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IB Chemistry S3.2: Functional groups worked solutions

Worked solutions to every practice question in S3.2: Functional groups of Photon Chemistry: Study Guide for the IB Diploma (SL & HL). Try each question, then compare your working with the full solution.

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Basic practice

Question B1

[1 mark]
A trial adhesive contains the fragment \mathrm{CH_3CH_2C(=O)NHCH_3}. Which local feature determines its classification as an amide?
A   A carbonyl carbon bonded directly to nitrogen
B   An oxygen atom bonded to two alkyl groups
C   A nitrogen atom bonded only to saturated carbon atoms
D   A hydroxyl group on the carbonyl carbon
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Trace the nitrogen bond: it leads directly to the carbonyl carbon. This C(=O)-N connection identifies amido [1]. Option B describes ether-like connectivity, C omits carbonyl attachment, and D describes carboxyl.

Final answer

A

Question B2

[1 mark]
A catalogue lists only acyclic molecules with one \mathrm{C\equiv C} bond and no other multiple bonds. For n carbon atoms, which expression gives the hydrogen count?
A   2n+2
B   2n
C   n+2
D   2n-2
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An acyclic carbon skeleton with only single bonds has 2n+2 H sites. Replacing one single C-C bond by a triple uses four additional valences, giving 2n-2 H [1].

Final answer

D

Question B3

[1 mark]
Two pure acyclic alcohol homologues have the same number of hydroxyl groups. The longer-chain member has the higher boiling point. Which explanation is best?
A. Its carbon-carbon covalent bonds must be broken during boiling. B. Its O-H bonds are replaced by stronger carbon-carbon bonds. C. Its molecules lose their functional groups as they vaporize. D. Its larger electron clouds give stronger London attractions between molecules.
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Both compounds retain their hydroxyl groups, so both can hydrogen-bond. The longer chain has a larger, more polarizable electron cloud and stronger London attractions [1]. Boiling separates molecules without breaking their internal covalent bonds.

Final answer

D

Question B4

[1 mark]
An inventory is to contain every acyclic saturated monohydric alcohol with formula \mathrm{C_4H_{10}O}. Count different connectivities only; mirror-image forms share one entry. How many entries are needed?
A   2
B   3
C   4
D   5
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On an unbranched four-carbon skeleton, inequivalent OH sites give butan-1-ol and butan-2-ol. On the branched skeleton, attachment at an end or the central carbon gives 2-methylpropan-1-ol and 2-methylpropan-2-ol. These four entries exhaust the possible skeleton/site combinations [1]. Reverse-equivalent and mirror drawings add no entry under the stated rule.

Final answer

C; four connectivities.

Question B5

[1 mark]
A methyl ester gives a molecular ion at m/z=102 and a fragment at m/z=71. Which proposed neutral loss has the required mass? Use nominal masses \mathrm{C} = 12, \mathrm{H} = 1 and \mathrm{O} = 16.
A   \mathrm{{}^{\bullet}OCH_3}
B   \mathrm{{}^{\bullet}CH_2CH_3}
C   \mathrm{H_2O}
D   \mathrm{{}^{\bullet}COOH}
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The neutral-loss number is 102-71=31. Methoxy has nominal mass 16+12+3=31, so A fits [1]. The other candidates have masses 29, 18 and 45 respectively. Agreement of mass alone does not prove a unique fragmentation route.

Final answer

A; methoxy radical.

Question B6

[1 mark]
A two-proton signal is a quartet in a simple first-order proton NMR spectrum. Assume coupling only to one set of equivalent adjacent protons. How many protons are in that neighbouring set?
A   1
B   2
C   3
D   4
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For one equivalent neighbouring set, n+1=4, so n=3 [1]. The observed group integrates to two H; that population is independent of its neighbour count.

Final answer

C; three neighbouring H.

