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What will be the product of this reaction?

\mathrm{C}_{3} \mathrm{F}_{7} \mathrm{CONH}_{2} \xrightarrow[{\mathrm{NaOH}, \Delta}]{{\mathrm{Br}_{2}}}

Option: 1

\mathrm{O=C=N-C_3F_7}


Option: 2

\mathrm{O=C(OH)-C_3F_7}


Option: 3

\mathrm{C_3F_7-Br}


Option: 4

None of these.


Answers (1)

best_answer

 The reaction occurs as follows:

\mathrm{C}_{3} \mathrm{F}_{7} \mathrm{CONH}_{2} \xrightarrow[{\mathrm{NaOH}, \Delta}]{{\mathrm{Br}_{2}}} \mathrm{C}_{3} \mathrm{F}_{7} \mathrm{Br}

If the amide contains an electron-withdrawing group (-I), the product is bromide.

Therefore the correct option is (3).

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mansi

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