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The compound C 7 _7 7 ā H 8 _8 8 ā undergoes the following reactions: C 7 H 8 ā 3 C l 2 / Ī A ā B r 2 / F e B ā Z n / H C l C \newline C_7H_8 \xrightarrow{3 Cl_2 / \Delta} A \xrightarrow{Br_2/Fe} B \xrightarrow{Zn/HCl} C \newline C 7 ā H 8 ā 3 C l 2 ā /Ī ā A B r 2 ā / F e ā B Z n / H Cl ā C The compound C is
3-bromo-2,4,6-trichlorotoluene
Explanation To solve this problem, we need to determine the structure of compound C. Let's analyze the reactions step-by-step. \\
1. Starting compound: C 7 H 8 C_7H_8 C 7 ā H 8 ā is toluene, which is C 6 H 5 C H 3 . C_6H_5CH_3. \\
C 6 ā H 5 ā C H 3 ā .
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2. Reaction with
3 Ā C l 2 / Ī : 3 \ Cl_2 / \Delta: \\
3 Ā C l 2 ā /Ī : ā¢ Toluene undergoes free radical chlorination at the methyl group.
ā¢ The methyl group
is converted to a trichloromethyl group
( C C l 3 ) . (CCl_3). \\
( CC l 3 ā ) . ā¢ Therefore, compound A is
C 6 H 5 C C l 3 C_6H_5CCl_3 C 6 ā H 5 ā CC l 3 ā (benzotrichloride).
3. Reaction with
B r 2 / F e : Br_2/Fe: \\
B r 2 ā / F e : ā¢ Benzotrichloride undergoes electrophilic aromatic substitution.
ā¢ The bromine substitutes at the meta position relative to the
group,
because the
group is a strong electron-withdrawing group,
directing the incoming electrophile to the meta position.
ā¢ Therefore, compound B is
-bromo-benzotrichloride
(3-bromo-1,1,1-trichlorotoluene).
4. Reaction with
Z n / H C l : Zn/HCl: \\
Z n / H Cl : ā¢ The
group is reduced to a methyl group
by
Z n / H C l . Zn/HCl. \\
Z n / H Cl . ā¢ This reduction converts the trichloromethyl group back to a methyl group.
ā¢ Therefore, compound C is
-bromotoluene
Thus, the compound C is
-bromotoluene, \ which corresponds to Option 3.