This question involves free radical halogenation of alkanes, specifically focusing on the selectivity of bromination versus chlorination and the stability of free radicals.

The given reaction is: ${CH}_{3} {CH}_{2} {CH}_{3} \underset{hν}{\overset{{Cl}_{2}}{\to}} A \underset{hν}{\overset{{Br}_{2}}{\to}} B$

Step 1: Chlorination of Propane (Formation of A)

Propane ( ${CH}_{3} {CH}_{2} {CH}_{3}$ ) has two types of hydrogen atoms:

6 primary hydrogens (on the two methyl groups, $- {CH}_{3}$ )
2 secondary hydrogens (on the methylene group, $- {CH}_{2} -$ )

Chlorination is a relatively unselective free radical substitution reaction. The product distribution is governed by the probability factor (number of each type of H) and the slight difference in reactivity. The reactivity ratio for chlorination is approximately: 1° H : 2° H = 1 : 3.5.

Therefore, the major product of chlorination will be 1-chloropropane ( ${CH}_{3} {CH}_{2} {CH}_{2} Cl$ ) because there are many more primary hydrogens. However, a significant amount of 2-chloropropane ( $({CH}_{3}) 2 CHCl$ ) is also formed. For the purpose of this sequential reaction, we must consider that A is the mixture of alkyl chlorides formed. The next step (bromination) will act on this mixture.

Step 2: Bromination of the Alkyl Chloride Mixture (Formation of B)

Bromination is a highly selective free radical substitution reaction. The reactivity ratio for bromination is approximately: 1° H : 2° H : 3° H = 1 : 80 : 1600. It strongly prefers to substitute the hydrogen on the carbon atom that will form the most stable free radical.

Let's analyze the bromination of each component in mixture A:

1-Chloropropane ( ${CH}_{3} {CH}_{2} {CH}_{2} Cl$ ): This molecule has three types of hydrogens:
1. Hydrogens on the $- {CH}_{2} Cl$ group (2° but deactivated by the electron-withdrawing Cl atom).
2. Hydrogens on the $- {CH}_{2} -$ group (2°).
3. Hydrogens on the terminal $- {CH}_{3}$ group (1°).
Bromination will occur almost exclusively on the carbon adjacent to the chlorine, the -CH2- group, to form ClCH2CH2CH2Br. The radical formed here is a 2° radical, and while the Cl atom is electron-withdrawing, the stability of the secondary radical outweighs the deactivating effect in this case for bromination's high selectivity.
2-Chloropropane ( $({CH}_{3}) 2 CHCl$ ): This molecule has only one type of hydrogen: nine equivalent 1° hydrogens on the methyl groups. Bromination will substitute one of these to form ${BrCH}_{2} CH (Cl) {CH}_{3}$ . The radical site is primary, but it is adjacent to the chlorine-substituted carbon. The electron-withdrawing chlorine atom stabilizes the forming primary radical through a phenomenon called alpha-halogen stabilization, making this substitution favorable.

Looking at the options, the correct pair A and B must show this progression. The first molecule in the sequence (A) is an alkyl chloride, and the second (B) is a dibromo compound derived from it.

Final Answer: The correct option is the one where:
A is ${CH}_{3} {CH}_{2} {CH}_{2} Cl$ (1-chloropropane) and
B is ${BrCH}_{2} {CH}_{2} {CH}_{2} Br$ (1,3-dibromopropane).
This corresponds to the first image option.

Related Topics & Formulae

Free Radical Halogenation: A reaction where a halogen ( $Cl 2$ or $Br 2$ ) replaces a hydrogen atom in an alkane under UV light ( $hν$ ) initiation. The mechanism involves initiation, propagation, and termination steps.

Reactivity and Selectivity:
The relative rate of abstraction for different hydrogens is given by:
$Relative Rate = Number of H \times Reactivity$
Chlorination reactivity: 1° : 2° : 3° ≈ 1 : 3.5 : 5
Bromination reactivity: 1° : 2° : 3° ≈ 1 : 80 : 1600

Radical Stability: The stability of alkyl radicals follows the order: Tertiary (3°) > Secondary (2°) > Primary (1°) > Methyl. More substituted radicals are more stable due to hyperconjugation and inductive effects.