Question B7

[1 mark]
Name the compound \ce{CH3CH(CH3)CH2CHO}.
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Include the aldehyde carbon in the parent: there are four carbons in that chain. Number CHO as C1; the methyl branch is on C3. The name is 3-methylbutanal [1].

Final answer

3-methylbutanal.

Question B8

[1 mark]
Give the empirical formula corresponding to \ce{C6H12O2}.
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The common divisor of 6, 12 and 2 is 2. Divide every subscript by 2 to obtain \mathrm{C_3H_6O} [1].

Final answer

\mathrm{C_3H_6O}.

Question B9

[2 marks]
Classify the amine \ce{CH3CH2NHCH2CH3} as primary, secondary or tertiary. Justify your answer.
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The nitrogen has two ethyl groups and one H attached. It is a secondary amine [1] because two carbon-containing groups attach directly to N [1]. Count at nitrogen, rather than at either adjacent carbon.

Final answer

Secondary; N is bonded to two carbon groups.

Question B10

[2 marks]
An IR trace has a strong band at 1720\,\mathrm{cm^{-1}} and a very broad band spanning 2500–3000\,\mathrm{cm^{-1}}. Identify the compound class and the bond responsible for the very broad band.
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The carbonyl band together with very broad acid OH absorption identifies a carboxylic acid [1]. The broad band is due to O-H stretching in hydrogen-bonded acid molecules [1]. These assignments use booklet section 20.

Final answer

Carboxylic acid; hydrogen-bonded O-H.

Tutorial practice

Question U3

[6 marks]
A model-building exercise uses hex-3-ene and 1,2-dimethylcyclopropane.
(a) Draw and label the cis and trans arrangements of hex-3-ene. Show both groups attached to each double-bond carbon.   [2]
(b) Draw wedge-dash structures of cis- and trans-1,2-dimethylcyclopropane. Hydrogens on the substituted carbons may be omitted.   [2]
(c) Explain why the cyclopropane ring permits distinct cis and trans forms although its carbon-carbon bonds are single bonds.   [2]
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(a) Put an ethyl group and H on each end of C=C. Ethyls on the same side give cis [1]; ethyls on opposite sides give trans [1]. (b) Draw a three-carbon ring with adjacent methyl substituents. Two solid wedges (or two hashed bonds) give cis [1]; one solid and one hashed gives trans [1]. (c) A small ring constrains rotation and face exchange [1]; converting the configurations requires disrupting that constraint rather than simply turning one freely rotatable chain bond [1]. Fresh completed drawings are supplied in the local solution diagrams.

Final answer

(a) cis/trans hex-3-ene. (b) Same-face and opposite-face methyl groups. (c) Ring constraints prevent free face exchange.

Question U4

[5 marks]
A sample has connectivity \mathrm{CH_3CH(Br)CH_2OH}.
(a) Draw a pair of enantiomers using tapered solid wedges and hashed bonds. Label all four groups around the chiral carbon.   [3]
(b) Explain the structural relationship between the two drawings.   [2]
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(a) The chiral carbon bears H, Br, methyl and CH2OH, four distinct groups [1]. Draw two in-plane bonds, one tapered forward wedge and one backward hashed bond [1]. Reflect all group positions to produce opposite handedness [1]. (b) The two forms retain the same atom connections [1] but are nonsuperimposable mirror images [1]. A rotated drawing alone is not a second enantiomer.

Final answer

Mirror forms of CH3CH(Br)CH2OH; nonsuperimposable enantiomers.

Question U1

[7 marks]
Read drawing Q1. (a) Give its molecular formula and empirical formula. [2] (b) Supply a systematic name. [1] (c) Classify its alcohol centre and justify your choice. [2] (d) Write a condensed formula for 3-bromopent-1-ene. [2]
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Count four carbons on the drawn path and the one-carbon branch; valence supplies twelve H [1]. The subscripts share no factor greater than one [1]. The OH end is C1 and the methyl branch is C2 [1]. Primary [1], because the OH-bearing carbon has one carbon neighbour [1]. A five-carbon path with C1=C2 [1] and Br at C3 with valid valences [1].

