The problems are on the organic nomenclature worksheet. Try them on your own first, then compare with the answers below.
Quick answer: IUPAC nomenclature names an organic compound from its longest carbon chain, its functional group and its substituents, with numbers giving their positions. These worked answers correct the 20 problems of the Chemistry 11-12 worksheet on organic nomenclature, from alkane names and formulae to name complex structures: each solution shows the formula, the substitution and the final answer with its unit.
Each answer gives the reasoning, the correct name or number in bold, and the rule used. When naming, check the longest chain, then the lowest locants, then alphabetical order of the prefixes.
Level: Chemistry grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB S3.2
Problem 1 Alkane names and formulae
1. methane, ethane, propane, butane, pentane, hexane.2. CₙH₂ₙ₊₂.
3. n = 7 gives C₇H₁₆.
Problem 2 Name the straight chains
1. pentan-1-ol2. propanal
3. butanone
4. propanoic acid
Problem 3 Condensed formulae
1. CH₃CH(CH₃)CH₂CH₃2. (CH₃)₄C or C(CH₃)₄
3. Both are C₅H₁₂.
Problem 4 Identify functional groups
alkene and alcohol; ester; amine; halogenoalkane.Problem 5 Which type of isomerism
1. position2. chain
3. functional group
4. functional group
Problem 6 Branched names
1. 2,2-dimethylbutane2. 2-chlorobutane
3. 2,3-dimethylpentane
4. 3-bromopentane
Problem 7 Isomers of C₄H₉Br
1. 1-bromobutane, 2-bromobutane, 1-bromo-2-methylpropane, 2-bromo-2-methylpropane.2. Primary, secondary, primary, tertiary respectively (the carbon bonded to Br has 1, 2, 1 and 3 carbon neighbours).
Problem 8 Boiling points of C₅H₁₂ isomers
1. 36 − 10 = 26 °C.2. More branching makes the molecule more compact with a smaller surface area for contact, so the London forces are weaker and the boiling point is lower.
Problem 9 E and Z but-2-ene
1. Left is Z (both CH₃ on the same side); right is E.2. In but-1-ene, one double-bond carbon has two H atoms, so it lacks two different groups.
3. The π bond needs overlapping p orbitals, which would be broken by twisting the bond.
Problem 10 Chiral or not
3-methylhexane (carbon 3 has H, CH₃, C₂H₅, C₃H₇) and 2-bromopentane (carbon 2 has H, Br, CH₃, C₃H₇) are chiral. The others have a carbon with two identical groups or no sp³ carbon with four different groups.Problem 11 Combustion analysis
1. n(CO₂) = 0.6600 ÷ 44.0 = 0.01500 mol so n(C) = 0.0150 mol; n(H₂O) = 0.3600 ÷ 18.0 = 0.0200 mol so n(H) = 0.0400 mol.2. C : H = 0.015 : 0.040 = 3 : 8, so C₃H₈.
3. Mr = 44, so the molecular formula is C₃H₈ and there is one structure (propane).
Problem 12 Enantiomeric excess
1. ee = 5.0 ÷ 20.0 × 100 = 25 %.2. (+) = (100 + 25) ÷ 2 = 62.5 %; (−) = 37.5 %.
Problem 13 Formula from percentages
1. C 54.5 ÷ 12.0 = 4.54; H 9.1 ÷ 1.0 = 9.1; O 36.4 ÷ 16.0 = 2.275. Dividing by 2.275 gives 2 : 4 : 1, so C₂H₄O.2. Empirical mass 44; 88 ÷ 44 = 2, so C₄H₈O₂.
3. For example butanoic acid and ethyl ethanoate (also 2-methylpropanoic acid, methyl propanoate).
Problem 14 Alkenes and alkynes
1. but-2-yne2. 2-methylbut-1-ene
3. pent-2-ene (shows E/Z)
4. but-1-yne
Problem 15 Spot the errors
1. The longest chain has five carbons: 3-methylpentane.2. Number from the end giving the lowest locant to OH: butan-2-ol.
3. In a Z isomer they are on the same side.
Problem 16 Assign E or Z
1. Left carbon: Cl (atomic number 17) > CH₃. Right carbon: CH₃ > H.2. Cl is top-left and CH₃ is top-right, on the same side: Q is Z.
3. The other isomer has CH₃ on the right carbon at the bottom, giving E.
Problem 17 Lactic acid
1. Carbon 2 with H, OH, COOH and CH₃.2. Same melting point and boiling point; they rotate plane-polarised light in opposite directions.
3. The reaction can attack a flat intermediate from either face with equal chance, forming equal amounts of both enantiomers.
Problem 18 Isomers of C₄H₈
1. but-1-ene, but-2-ene, 2-methylpropene and cyclobutane (also methylcyclopropane).2. But-2-ene.
3. Five structural isomers; but-2-ene has two stereoisomers, so 6 in total.
Problem 19 Isomers of C₃H₆O₂
1. propanoic acid, CH₃CH₂COOH.2. methyl ethanoate and ethyl methanoate: functional group (and position) isomers.
3. Add Na₂CO₃: an acid releases CO₂ and fizzes, whereas esters do not.
Problem 20 Name complex structures
1. 4-chloro-2-methylpentan-1-ol2. 2-methylpent-2-ene
3. 3-methylbutanoic acid
4. CH₃CHClC(CH₃)ClCH₂CH₃
5. Two chiral centres, so 2² = 4 stereoisomers.
