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Match List-I with List-II and select the correct answer using the codes given below the lists:
\begin{tabular}{|l|l|l|l|}
\hline \multicolumn{2}{|c|}{\begin{tabular}{c}
List I \\
(Reaction)
\end{tabular}} & \multicolumn{2}{c|}{\begin{tabular}{c}
Column II \\
(Reagents)
\end{tabular}} \\
\hline (A) & \( \mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2} \rightarrow \mathrm{CH}_{3}-\mathrm{CHBr}^{-} \mathrm{CH}_{3} \) & (P) & \( \mathrm{HBr} \) \\
\hline (B) & \( \mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2} \rightarrow \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2} \mathrm{Br} \) & (Q) & \( \mathrm{Br} / \mathrm{CCl}_{4} \) \\
\hline (C) & \( \mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2} \rightarrow \mathrm{BrCH}_{2}-\mathrm{CH}=\mathrm{CH}_{2} \) & (R) & \( \mathrm{HBr} / \mathrm{Peroxide} \) \\
\hline (D) & \( \mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2} \rightarrow \mathrm{CH}_{3}-\mathrm{CHBr}-\mathrm{CH}_{2} \mathrm{Br} \) & (S) & \( \mathrm{NBS} \) \\
\hline
\end{tabular}
(1) A-(R); B-(S); C-(Q); D-(P)
(2) \( \mathrm{A}-(\mathrm{S}) ; \mathrm{B}-(\mathrm{P}) ; \mathrm{C}-(\mathrm{R}) ; \mathrm{D}-(\mathrm{Q}) \)
(3) \( \mathrm{A}-(\mathrm{P}) ; \mathrm{B}-(\mathrm{R}) ; \mathrm{C}-(\mathrm{S}) ; \mathrm{D}-(\mathrm{Q}) \)
(4) \( \mathrm{A}-(\mathrm{S}) ; \mathrm{B}-(\mathrm{Q}) ; \mathrm{C}-(\mathrm{P}) ; \mathrm{D}-(\mathrm{R}) \)


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