Q.Which reagent will you use for the following reaction?
$\mathrm{CH_3CH_2CH_2CH_3 \rightarrow CH_3CH_2CH_2CH_2Cl + CH_3CH_2CHClCH_3}$

The reaction converts butane into a mixture of 1-chlorobutane and 2-chlorobutane. This is a free-radical chlorination, which requires homolytic cleavage of the $\mathrm{Cl_2}$ bond.

Reasoning:

1. Alkanes do not react with $\mathrm{Cl_2}$ in the dark or with ionic reagents like $\mathrm{NaCl + H_2SO_4}$.
2. $\mathrm{Cl_2}$ in the presence of iron in dark promotes electrophilic aromatic substitution (not alkane substitution). …

This is a classic example of free‑radical halogenation of alkanes. Alkanes are generally unreactive because their C–H and C–C bonds are strong and non‑polar. To break a C–H bond and replace H with Cl, you need to generate chlorine radicals — highly reactive species that can abstract a hydrogen atom from the alkane.

The key idea: chlorine radicals are produced when $\mathrm{Cl_2}$ is exposed to UV light (or heat, around 250–400 °C). In the dark, $\mathrm{Cl_2}$ does not dissociate into radicals, so no reaction occurs with an alkane. The presence of iron (Fe) is used for electrophilic aromatic substitution (chlorination of benzene), not for alkanes. And $\mathrm{NaCl + H_2SO_4}$ generates HCl, not chlorine radicals — that mixture won’t chlorinate an alkane.

Now let’s walk through the reasoning step by step.

1. Identify the reaction type.

   The starting material is butane ($\mathrm{CH_3CH_2CH_2CH_3}$), a straight‑chain alkane. The products are two monochlorinated isomers: 1‑chlorobutane and 2‑chlorobutane. This is a substitution reaction where one hydrogen is replaced by chlorine. For alkanes, such substitution occurs only via a free‑radical chain mechanism.

2. What generates chlorine radicals?

   The chlorine molecule ($\mathrm{Cl_2}$) has a Cl–Cl bond that can be homolytically cleaved by absorbing energy.

UV light provides the necessary energy (about 243 kJ/mol). Heat can also do it, but the question specifically mentions UV light in option (i). Without UV or heat, no radicals form.

3. Why not the other options?

   • (ii) $\mathrm{NaCl + H_2SO_4}$: This mixture produces HCl gas (and $\mathrm{NaHSO_4}$). HCl does not dissociate into chlorine radicals under normal conditions; it’s a source of $\mathrm{H^+}$ and $\mathrm{Cl^-}$, not $\mathrm{Cl^\bullet}$. No radical chlorination occurs.
   • (iii) $\mathrm{Cl_2}$ gas in dark: In the dark, $\mathrm{Cl_2}$ molecules remain intact. Without radical initiation, the alkane and chlorine simply mix without reacting.
   • (iv) $\mathrm{Cl_2}$ gas in the presence of iron in dark: Iron (Fe) is a Lewis acid that polarises $\mathrm{Cl_2}$ to generate $\mathrm{Cl^+}$ (electrophilic chlorine), used for aromatic chlorination (e.g., benzene → chlorobenzene). But alkanes do not undergo electrophilic substitution — they lack a π‑electron system to attack. So this reagent is ineffective for butane.

4. Why does the reaction give two products? …

Concept: Free Radical Halogenation of Alkanes

The reaction shown is the chlorination of n-butane to give a mixture of 1-chlorobutane and 2-chlorobutane. This is a classic example of free radical substitution.

Method: Free Radical Substitution (Chain Mechanism)

Steps:

1. Identify the type of reaction:

   An alkane ($\mathrm{CH_3CH_2CH_2CH_3}$) reacting with chlorine to form two isomeric monochlorinated products. This is not an electrophilic substitution (which requires an aromatic ring) or an ionic addition.

2. Recall the conditions for free radical chlorination:

   • Requires UV light or high temperature to break the $\mathrm{Cl-Cl}$ bond homolytically.
   • In darkness, no reaction occurs at room temperature.
   • Presence of iron (Fe) in dark promotes electrophilic chlorination (for aromatic compounds), not alkanes.

3. Analyze each option:

   • (i) $\mathrm{Cl_2}$/UV light → Provides energy for homolytic cleavage → correct for free radical substitution. …

Here are the common mistakes students make on this question, along with how to avoid each.

Mistake 1: Confusing Free Radical Halogenation with Electrophilic Substitution

• The Mistake: Students see $\mathrm{Cl_2}$ and a hydrocarbon and immediately think of the addition reaction (like with alkenes) or electrophilic substitution (like with benzene). They then pick option (iii) or (iv), which are conditions for those other reactions.
• Why it’s Wrong: The reactant is butane ($\mathrm{CH_3CH_2CH_2CH_3}$), a saturated alkane. Alkanes do not undergo addition or electrophilic substitution. The reaction shown is a substitution (one H is replaced by Cl), producing two isomers.
• How to Avoid: Always identify the functional group first.
  • Alkane → Free radical substitution (needs UV light or heat).
  • Alkene/Alkyne → Addition (needs dark, no UV).
  • Benzene → Electrophilic substitution (needs a Lewis acid catalyst like $\mathrm{FeCl_3}$ or iron).

Mistake 2: Ignoring the "UV Light" Requirement for Free Radical Mechanism

• The Mistake: Students pick option (i) correctly but don't understand why UV light is essential. They might think $\mathrm{Cl_2}$ alone is enough.
• Why it’s Wrong: The $\mathrm{Cl-Cl}$ bond is strong. Without UV light (or heat $\approx$ 250°C), the bond does not break homolytically to form chlorine free radicals ($\mathrm{Cl}^\bullet$). No radicals → no chain reaction → no product.
• How to Avoid: Memorize the initiation step of free radical halogenation:

The symbol $h\nu$ (UV light) is non-negotiable for this reaction at room temperature.

Mistake 3: Thinking $\mathrm{NaCl + H_2SO_4}$ Produces $\mathrm{Cl_2}$ for Substitution

• The Mistake: Students recall that $\mathrm{NaCl + H_2SO_4}$ produces $\mathrm{HCl}$ gas (or $\mathrm{Cl_2}$ with an oxidizer). They assume this $\mathrm{Cl_2}$ can then substitute into the alkane.
• Why it’s Wrong: This mixture is used to generate $\mathrm{HCl}$ gas in situ, not free chlorine radicals. Even if some $\mathrm{Cl_2}$ is formed, there is no UV light to initiate the radical chain. This reagent is used for converting alcohols to alkyl chlorides (via $\mathrm{SN1}$ or $\mathrm{SN2}$), not for alkanes.
• How to Avoid: Know the specific use of common reagents.
  • $\mathrm{NaCl + H_2SO_4}$ → For alcohols (to make alkyl halides).
  • $\mathrm{Cl_2}$/UV → For alkanes (free radical substitution).

Mistake 4: Misinterpreting the Role of Iron in Halogenation …