Final answer

(a) C5H12O; C5H12O. (b) 2-methylbutan-1-ol. (c) Primary: one carbon neighbour at the OH-bearing carbon. (d) CH2=CHCH(Br)CH2CH3.

Question U2

[6 marks]
Two register entries are L: CH3CH2CH2OH and M: CH3CH(OH)CH3. (a) Give their common molecular formula and classify their isomer relationship. [2] (b) Write an ether structure with that formula and identify the change in connectivity that makes it a different compound class. [2] (c) Draw the two carbon frameworks possible for C4H10 and name their isomer relationship. [2]
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Both inventories give C3H8O [1]; OH changes its position on an unchanged three-carbon skeleton [1]. CH3OCH2CH3 [1]; its C–O–C link replaces C–O–H [1]. Both correct four-carbon frameworks [1]; chain isomerism [1]. Reversed drawings are one entry.

Final answer

(a) C3H8O; position isomers. (b) C2H5OCH3, with a C–O–C oxygen link. (c) The linear four-carbon chain versus its centrally branched framework; chain isomers.

Question U5

[7 marks]
For CH3CH2C(=O)CH2CH3, use a first-order proton NMR model: each CH2 pair is equivalent and long-range coupling is ignored. (a) Predict the number of signals and their proton integrals. [2] (b) Give each multiplicity with its neighbour count. [2] (c) Explain why all TMS protons make one reference signal and state its assigned shift. [2] (d) Use booklet section 21 to give a suitable shift range for the ketone-adjacent CH2 signal. [1]
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The two halves exchange under molecular symmetry, producing two environments [1] with combined populations 6 and 4 [1]. Each methyl couples to two adjacent CH2 protons: triplet [1]. Each CH2 couples to its three methyl neighbours: quartet [1]. The four equivalent TMS methyl groups give twelve equivalent protons and one signal [1], assigned zero ppm [1]. The CH2 protons next to ketone carbonyl lie in the supplied 2.2–2.7 ppm range [1]. Seven resolved lines here do not mean seven environments.

Final answer

(a) Two; 6H and 4H. (b) CH3 triplet, two neighbours; CH2 quartet, three neighbours. (c) Twelve equivalent TMS protons; 0 ppm by definition. (d) 2.2–2.7 ppm.

Question U6

[4 marks]
Compare pent-1-ene with cyclopentane. (a) Write their molecular formulas and identify the isomer relationship. [2] (b) Explain why a common general formula does not put them in the same homologous series. [2]
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(a) Both have \mathrm{C_5H_{10}} [1]. They are structural isomers: an open chain with C=C versus a saturated ring [1]. (b) Pent-1-ene belongs to the alkene series, with a carbon-carbon double bond [1]; cyclopentane is a cycloalkane, with only single bonds in its ring [1]. Shared \mathrm{C_nH_{2n}} does not establish shared connectivity or functional group.

Final answer

Both \mathrm{C_5H_{10}}; structural (ring-chain) isomers; alkene versus cycloalkane.

Question U7

[5 marks]
A CHO compound gives a singly charged molecular ion at m/z=86 and a fragment at m/z=43. Its IR trace has a strong band at 1718\,\mathrm{cm^{-1}} and reliably no broad acid O–H band. Use nominal C, H and O masses of 12, 1 and 16. (a) Calculate the neutral-loss mass and give one possible neutral composition. [2] (b) Identify the bond causing the 1718\,\mathrm{cm^{-1}} band and assess a carboxylic-acid proposal. [2] (c) Explain why these observations do not prove a unique molecular structure. [1]
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(a) The neutral loss is 86-43=43 [1]. One possible neutral is \mathrm{C_3H_7^{\bullet}}, with nominal mass 3(12)+7=43 [1]. This is a possible mass assignment, not proof of a mechanism. (b) The absorption supports C=O [1]; reliable absence of broad acid O-H evidence contradicts an acid assignment [1]. (c) Mass and bond evidence do not fix all atom connections; alternative carbonyl isomers can remain [1].

