Organic Chemistry I Problem Solving

This final lesson brings together everything from the course. The key to success in A-Level organic chemistry is not just knowing individual reactions, but being able to link them together — predicting products, identifying reagents, drawing mechanisms, and planning multi-step syntheses.

---

Reaction Pathway Map

The following diagram shows the key interconversions you have learned. Being able to navigate this map is essential for synthesis questions.

graph TD
    A["Alkane"] -->|"Free radical sub. (X₂, UV)"| B["Halogenoalkane"]
    C["Alkene"] -->|"+ HX, room temp"| B
    C -->|"+ H₂O, H₃PO₄, 300 °C"| D["Alcohol"]
    C -->|"+ H₂, Ni, 150 °C"| A
    C -->|"+ Br₂, room temp"| E["Dihalogenoalkane"]
    C -->|"Polymerisation"| F["Addition Polymer"]
    B -->|"NaOH(aq), reflux"| D
    B -->|"KCN, ethanol/water, reflux"| G["Nitrile (chain +1C)"]
    B -->|"Excess NH₃, ethanol, sealed tube"| H["Amine"]
    B -->|"NaOH in ethanol, reflux"| C
    D -->|"K₂Cr₂O₇/H₂SO₄, distil (1° only)"| I["Aldehyde"]
    D -->|"K₂Cr₂O₇/H₂SO₄, reflux (1° only)"| J["Carboxylic Acid"]
    D -->|"K₂Cr₂O₇/H₂SO₄ (2° only)"| K["Ketone"]
    D -->|"Conc. H₂SO₄, 170 °C"| C
    D -->|"+ Carboxylic acid, conc. H₂SO₄, reflux"| L["Ester"]
    I -->|"K₂Cr₂O₇/H₂SO₄, reflux"| J

Notice the interconnections: alkenes can be converted to halogenoalkanes, which can be converted to alcohols, which can be dehydrated back to alkenes. These interconversions allow you to plan synthetic routes.

---

Predicting Products

To predict the product of a reaction, identify:

1. The functional group in the starting material (this determines which reactions are possible)
2. The reagent and conditions (this tells you which specific reaction occurs)
3. Apply the reaction to determine the product

Worked Example 1

Starting material: But-1-ene. Reagent: HBr at room temperature.

1. But-1-ene is an alkene (C=C present)
2. HBr adds to alkenes (electrophilic addition)
3. Apply Markovnikov's rule: H adds to C1 (more H atoms), Br to C2
4. Product: 2-bromobutane

Worked Example 2

Starting material: 2-bromobutane. Reagent: NaOH(aq), heat under reflux.

1. 2-bromobutane is a halogenoalkane
2. Aqueous NaOH acts as a nucleophile → nucleophilic substitution
3. OH⁻ replaces Br⁻
4. Product: butan-2-ol

Worked Example 3

Starting material: Butan-2-ol. Reagent: K₂Cr₂O₇/H₂SO₄, heat.

1. Butan-2-ol is a secondary alcohol
2. Secondary alcohols are oxidised to ketones
3. Product: butanone (whether distilled or refluxed)
4. Observation: solution changes from orange to green

---

Identifying Reagents and Conditions

Given a starting material and a product, work backwards to identify what reagent and conditions are needed.

Worked Example 4

Convert propan-1-ol to propanal.

1. Propan-1-ol is a primary alcohol; propanal is an aldehyde
2. This is oxidation of a primary alcohol (one stage only)
3. Reagent: acidified potassium dichromate, K₂Cr₂O₇ / H₂SO₄
4. Conditions: heat and distil (to collect aldehyde before further oxidation)

Worked Example 5

Convert propene to propan-2-ol.

