QUESTIONS
1. A student is given an unlabelled bottle containing a colourless liquid known only to be a member of the alkanol series with molecular mass 46. Using this information alone:
(a) deduce the molecular formula of the compound
(b) deduce its empirical formula
(c) state its IUPAC name and one everyday use of this specific compound
2. Match each reagent in Column I to the correct organic transformation it brings about in Column II (write only the matching letter beside each number):
Column I
i) Hot concentrated H₂SO₄
ii) I₂/NaOH (iodoform test conditions)
iii) NH₂OH
iv) Ozone (O₃) followed by Zn/H₂O
v) Conc. HNO₃ alone (no H₂SO₄), cold, dilute
Column II
A. Oxidative cleavage of a C=C double bond
B. Detects the presence of a CH₃CO− or CH₃CH(OH)− group
C. Sulphonation of an aromatic ring
D. Condensation with a carbonyl group to form an oxime
E. Mild nitration, generally slower than the mixed-acid method
3. State whether each of the following statements is TRUE or FALSE. Where a statement is FALSE, rewrite it correctly.
a) Members of a homologous series can be separated from one another by a constant unit of CH₂.
b) The alkyne general formula CₙH₂ₙ₋₂ is identical to that of a cycloalkene.
c) A compound and its next homologue always have identical chemical reactivity but different physical properties.
d) Ethyne (C₂H₂) is the first member of the alkyne series.
4. Threonine and aspartic acid are two amino acids whose side chains are −CH(OH)CH₃ and −CH₂COOH respectively.
a) i) Without drawing the full structures, state which functional group each side chain adds to the basic amino acid skeleton, and hence classify each amino acid as acidic, basic or neutral.
ii) An amino acid solution is placed in an electric field at a pH exactly equal to its isoelectric point. Predict and explain what is observed.
b) Draw the full structural formula of the dipeptide formed when the amino group of threonine reacts with the carboxyl group of aspartic acid.
5. Propane is subjected to monochlorination under UV light.
a) List, in order, the three stages of the mechanism involved, briefly describing what happens at each stage.
b) State the type of bond fission occurring at the very first stage.
c) Two structural isomers of C₃H₇Cl are possible from this reaction. Draw both and explain, using the number of equivalent hydrogen environments in propane, why exactly two (and not three) isomers result.
6. Complete the table below by inserting the correct number of pi (π) electrons present in each molecule listed:
| Molecule | Structural feature | Number of π electrons |
|---|---|---|
| But-2-ene | one C=C | ? |
| Hexa-2,4-diyne | two C≡C | ? |
| Benzene | three alternating C=C | ? |
| Cyclohexane | none | ? |
| Butanal | one C=O | ? |
7. "The petrochemical industry begins with crude oil and ends with everyday plastic products in our homes." Using this statement as a guide, briefly trace the journey of crude oil through THREE (3) key stages from extraction to a finished polymer product, naming a process involved at each stage.
8. For each pair of substituted benzene derivatives below, state which member of the pair would react FASTER with an electrophile (E⁺) in electrophilic aromatic substitution, and give a brief reason:
a) Phenol vs. Nitrobenzene
b) Toluene vs. Benzoic acid
c) Aniline vs. Chlorobenzene
9. An examiner accidentally wrote the formula "C₅H₁₂O₂" for a member of the alkanol homologous series. Explain, with a calculation, why this formula CANNOT represent a simple (monohydric) alkanol, and state what class of compound it could instead represent.
10. Arrange the following five compounds in order of INCREASING boiling point, and briefly justify your ranking based on their functional groups and intermolecular forces: butane, butan-1-ol, butanal, butanoic acid, 1-chlorobutane.
SECTION B
11. A laboratory technician has four unlabelled bottles containing: an alkene, an alkanal, an alkanoic acid, and a haloalkane.
a) Design a systematic testing scheme (a flowchart or step-by-step sequence of tests) that would allow the technician to identify all four compounds unambiguously, stating the reagent, expected observation, and conclusion at each step. (18 marks)
b) Explain briefly why it would NOT be possible to distinguish an alkanal from an alkanone using Tollens' reagent alone if only the observation "no reaction" is obtained with an unknown sample — what further test would you recommend? (7 marks)
12. A hydrocarbon X, on complete combustion, produces only CO₂ and H₂O. When 2.8g of X was burnt completely, 8.8g of CO₂ and 3.6g of H₂O were obtained. The vapour density of X was found to be 28.