Final answer

Loss 43; e.g. \mathrm{C_3H_7^{\bullet}}; C=O; acid rejected; several connectivities remain.

Question U8

[5 marks]
A compound has formula \ce{C3H8O}. Its first-order proton NMR contains a 3H triplet, a 2H quartet and a 3H singlet. Ignore long-range coupling. Reliable IR data show no O–H absorption. (a) Deduce a condensed structure and identify its compound class. [2] (b) Assign the triplet and quartet, stating the neighbouring proton count in each case. [2] (c) Use the IR evidence to reject the alcohol alternatives. [1]
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(a) The structure is \mathrm{CH_3OCH_2CH_3} [1], an ether [1]. The methoxy protons form the singlet because no adjacent proton set couples across O under the stated approximation. (b) The ethyl CH3 is a triplet from two adjacent CH2 protons [1]; CH2 is a quartet from the three CH3 protons [1]. (c) Both propan-1-ol and propan-2-ol contain OH, so reliable absence of its IR band rejects both [1].

Final answer

\mathrm{CH_3OCH_2CH_3}, ether; ethyl 3H triplet/2H quartet; no alcohol O-H.

Advanced practice

Question A1

[7 marks]
An unknown compound contains only \mathrm{C}, \mathrm{H} and \mathrm{O}. Its singly charged molecular ion occurs at m/z=88. IR shows strong absorption at 1736\,\mathrm{cm^{-1}} and no broad \mathrm{O-H} band. Proton NMR gives \begin{array}{c|c|c}\delta/\mathrm{ppm}&\text{integral}&\text{multiplicity}\\\hline1.00&3&\text{triplet}\\2.35&2&\text{quartet}\\3.70&3&\text{singlet}\end{array}. Use nominal masses \mathrm{C} = 12, \mathrm{H} = 1, \mathrm{O} = 16.
(a) Deduce a condensed structure, assigning all three proton signals.   [4]
(b) Show that your molecular formula agrees with the molecular-ion mass.   [1]
(c) Explain how the IR observations discriminate your structure from a carboxylic acid isomer.   [2]
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(a) Assemble CH3CH2C(=O)OCH3 [1]. The terminal CH3 is a triplet from two neighbouring H [1]; the carbonyl-adjacent CH2 is a quartet from three neighbouring H [1]; OCH3 is a singlet because the adjacent oxygen does not provide a neighbouring proton set [1]. (b) The formula is C4H8O2 and 4(12)+8+2(16)=88 [1]. (c) Carbonyl absorption fits both ester and acid [1], but reliable absence of the very broad acid OH band contradicts the acid alternative [1]. The shifts are supplied typical values, not measurements predicted from first principles.

Final answer

CH3CH2C(=O)OCH3, C4H8O2.

Question A2

[7 marks]
A greenhouse-gas absorption band is centred at wavelength 4.30\,\mathrm{\mu m}. Wavenumber is reciprocal wavelength, expressed in \mathrm{cm^{-1}}.
(a) Calculate the band wavenumber to three significant figures.   [2]
(b) State the requirement for a molecular vibration to absorb IR radiation.   [1]
(c) Explain why the fact that \mathrm{CO_2} has no permanent molecular dipole does not prevent it absorbing IR.   [2]
(d) Outline how absorption of outgoing IR contributes to warming near the surface.   [2]
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(a) Convert 4.30\,\mu\mathrm m=4.30\times10^{-4}\,\mathrm{cm} [1]. Its reciprocal is 2325.581\ldots\,\mathrm{cm^{-1}}, or 2.33\times10^3\,\mathrm{cm^{-1}} to three significant figures [1]. (b) The vibration must change the molecular dipole [1]. (c) CO2 has cancelling dipoles at its equilibrium linear geometry [1], but bending and asymmetric stretching change that cancellation [1]. (d) Greenhouse molecules absorb surface-emitted IR [1]; subsequent energy exchange and emission include downward radiation, reducing net surface cooling at a given surface temperature [1].

