Q.Given, two haloalkanes A and B :
A : CH3 – CH2 – CH2 – Br B : CH3 – CH(Br) – CH3

Alcohol dehydration is a classic example of an elimination reaction, specifically an E1 mechanism when catalyzed by acid. Understanding the mechanism is key to identifying the intermediates. The core idea here is that a poor leaving group (the hydroxyl group, $-\text{OH}$) must be converted into a good leaving group (water, $-\text{OH}_2^+$) before it can depart. The departure of this good leaving group is the slowest step, as it involves breaking a bond and forming a high-energy, electron-deficient species.

Let's break down the process:

1. Protonation of the Alcohol:

   Alcohols are not good substrates for direct elimination because the hydroxyl group ($\text{OH}^-$) is a strong base and thus a poor leaving group. In the presence of an acid (like $\text{H}_2\text{SO}_4$ or $\text{H}_3\text{PO}_4$), the oxygen atom of the alcohol, with its lone pairs, acts as a Lewis base and gets protonated. This step is fast and reversible.

   $\text{R-CH}_2\text{-CH}_2\text{-OH} + \text{H}^+ \rightleftharpoons \text{R-CH}_2\text{-CH}_2\text{-OH}_2^+$

   The protonated alcohol, $\text{R-OH}_2^+$, now has a good leaving group: a neutral water molecule ($\text{H}_2\text{O}$).

2. Formation of the Carbocation:

   This is the crucial step and the rate-determining step (slowest step) of the reaction. The protonated hydroxyl group departs as a neutral water molecule, leaving behind a positively charged carbon atom. This positively charged carbon species is called a carbocation.

   $\text{R-CH}_2\text{-CH}_2\text{-OH}_2^+ \xrightarrow{\text{slow}} \text{R-CH}_2\text{-CH}_2^+ + \text{H}_2\text{O}$

   This step is slow because it involves breaking a strong carbon-oxygen bond and forming a high-energy, unstable carbocation. The activation energy for this step is high. The stability of the carbocation formed directly influences the rate of this step; more stable carbocations form faster.

   The general order of carbocation stability is:

   $\text{Tertiary} (3^\circ) > \text{Secondary} (2^\circ) > \text{Primary} (1^\circ) > \text{Methyl}$

   This stability is primarily due to hyperconjugation and inductive effects from alkyl groups.

3. Deprotonation and Alkene Formation: …

When an ether reacts with a hydrogen halide, the first step is always protonation of the ether oxygen, which converts it into a good leaving group. The regiochemistry — which fragment becomes the iodide and which becomes the alcohol — is decided by the mechanism.

1. Protonation. The oxygen lone pair takes a proton from HI:

2. Choose the mechanism. An $S_N1$ (carbocation) route operates only when one alkyl group can form a stable (tertiary or benzylic/allylic) cation. Here the choices are methyl (primary) and isopropyl (secondary) — neither gives a stable enough cation, so cleavage goes by $S_N2$.

3. $S_N2$ attacks the less hindered carbon. Iodide approaches the carbon with least steric crowding. The methyl carbon is far less hindered than the secondary isopropyl carbon, so $I^-$ attacks the methyl group:

The isopropyl group departs as isopropyl alcohol. …

The key here is the reagent: alcoholic KOH. This is a classic strong base in a polar, protic solvent (ethanol). Unlike aqueous KOH (which promotes substitution), the alcoholic medium suppresses the nucleophilic character of the hydroxide ion and enhances its basicity. So the reaction follows an E2 elimination pathway — not SN1 or SN2.

Why? Because the alkoxide ion (from ethanol) is a weaker nucleophile but a strong base, and the high temperature of the reaction (usually reflux) favours elimination over substitution. The substrate is a secondary alkyl halide, which can undergo both E1 and E2, but with a strong base like KOH, E2 dominates.

Now, the molecule is 2-Bromopentane:

The bromine is on carbon 2. In an E2 elimination, the base abstracts a proton from a β-carbon (adjacent to the carbon bearing the leaving group), while the leaving group departs simultaneously. The question is: which β-hydrogen is removed?

There are two possible β-carbons:

1. β-carbon 1 (C1) — gives a terminal alkene: Pent-1-ene
2. β-carbon 3 (C3) — gives an internal alkene: Pent-2-ene

The Saytzeff rule (Zaitsev’s rule) tells us that the more substituted alkene is more stable (due to hyperconjugation and inductive effects). Pent-2-ene is disubstituted (two alkyl groups on the double bond), while Pent-1-ene is monosubstituted. So Pent-2-ene is the major product.

Let’s walk through the steps:

1. Identify the substrate and reagent.

   2-Bromopentane is a secondary alkyl halide. Alcoholic KOH is a strong base in ethanol. The reaction conditions (heat, base, alcohol solvent) scream E2 elimination.

2. Locate the β-hydrogens.

   The α-carbon (C2) has two β-carbons:

   • C1 (methyl group) has 3 hydrogens.
   • C3 (methylene group) has 2 hydrogens. Both are accessible to the base.

3. Apply the Saytzeff rule. …