a) Determine the empirical formula of X. (8 marks)
b) Determine the molecular formula of X. (4 marks)
c) Suggest a possible structure for X and name it, given that X decolourises bromine water rapidly. (5 marks)
d) Write a balanced equation for the reaction of X with hydrogen bromide (HBr), stating and justifying which product predominates using an appropriate rule. (8 marks)
ANSWERS
Question 1
Alkanol general formula: CₙH₂ₙ₊₁OH, molar mass = 14n+18
For M=46: 14n+18=46 → 14n=28 → n=2
a) Molecular formula: C₂H₅OH (C₂H₆O)
b) Empirical formula: C₂H₆O (already in lowest terms)
c) IUPAC name: Ethanol. Everyday use: used as the alcohol in alcoholic beverages / as a solvent / as a fuel additive (e.g., in petrol blends)
Question 2
i) → C (hot conc. H₂SO₄ sulphonates aromatic rings)
ii) → B (iodoform test detects CH₃CO− or CH₃CH(OH)− groups)
iii) → D (hydroxylamine condenses with C=O to form an oxime)
iv) → A (ozonolysis cleaves C=C oxidatively)
v) → E (dilute HNO₃ alone gives slower, milder nitration than the mixed acid method)
Question 3
a) TRUE.
b) FALSE. Correction: The alkyne general formula CₙH₂ₙ₋₂ is the same as that of a cycloalkene by formula only — but they are NOT chemically identical; this is a case of isomerism (same general formula, different structure/functional group), not a true match of the series. (Statement is best marked FALSE because it implies the two are "identical" rather than merely sharing the same general formula.)
c) FALSE. Correction: A compound and its next homologue have similar (not identical) chemical reactivity, along with a gradual (not identical) trend in physical properties such as boiling point and density.
d) TRUE.
Question 4
a) i) Threonine's side chain (−CH(OH)CH₃) adds a hydroxyl (alcohol) group — this makes threonine a neutral amino acid, since −OH does not significantly ionise to acid or base at physiological pH. Aspartic acid's side chain (−CH₂COOH) adds an additional carboxyl group — this makes aspartic acid an acidic amino acid, since it has two acidic (−COOH) groups against one basic (−NH₂) group.
ii) At the isoelectric point, the amino acid exists predominantly as its zwitterion with a net charge of zero; placed in an electric field, it will show no net migration toward either electrode (it remains stationary, since there is no net charge to be attracted to either the anode or cathode).
b) Dipeptide (threonine's −NH₂ + aspartic acid's −COOH, condensation losing H₂O):
CH₃CH(OH)CH(NH₂)CO—NH—CH(CH₂COOH)COOH
(Threonyl-aspartic acid dipeptide, with the new peptide bond, −CO−NH−, shown linking the two residues)
Question 5
a) Three stages of the free radical mechanism:
- Initiation: UV light supplies energy to homolytically split the Cl–Cl bond, generating two chlorine free radicals: Cl₂ → 2Cl·
- Propagation: A chlorine radical abstracts a hydrogen atom from propane, forming HCl and a propyl radical, which then reacts with another Cl₂ molecule to form the chloropropane product and regenerate a Cl radical (chain continues): C₃H₈+Cl·→C₃H₇·+HCl; C₃H₇·+Cl₂→C₃H₇Cl+Cl·
- Termination: Two radicals combine to form a stable molecule, ending the chain: e.g., Cl·+Cl·→Cl₂, or C₃H₇·+Cl·→C₃H₇Cl
b) Homolytic bond fission (each atom retains one electron of the broken bond, forming radicals).
c) Propane (CH₃CH₂CH₃) has two distinct types of hydrogen environment by symmetry: the 6 equivalent hydrogens on the two terminal (C1 and C3) carbons, and the 2 equivalent hydrogens on the central (C2) carbon.
Isomer 1 — 1-chloropropane: CH₃—CH₂—CH₂—Cl
Isomer 2 — 2-chloropropane: CH₃—CHCl—CH₃
Only two isomers result (not three) because C1 and C3 are chemically equivalent by the molecule's symmetry, so substitution at either end gives the same product (1-chloropropane); only substitution at the unique central carbon gives a distinct second product.
Question 6
| Molecule | Structural feature | Number of π electrons |
|---|---|---|
| But-2-ene | one C=C | 2 |
| Hexa-2,4-diyne | two C≡C | 8 |
| Benzene | three alternating C=C | 6 |
| Cyclohexane | none | 0 |
| Butanal | one C=O | 2 |
Question 7
- Extraction: Crude oil is extracted from underground reservoirs by drilling.
- Refining (fractional distillation): Crude oil is separated into fractions (e.g., naphtha) based on differing boiling points at a refinery.
- Cracking and polymerisation: The naphtha fraction undergoes catalytic/thermal cracking to produce small alkene monomers (e.g., ethene), which are then polymerised (addition polymerisation) to form the finished plastic (e.g., polyethene).
Question 8
a) Phenol reacts faster. The −OH group is strongly activating (electron-donating via resonance/lone pair donation into the ring), increasing electron density and reactivity toward electrophiles, whereas −NO₂ is strongly deactivating (electron-withdrawing), making nitrobenzene far less reactive.
b) Toluene reacts faster. The −CH₃ group is weakly activating (electron-donating via the inductive/hyperconjugation effect), while −COOH is deactivating (electron-withdrawing), making benzoic acid less reactive than toluene.
c) Aniline reacts faster. The −NH₂ group is strongly activating (lone pair on nitrogen donates into the ring by resonance), while −Cl, though weakly deactivating overall, withdraws electron density inductively, making chlorobenzene less reactive than aniline.
