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JEE CHEMISTRY MCQ: HYDROCARBONS

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Application +4 | -1
The principal organic product formed in the reaction,

CH2=CH(CH2)8COOH+HBr→Peroxide\mathrm{C}{\mathrm{H}_2} = \mathrm{CH}{\left( {\mathrm{C}{\mathrm{H}_2}} \right)_8}\mathrm{COOH} + \mathrm{HBr}\xrightarrow{{\mathrm{Peroxide}}}…..is:
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Correct option: C
CH2BrCH2(CH2)8COOH\mathrm{C}{\mathrm{H}_2}\mathrm{BrC}{\mathrm{H}_2}{\left( {\mathrm{C}{\mathrm{H}_2}} \right)_8}\mathrm{COOH}
The correct answer is C because in the presence of peroxide, HBr adds to the alkene via anti-Markovnikov's rule.

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