Q.(a) Write the mechanism of the following reaction: Ethanol $CH_3CH_2OH \xrightarrow[443\,K]{H^+} CH_2=CH_2$ Ethene

1. Mechanism of ethanol dehydration to ethene This is an acid-catalysed elimination (E1 mechanism). Step 1: Protonation of the –OH group by $H^+$ forms a good leaving group ($H_2O$). Step 2: Loss of water generates a primary carbocation (unstable, but rearrangement is not possible here). Step 3: Deprotonation at the $\beta$-carbon by a base (e.g., $HSO_4^-$) gives the alkene.

   $CH_3CH_2OH \xrightarrow{H^+} CH_3CH_2OH_2^+ \rightarrow CH_3CH_2^+ + H_2O \rightarrow CH_2=CH_2 + H_3O^+$

2. Main products

(i) Hydroboration–oxidation of but-1-ene

This adds water across the double bond with anti-Markovnikov regiochemistry.

The boron attaches to the less substituted carbon; oxidation replaces B with –OH.

Product: Butan-1-ol ($CH_3CH_2CH_2CH_2OH$). …

(a) Mechanism of Ethanol Dehydration to Ethene

The reaction $CH_3CH_2OH \xrightarrow[H^+]{443\,K} CH_2=CH_2$ is an acid‑catalysed elimination (E1 mechanism). The key idea: the alcohol is first protonated to turn the poor leaving group (–OH) into a good one (–OH$_2^+$), then water leaves to form a carbocation, which finally loses a proton to give the alkene.

Step 1: Protonation of the –OH group

The lone pair on oxygen of ethanol attacks a proton from the acid ($H^+$), forming an oxonium ion.

This step is fast and reversible. The oxygen now carries a positive charge, making the C–O bond much weaker.

Step 2: Loss of water to form a carbocation

The C–O bond breaks heterolytically, and water leaves. This is the rate‑determining step (slow).

A primary carbocation ($CH_3CH_2^+$) is formed. It is unstable, but the high temperature (443 K) provides enough energy for this step to occur.

Step 3: Deprotonation to form ethene

A water molecule (or the conjugate base of the acid) abstracts a proton from the $\beta$‑carbon (the carbon adjacent to the carbocation). The electron pair shifts to form the C=C double bond.

The $H_3O^+$ regenerates the acid catalyst.

(b) Main Products of the Given Reactions

(i) But‑1‑ene with Hydroboration‑Oxidation

Reaction:

$CH_3CH_2CH=CH_2 \xrightarrow{\text{a) } B_2H_6 \text{ b) } 3H_2O_2/OH^-}$

Concept: Hydroboration‑oxidation is a two‑step method that adds water across a double bond anti‑Markovnikov — the hydrogen goes to the more substituted carbon, and the –OH goes to the less substituted carbon. The reaction is syn addition (both H and OH add from the same face), but that stereochemistry is not relevant here because the product is an achiral alcohol.

Step 1: Hydroboration

Diborane ($B_2H_6$) dissociates into $BH_3$, which adds to the double bond. Boron attaches to the less hindered (less substituted) carbon — in but‑1‑ene, that is the terminal carbon (C1). Hydrogen attaches to C2.

This step repeats until all three hydrogens on boron are replaced, forming a trialkylborane.

Step 2: Oxidation

Alkaline hydrogen peroxide ($H_2O_2/OH^-$) oxidises the C–B bond to a C–OH bond, replacing boron with a hydroxyl group. The boron is removed as boric acid.

Main product: Butan‑1‑ol ($CH_3CH_2CH_2CH_2OH$). No rearrangement occurs because the addition is concerted.

(ii) Phenol with NaOH and then CO$_2$ (Kolbe’s Reaction)

Reaction:

$C_6H_5OH \xrightarrow{\text{a) aq. NaOH b) } CO_2, H^+}$ …