when we react benzene with excess cl2 / hv why do we get lindane and not benzenering with cl in every corner

Key Concepts Involved

1. Reaction Conditions Control Mechanism
   • Electrophilic Substitution (EAS) → Requires a Lewis acid such as FeCl₃/AlCl₃, no light. Keeps aromaticity.
   • Free-Radical Addition → Needs radical initiators: hν, heat, peroxides. Breaks aromaticity and adds across C=C.

2. Why Does Light Promote Radical Path?
   • hν splits Cl₂ → 2 Cl· (homolysis).
   • Cl· is a very reactive electrophilic radical. It will attack a π-bond to form a new σ-bond, giving a cyclohexadienyl radical. Repetition gives HCH.

3. Energy Considerations
   • Destroying aromaticity costs ≈151 kJ mol⁻¹ (benzene resonance energy).
   • Each C–Cl bond formed releases ≈327 kJ mol⁻¹, and each H–Cl bond ≈431 kJ mol⁻¹. The overall heat released by six additions more than compensates for the loss of aromaticity, so the pathway becomes favourable under radical conditions.

4. Why Not Hexachlorobenzene?
   • Formation of hexachlorobenzene needs six consecutive EAS steps. Radical conditions do not generate the electrophile Cl⁺ required for EAS; they generate Cl· instead.
   • In the absence of FeCl₃, the Cl₂ molecule is not activated for EAS.
   • Therefore radical addition dominates, giving HCH (lindane) rather than C₆Cl₆.

Logical Chain A Student Should Follow

1. Identify Reagents → Cl₂ + hν → radicals.
2. Recall Mechanism → radicals mean addition to double bonds or abstraction from alkanes.
3. Apply To Benzene → radical adds, breaks aromaticity, stepwise addition gives HCH.
4. Contrast With FeCl₃ Case → electrophile, substitution, hexachlorobenzene possible.
5. Conclude → HCH (lindane) is formed, not fully substituted benzene.

Imagine Benzene As A Strongly Glued Wheel

• Think of benzene as a strongly glued 6-spoke wheel. The glue (called aromaticity) makes the wheel very hard to open up.
• When you add Cl₂ with help of FeCl₃ (no light) you just paste one Cl on each spoke and keep the wheel intact. That is called substitution.
• But when you shine bright light (hν) on Cl₂, you break Cl₂ into angry single-Cl radicals. These radicals don’t politely replace hydrogen; they attack the wheel itself, prying open each double bond and snapping it into an ordinary 6-corner ring (cyclohexane type).
• Finally every corner gets one Cl and one H → you obtain hexachlorocyclohexane (HCH). One of its forms is called lindane.
• So, with light, the chemistry route is different – the radicals do addition (open the wheel) instead of substitution. Therefore you don’t get a flat Cl-decorated wheel (hexachlorobenzene); you get a puckered chair-like ring with Cl’s all around – lindane.

Complete Step-By-Step Solution

1. Reagents & Conditions
   $\text{Benzene} + \text{excess } Cl_2 \xrightarrow{h\nu} \text{?}$

2. Initiation (light)
   $Cl_2 \xrightarrow{h\nu} 2\,Cl\cdot$

3. Propagation (first addition)
   $Cl\cdot + C_6H_6 \;\to\; C_6H_6Cl\cdot$ (radical adds to one C of the ring)
   $C_6H_6Cl\cdot + Cl_2 \;\to\; C_6H_6Cl\_2 + Cl\cdot$

4. Repeat Propagation
   Continue the above two steps five more times; each turn adds one Cl and generates another Cl·, until we have:
   $C_6H_6 + 6\,Cl_2 \;\to\; C_6H_6Cl_6$

5. Product Structure
   • The resulting product is hexachlorocyclohexane (HCH).
   • Multiple stereoisomers form; the γ-isomer is called lindane.

6. Why Hexachlorobenzene (C₆Cl₆) Is Not Formed
   • C₆Cl₆ would require six electrophilic substitutions with Cl⁺.
   • No FeCl₃/AlCl₃ present → Cl⁺ not generated.
   • Radical mechanism is faster under these conditions and dominates.

7. Final Answer
   $\boxed{\text{Hexachlorocyclohexane (lindane) is obtained, not hexachlorobenzene}.}$