The number of chlorine atoms present in the organic products X and Y of the following reactions, respectively, are :
Shows benzene undergoing two distinct halogenation reactions to produce X and Y.
A.3 and 6
B.6 and 6
C.6 and 3
D.3 and 3
Solution & Explanation
Core Logic
Reaction 1 (Formation of X): Electrophilic aromatic substitution.
Benzene + 6Cl₂ Anhydr. AlCl₃, dark, cold C₆Cl₆ + 6HCl$+ 6Cl_2 \xrightarrow{\text{Anhydr. } AlCl_3, \text{dark, cold}} C_6Cl_6 + 6HCl$
Excess chlorine in the presence of a Lewis acid substitutes all six hydrogen atoms on the benzene ring to form Hexachlorobenzene (C₆Cl₆$C_6Cl_6$).
Shows benzene undergoing two distinct halogenation reactions to produce X and Y.
Number of chlorine atoms in X = 6.
Reaction 2 (Formation of Y): Free radical addition.
Benzene + 3Cl₂ UV, 500K C₆H₆Cl₆$+ 3Cl_2 \xrightarrow{UV, 500K} C_6H_6Cl_6$
Chlorine adds across all three double bonds in the presence of UV light to form Benzene hexachloride (BHC), also known as Gammexane.
Shows benzene undergoing two distinct halogenation reactions to produce X and Y.
Number of chlorine atoms in Y = 6.
Step 1: Final Conclusion
Both products X (hexachlorobenzene) and Y (benzene hexachloride) contain exactly 6 chlorine atoms.
Pattern Recognition
Lewis acid (AlCl₃$AlCl_3$) = Substitution (retains aromaticity). UV light/Heat = Addition (destroys aromaticity). In both exhaustive cases, 6 chlorines end up on the carbon ring.
Chapter Mix
Class 11 Chemistry: Hydrocarbons
Class 12 Chemistry: Haloalkanes and Haloarenes
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
