A) \[C{{H}_{3}}CHBr{{(C{{H}_{2}})}_{8}}COOH\]
B) \[C{{H}_{2}}=CH{{(C{{H}_{2}})}_{8}}COBr\]
C) \[C{{H}_{2}}BrC{{H}_{2}}{{(C{{H}_{2}})}_{8}}COOH\]
D) \[C{{H}_{2}}=CH{{(C{{H}_{2}})}_{7}}CHBrCOOH\]
Correct Answer: C
Solution :
[c] Anti Markovnikov's addition
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