Q.Predict the major product of acid catalysed dehydration of

Concept: Acid-catalysed dehydration of alcohols follows Zaitsev’s rule — the more substituted alkene is the major product, formed via carbocation rearrangement when a more stable carbocation is possible.

1. 1-Methylcyclohexanol Protonation and loss of water gives a tertiary carbocation at the 1-position. No rearrangement is needed. Elimination of a proton from the adjacent carbon (C-2) yields the more substituted alkene — 1-methylcyclohexene.
2. Butan-1-ol Protonation and loss of water gives a primary carbocation (unstable). A 1,2-hydride shift converts it to the more stable secondary carbocation at C-2. Elimination then gives but-2-ene (major), which is more substituted than but-1-ene.
   ✓Final answer

   The major products are (i) 1-methylcyclohexene and (ii) but-2-ene.

The Concept: Why Dehydration Works This Way

Acid-catalysed dehydration of alcohols is an elimination reaction (E1 mechanism under these conditions). The alcohol’s –OH group is a poor leaving group, so we first protonate it with acid (usually H₂SO₄ or H₃PO₄) to turn it into –OH₂⁺, which leaves as water. This creates a carbocation intermediate.

Once the carbocation forms, two things can happen:

• Elimination: A base (often water or the conjugate base of the acid) removes a β-hydrogen, forming a double bond.
• Rearrangement: If the carbocation can shift to a more stable position (via a hydride or alkyl shift), it will do so before elimination.

The major product is the most substituted alkene (Zaitsev’s rule) — unless rearrangement leads to an even more stable alkene.

Let’s apply this to each case.

(i) 1-Methylcyclohexanol

Step 1: Protonation and loss of water

The –OH group gets protonated by H⁺ from the acid:

Water leaves, forming a tertiary carbocation at the carbon that originally held the –OH (the 1-position of the ring, which also bears the methyl group).

This carbocation is tertiary — already quite stable. No rearrangement is needed because a tertiary carbocation is more stable than any secondary or primary alternative.

Step 2: Elimination of a β-hydrogen

Now a base (water or HSO₄⁻) removes a hydrogen from a carbon adjacent to the carbocation. There are two possible β-positions:

• Removing a hydrogen from the ring carbon next to the carbocation (say, C2) gives 1-methylcyclohexene (the double bond between C1 and C2).
• Removing a hydrogen from the methyl group itself would give methylenecyclohexane (double bond exocyclic to the ring).

Which is major? Zaitsev’s rule says: the more substituted alkene is favoured. 1-Methylcyclohexene is trisubstituted (three alkyl groups on the double bond carbons), while methylenecyclohexane is disubstituted. So 1-methylcyclohexene is the major product.

Final product for (i): 1-methylcyclohexene.

(ii) Butan-1-ol

Step 1: Protonation and loss of water

Butan-1-ol is a primary alcohol. Protonation gives butan-1-olium ion, and water leaves to form a primary carbocation (CH₃CH₂CH₂CH₂⁺).

A primary carbocation is very unstable. So before elimination can occur, the carbocation will rearrange via a 1,2-hydride shift to form a more stable secondary carbocation.

This secondary carbocation (butan-2-ylium) is more stable.

Step 2: Elimination from the secondary carbocation

Now we have a secondary carbocation at C2. Elimination of a β-hydrogen can occur in two directions:

• Remove a hydrogen from C1 (the end carbon) → gives but-1-ene (CH₂=CH–CH₂–CH₃).
• Remove a hydrogen from C3 → gives but-2-ene (CH₃–CH=CH–CH₃).

But-2-ene is more substituted (disubstituted) than but-1-ene (monosubstituted), so Zaitsev’s rule predicts but-2-ene as the major product. However, but-2-ene exists as two stereoisomers: cis and trans. The trans isomer is more stable (less steric hindrance) and is usually the major stereoisomer.

So the major product from butan-1-ol is trans-but-2-ene, with some cis-but-2-ene and a little but-1-ene.

Final product for (ii): trans-but-2-ene (or simply but-2-ene as the major alkene).

Summary Table

Method: Zaitsev’s Rule (Saytzeff’s Rule) for Acid-Catalysed Dehydration of Alcohols

This is an E1 elimination reaction (unimolecular elimination) under acidic conditions. The key idea: the more substituted alkene (more alkyl groups on the double-bond carbons) is the major product.

General Steps

1. Protonation of the –OH group by acid ($H^+$).
2. Loss of water to form a carbocation intermediate.
3. Rearrangement (if a more stable carbocation is possible via hydride or alkyl shift).
4. Deprotonation (loss of $H^+$ from a neighbouring carbon) to form the alkene.
5. Apply Zaitsev’s Rule: the major alkene is the one with the most substituted double bond (more alkyl groups = more stable).

(i) 1-Methylcyclohexanol

Starting molecule:

A cyclohexane ring with an –OH and a –CH$_3$ on the same carbon (C1).

Step-by-step:

• Protonation of –OH → $OH_2^+$ leaves as water → forms a tertiary carbocation at C1 (already stable, no rearrangement needed).
• Deprotonation can occur from:
  • C2 (gives 1-methylcyclohexene — double bond between C1 and C2)
  • C6 (gives methylenecyclohexane — double bond between C1 and C6, exocyclic)

Apply Zaitsev’s Rule:

1-methylcyclohexene has a trisubstituted double bond (C1 has –CH$_3$, C2 has two ring carbons).

Methylenecyclohexane has a disubstituted double bond (exocyclic, less substituted).

Major product: 1-methylcyclohexene

(ii) Butan-1-ol

Starting molecule:

$CH_3CH_2CH_2CH_2OH$ (primary alcohol).

