Silver benzoate reacts with bromine to give:

C_{6} H_{6}

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C_{6} H_{5} C O O B r

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m - B r - C_{6} H_{4} - C O O A g

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C_{6} H_{5} B r

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Solution

The correct option is D $C_{6} H_{5} B r$
d is the correct option because it correctly shows the product formed

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Similar questions

o - B r - C_{6} H_{4} - {N H}_{2}

C_{6} H_{6} \to C_{6} H_{5} {C H}_{3}

C_{6} H_{6} + A + A l {C l}_{3} ⟶ C_{6} H_{5} C O {N H}_{2}

I C_{6} H_{5} N H_{2}

II p - N O_{2} C_{6} H_{4} N H_{2}

III m - N O_{2} C_{6} H_{4} N H_{2}

IV m - C H_{3} O C_{6} H_{4} N H_{2}

C_{6} H_{5} {N H}_{2}, C_{6} H_{5} N H M e, C_{6} H_{5} N M e_{2}, o - M e C_{6} H_{4} {N H}_{2}

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