1. Propene is an alkene; propan-2-ol is an alcohol
2. This is hydration (addition of water)
3. Reagent: steam (H₂O) with H₃PO₄ catalyst
4. Conditions: 300 °C, 60 atm
5. (Markovnikov's rule: H adds to CH₂, OH adds to CH giving propan-2-ol)

---

Drawing Mechanisms

In exams, you will be asked to draw mechanisms for:

• Free radical substitution (single-headed curly arrows, three stages)
• Electrophilic addition (double-headed curly arrows, two steps via carbocation)
• Nucleophilic substitution — SN1 (two steps via carbocation) or SN2 (one step, backside attack)

Common Mechanism Errors to Avoid

1. Wrong type of curly arrow: Use fish-hook (single-headed) only for radical mechanisms; use standard (double-headed) for all ionic mechanisms
2. Arrows starting from the wrong place: Curly arrows must start from an electron source (bond or lone pair), never from an atom or a positive charge
3. Missing charges: Always show the + on carbocations and the charges on ions
4. Missing dipoles: Show δ+ and δ− on polar bonds where relevant
5. Incomplete mechanisms: Show all steps — do not skip the carbocation intermediate in electrophilic addition or SN1

Mechanism Summary Table

Mechanism	Arrow Type	Steps	Key Intermediate	Favoured By
Free radical substitution	Fish-hook (½-headed)	3 stages (I, P, T)	Free radicals	UV light + halogen
Electrophilic addition	Double-headed	2 steps	Carbocation	C=C + electrophile
SN2	Double-headed	1 step	Transition state only	Primary RX + strong Nu⁻
SN1	Double-headed	2 steps	Carbocation	Tertiary RX

---

Multi-Step Synthesis

Sometimes you need to convert one compound to another that cannot be achieved in a single reaction. You must plan a synthetic route using two or more steps.

Worked Example 6

Convert ethane to ethanol.

There is no direct one-step conversion. But consider:

Step 1: Ethane → bromoethane (free radical substitution with Br₂, UV light) Step 2: Bromoethane → ethanol (nucleophilic substitution with NaOH(aq), reflux)

Worked Example 7

Convert propan-1-ol to propanoic acid.

Method: Oxidise with excess acidified K₂Cr₂O₇ under reflux. This takes propan-1-ol → propanal → propanoic acid in a single experimental procedure (though two oxidation stages occur).

Worked Example 8

Convert but-1-ene to butanenitrile.

Step 1: But-1-ene + HBr → 2-bromobutane (electrophilic addition, Markovnikov's rule)

But wait — this gives 2-bromobutane, and we need a CN on C1 for butanenitrile. The Markovnikov product places Br on C2, not C1. We need an alternative route.

Better route: Step 1: But-1-ene + H₂O (steam, H₃PO₄, 300 °C) → butan-2-ol (hydration)

This still places the functional group on C2. To get to butanenitrile (CN on C1), we need a C1 halogenoalkane.

Best route (starting from a different starting material): Start from 1-bromobutane. Treat with KCN in ethanol/water, reflux → pentanenitrile (5 carbons). But that is not butanenitrile either.

Actually, for butanenitrile (CH₃CH₂CH₂CN, 4 carbons including the CN carbon), we need a 3-carbon halogenoalkane: 1-bromopropane + KCN → butanenitrile.

This illustrates why thinking about the pathway holistically matters. The CN⁻ reaction adds one carbon — so to get a 4-carbon nitrile, start from a 3-carbon halogenoalkane.

Worked Example 9

Convert ethanol to ethyl ethanoate (an ester).

Step 1: Ethanol → ethanoic acid (oxidise with excess K₂Cr₂O₇/H₂SO₄, reflux). This goes through ethanal as intermediate. Step 2: Ethanoic acid + ethanol → ethyl ethanoate (esterification with conc. H₂SO₄ catalyst, reflux)

Note: you need both ethanol and ethanoic acid for the ester. You could oxidise some of the ethanol to make ethanoic acid, then react it with the remaining ethanol.

---

Isomer Identification in Synthesis

When planning a synthesis, remember that some reactions produce mixtures of isomers:

• Free radical substitution gives mixtures of position isomers (and multiple substitution products)
• Electrophilic addition to unsymmetrical alkenes gives major and minor products (Markovnikov)
• SN1 reactions on chiral substrates give racemic mixtures

Always consider whether the desired product is the major or minor product and whether purification will be needed.

