Question 1 of 5: Functional-Group Structural Isomers

Question 1: Functional-Group Structural Isomers (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Each part fixes a molecular formula and a functional-group family, and asks for several different carbon skeletons/substitution patterns that both satisfy the formula exactly. The degree of unsaturation of every formula below is exactly 1 (one C=O for the aldehyde/ketone, one C=O for the acid's carbonyl with the second oxygen as –OH, and zero for the saturated amines — consistent with C_nH_2n+3N carrying no ring or π-bond at all), so the only freedom in each part is where the functional group sits on the chain or how the chain branches.

(i) Three secondary amines, C₄H₁₁N (6 points). A secondary amine has nitrogen bonded to exactly two carbon groups and one H; the two carbon groups must together contain all four carbons of the formula, split any way that gives two distinct alkyl fragments (4+0 is a primary amine, not allowed here, so the real choices are 3+1 and 2+2 in various isomeric forms):

These are diethylamine (Et–NH–Et, a symmetric 2+2 split), N-methylpropan-1-amine and N-methylpropan-2-amine (both a 1+3 split, differing in whether the propyl group is straight-chain or branched at the point of attachment) — three genuinely different secondary amines, all C₄H₁₁N.

(ii) Two aldehydes, C₄H₈O (4 points). An aldehyde's carbonyl must sit at the end of the chain (C1); with four carbons the only freedom left is whether the remaining three-carbon tail is straight or branched at C2:

Butanal (straight chain) and 2-methylpropanal/isobutyraldehyde (branched at the carbon next to the carbonyl) — both C₄H₈O, constitutional isomers of each other.

(iii) Two carboxylic acids, C₄H₈O₂ (4 points). Same logic as the aldehyde, but the terminal group is –COOH instead of –CHO (using both oxygens of the formula):

Butanoic acid and 2-methylpropanoic acid (isobutyric acid) — both C₄H₈O₂.

(iv) Three ketones, C₅H₁₀O (6 points). A ketone's carbonyl must sit on an internal carbon (never C1), so with five carbons there are two straight-chain placements (C2 or C3) plus one branched skeleton:

Pentan-2-one and pentan-3-one (the carbonyl at two different internal positions of the same straight chain) and 3-methylbutan-2-one (a branched skeleton with the carbonyl next to the branch point) — three distinct ketones, all C₅H₁₀O.