Q.Arrange each set of compounds in order of increasing boiling points.

Concept: Boiling Point Trends — Boiling point depends on molecular mass (higher mass → stronger London forces) and branching (more branching → lower surface area → weaker forces).

Reasoning:

1. Set (i): All are haloalkanes. Boiling point increases with molecular mass.

   • Chloromethane ($\ce{CH3Cl}$, $M \approx 50.5$) < Bromomethane ($\ce{CH3Br}$, $M \approx 95$) < Dibromomethane ($\ce{CH2Br2}$, $M \approx 174$) < Bromoform ($\ce{CHBr3}$, $M \approx 253$).

2. Set (ii): Same molecular formula ($\ce{C3H7Cl}$) for 1-chloropropane and isopropyl chloride; 1-chlorobutane ($\ce{C4H9Cl}$) has higher mass. …

The Core Idea: What Controls Boiling Point in Haloalkanes?

Boiling point is the temperature at which the vapour pressure of a liquid equals the surrounding pressure. To boil, molecules must overcome the intermolecular forces holding them together. In haloalkanes, the dominant forces are van der Waals forces (London dispersion forces) and, to a lesser extent, dipole-dipole interactions from the polar C–X bond.

The key factors are:

1. Molecular mass and size: Heavier, larger molecules have more electrons, leading to stronger temporary dipoles and greater dispersion forces. This usually raises the boiling point.
2. Surface area and branching: A straight chain has a larger surface area for contact than a branched isomer. More branching means a more compact shape, weaker intermolecular forces, and a lower boiling point.
3. Halogen identity: For the same alkyl group, boiling point increases with the size and polarizability of the halogen: $I > Br > Cl > F$. A larger halogen atom contributes more to dispersion forces.

Now, let's apply these principles to each set.

(i) Bromomethane, Bromoform, Chloromethane, Dibromomethane

Step 1: Identify the compounds and their molecular masses.

We have four methane derivatives. Let's list them with their molecular formulas and approximate molar masses.

Step 2: Apply the mass-size rule.

The most immediate trend is the increase in molecular mass. As we go from one halogen to two, then three, the molecule gets much heavier and larger. This dramatically increases the strength of London dispersion forces.

Therefore, the boiling points should increase with the number of halogen atoms (and thus mass). The lightest is Chloromethane, then Bromomethane, then Dibromomethane, and the heaviest is Bromoform.

Step 3: Compare Chloromethane and Bromomethane.

Both are monohalomethanes. Bromomethane ($CH_3Br$) has a bromine atom (atomic mass 80), while Chloromethane ($CH_3Cl$) has a chlorine atom (atomic mass 35.5). Bromomethane is nearly twice as heavy. Its larger electron cloud makes it much more polarizable, leading to stronger dispersion forces. So, Bromomethane has a higher boiling point than Chloromethane.

Step 4: Assemble the order.

Combining the mass trend, the order of increasing boiling points is:

Chloromethane < Bromomethane < Dibromomethane < Bromoform

(ii) 1-Chloropropane, Isopropyl chloride, 1-Chlorobutane

Step 1: Identify the compounds and their structures.

All three are monochlorinated alkanes. Let's draw their structures.

• 1-Chloropropane ($n$-propyl chloride): $CH_3-CH_2-CH_2-Cl$ (a straight chain of 3 carbons).
• Isopropyl chloride (2-Chloropropane): $CH_3-CH(Cl)-CH_3$ (a branched chain of 3 carbons; the chlorine is on the middle carbon).
• 1-Chlorobutane ($n$-butyl chloride): $CH_3-CH_2-CH_2-CH_2-Cl$ (a straight chain of 4 carbons). …

Method: Intermolecular Force Analysis (Van der Waals & Dipole-Dipole)

This method uses two key factors that determine boiling point:

• Molecular mass (higher mass → stronger London dispersion forces → higher boiling point)
• Molecular shape (more branching → lower surface area → weaker dispersion forces → lower boiling point)

