A) 3-chloropentane
B) 2-chloropentane
C) 1,2-dichloropentane
D) 1-chloropentane
Correct Answer: B
Solution :
The addition of\[HCl=+92\text{ }kJ\]to pent-1-ene takes place according to Markownikoffs rule, i.e., the negative part of the reagent will be attached to the carbon atom which contains lesser number of H-atom. \[\underset{pent-1-ene}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}CH=C{{H}_{2}}+HCl}}\,\]\[\xrightarrow[{}]{{}}\underset{2-chloropentane}{\mathop{C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}-\underset{\begin{smallmatrix} | \\ Cl \end{smallmatrix}}{\mathop{CH}}\,-C{{H}_{3}}}}\,\]
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