Step-by-step:

• Protonation and loss of water → forms a primary carbocation (very unstable).
• Hydride shift occurs: a hydrogen from C2 moves to C1 → forms a secondary carbocation at C2 (more stable).
• Deprotonation can occur from:
  • C1 (gives but-1-ene — double bond between C1 and C2)
  • C3 (gives but-2-ene — double bond between C2 and C3)

Apply Zaitsev’s Rule:

But-2-ene has a disubstituted double bond (both C2 and C3 have one alkyl group each).

But-1-ene has a monosubstituted double bond (terminal).

Major product: But-2-ene (exists as cis/trans isomers, but both are more substituted than but-1-ene)

Final Answer Summary

Key takeaway: Always check for carbocation rearrangement first, then apply Zaitsev’s rule to find the most substituted alkene.

Here are the common mistakes students make with acid-catalysed dehydration of alcohols, specifically for the two examples you gave, along with how to avoid each.

Mistake 1: Forgetting Carbocation Rearrangement (1-methylcyclohexanol)

The Mistake: Students often simply remove water from 1-methylcyclohexanol and draw the direct alkene (1-methylcyclohexene). They miss the possibility of a hydride shift or methyl shift to form a more stable carbocation.

Why it happens: The reaction starts with protonation of the OH group, then loss of water to form a carbocation. In 1-methylcyclohexanol, the initial carbocation is secondary (on the carbon that had the OH). Students assume this is the final intermediate.

How to avoid it:

• Always check carbocation stability. A secondary carbocation can rearrange to a tertiary carbocation via a 1,2-hydride shift (or methyl shift) if a more substituted carbon is adjacent.
• For 1-methylcyclohexanol, the initial secondary carbocation can undergo a hydride shift from the adjacent tertiary carbon (the one with the methyl group) to give a tertiary carbocation.
• The major product is then 1-methylcyclohexene (the more substituted alkene, obeying Zaitsev’s rule) — but only after considering that the tertiary carbocation leads to the same alkene as the direct product. However, if a different shift were possible (e.g., ring expansion), the product would change. Always draw the carbocation and check for shifts.

Correct approach:

• Step 1: Protonate OH → $\ce{OH2^+}$ leaves → secondary carbocation.
• Step 2: Check adjacent carbons for more stable carbocation. Here, a hydride shift from the tertiary carbon (bearing the methyl) gives a tertiary carbocation.
• Step 3: Deprotonation from the tertiary carbocation gives the most substituted alkene: 1-methylcyclohexene.

Mistake 2: Ignoring Zaitsev’s Rule (butan-1-ol)

The Mistake: For butan-1-ol (a primary alcohol), students often draw but-1-ene as the major product, thinking the OH leaves from C1 and the double bond forms between C1 and C2.

Why it happens: They forget that acid-catalysed dehydration of primary alcohols proceeds via an E1 mechanism (not E2) under these conditions, and that the carbocation formed (if any) is unstable. Actually, for primary alcohols, the mechanism is often E2-like or involves a carbocation rearrangement to a more stable one.

How to avoid it:

• Primary alcohols do not form stable primary carbocations. Instead, the reaction proceeds via a concerted E2 mechanism (with acid catalysis) or via a rearrangement to a secondary carbocation.
• For butan-1-ol, the initial protonation and loss of water would give a primary carbocation (very unstable). Instead, a 1,2-hydride shift occurs to form a secondary carbocation (more stable).
• Then, deprotonation follows Zaitsev’s rule: the more substituted alkene is but-2-ene (as a mixture of cis/trans), not but-1-ene.

Correct approach:

• Step 1: Protonate OH → loss of water → primary carbocation (unstable).
• Step 2: Hydride shift from C2 to C1 → secondary carbocation at C2.
• Step 3: Deprotonation from C3 (adjacent to C2) gives but-2-ene (major), not but-1-ene.

Mistake 3: Forgetting Stereochemistry (cis/trans in but-2-ene)

The Mistake: Students write "but-2-ene" without specifying the stereoisomers, or they assume only one isomer forms.

Why it happens: They focus only on the carbon skeleton and ignore that the double bond can have E/Z (cis/trans) isomers.

How to avoid it:

• In acid-catalysed dehydration, the deprotonation step is not stereoselective — both cis and trans isomers form.
• For but-2-ene, the trans isomer is more stable (less steric hindrance) and is the major product.
• Always indicate the major stereoisomer if the question asks for the major product.

Correct answer: Major product is trans-but-2-ene (along with some cis-but-2-ene and a little but-1-ene).

Mistake 4: Not Considering Ring Expansion (for cyclic alcohols)

The Mistake: For 1-methylcyclohexanol, students don’t check if a ring expansion (e.g., from a six-membered ring to a seven-membered ring) is possible.

Why it happens: They assume the ring stays the same size.

How to avoid it:

• In cyclic alcohols, if a carbocation forms adjacent to a ring carbon, a 1,2-alkyl shift can expand the ring (e.g., cyclopentyl to cyclohexyl).
• For 1-methylcyclohexanol, the initial secondary carbocation is adjacent to a tertiary carbon (with the methyl). A methyl shift would give a tertiary carbocation but no ring expansion because the methyl is a side group, not part of the ring. However, if the OH were on a different carbon (e.g., 2-methylcyclohexanol), ring expansion to a cycloheptane derivative is possible.
• Always check for ring expansion when a carbocation is on a ring carbon adjacent to a ring junction.

Summary Table of Mistakes and Fixes

Final Correct Answers (for your reference)

1. 1-methylcyclohexanol → Major product: 1-methylcyclohexene (via tertiary carbocation, no ring expansion here)
2. butan-1-ol → Major product: trans-but-2-ene (via hydride shift to secondary carbocation, then Zaitsev deprotonation)