Final answer

2.33\times10^3\,\mathrm{cm^{-1}}; changing dipole; CO2 has active vibrations.

Question A4

[7 marks]
A sample is known to be one of the acyclic saturated alcohol isomers of \mathrm{C_4H_{10}O}. A supplied test rule states: under these mild oxidation conditions, a secondary alcohol gives a ketone, a primary alcohol gives an aldehyde or acid, and a tertiary alcohol shows no reaction. The sample forms a ketone.
(a) Use the supplied rule to determine the alcohol class.   [1]
(b) Deduce and name the only alcohol connectivity that fits the formula and this class.   [2]
(c) Write a condensed structural formula for the ketone product and identify its functional group by name.   [2]
(d) Explain why 2-methylpropan-2-ol does not fit the observation.   [1]
(e) Write the balanced oxidation equation for your alcohol, using [\mathrm{O}] for one oxygen atom supplied by the oxidizing agent.   [1]
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(a) The supplied rule selects a secondary alcohol [1]. (b) C2H5CH(OH)CH3 [1] is butan-2-ol [1]; it is the sole secondary alcohol connectivity with four carbons. (c) Oxidation gives C2H5C(=O)CH3 [1], containing carbonyl [1]. (d) 2-Methylpropan-2-ol is tertiary, so the supplied mild test predicts no ketone-producing reaction [1]. (e) CH3CH(OH)CH2CH3 + [O] -> CH3C(=O)CH2CH3 + H2O [1]. Both sides have four C, ten H and two O when the supplied oxygen atom is included.

Final answer

Secondary; butan-2-ol; butan-2-one; tertiary alternative fails.

Question A3

[6 marks]
S3.2 + S2.2 + Inquiry. A model has connectivity CH3CH(OH)CH(Br)CH2CH3. Treat its two substituted tetrahedral carbons as independent stereocentres; no other stereogenic feature is present. (a) Locate the centres and determine the number of configurations, checking whether internal symmetry reduces it. [2] (b) Equal quantities of one configuration and its mirror image are mixed. Predict their net optical rotation under identical conditions. [1] (c) Explain why a zero reading from an unspecified solution does not establish that same composition. [1] (d) A chiral enzyme binds the two mirror forms differently. Give a structural explanation. [2]
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C2 and C3 each bear four distinct groups [1]. Two independent binary arrangements give four, with chemically unequal ends preventing meso reduction [1]. Equal opposite rotations cancel [1]. An achiral solute is an alternative explanation [1]. The binding site has a fixed chiral arrangement [1]; exchanging the molecular handedness changes the relative placement of interacting groups and hence binding [1].

Final answer

(a) C2, C3; four configurations, no internal symmetry reduction. (b) Zero. (c) An achiral solute can also give zero. (d) Different three-dimensional fit to a chiral binding site gives different interactions.

Connected questions

Question C1

[8 marks]
A fuel contains only C, H and O, with one O per molecule, and is an acyclic saturated alcohol. Complete combustion of 0.00200\,\mathrm{mol} fuel gives 0.00800\,\mathrm{mol} \ce{CO2} and 0.0100\,\mathrm{mol} \ce{H2O}. A mild oxidation test produces a ketone. (a) Deduce the molecular formula from the product amounts and the stated alcohol class. [2] (b) Deduce the alcohol connectivity and give its systematic name. [2] (c) Write a balanced equation for complete combustion of this alcohol. [2] (d) Calculate the amount of oxygen consumed by the original fuel sample. [2]
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(a) Each fuel molecule has 0.00800/0.00200=4 carbon atoms [1] and 2(0.0100)/0.00200=10 H atoms; one O is specified, hence \mathrm{C_4H_{10}O} [1]. (b) Ketone formation selects the secondary alcohol \mathrm{CH_3CH(OH)CH_2CH_3} [1], butan-2-ol [1]. (c) \mathrm{C_4H_{10}O+6O_2\to4CO_2+5H_2O}: C/H products [1] and oxygen coefficient 6 [1]. (d) Use the 1:6 fuel:oxygen ratio [1]; n(\mathrm{O_2})=6(0.00200)=0.0120\,\mathrm{mol} to three significant figures [1].

