Q.A) Arrange the following alkyl halides in ascending order of their reactivity towards SN1 reaction. CH3X, CH3-CH2-X, CH3-CH(X)-CH3, CH3-C(CH3)2-X (X = Same) B) Complete the following chemical reactions and write the products. i) CH3-CH2-Cl + KOH (Alc.) --> ii) R-CH=CH2 + HBr --Peroxide-->

Why this approach works

Nucleophilic substitution reactions (SN1 and SN2) both require the leaving group (here, Cl⁻) to depart. The ease of this departure depends on:

• For SN1: The stability of the carbocation formed after Cl⁻ leaves. More stable carbocations (tertiary, allylic, benzylic) react faster.
• For SN2: The steric hindrance around the carbon bearing the leaving group. Less hindered carbons (methyl > primary > secondary) react faster.

But there is a special case: when chlorine is directly bonded to an aromatic ring (an aryl halide), the C–Cl bond gains partial double-bond character due to resonance. This makes it much stronger and harder to break — so aryl halides are notoriously unreactive in typical nucleophilic substitutions unless special conditions (like very strong nucleophiles or high temperatures) are used.

Let’s examine each option.

Step-by-step reasoning

1. Option (A): Benzyl chloride ($C_6H_5CH_2Cl$)

Here, chlorine is on a carbon next to the benzene ring, not directly on it. The benzylic carbocation ($C_6H_5CH_2^+$) is highly stabilised by resonance with the ring. So SN1 is very fast. SN2 is also possible because the benzylic carbon is primary and not too hindered. This compound is highly reactive.

2. Option (B): 1-chloro-4-methylbenzene (4-methylchlorobenzene)

Chlorine is directly attached to the benzene ring. The lone pairs on chlorine participate in resonance with the aromatic π-system, giving the C–Cl bond partial double-bond character. This bond is very strong — about 30–40 kJ/mol stronger than a typical alkyl C–Cl bond. Neither SN1 (no stable carbocation — aryl cations are extremely unstable) nor SN2 (the carbon is sp² hybridised and the backside is blocked by the ring) works under normal conditions. This is the least reactive.

Understanding SN1 Reactivity and Carbocation Stability

Hydrolysis by aqueous KOH can proceed through two pathways: SN2 (bimolecular substitution) or SN1 (unimolecular, carbocation-mediated). When we ask which compound hydrolyses "most easily," we're really asking which forms the most stable intermediate or transition state.

The key insight: carbocation stability dictates SN1 reactivity, and resonance stabilization trumps inductive effects. Let's examine each structure.

Step-by-Step Analysis

1. Identify the type of halide in each compound

• (A) $CH_2 = CH - Br$: Vinyl halide (Br directly on $sp^2$ carbon)
• (B) $CH_3 - CH_2 - Br$: Primary alkyl halide
• (C) $CH_3 - CH(Br) - CH_3$: Secondary alkyl halide
• (D) $CH_2 = CH - CH_2 - Br$: Allylic halide (Br on carbon adjacent to C=C)

2. Evaluate vinyl halide (A)

Vinyl halides are notoriously unreactive toward both SN1 and SN2. The C–Br bond has significant $sp^2$ character (shorter, stronger), and the hypothetical vinyl cation would be extremely unstable due to the electron-withdrawing effect of the $sp^2$ hybridized carbon. This compound is essentially inert under typical hydrolysis conditions.

3. Compare primary (B) vs. secondary (C) alkyl halides

• Primary carbocation: highly unstable, so (B) proceeds mainly via SN2 (slow with weak nucleophile in aqueous medium)
• Secondary carbocation: more stable than primary due to hyperconjugation from two adjacent alkyl groups, so (C) can proceed via SN1, but still not particularly fast

The order so far: (C) > (B) >> (A).

4. Recognize the allylic system in (D)

When $CH_2 = CH - CH_2 - Br$ ionizes, it forms the allyl cation $CH_2 = CH - CH_2^+$. This cation is resonance-stabilized: …

Why $S_N1$ Reactivity Order Matters

$S_N1$ reactions proceed through a two-step mechanism: first, the leaving group departs to form a carbocation intermediate; second, the nucleophile attacks this carbocation. The rate-determining step is the first step — carbocation formation. So the faster a molecule can form a stable carbocation, the faster it undergoes $S_N1$ reaction.

This means we don't look at steric hindrance or nucleophile strength here. We look purely at carbocation stability. The more stable the carbocation intermediate, the lower the activation energy for its formation, and the faster the reaction.

Let's examine each compound.

1. Compound (A): 3-Bromocyclohex-1-ene

   The bromine is on the allylic carbon — the carbon adjacent to the double bond. When the Br leaves, the resulting carbocation is allylic.

   An allylic carbocation is resonance-stabilized: the positive charge can be delocalized into the adjacent $\pi$ bond. This gives two resonance structures, spreading the charge over two carbons.

   Resonance stabilization of an allylic carbocation:

   $\ce{CH2=CH-CH2+ <-> +CH2-CH=CH2}$

   This delocalization significantly lowers the energy of the carbocation, making it much more stable than a simple secondary or tertiary alkyl carbocation.

2. Compound (B): (Bromomethyl)cyclohexane

   Here the bromine is on a primary carbon (the CH$_2$ group attached to the ring). If Br leaves, we get a primary carbocation — $\ce{RCH2+}$.

   Primary carbocations are highly unstable (no alkyl groups to donate electron density via hyperconjugation, no resonance). They are so unstable that $S_N1$ reactions on primary substrates are essentially impossible under normal conditions. This compound would react via $S_N2$, not $S_N1$.

