Animated Solution for Chemistry - Organic Chemistry: The increasing order of the boiling points of the major products A, B and C of the following reactions will be

We start by identifying the major products of the three given reactions.

Reaction I involves the addition of $\text{HBr}$ to 1-hexene in the presence of benzoyl peroxide. The peroxide initiates a free radical mechanism, leading to an anti-Markovnikov addition. The bromine atom attaches to the less substituted terminal carbon, yielding 1-bromohexane (Product A).

Reaction II is the addition of $\text{HBr}$ to 1-pentene without any peroxide. This proceeds via a standard electrophilic addition, following Markovnikov's rule. The intermediate is a stable secondary carbocation, resulting in the formation of 2-bromopentane (Product B).

Reaction III features 2-pentene reacting with $\text{HBr}$. Protonation of the double bond can form a secondary carbocation at either C2 or C3. Both lead to secondary alkyl halides, but for the purpose of comparing distinct isomers, we identify the product as 3-bromopentane (Product C).

Now that we have our products, we need to compare their boiling points. The primary factor determining the boiling point of organic compounds is the strength of the intermolecular van der Waals forces.

These forces increase with the size of the electron cloud, which is directly proportional to the molecular mass. Product A (1-bromohexane) contains 6 carbon atoms, whereas Products B and C contain only 5 carbon atoms.

Because of its larger molecular mass and greater surface area, 1-bromohexane will have the strongest intermolecular forces. Therefore, Product A has the highest boiling point.

The real challenge lies in comparing Product B (2-bromopentane) and Product C (3-bromopentane). Both are isomers with the exact same molecular mass.

For isomeric alkyl halides, the boiling point decreases as branching increases. Branching makes the molecule more compact and spherical. A spherical shape minimizes the surface area available for intermolecular contact, thereby weakening the van der Waals forces.

When we compare 2-bromopentane and 3-bromopentane, 2-bromopentane is effectively more sterically hindered or "branched" at the end of the chain, making it slightly more compact than the more symmetrical 3-bromopentane.

Because 2-bromopentane has a smaller effective surface area, its intermolecular forces are slightly weaker. Consequently, the boiling point of 2-bromopentane (B) is lower than that of 3-bromopentane (C).

Putting it all together, the boiling point of B is less than C, and both are significantly less than A.

This corresponds to the reaction order:

Thus, the correct option is (a).