building Carbon Skeletons with Grignard Reagents

Step-by-Step Solution

Method: Grignard Reaction Followed by Acidic Hydrolysis

Why this works (the concept)

Step-by-step procedure

Key points for exams …

Common Mistakes: Propan-2-one → tert-Butyl Alcohol

✓ The Correct Approach

✗ Mistake #1: Using the Wrong Reagent

✗ Mistake #2: Forgetting the Second Equivalent

✗ Mistake #3: Skipping the Acidic Workup

✗ Mistake #4: Using Aqueous Conditions for Grignard

Q.How can propan-2-one be converted into tert-butyl alcohol?

The Concept: Building Carbon Skeletons with Grignard Reagents

Step-by-Step Solution

🎓Unlock everything free for 14 days

Unlock everything free for 14 days