When methylbenzene is treated with bromine in the presence of a catalyst, a mixture of two monobromo isomers is formed. What are the structures of these two isomers?
Correct answer: C. Ortho-bromotoluene and para-bromotoluene
- A. Ortho-bromotoluene and meta-bromotoluene
- B. Meta-bromotoluene and para-bromotoluene
- C. Ortho-bromotoluene and para-bromotoluene
- D. Ortho-bromotoluene and ortho-bromotoluene
- E. Para-bromotoluene and para-bromotoluene
Explanation
The correct answer is Ortho-bromotoluene and para-bromotoluene. The methyl group (-CH3) on the benzene ring acts as an ortho-para directing group, meaning it directs incoming substituents (like bromine in this case) to the ortho (positions 2 and 6) and para (position 4) positions on the ring. Thus, when methylbenzene (toluene) is treated with bromine in the presence of a catalyst, bromine substitutes at these positions, forming ortho-bromotoluene and para-bromotoluene as the two monobromo isomers.Other options are incorrect because they either suggest meta substitution, which is not favored by the -CH3 group, or imply duplication of the same isomer, which does not align with the formation of two distinct isomers.
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