Q.Write the mechanism of hydration of ethene to yield ethanol.

The key idea is electrophilic addition across the carbon-carbon double bond, where water adds in the presence of an acid catalyst.

Step 1: Protonation of ethene

The $\pi$ bond of ethene attacks a proton ($\ce{H+}$) from the acid catalyst (e.g., $\ce{H2SO4}$), forming a carbocation intermediate (ethyl cation).

Step 2: Nucleophilic attack by water …

The hydration of ethene is a classic example of an electrophilic addition reaction — the most characteristic reaction of alkenes. The key idea is that the $\pi$ bond in ethene is electron-rich and acts as a nucleophile, attacking an electrophile. Here, the electrophile is a proton ($\text{H}^+$) from the acid catalyst. Once the proton adds, a carbocation intermediate forms, which is then attacked by water (a nucleophile). A final deprotonation step gives the alcohol.

Why does this need an acid catalyst? Water alone is a poor electrophile — the $\text{O–H}$ bond is too strong to break easily. The acid provides the $\text{H}^+$ that initiates the attack. The catalyst is regenerated at the end, so it is not consumed.

Let’s walk through the mechanism step by step.

1. Protonation of the double bond The acid catalyst (typically dilute $\text{H}_2\text{SO}_4$ or $\text{H}_3\text{PO}_4$) donates a proton to the $\pi$ bond of ethene. The $\pi$ electrons attack the $\text{H}^+$, forming a $\sigma$ bond between carbon and hydrogen. This leaves the other carbon with a positive charge — a carbocation intermediate.

The carbocation is a primary carbocation (ethyl cation). It is relatively unstable but is formed here because ethene is symmetrical and the alternative (a secondary carbocation) is not possible.

2. Nucleophilic attack by water The carbocation is electron-deficient. A water molecule (which has lone pairs on oxygen) acts as a nucleophile and donates a pair of electrons to the positively charged carbon. This forms a new $\text{C–O}$ bond, giving a protonated alcohol (an oxonium ion).

3. Deprotonation to form ethanol The oxonium ion is acidic — the oxygen now has a positive charge and a hydrogen can be lost. A water molecule (or the conjugate base of the acid) abstracts a proton from the oxygen, regenerating the acid catalyst and yielding neutral ethanol.

The $\text{H}_3\text{O}^+$ can then dissociate to give back $\text{H}^+$, completing the catalytic cycle. …

Method: Acid-Catalysed Electrophilic Addition (via Carbocation Intermediate)

This is the standard mechanism for hydration of alkenes — converting ethene ($\ce{CH2=CH2}$) to ethanol ($\ce{CH3CH2OH}$) using dilute acid ($\ce{H3O+}$).

Why this happens (the concept)

The $\pi$ bond in ethene is electron-rich and acts as a nucleophile. It attacks a proton ($\ce{H+}$) from the acid — an electrophile. This forms a carbocation, which is then attacked by water (a nucleophile). Finally, deprotonation gives the alcohol.

Step-by-step mechanism

Step 1: Protonation of the $\pi$ bond (slow, rate-determining step)

The $\pi$ electrons attack $\ce{H+}$ from $\ce{H3O+}$, forming a primary carbocation and a water molecule.

Key point: The carbocation is the reactive intermediate. It is unstable (primary) but formed here because ethene is symmetrical — no regioselectivity issue.

Step 2: Nucleophilic attack by water (fast)

Water (a lone-pair donor) attacks the positively charged carbon, forming a protonated alcohol (oxonium ion).

Step 3: Deprotonation (fast) …

Common Mistakes in Hydration of Ethene (Electrophilic Addition)

Students often lose marks on this mechanism despite it being straightforward. Here are the most frequent errors and how to avoid each.

1. Writing the Wrong Reagent or Conditions

Mistake:

Writing "H₂O" alone, or using dilute acid at room temperature.

Why it's wrong:

Ethene does not react with neutral water. The reaction requires:

• Concentrated H₂SO₄ (acts as catalyst and proton source)
• Followed by hydrolysis with water

How to avoid:

Always write:

Step 1: Ethene + conc. H₂SO₄ → ethyl hydrogen sulphate

Step 2: Ethyl hydrogen sulphate + H₂O (warm) → ethanol + H₂SO₄ (regenerated)

2. Forgetting the Carbocation Intermediate

Mistake:

Showing direct addition of H⁺ and OH⁻ in one step.

Why it's wrong:

Electrophilic addition proceeds via a carbocation intermediate. The mechanism is two-step:

1. Slow step: Electrophile (H⁺) attacks π-bond → forms carbocation
2. Fast step: Nucleophile (HSO₄⁻ or H₂O) attacks carbocation

How to avoid:

Draw the curly arrow from the π-bond to H⁺, then show the secondary carbocation (CH₃–CH₂⁺) clearly.

3. Showing Markovnikov Addition Incorrectly

Mistake:

Adding H⁺ to the wrong carbon in substituted alkenes.

Why it's wrong:

For ethene, both carbons are identical — Markovnikov rule doesn't apply. But students often misapply it anyway.

How to avoid:

For ethene, either carbon gets the H⁺ — the product is the same. But for propene or higher alkenes, remember:

H⁺ goes to the carbon with more H atoms already (to form the more stable carbocation).

4. Writing the Final Product Incorrectly

Mistake:

Writing CH₃CH₂OH as CH₃CH₂O⁻ or CH₃CH₂⁺.

Why it's wrong:

Ethanol is a neutral molecule, not an ion. The final step involves deprotonation (loss of H⁺) to give the alcohol.

How to avoid:

After the nucleophile attacks, show a proton transfer (e.g., from H₂O⁺ to HSO₄⁻) to yield neutral ethanol.

5. Omitting the Regeneration of Catalyst

Mistake:

Writing H₂SO₄ as a reactant that gets consumed.

Why it's wrong:

H₂SO₄ is a catalyst — it is regenerated at the end. The net reaction is:

How to avoid: …