3. Compound (C): Bromocyclohexane

   The bromine is on a secondary carbon of the cyclohexane ring. Loss of Br gives a secondary carbocation.

   Secondary carbocations are moderately stable — they have two alkyl groups providing hyperconjugative stabilization — but they are far less stable than an allylic carbocation. No resonance is possible here.

4. Compound (D): Bromobenzene

   The bromine is directly attached to an aromatic ring. If Br leaves, we would get a phenyl carbocation — a positive charge on an $sp^2$ carbon of the benzene ring. …

Understanding the Lucas Test

The reaction you're looking at is the Lucas test, a classic qualitative test to distinguish between primary, secondary, and tertiary alcohols. The reagent is a mixture of concentrated hydrochloric acid and zinc chloride ($\ce{ZnCl2}$), which acts as a Lewis acid catalyst.

The key to predicting reactivity lies in understanding how this reaction proceeds. Zinc chloride coordinates with the oxygen of the alcohol, making it a better leaving group. For secondary and tertiary alcohols, the mechanism is predominantly $\mathrm{S_N1}$: the protonated alcohol loses water to form a carbocation, which then captures chloride ion. For primary alcohols, the mechanism leans toward $\mathrm{S_N2}$ because primary carbocations are too unstable to form.

Since carbocation stability increases in the order $1^\circ < 2^\circ < 3^\circ$, and the rate-determining step in $\mathrm{S_N1}$ is carbocation formation, tertiary alcohols react fastest, followed by secondary, then primary.

Step-by-Step Analysis

1. Tertiary alcohols ($3^\circ$)

   When a tertiary alcohol is treated with Lucas reagent, the $\ce{-OH}$ group is protonated and $\ce{ZnCl2}$ coordinates to facilitate departure of water. The resulting tertiary carbocation ($\ce{R3C+}$) is highly stable due to hyperconjugation and inductive effects from three alkyl groups. This carbocation forms rapidly, so the reaction is immediate — you see cloudiness (the insoluble alkyl chloride) within seconds at room temperature.

2. Secondary alcohols ($2^\circ$)

   Secondary alcohols also proceed via $\mathrm{S_N1}$, but the secondary carbocation ($\ce{R2CH+}$) is less stable than a tertiary one (only two alkyl groups stabilizing the positive charge). The reaction is slower, typically requiring 5–10 minutes of warming to produce visible turbidity.

3. Primary alcohols ($1^\circ$) …

The question asks us to identify which substitution mechanism leads to racemisation—the formation of a 50:50 mixture of enantiomers from an optically active starting material.

To understand this, we need to examine what happens at the chiral centre during each mechanism.

Understanding the $S_N1$ Mechanism

In an $S_N1$ reaction, the mechanism proceeds in two distinct steps:

1. Formation of a carbocation intermediate

   The leaving group departs first, creating a carbocation. This is the rate-determining step (hence "1" in $S_N1$—the rate depends only on the substrate concentration).

2. Nucleophilic attack on the carbocation

   Here's the crucial point: a carbocation is $sp^2$ hybridised and planar. The three substituents lie in a plane, with an empty $p$-orbital perpendicular to that plane.

When the nucleophile approaches this planar carbocation, it can attack from either the front face or the back face with equal probability. If the starting material was a pure enantiomer (say, $R$-configuration), attack from one side regenerates the $R$-enantiomer while attack from the other side produces the $S$-enantiomer.

The result? A racemic mixture—equal amounts of both enantiomers. This is racemisation.

Understanding the $S_N2$ Mechanism

The $S_N2$ mechanism is fundamentally different:

1. Single concerted step

   The nucleophile attacks from the backside (opposite to the leaving group) in a single step. There is no intermediate.

2. Inversion of configuration

   Because the attack must occur from the back, the stereochemistry at the chiral centre is inverted—like an umbrella flipping inside-out. This is called Walden inversion. …

This question tests your understanding of SN2 reactivity — specifically, how the structure of the alkyl halide and the nature of the nucleophile determine whether substitution actually happens. The key idea is simple: SN2 requires a sterically accessible carbon atom for the backside attack. A bulky substrate or a bulky nucleophile can shut it down.

Let’s examine each reaction step by step.

1. Reaction (A): $CH_3CH_2Br + Na^+\,{}^-O-C(CH_3)_3 \rightarrow CH_3CH_2-O-C(CH_3)_3$

   Here, the substrate is ethyl bromide — a primary alkyl halide. Primary halides are excellent for SN2 because the carbon is wide open for attack. The nucleophile is the tert-butoxide ion $^-O-C(CH_3)_3$, which is a very strong base and a bulky nucleophile.

   Now, here’s the nuance: tert-butoxide is indeed bulky, but it can still perform SN2 on a primary substrate because the steric hindrance around the reacting carbon is minimal. The backside approach is not blocked. So this reaction proceeds via SN2, giving the ether product.

   Tip

   A common trap is to think tert-butoxide always does elimination. That’s true for secondary and tertiary halides, but on primary halides, SN2 is still the dominant pathway — the substrate’s openness outweighs the nucleophile’s bulk.

2. Reaction (B): $(CH_3)_3C-Cl + Na^+\,{}^-O-CH_2CH_3 \rightarrow CH_3CH_2-O-C(CH_3)_3$

   Here, the substrate is tert-butyl chloride — a tertiary alkyl halide. Tertiary carbons are sterically crowded; the three methyl groups block the backside approach completely. SN2 is impossible. …