---

Linking Mechanisms to Observations

In exam questions, you may be asked to explain observations in terms of mechanism:

Observation	Mechanistic Explanation
Reaction rate doubles when [OH⁻] doubles	OH⁻ is in the rate-determining step → SN2
Reaction rate unchanged when [OH⁻] doubles	OH⁻ not in the rate-determining step → SN1
Product is a racemic mixture	Planar carbocation intermediate → SN1
Product shows inversion of configuration	Backside attack → SN2
Bromine water decolourised	C=C present → electrophilic addition
Dichromate stays orange	Tertiary alcohol (or not an alcohol at all)
Dichromate turns green, Tollens gives silver mirror	Primary alcohol → aldehyde formed
Dichromate turns green, Tollens gives no reaction	Secondary alcohol → ketone formed

---

Strategy for Problem Solving

1. Identify the functional groups in the starting material and target molecule
2. Map the transformation: What bonds need to be made or broken?
3. Check your reaction map: Is there a direct route? If not, what intermediate is needed?
4. Write reagents and conditions for each step
5. Consider selectivity: Will the desired product be the major product?
6. Draw mechanisms if asked — check arrow types, charges, intermediates
7. Review your answer: Does the product have the correct molecular formula? Are the reagents and conditions matched correctly?

Being systematic is more important than being fast. Work through the logic step by step, and the pathway will reveal itself.

---

A-Level Deep Dive: Organic Chemistry I Problem Solving

Spec mapping

Edexcel 9CH0 specification Topics 6 and 7 — Organic Chemistry I (synoptic across both) assesses the integrated application of nomenclature, isomerism, alkane and alkene reactivity, free-radical and electrophilic mechanisms, halogenoalkane substitution and elimination, and alcohol oxidation. The synoptic problem-solving sub-topic appears across Paper 2 (multi-step synthesis questions, identification of unknown compounds from a battery of test data) and especially on Paper 3 (general and practical principles, where 12-mark synoptic questions ask candidates to identify an unknown organic compound from IR, NMR and chemical-test data, then propose a synthesis route from a given starting material) — refer to the official specification document for exact wording. The data booklet provides characteristic IR ranges, $^{1}H$1H NMR chemical-shift table, and $^{13}C$13C NMR chemical-shift table; candidates apply these alongside organic mechanism knowledge.

---

Worked example with full mark scheme

Question (12 marks): Compound X has molecular formula $C_4H_8O$C4​H8​O.

• IR: broad absorption at 3300 cm$^{-1}$−1; absorption at 1640 cm$^{-1}$−1.
• $^{1}H$1H NMR: peaks at δ 5.8 (multiplet, 1H), 5.2 (multiplet, 2H), 4.1 (broad singlet, 1H, exchangeable with D2O), 2.4 (multiplet, 2H), 0.9 (triplet, 3H).
• Chemical tests: gives orange precipitate with 2,4-DNPH? No. Decolourises Br2 water? Yes.

(a) Identify the functional groups present in X, giving evidence from each piece of data. (4)

(b) Propose a structure for X, naming the compound. (3)

(c) X can be converted to butanoic acid in two steps. Suggest the two-step synthesis, with reagents and conditions. (5)

Solution with mark scheme:

(a) Step 1 — IR analysis.

• Broad 3300 cm$^{-1}$−1 → O-H stretch (alcohol or carboxylic acid).
• 1640 cm$^{-1}$−1 → C=C stretch (alkene).

M1 — both functional groups (alcohol/acid + alkene) identified.

Step 2 — $^{1}H$1H NMR.

• δ 5.8 (1H, multiplet) → vinylic H (-CH=).
• δ 5.2 (2H, multiplet) → terminal =CH2 (two H on the same C of a C=C).
• δ 4.1 (1H, broad, D2O exchangeable) → -OH (the broad singlet is characteristic, and D2O exchange confirms the OH).
• δ 2.4 (2H, multiplet) → -CH2- adjacent to both an alkene and an OH.
• δ 0.9 (3H, triplet) → -CH3 adjacent to a -CH2-.

M1 — 5.2 and 5.8 ppm assigned to alkene H atoms (1H + 2H = 3H total); 4.1 ppm as -OH; 0.9 ppm as -CH3.

Step 3 — chemical tests.

• 2,4-DNPH negative → no carbonyl (C=O). Rules out aldehyde and ketone. Combined with broad 3300 cm$^{-1}$−1, this means -OH is alcohol (not carboxylic acid; acid would give C=O at ~1710 cm$^{-1}$−1 which is absent).
• Br2 water decolourisation → alkene confirmed.

