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Questions on naming and representing organic molecules

Empirical, molecular, general, structural, displayed and skeletal formulae; homologous series, functional groups and alkyl groups; the thirteen functional groups of A level with their suffixes and prefixes; and the IUPAC rules for naming chains to six carbons, including locants, alphabetical order, di and tri, rings, esters, and molecules with two functional groups; hydrocarbons, saturated and unsaturated; straight, branched and cyclic skeletons; and the shape round each carbon with sp³, sp² and sp hybridisation.

18 original questions · 64 marks · the naming and representing organic molecules notes · Introduction to organic chemistry · Flashcards for this unit

To print: the written questions (PDF, 76 KB) · the mark schemes (PDF, 85 KB)

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• 6 marks · Explain · NRO-Q23

  Explain, step by step, how the IUPAC name of the compound HOCH₂CH₂CH(CH₃)CH(Br)CH₃ is worked out, and give the name.

  Mark scheme

  Indicative content

  1. the functional groups are OH and Br; halogens are only ever prefixes, so the alcohol gives the suffix -ol
  2. the longest chain containing the carbon that carries the OH runs from the CH₂OH carbon to the CH₃ beyond the CHBr, five carbons, so the stem is pentan
  3. the CH₃ on the third carbon of that chain is a branch, because counting through it gives a chain of only four carbons
  4. the chain is numbered from the CH₂OH end so that the suffix group has the lowest locant, 1, rather than 5
  5. with that numbering the methyl branch is on carbon 3 and the bromine on carbon 4, and the prefixes go in alphabetical order, bromo before methyl
  6. the e of pentane is dropped before -ol, giving 4-bromo-3-methylpentan-1-ol, C₆H₁₃BrO, with hyphens between numbers and letters

  Levels

  • Level 3 (5–6 marks) every step is present and justified: the suffix chosen from the OH, the longest chain through the OH carbon, the methyl identified as a branch, the numbering from the OH end for the lowest locant, the alphabetical order of prefixes, and the correct name with its punctuation
  • Level 2 (3–4 marks) the name is right and most steps are explained; why the chain is numbered from the OH end, why that chain is the longest, or why bromo comes before methyl is asserted
  • Level 1 (1–2 marks) the -ol suffix or the five-carbon chain is found; the chain is usually numbered from the bromine end, which gives the wrong locants

• 5 marks · Deduce · NRO-Q22

  Deduce the molecular formula, the structural formula and the name of compound Q, an alcohol with a branched carbon chain whose OH group is on a carbon atom at the end of the chain. Its composition by mass and relative molecular mass are given in the table.

  Element or quantity	Value
  Carbon	64.9%
  Hydrogen	13.5%
  Oxygen	21.6%
  Relative molecular mass	74.0

  Mark scheme

  1. M1 moles in 100 g: C 64.9 ÷ 12.0 = 5.41, H 13.5 ÷ 1.0 = 13.5, O 21.6 ÷ 16.0 = 1.35
  2. M1 dividing by 1.35 gives 4.01 : 10.0 : 1, so the empirical formula is C₄H₁₀O
  3. B1 C₄H₁₀O has Mr 74.0, equal to the relative molecular mass, so the molecular formula is also C₄H₁₀O
  4. B1 the branched four-carbon skeleton with OH on an end carbon is (CH₃)₂CHCH₂OH; allow a displayed or skeletal formula
  5. B1 Q is 2-methylpropan-1-ol; error carried forward from a wrong structure only if it is a branched C₄ alcohol

• 4 marks · Give · NRO-Q1

  Give the molecular formula and the empirical formula of hexane, CH₃(CH₂)₄CH₃, the general formula of the homologous series it belongs to, and the molecular formula of the member of that series with eight carbon atoms.

  Mark scheme

  1. B1 the molecular formula of hexane is C₆H₁₄
  2. B1 the empirical formula is C₃H₇, the ratio 6 : 14 divided by the highest common factor, 2
  3. B1 hexane is an alkane, and the general formula of the alkanes is CₙH₂ₙ₊₂
  4. B1 with n = 8 the molecular formula is C₈H₁₈ (octane); error carried forward from a wrong general formula

• 4 marks · Name · NRO-Q3

  Name the four compounds CH₃CH₂CH(CH₃)CH₂OH, CH₃COCH₂CH₂CH₃, CH₃CH₂COOCH₃ and CH₃CH₂CONH₂.

