Q.Arrange the following compounds in increasing order of boiling point.
Propan-1-ol, butan-1-ol, butan-2-ol, pentan-1-ol

Concept

Alkanes are held together by London dispersion forces, which strengthen with (i) more electrons — longer chains — and (ii) larger surface contact — straighter chains. So for a fixed carbon count, more branching lowers the boiling point because the molecule becomes more compact.

Solution

Identify each compound:

• D — n-butane: $\mathrm{C_4}$, straight; fewest electrons $\Rightarrow$ lowest boiling point overall ($\approx -0.5^\circ\mathrm{C}$).
• A — 2,2-dimethylpropane (neopentane): $\mathrm{C_5}$, most branched $\Rightarrow$ lowest among the $\mathrm{C_5}$'s ($\approx 9.5^\circ\mathrm{C}$).
• B — 2-methylbutane (isopentane): $\mathrm{C_5}$, one branch ($\approx 28^\circ\mathrm{C}$). …

Concept and intuition

Vapour pressure is a measure of how readily molecules escape from the liquid into the gas phase. For a pure liquid, stronger intermolecular forces (like hydrogen bonding or dipole-dipole interactions) mean molecules are held more tightly, so fewer escape — giving a lower vapour pressure at any given temperature. For a solution of two volatile liquids, Raoult’s law tells us that the total vapour pressure is a weighted average of the pure vapour pressures (assuming ideal behaviour). Therefore, the solution’s vapour-pressure curve must lie between the curves of the two pure components. In the graph, the three curves are ordered from highest vapour pressure (curve A) to lowest (curve C) at any fixed temperature. So:

• Curve A: highest vapour pressure → weakest intermolecular forces.
• Curve C: lowest vapour pressure → strongest intermolecular forces.
• Curve B: intermediate → must be the solution.

Now let’s confirm step by step.

1. Identify which curve is the solution For an ideal solution of two volatile liquids, the total vapour pressure at a given temperature is

where $P_A^*$ and $P_B^*$ are the vapour pressures of the pure liquids and $\chi$ are mole fractions (which sum to 1). Since $P_{\text{soln}}$ is a weighted average, it always lies between $P_A^*$ and $P_B^*$. Therefore, on the graph, the solution’s curve must be the one that sits between the other two at every temperature. The curves are labelled A (leftmost, highest vapour pressure), B (middle), and C (rightmost, lowest vapour pressure). So curve B is the solution.

2. Identify which pure liquid has the strongest intermolecular forces …

Step 1 — Classify polymers by intermolecular force (the concept).

The molecular-force classification explains a polymer's entire mechanical behaviour:

Step 2 — Why Terylene sits at the top.

Terylene is a polyester made from ethylene glycol and terephthalic acid. Its backbone carries repeating polar ester groups ($-\!COO\!-$), whose $C{=}O$ dipoles set up strong dipole–dipole attractions (and hydrogen-bonding-type interactions) between neighbouring chains. These forces let the chains pack into a close, crystalline arrangement — giving the high tensile strength, high melting point and thread-forming ability that define a fibre.

Step 3 — Why the other three are weaker. …

The key here is that both phenol and ethanol have an —OH group, so they share some reactions (like with sodium metal). But phenol is an aromatic alcohol — the —OH is attached directly to a benzene ring. That changes its chemistry dramatically. The ring activates the ortho and para positions toward electrophilic substitution, which ethanol, being aliphatic, cannot do.

Bromine water is a classic test for phenols because it exploits this difference. Let's see why each option works or fails.

1. Bromine water (Option A) — This is the correct choice. Phenol reacts instantly with bromine water at room temperature to form a white precipitate of 2,4,6-tribromophenol. The reaction is:

Ethanol does not react with bromine water under these conditions — no decolourisation, no precipitate. So you get a clear visual distinction.

2. Sodium metal (Option B) — Both phenol and ethanol react with sodium metal to liberate hydrogen gas:

This test cannot distinguish between them — both give the same observation.

3. Iron metal (Option C) — Iron does not react with either phenol or ethanol under ordinary conditions. No useful test here. …