Q.Draw and name all the chain (structural) isomers of $\text{C}_5\text{H}_{12}$, and state how their boiling points compare.

$\text{C}_5\text{H}_{12}$ can be arranged in three distinct ways: a straight chain, $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_3$ (n-pentane); a singly-branched chain, $(\text{CH}_3)_2\text{CHCH}_2\text{CH}_3$ (2-methylbutane, or isopentane); and a doubly-branched, highly compact chain, $(\text{CH}_3)_4\text{C}$ (2,2-dimethylpropane, or neopentane). Boiling point is set by the strength of London dispersion forces between molecules, which depend on how much surface area is available for intermolecular contact -- a long, thin, straight-chain molecule has more surface contact than a compact, near-spherical branched one, so boiling point falls steadily as branching increases: n-pentane ($36\,°\text{C}$) $>$ isopentane ($28\,°\text{C}$) $>$ neopentane ($9.5\,°\text{C}$).

Systematically place the five carbons as (a) one unbranched chain,

(b) a 4-carbon chain with one methyl branch,

(c) a 3-carbon chain with two methyl branches, checking each is a genuinely distinct connectivity; then rank boiling points by surface-area/dispersion-force reasoning.

Missing neopentane because it looks unusual (four methyls on one carbon); assuming boiling point depends on molecular weight alone rather than shape/surface area, since all three isomers have the identical molecular formula and mass.