Q.(a) Arrange the following compounds in order of their increasing boiling points (1 mark): Bromomethane, Bromoform, Chloromethane, Dibromomethane

Boiling point of haloalkanes increases with molecular mass/surface area and decreases with branching among isomers, fixing the order among these three compounds.

• 1-chloropropane, $CH_3CH_2CH_2Cl$ ($C_3H_7Cl$): a straight chain, $\text{bp}\approx46.6\,^{\circ}C$.
• Isopropyl chloride (2-chloropropane), $(CH_3)_2CHCl$ ($C_3H_7Cl$): the same molecular formula as 1-chloropropane but branched — the more compact, nearly spherical shape has less surface area available for van der Waals (London dispersion) contact with neighbouring molecules, so it boils lower, $\text{bp}\approx35.7\,^{\circ}C$. …

Boiling points of alkyl halides depend primarily on molecular mass and van der Waals forces, not bond polarity. n-Butyl bromide (heavier) boils higher than n-Butyl chloride, making the assertion false; the C–Cl bond is more polar than C–Br, making the reason true.

The boiling point trend RI > RBr > RCl > RF is driven by increasing molecular size and polarizability, which strengthen van der Waals forces. The Reason correctly identifies this cause, so both statements are true and the Reason is the correct explanation.

For the same alkyl group R, boiling point of alkyl halides increases as the halogen gets heavier/bigger: RF < RCl < RBr < RI.

Boiling point of a covalent molecular compound depends mainly on the strength of the intermolecular (van der Waals/London dispersion) forces, which in turn depend on molecular mass and polarisability of the atoms. Going down the halogen group F → Cl → Br → I, both atomic size and the number of electrons increase sharply, making the electron cloud more polarisable — this produces much stronger London dispersion forces between molecules, and also increases molecular mass, both of which raise the boiling point …

Among isomeric haloalkanes, boiling point decreases as branching increases, because branching reduces the molecular surface area available for van der Waals (London dispersion) contact between molecules.

The three isomers (all C4H9Br):

1. isobutyl bromide, (CH3)2CH-CH2-Br - one branch
2. n-butyl bromide, CH3-CH2-CH2-CH2-Br - straight chain, no branching
3. tert-butyl bromide, (CH3)3C-Br - most branched (three methyl groups on the C-Br carbon) …

Both statements are individually correct chemistry facts, but [R] compares halides to hydrocarbons, while [A] is about comparing halides to EACH OTHER — so [R] doesn't actually explain the trend stated in [A].

[A] For a given alkyl group, boiling point order $RI > RBr > RCl > RF$ is TRUE — boiling point rises with the size/polarizability of the halogen (van der Waals forces increase down the halogen group despite similar dipole moments), and with increasing molar mass.

[R] Alkyl halides (Cl, Br, I) do have higher boiling points than hydrocarbons of comparable molecular mass — TRUE, because of their greater polarizability and the presence of a permanent C–X dipole (dipole-dipole interactions) in addition to van der Waals forces present in hydrocarbons.

…

Among alkyl halides, boiling point rises with molecular mass and with the number of heavier halogen atoms present, since both increase the strength of van der Waals (London dispersion) forces between molecules.

Reasoning from structure to boiling point

All four compounds are small, only weakly polar haloalkanes, so intermolecular attraction is dominated by van der Waals (dispersion) forces, which grow stronger as molecular size/mass and the number of polarisable halogen atoms increase:

• Chloromethane, $CH_3Cl$ (M $\approx$ 50.5 g/mol) — smallest, only one Cl atom $\to$ weakest van der Waals forces $\to$ lowest boiling point ($\approx$ –24 °C).
• Bromomethane, $CH_3Br$ (M $\approx$ 95 g/mol) — one Br atom, heavier and more polarisable than Cl $\to$ stronger forces than chloromethane (bp $\approx$ 3.6 °C). …

Boiling point of methyl halides rises with the size/polarizability of the halogen, not with dipole moment.

Although C–F and C–Cl bonds are more polar, boiling point of haloalkanes is governed mainly by the strength of van der Waals (London dispersion) forces, which increase with the size and polarizability of the halogen atom and with the molecular mass. Order of at …

Boiling points of alkyl halides increase with halogen size (R-I > R-Br > R-Cl > R-F) because larger halogens create stronger van der Waals forces, not weaker ones. The assertion is true but the reason contradicts it. The correct option is (C).

Among structural isomers, the most branched (most spherical) one has the least surface area for intermolecular contact and hence the lowest boiling point.

$C_4H_9Cl$ has four structural isomers: 1-chlorobutane ($n$-butyl chloride, straight chain), 2-chlorobutane ($sec$-butyl chloride), 1-chloro-2-methylpropane (isobutyl chloride), and 2-chloro-2-methylpropane ($tert$-butyl chloride, most branched). Boiling point falls as branching increases, because a more branched (more compact/spherical) molecule has less surface area available for van der Waals contact with neighbouring molecules than an elongated stra …

Boiling point depends on the strength of van der Waals forces, which in turn depend on the surface area available for intermolecular contact; branching reduces that surface area, so more-branched isomers boil at lower temperatures.

For isomeric haloalkanes (same molecular formula, same molar mass), the dominant intermolecular force is the van der Waals (London dispersion) force, whose strength depends heavily on the total surface area of contact between neighbouring molecules — the larger the contact area, the stronger the net attractive force, and the higher the boiling point.

• A straight-chain (unbranched) haloalkane has an elongated shape that allows extensive, close, side-by-side surface contact between neighbouring molecules — maximising van der Waals attraction.
• As branching increases, the molecule becomes more compact and closer to a spherical shape. A sphere has the minimum possible surface area for a given volume, so branched molecules present much less surface area for intermolecular contact, and can also approach each other less closely due to the bulky branches getting in the way. …