  Mark scheme

  1. B1 CH₃CH₂CH(CH₃)CH₂OH is 2-methylbutan-1-ol; reject 3-methylbutan-4-ol, which numbers from the wrong end
  2. B1 CH₃COCH₂CH₂CH₃ is the ketone pentan-2-one
  3. B1 CH₃CH₂COOCH₃ is the ester methyl propanoate; reject propyl methanoate, which reverses the two parts
  4. B1 CH₃CH₂CONH₂ is the amide propanamide, the C=O carbon counting in the three-carbon stem

• 4 marks · Draw · NRO-Q5

  Draw the skeletal formula of each of the compounds (CH₃)₂CHCH₂CHO and CH₂=C(CH₃)CH₂CH₃, and name each one.

  Mark scheme

  1. B1 skeletal formula of the aldehyde: a four-carbon zigzag with a branch line on the third carbon from the CHO end, and the end carbon drawn with a double line to O; allow the aldehyde H written in
  2. B1 the aldehyde is 3-methylbutanal, numbered from the CHO carbon, with no locant for the aldehyde group
  3. B1 skeletal formula of the alkene: a four-carbon chain with a double line between carbons 1 and 2 and a branch line on carbon 2
  4. B1 the alkene is 2-methylbut-1-ene; reject 2-ethylprop-1-ene, which misses the longest chain

• 4 marks · Explain · NRO-Q7

  Explain, in terms of the homologous series they belong to, why ethanol, CH₃CH₂OH, and propan-1-ol, CH₃CH₂CH₂OH, are both oxidised by acidified potassium dichromate(VI), but propan-1-ol boils about 19 °C higher.

  Mark scheme

  1. B1 both are alcohols, members of a homologous series with the general formula CₙH₂ₙ₊₁OH, each member differing from the next by CH₂
  2. B1 they have the same functional group, OH, which is responsible for their characteristic reactions
  3. B1 the reaction with dichromate takes place at the OH and the carbon joined to it, so the length of the chain makes little difference to it
  4. B1 propan-1-ol has one more CH₂ and so more electrons, giving stronger van der Waals forces between molecules and a higher boiling point; ignore references to hydrogen bonding if both are said to have it

• 4 marks · Evaluate · NRO-Q17

  Evaluate a student's name of 2-methylpentan-4-ol for the alcohol (CH₃)₂CHCH₂CH(OH)CH₃.

  Mark scheme

  1. B1 the student has found the correct chain of five carbons, with an OH group and a methyl branch
  2. B1 the chain must be numbered from the end that gives the suffix group, OH, the lowest number
  3. B1 numbered from the other end, the OH is on carbon 2 and the methyl group on carbon 4
  4. B1 the correct name is 4-methylpentan-2-ol, so the student's name is wrong only in its numbering

• 3 marks · Explain · NRO-Q6

  Explain why the compound CH₃CH(C₂H₅)CH₃ is named 2-methylbutane and not 2-ethylpropane.

  Mark scheme

  1. B1 the longest continuous carbon chain runs through the ethyl group and contains four carbons, not three
  2. B1 so the stem is butane and the remaining CH₃ is a methyl branch
  3. B1 numbering from the end nearer the branch puts the methyl group on carbon 2, giving 2-methylbutane

• 3 marks · Explain · NRO-Q9

  Explain why the molecular formula C₄H₈O₂ is not enough to identify a compound, but a structural formula is.

  Mark scheme

  1. B1 a molecular formula gives only the number of atoms of each element in a molecule
  2. B1 more than one compound has the formula C₄H₈O₂, for example butanoic acid and ethyl ethanoate; accept any two correct isomers
  3. B1 a structural formula shows how the atoms are arranged, so it fits only one of these compounds

• 3 marks · Calculate · NRO-Q10

  Calculate the empirical formula and the molecular formula of a compound whose composition by mass and relative molecular mass are given in the table.

  Element or quantity	Value
  Carbon	54.5%
  Hydrogen	9.1%
  Oxygen	36.4%
  Relative molecular mass	88.0

  Mark scheme

  1. M1 moles in 100 g: C 54.5 ÷ 12.0 = 4.54, H 9.1 ÷ 1.0 = 9.1, O 36.4 ÷ 16.0 = 2.275
  2. M1 dividing by the smallest, 2.275, gives the ratio 2 : 4 : 1, so the empirical formula is C₂H₄O
  3. A1 C₂H₄O has Mr 44.0, and 88.0 ÷ 44.0 = 2, so the molecular formula is C₄H₈O₂

• 3 marks · Calculate · NRO-Q11

  Calculate the percentage by mass of oxygen in propan-2-ol, CH₃CH(OH)CH₃. Use Ar values H 1.0, C 12.0, O 16.0.