Question 9
For a simple monohydric alkanol, CₙH₂ₙ₊₁OH = CₙH₂ₙ₊₂O (general formula has only ONE oxygen atom).
C₅H₁₂O₂ has two oxygen atoms, which is inconsistent with the alkanol general formula (CₙH₂ₙ₊₂O, n=5 would give C₅H₁₂O, not C₅H₁₂O₂).
Explanation: Since the formula contains 2 oxygen atoms rather than 1, it cannot be a simple monohydric alkanol. It could instead represent a diol (dihydric alcohol, CₙH₂ₙ(OH)₂) or possibly an ether-alcohol / hydroxy-ether, since these classes accommodate two oxygen atoms within a similar hydrogen count.
Question 10
Increasing order of boiling point:
Butane < 1-chlorobutane < Butanal < Butan-1-ol < Butanoic acid
Justification:
- Butane (weakest — only van der Waals/London dispersion forces, non-polar molecule).
- 1-Chlorobutane (slightly stronger — polar C−Cl bond gives permanent dipole-dipole forces in addition to dispersion forces).
- Butanal (stronger still — polar C=O gives dipole-dipole forces, generally somewhat stronger than a single C−Cl dipole due to the carbonyl's larger dipole moment).
- Butan-1-ol (stronger — capable of hydrogen bonding via the −OH group, a much stronger intermolecular force than dipole-dipole alone).
- Butanoic acid (strongest — capable of hydrogen bonding AND forms stable dimers via two hydrogen bonds between pairs of −COOH groups, requiring the most energy to separate molecules, giving the highest boiling point).
Question 11
a) Systematic identification scheme:
Step 1 — Test with bromine water (added dropwise, shaken):
- If the orange colour is decolourised rapidly → sample is the alkene (addition reaction across C=C). (Identified — stop testing this sample.)
- If no immediate decolourisation → proceed to Step 2 for remaining three samples.
Step 2 — Test with Tollens' reagent (ammoniacal AgNO₃), warm gently:
- If a silver mirror forms on the test tube walls → sample is the alkanal (oxidised to carboxylic acid, Ag⁺ reduced to Ag). (Identified — stop testing this sample.)
- If no silver mirror forms → proceed to Step 3 for remaining two samples.
Step 3 — Test with sodium hydrogen carbonate (NaHCO₃) solution:
- If effervescence (gas bubbles, CO₂) is observed → sample is the alkanoic acid (acid-carbonate reaction releasing CO₂). (Identified — stop testing this sample.)
- If no effervescence → sample is the haloalkane (by elimination, confirmed as it does not react with any of the above reagents).
Optional confirmation for haloalkane: Warm with aqueous AgNO₃/ethanol; formation of a precipitate (white/cream/yellow depending on halogen) confirms a haloalkane.
b) Both alkanals and alkanones contain a carbonyl group, but Tollens' reagent specifically oxidises only aldehydes (which have a hydrogen directly attached to the carbonyl carbon, making them susceptible to oxidation), while ketones lack this hydrogen and cannot be oxidised in the same way, giving "no reaction" for ketones. However, "no reaction" alone cannot distinguish a ketone from a compound that simply has no carbonyl at all, so a positive test for the carbonyl group itself is needed first, such as 2,4-dinitrophenylhydrazine (Brady's reagent), which forms an orange/yellow precipitate with BOTH aldehydes and ketones (confirming presence of C=O), after which a negative Tollens' result would then specifically confirm the sample is a ketone rather than simply "no carbonyl present."
Question 12
a) Moles CO₂ = 8.8/44 = 0.2 mol → moles C = 0.2 mol → mass C = 0.2×12 = 2.4g
Moles H₂O = 3.6/18 = 0.2 mol → moles H = 0.4 mol → mass H = 0.4×1 = 0.4g
Total mass C+H = 2.4+0.4 = 2.8g = mass of X (matches given 2.8g, confirming X contains only C and H — a pure hydrocarbon)
Ratio C:H = 0.2 : 0.4 = 1 : 2
Empirical formula: CH₂
b) Molar mass M = 2×VD = 2×28 = 56
Empirical formula mass (CH₂) = 14
Molecular mass/empirical mass = 56/14 = 4
Molecular formula: C₄H₈
c) Since X decolourises bromine water rapidly, X contains a C=C double bond (an alkene). A possible structure consistent with C₄H₈:
But-1-ene: CH₂=CH—CH₂—CH₃
d) Reaction with HBr:
CH₂=CHCH₂CH₃ + HBr → CH₃CHBrCH₂CH�3 (major product)
By Markovnikov's rule, the hydrogen atom of HBr adds to the carbon of the double bond already bearing the greater number of hydrogen atoms (C1), while the bromine adds to the more substituted carbon (C2), because this pathway proceeds via the more stable secondary carbocation intermediate rather than the less stable primary carbocation, making 2-bromobutane the major (predominant) product.