Alkyl halide
The reactivity of alkyl halides is influenced by their chemical structure and the reaction conditions, leading to different mechanistic pathways. A chemist was studying the hydrolysis of the different bromo-alkanes (P) CH3Br (Q) (CH3)2CHBr and (R) (CH3)3C-Br. When reacted at a low concentration of NaOH in aqueous acetone medium, the rate of reaction was found to be P > Q > R. However, when the same reaction was carried out with a high conc. NaOH in purely aqueous solution, the order of reactivity reversed, becoming R > Q > P. Furthermore, the hydrolysis of (R) resulted in the formation of a racemic mixture while (P) led to the inversion of configuration.
Methyl chloride, methyl bromide, ethyl chloride and some chlorofluoromethanes are gases at room temperature. Higher members are liquids or solids. Molecules of organic halogen compounds are generally polar. Due to greater polarity as well as higher molecular mass as compared to the parent hydrocarbon, the intermolecular forces of attraction (dipole-dipole and van der Waals) are stronger in the halogen derivatives. This is the reason, the boiling points of chlorides, bromides and iodides are considerably higher than those of the hydrocarbons of comparable molecular mass.
The attractions get stronger as the molecules get bigger in size and have more electrons. For the same alkyl group, the boiling points of alkyl halides decrease in the order: RI > RBr > RCl > RF. This is because with the increase in size and mass of halogen atom, the magnitude of van der Waals forces increases.
Which of the following will have highest boiling point?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → CH3CH2CH2CH2CH2Br
The boiling points of alkyl halides depend on the strength of the van der Waals forces between molecules. For the same halogen, these forces grow as the molecule gets bigger (more carbon atoms, more electrons) and weaken with branching, which gives a smaller surface area.
Options (1), (2) and (3) are \( \ce{C4} \) bromides; Option (4) is a \( \ce{C5} \) bromide and has the largest molecules:
1-bromopentane > 1-bromobutane > 2-bromobutane > 2-bromo-2-methylpropane
So the highest boiling point is that of CH3CH2CH2CH2CH2Br (1-bromopentane).
Why the other options are incorrect
Option (1) 2-bromo-2-methylpropane — Incorrect. The most branched \( \ce{C4} \) isomer (3°), nearly spherical; it has the lowest boiling point of the four.
Option (2) CH3CH2CH2CH2Br — Incorrect. The straight-chain \( \ce{C4} \) bromide; highest of the \( \ce{C4} \) isomers, but it has one carbon fewer than 1-bromopentane.
Option (3) 2-bromobutane — Incorrect. A branched (2°) \( \ce{C4} \) isomer, so it boils below 1-bromobutane.
Note: It seems that there is a mistake in the original NTA paper, which says that the hydrolysis of (R) (CH3)3C-Br gave a racemic mixture and that (P) CH3Br showed inversion of configuration; neither compound is chiral, so neither observation can be made on them.
NCERT (2026-27 edition):
Physical Properties – …
