The question revolves around the concept of halogenation in alkanes. In the example provided:
\(\mathrm{R-H} + \mathrm{Br}^- \longrightarrow \mathrm{R}^- + \mathrm{Br-H}\)
This reaction is essentially a free radical substitution reaction. The halogenation process, specifically with bromine, exhibits low reactivity and high selectivity. Let's break down these terms:
The option "Low reactivity and high selectivity" explains the nature of bromination (using Br2) in alkanes. This results in low activation energy requirement and the formation of the most stable radical intermediate, thus giving highly selective product formation. Therefore, the correct answer is Low reactivity and high selectivity.
Ruling out other options:
Hence, the bromination of alkanes is characterized as having low reactivity and high selectivity.
Given below are the four isomeric compounds \(P, Q, R, S\): 
\(P\): Aromatic compound containing an \(-\mathrm{OH}\) group
\(Q\): Aromatic compound containing an \(-\mathrm{CHO}\) group (aldehyde)
\(R\): Aromatic compound containing a ketone group
\(S\): Aromatic compound containing a ketone group Identify the correct statements from below:
[A.] \(Q, R\) and \(S\) will give precipitate with \(2,4\)-DNP.
[B.] \(P\) and \(Q\) will give positive Baeyer’s test.
[C.] \(Q\) and \(R\) will give sooty flame.
[D.] \(R\) and \(S\) will give yellow precipitate with \(I_2/\mathrm{NaOH}\).
[E.] \(Q\) alone will deposit silver with Tollens’ reagent. Choose the correct option.
Match the LIST-I with LIST-II 
Choose the correct answer from the options given below:

Match the following:
(P) Schedule H
(Q) Schedule G
(R) Schedule P
(S) Schedule F2
Descriptions:
(I) Life period of drugs
(II) Drugs used under RMP
(III) List of Prescription Drugs
(IV) Standards for surgical dressing