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NEET CHEMISTRYHydrocarbonsMedium

Question

The compound C7H8C_7H_8 undergoes the following reactions: C7H8→3Cl2/ΔA→Br2/FeB→Zn/HClCC_7H_8 \xrightarrow{3Cl_2/\Delta} A \xrightarrow{Br_2/Fe} B \xrightarrow{Zn/HCl} C The product 'C' is:

A

m–Bromotoluene

B

o–Bromotoluene

C

3–Bromo–2,4,6–trichlorotoluene

D

p–Bromotoluene

Step-by-Step Solution

  1. Identify Reactant: The compound C7H8C_7H_8 is Toluene (C6H5CH3C_6H_5CH_3).
  2. Step 1 (Side Chain Halogenation): Reaction with 3Cl23Cl_2 in the presence of heat/light (hνh\nu) is a free radical substitution reaction. All three hydrogen atoms of the methyl group are replaced by chlorine atoms [NCERT 12th, Ch 10, Sec 10.6]. C6H5CH3+3Cl2→hνC6H5CCl3 (A: Benzotrichloride)C_6H_5CH_3 + 3Cl_2 \xrightarrow{h\nu} C_6H_5CCl_3 \text{ (A: Benzotrichloride)}
  3. Step 2 (Electrophilic Aromatic Substitution): Reaction of 'A' with Br2/FeBr_2/Fe is a ring substitution (halogenation). The −CCl3-CCl_3 group is a strong electron-withdrawing group (due to -I effect of three Cl atoms) and acts as a meta-directing group [NCERT 11th, Ch 13, Sec 13.5.6]. C6H5CCl3+Br2→Fem-Bromo-benzotrichloride (B)C_6H_5CCl_3 + Br_2 \xrightarrow{Fe} m\text{-Bromo-benzotrichloride (B)}
  4. Step 3 (Reduction): Reaction with Zn/HClZn/HCl reduces the trichloromethyl group (−CCl3-CCl_3) back to the methyl group (−CH3-CH_3). The aromatic C-Br bond is strong and generally resists this reduction condition. m-Br-C6H4CCl3→Zn/HClm-Br-C6H4CH3 (C: m-Bromotoluene)m\text{-Br-}C_6H_4CCl_3 \xrightarrow{Zn/HCl} m\text{-Br-}C_6H_4CH_3 \text{ (C: m-Bromotoluene)}

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Hydrocarbons. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYHydrocarbonscompoundundergoesfollowingreactionsxrightarrowcldelta

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