(i) Bromomethane, Bromoform, Chloromethane, Dibromomethane

Step 1: Identify molecular masses

• Chloromethane ($\text{CH}_3\text{Cl}$): $M \approx 50.5$
• Bromomethane ($\text{CH}_3\text{Br}$): $M \approx 95$
• Dibromomethane ($\text{CH}_2\text{Br}_2$): $M \approx 174$
• Bromoform ($\text{CHBr}_3$): $M \approx 253$

Step 2: Arrange by increasing mass

Chloromethane < Bromomethane < Dibromomethane < Bromoform

Step 3: Check for shape effects — all are small, compact molecules with no branching, so mass dominates.

Final order (increasing boiling point):

Chloromethane < Bromomethane < Dibromomethane < Bromoform

(ii) 1-Chloropropane, Isopropyl chloride, 1-Chlorobutane

Step 1: Identify molecular masses

• 1-Chloropropane ($\text{C}_3\text{H}_7\text{Cl}$): $M \approx 78.5$
• Isopropyl chloride ($\text{C}_3\text{H}_7\text{Cl}$): $M \approx 78.5$ (same formula!)
• 1-Chlorobutane ($\text{C}_4\text{H}_9\text{Cl}$): $M \approx 92.5$

Step 2: Compare by mass first …

Here are the common mistakes students make when solving boiling point trend questions, specifically for the two sets you provided, along with how to avoid each.

General Mistake #1: Confusing Boiling Point with Melting Point or Solubility

The Mistake: Students often think "more polar = always higher boiling point" without considering molecular mass or surface area. They might rank Chloromethane above Bromomethane because C–Cl is more polar than C–Br.

Why it’s wrong: Boiling point depends on strength of intermolecular forces (van der Waals forces + dipole-dipole). For similar molecules, molecular mass (and thus number of electrons) is the dominant factor. Heavier molecules have stronger London dispersion forces.

How to avoid: Always check molecular mass first. For haloalkanes with the same carbon skeleton, heavier halogen = higher boiling point.

• Correct order for (i) part: Chloromethane ($CH_3Cl$, ~50.5 g/mol) < Bromomethane ($CH_3Br$, ~95 g/mol) < Dibromomethane ($CH_2Br_2$, ~174 g/mol) < Bromoform ($CHBr_3$, ~253 g/mol)

Mistake #2: Ignoring the Effect of Branching (Surface Area)

The Mistake: In set (ii), students rank 1-Chloropropane and Isopropyl chloride based only on molecular mass (same formula $C_3H_7Cl$) and guess they are equal, or they rank Isopropyl chloride higher because it has more branching (thinking branching increases boiling point).

Why it’s wrong: Branching decreases surface area, which weakens London dispersion forces. More branching = lower boiling point for isomers.

How to avoid: Draw the structures. A straight chain has more surface contact between molecules than a branched one.

• Correct order for (ii): Isopropyl chloride (branched) < 1-Chloropropane (straight chain) < 1-Chlorobutane (longer chain, higher mass)

Mistake #3: Forgetting to Compare Chain Length First

The Mistake: In set (ii), students compare 1-Chloropropane and 1-Chlorobutane only by the halogen, ignoring that 1-Chlorobutane has a longer carbon chain.

Why it’s wrong: For the same functional group, longer carbon chain = larger surface area + higher molecular mass = stronger van der Waals forces = higher boiling point.

How to avoid: Always compare carbon chain length before looking at branching or halogen type.

Mistake #4: Misapplying the "More Halogen Atoms = Higher BP" Rule

The Mistake: In set (i), students think that because Bromoform has three Br atoms, it must have the highest boiling point — which is correct — but they then incorrectly place Dibromomethane below Bromomethane because "two Br atoms are less than three."

Why it’s wrong: The trend is monotonic: more halogen atoms (especially heavier ones) increase molecular mass and polarizability, raising boiling point. But you must compare each compound individually.

How to avoid: List molecular masses explicitly: