Q.Write the mechanism of acid dehydration of ethanol to yield ethene.

The key idea is acid-catalysed dehydration of an alcohol — an elimination reaction where concentrated sulphuric acid acts as both a dehydrating agent and a catalyst.

Step 1: Protonation of the hydroxyl group

Ethanol reacts with $\ce{H2SO4}$. The lone pair on the oxygen of the $\ce{-OH}$ group abstracts a proton ($\ce{H+}$) from the acid, forming a protonated ethanol (oxonium ion). This converts the poor leaving group ($\ce{-OH}$) into a good one ($\ce{H2O}$).

Step 2: Formation of the carbocation

The $\ce{C-O}$ bond in the oxonium ion breaks heterolytically, expelling a water molecule and generating a primary ethyl carbocation ($\ce{CH3CH2+}$).

Step 3: Elimination of a proton to form the alkene …

This is a classic example of an elimination reaction — specifically, an E1 mechanism (unimolecular elimination). The key idea is that the hydroxyl group ($-OH$) is a poor leaving group on its own. By protonating it with a strong acid (like concentrated $H_2SO_4$ or $H_3PO_4$), we convert it into water ($H_2O$), which is an excellent leaving group. Once water leaves, a carbocation forms, and then a nearby base (often the conjugate base of the acid, like $HSO_4^-$) removes a proton to form the double bond.

Let’s walk through the mechanism step by step.

1. Protonation of the hydroxyl group Ethanol ($CH_3CH_2OH$) has a lone pair on oxygen. In the presence of a strong acid like sulfuric acid ($H_2SO_4$), the oxygen gets protonated. This step is fast and reversible.

The product is an ethyloxonium ion — the oxygen now has a positive charge and is bonded to a water molecule waiting to leave.

2. Formation of the carbocation (rate-determining step) The $C-O$ bond breaks heterolytically, and water leaves as a neutral molecule. This step is slow because it involves breaking a bond and forming a high-energy carbocation.

The result is a primary carbocation ($CH_3CH_2^+$). Primary carbocations are relatively unstable, but under these strongly acidic conditions and at high temperatures (around 170°C), the reaction is driven forward.

3. Deprotonation to form the alkene The carbocation is highly electrophilic. A nearby base (here, the bisulfate ion $HSO_4^-$ or even a water molecule) abstracts a proton from the carbon adjacent to the carbocation (the $\beta$-carbon). This forms a $\pi$ bond between the two carbons, regenerating the acid catalyst.

The product is ethene ($CH_2=CH_2$), a colourless gas. …

Method: Acid-Catalyzed Dehydration of Alcohols (E1 Mechanism)

This reaction follows an E1 (unimolecular elimination) pathway. The key idea: the acid protonates the hydroxyl group, turning a poor leaving group ($-OH$) into a good one ($-OH_2^+$), which then leaves to form a carbocation. A base (here, water or bisulfate ion) then removes a beta-hydrogen to form the alkene.

Steps of the Mechanism

Step 1: Protonation of the hydroxyl group

Ethanol ($CH_3CH_2OH$) reacts with concentrated sulfuric acid ($H_2SO_4$). The acid donates a proton to the oxygen atom of the hydroxyl group.

• Why? The $-OH$ group is a poor leaving group. Protonating it converts it into water ($H_2O$), which is an excellent leaving group.

Step 2: Formation of a carbocation (rate-determining step)

The protonated ethanol loses a water molecule, forming a primary carbocation (ethyl carbocation).

• Key point: Primary carbocations are unstable. However, under strong acidic conditions and high temperature (around 170°C), this step is driven forward. The reaction is unimolecular — only the protonated alcohol is involved in this slow step.

Step 3: Deprotonation to form ethene

A base (either $HSO_4^-$ or $H_2O$) abstracts a beta-hydrogen (a hydrogen on the carbon adjacent to the carbocation). The electrons from the C–H bond move to form the C=C double bond.

• Result: Ethene ($CH_2=CH_2$) is formed, and the acid catalyst ($H_2SO_4$) is regenerated.

Overall Reaction …

Common Mistakes in Acid Dehydration of Ethanol → Ethene

Students often lose marks on this classic mechanism. Here are the most frequent errors and how to avoid each.

1. Forgetting the Protonation Step is First

Mistake: Starting directly with elimination of water from neutral ethanol.

Why it's wrong: Ethanol's $-\text{OH}$ group is a poor leaving group ($\text{OH}^-$ is a strong base). It must first be converted into a better leaving group.

How to avoid: Always begin with protonation of the hydroxyl oxygen by the acid ($\text{H}_2\text{SO}_4$ or $\text{H}_3\text{PO}_4$):

Key idea: The acid makes the leaving group water ($\text{H}_2\text{O}$), which is much more stable than $\text{OH}^-$.

2. Writing the Wrong Carbocation Intermediate

Mistake: Showing a primary carbocation ($\text{CH}_3\text{CH}_2^+$) without rearrangement.

Why it's wrong: Primary carbocations are highly unstable and rarely form. The actual mechanism involves a concerted (E1-like) elimination or a hydride shift if a more stable carbocation is possible.

How to avoid: For ethanol, the mechanism is E1 — after losing water, a primary carbocation does form briefly, but it immediately loses a proton to give ethene. Do not show a rearrangement here (no more stable carbocation exists). However, for higher alcohols (e.g., 2-butanol), always check for possible hydride or alkyl shifts.

Correct step:

Then:

3. Incorrect Arrow Pushing in the Elimination Step

Mistake: Showing the base (e.g., $\text{HSO}_4^-$) abstracting a proton from the wrong carbon or using a curly arrow that starts from a bond instead of a lone pair.

How to avoid:

• The base attacks a $\beta$-hydrogen (on the carbon adjacent to the carbocation).
• The arrow starts from the lone pair on the base (or from the $\text{C-H}$ bond if showing a concerted mechanism — but for E1, it's stepwise).
• The $\pi$ bond forms simultaneously as the $\text{C-H}$ bond breaks.

Correct arrow pushing:

Tip: Draw the carbocation with the positive charge on the correct carbon, then show the base removing a $\beta$-hydrogen.

4. Omitting the Regeneration of Acid Catalyst

Mistake: Showing the acid being consumed and not regenerated.

Why it's wrong: The acid is a catalyst — it is used in the first step and released in the last step.

How to avoid: The final step should show the proton transfer to the base (e.g., $\text{HSO}_4^-$), regenerating $\text{H}_2\text{SO}_4$:

Key idea: Net reaction: $\ce{CH3CH2OH -> CH2=CH2 + H2O}$ — the acid is unchanged.

5. Confusing Conditions (Temperature and Acid)

Mistake: Using the wrong temperature or acid.

Why it's wrong: Dehydration requires concentrated acid and high temperature ($\sim$170°C for ethanol). At lower temperatures ($\sim$140°C), ether forms instead.

How to avoid: Memorise the conditions: