Q.Draw structures and give IUPAC names for the three chain isomers of $\text{C}_5\text{H}_{12}$, and classify the carbon atoms in each as primary, secondary or tertiary.

n-Pentane, $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_3$, has its two end carbons bonded to only one other carbon each (primary, 1°) and its three middle carbons each bonded to two other carbons (secondary, 2°). 2-Methylbutane, $(\text{CH}_3)_2\text{CHCH}_2\text{CH}_3$, has three primary carbons (the two methyls on C2 and the terminal C4), one secondary carbon (C3), and one tertiary (3°) carbon (C2, bonded to three other carbons). 2,2-Dimethylpropane, $(\text{CH}_3)_4\text{C}$, has four equivalent primary (methyl) carbons and one central carbon bonded to four other carbons with no hydrogens of its own -- a quaternary carbon.

For each carbon, count how many OTHER carbon atoms it is directly bonded to: one = primary, two = secondary, three = tertiary, four = quaternary.

Counting hydrogens instead of carbon-carbon bonds to classify a carbon; forgetting that a quaternary carbon (bonded to four other carbons, like neopentane's centre) has no classification as 1°/2°/3° since those terms are defined by how many H's it could still lose, and it has none.