Final answer

\mathrm{C_4H_{10}O}; butan-2-ol; 6 mol O2 per mol fuel; 0.0120\,\mathrm{mol} O2.

Question C2

[8 marks]
Isomers X and Y have formula \ce{C4H10O}. X gives a 6H triplet and a 4H quartet in a first-order proton NMR spectrum, with no IR O–H band. Y is an alcohol whose non-exchangeable protons give one 9H singlet. (a) Deduce X and justify it using two independent spectroscopic observations. [3] (b) Deduce Y and classify its alcohol centre. [2] (c) Predict which isomer has the higher boiling temperature and explain through intermolecular attractions. [2] (d) Predict how Y responds to mild oxidation. [1]
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(a) X is \mathrm{CH_3CH_2OCH_2CH_3} [1]. Symmetry makes its ethyl groups equivalent, explaining the 6H triplet and 4H quartet [1]; no O-H band supports ether over alcohol [1]. (b) Y is \mathrm{(CH_3)_3COH} [1], a tertiary alcohol because its OH-bearing carbon bonds to three carbon groups [1]. The three equivalent methyl groups account for the 9H singlet. (c) Y has the higher boiling temperature [1] because its molecules hydrogen-bond to each other; X accepts H bonds from water but lacks an OH donor in the pure liquid [1]. (d) Y undergoes no oxidation under mild conditions [1]; there is no H on its OH-bearing carbon to remove in the usual carbonyl-forming oxidation.

Final answer

X = ethoxyethane; Y = 2-methylpropan-2-ol, tertiary; Y boils higher; no mild oxidation.

Question C3

[8 marks]
A synthetic ester gives four proton NMR signals: two triplets and two quartets, integrating to 3:3:2:2. Ignore long-range coupling. Candidate P is \ce{CH3CH2COOCH2CH3}; Q is \ce{CH3CH2CH2COOCH3}. In a separate reaction, 0.0200\,\mathrm{mol} of the identified ester reacts completely and warms 200.0\,\mathrm{g} solution by 0.600\,\mathrm{K}. Use c=4.18\,\mathrm{J\,g^{-1}\,K^{-1}}; neglect apparatus heat capacity and heat loss. (a) Select P or Q and assign the two quartet environments. [2] (b) Explain how the methyl signals reject the other candidate. [2] (c) Calculate the heat gained by the solution. [2] (d) Calculate the molar enthalpy change of the reaction, including its sign. [2]
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(a) P fits [1]. Both its carbonyl-adjacent CH2 and its OCH2 adjoin CH3, so each is a quartet [1]. (b) Both P methyl groups neighbour two H and form triplets [1]; Q instead requires an isolated OCH3 singlet, inconsistent with the observations [1]. (c) q=mc\Delta T=200(4.18)(0.600) [1] =501.6\,\mathrm J, reported as 502\,\mathrm J [1]. (d) The solution warms, so the reaction releases heat and has negative enthalpy [1]. \Delta H=-501.6/(1000\times0.0200)=-25.1\,\mathrm{kJ\,mol^{-1}} to three significant figures [1].

Final answer

P; both CH2 groups are quartets; Q has OCH3 singlet; q=502\,\mathrm J; \Delta H=-25.1\,\mathrm{kJ\,mol^{-1}}.

From the book

Photon Chemistry: Study Guide for the IB Diploma (SL & HL)

Functional groups from the core ideas to full exam questions, with question types, worked examples and graded practice.

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