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.
List I | List II | ||
|---|---|---|---|
| A | \(\Omega^{-1}\) | I | Specific conductance |
| B | \(∧\) | II | Electrical conductance |
| C | k | III | Specific resistance |
| D | \(\rho\) | IV | Equivalent conductance |
List I | List II | ||
|---|---|---|---|
| A | Constant heat (q = 0) | I | Isothermal |
| B | Reversible process at constant temperature (dT = 0) | II | Isometric |
| C | Constant volume (dV = 0) | III | Adiabatic |
| D | Constant pressure (dP = 0) | IV | Isobar |