A)
B)
C) \[C{{H}_{3}}-CH=C{{H}_{2}}\xrightarrow{NBS}C{{H}_{2}}=CH-C{{H}_{2}}-Br\]
D) \[C{{H}_{3}}CHO+N{{H}_{2}}OH\xrightarrow{{{H}^{+}}}C{{H}_{3}}-CH=N-OH\]
Correct Answer: C
Solution :
NBS form allylic free radical.
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