  Mark scheme

  1. M1 the molecular formula is C₃H₈O, so Mr = 3 × 12.0 + 8 × 1.0 + 16.0 = 60.0
  2. M1 percentage of oxygen = 16.0 ÷ 60.0 × 100
  3. A1 26.7% oxygen, to three significant figures; allow 27%

• 3 marks · Suggest · NRO-Q12

  Suggest why propanone and butanone are named without a number, the ketones with five carbons are named pentan-2-one and pentan-3-one, and butanal needs no number either.

  Mark scheme

  1. B1 in propanone and butanone the C=O can only be on carbon 2; on carbon 1 the compound would be an aldehyde, so no number is needed
  2. B1 in a five-carbon chain the C=O can be on carbon 2 or carbon 3, which are different compounds, so a locant is needed to say which
  3. B1 an aldehyde group can only be at the end of a chain, so its carbon is always carbon 1 and the locant is left out

• 3 marks · Suggest · NRO-Q13

  Suggest a name for CH₂=CHCH₂OH, which contains both a C=C bond and an OH group, and explain how each part of the name is chosen.

  Mark scheme

  1. B1 the alcohol gives the suffix, so the chain is numbered from the CH₂OH carbon and the OH is on carbon 1
  2. B1 the C=C starts at carbon 2 and stays in the stem as -en-, giving prop-2-en-1-ol
  3. B1 the e of -ene is dropped before the vowel of -ol; prop-2-ene-1-ol does not earn this mark

• 3 marks · Compare · NRO-Q14

  Compare the information shown by the displayed formula and by the skeletal formula of butan-2-ol, CH₃CH(OH)CH₂CH₃.

  Mark scheme

  1. B1 both show how the atoms are arranged, with the OH group on the second carbon of a four-carbon chain
  2. B1 the displayed formula shows every atom and every bond, including each C–H bond and the O–H bond
  3. B1 the skeletal formula shows the carbons only as line ends and corners and omits the hydrogens on carbon, but still shows the O and the H of the OH group

• 3 marks · Describe · NRO-Q15

  Describe how the chain of a branched alkane is chosen and numbered, and how the prefixes are written, using 3-ethyl-2,2-dimethylpentane as the example.

  Mark scheme

  1. B1 the stem comes from the longest continuous chain of carbon atoms, here five carbons, so pentane
  2. B1 the chain is numbered to give the lowest set of locants at the first point of difference, 2,2,3 rather than 3,4,4
  3. B1 the prefixes are in alphabetical order ignoring di, so ethyl before dimethyl, with commas between numbers and hyphens between numbers and letters

• 3 marks · Explain · NRO-Q19

  Explain why the H–C–H bond angle in ethane is about 109.5° but the H–C–H bond angle in ethene is about 120°.

  Mark scheme

  1. B1 each carbon in ethane has four regions of electron density, four single bonds, and each carbon in ethene has three, the C=C counting as one region
  2. B1 the regions of electron density repel each other and get as far apart as possible
  3. B1 four regions give a tetrahedral arrangement at 109.5°, and three give a trigonal planar arrangement at 120°; allow 107° to 112° and 118° to 122°

• 3 marks · Suggest · NRO-Q20 CAIE

  Suggest the bond angle around each carbon atom of ethanenitrile, CH₃CN, and the hybridisation of each carbon.

  Mark scheme

  1. B1 the carbon of the CH₃ group has four single bonds, so it is sp³ hybridised and tetrahedral, with bond angles of about 109.5°
  2. B1 the carbon of the C≡N has two regions of electron density, so it is sp hybridised and the C–C≡N angle is 180°
  3. B1 the C≡N bond is one σ bond and two π bonds, the π bonds formed from the two p orbitals left on the sp carbon

• 3 marks · Name · NRO-Q21

  Name the alcohol that is a six-carbon ring carrying one OH group, and the two cyclohexenes that carry a methyl group, one on a carbon of the C=C and one on the carbon next to the C=C.

  Mark scheme

  1. B1 the alcohol is cyclohexanol, with no locant because every carbon of the ring is equivalent
  2. B1 the methyl group on a carbon of the C=C gives 1-methylcyclohexene, the C=C carbons being 1 and 2
  3. B1 the methyl group on the carbon beside the C=C gives 3-methylcyclohexene; reject 6-methylcyclohexene, numbered the wrong way round the ring

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