A1 — alcohol + alkene are the two functional groups in X.

(b) Step 1 — assemble fragments.

From the NMR:

• =CH-CH2- (1H + 2H, allylic CH2)
• =CH2 (terminal alkene, 2H)
• -CH3 (terminal methyl, 3H, triplet)
• -OH (1H, broad)

But wait — the molecular formula is $C_4H_8O$C4​H8​O with degree of unsaturation $\frac{2 \cdot 4 + 2 - 8}{2} = 1$22⋅4+2−8​=1, consistent with one C=C and no rings. We need 4 carbons total.

M1 — count atoms: =CH2 (1C), =CH- (1C), -CH2- (1C), -CH3 (1C) = 4 carbons; 1 OH must be on one of these carbons.

The triplet at δ 0.9 (3H) tells us -CH3 is adjacent to one -CH2-. The fragment -CH2-CH3 is established (CH2 splits CH3 into a triplet). But the CH2 fragment from δ 2.4 (multiplet, 2H) is adjacent to the alkene H at δ 5.8. So there are two CH2 groups required? No — let me re-count.

If the structure is $CH_2=CH-CH(OH)-CH_3$CH2​=CH−CH(OH)−CH3​ (but-3-en-2-ol):

• =CH2 (2H, multiplet around 5.0–5.2 ppm) ✓
• =CH (1H, multiplet around 5.8 ppm) ✓
• -CH(OH) (1H, this would be a sextet ≈ 4.3 ppm)
• -OH (broad, ~3.5 ppm)
• -CH3 (3H, doublet, ~1.3 ppm)

But the NMR shows -CH3 as a triplet (next to -CH2-) and -OH at 4.1 ppm overlapping with -CH(OH)- region. Reconsidering with two CH2 groups visible:

Structure but-3-en-1-ol: $CH_2=CH-CH_2-CH_2-OH$CH2​=CH−CH2​−CH2​−OH. But this has 5H + 5H = wait, let me count. Formula $C_4H_8O$C4​H8​O. Atoms: =CH2 (2H), =CH (1H), -CH2- (2H), -CH2- (2H, with -OH), -OH (1H) = 8H ✓.

NMR check:

• =CH2 (2H, multiplet ~5.1 ppm) ✓
• =CH- (1H, multiplet ~5.7 ppm) ✓
• -CH2-CH2- (2H allylic ~2.3, 2H next to OH ~3.5) — two distinct CH2s.
• -OH (1H, broad)

But the NMR shows only one -CH2- multiplet at 2.4 (2H) and -CH3 at 0.9 (3H, triplet) — not two CH2s.

M1 — the triplet at 0.9 ppm (3H) clearly indicates -CH3 adjacent to -CH2-. The structure must contain -CH2-CH3 (an ethyl group).

Structure but-1-en-3-ol (more correctly but-3-en-2-ol) does not have an ethyl. Reconsider as 3-buten-1-ol? That has -CH2-CH2-OH + CH=CH2, with no -CH3 group at all.

The presence of the triplet at 0.9 ppm and only one CH2 multiplet at 2.4 ppm, with =CH2 (5.2) and =CH (5.8), along with -OH (4.1), strongly suggests but-1-en-3-ol is not correct either. Re-examining: $CH_2=CH-CH(OH)-CH_2-CH_3$CH2​=CH−CH(OH)−CH2​−CH3​? That is $C_5H_{10}O$C5​H10​O, not $C_4H_8O$C4​H8​O.

The most consistent structure for $C_4H_8O$C4​H8​O with the given data is indeed but-3-en-1-ol, $CH_2=CH-CH_2-CH_2-OH$CH2​=CH−CH2​−CH2​−OH, even though the methyl-as-triplet observation doesn't fit perfectly. (An alternative: reading the question carefully, the "0.9 (triplet, 3H)" might be a misprint or a CH3 group from a 5-C alcohol; either way, on the available evidence the structure is but-3-en-1-ol, accepting some NMR ambiguity.)

A1 — proposed structure: but-3-en-1-ol ($CH_2=CH-CH_2-CH_2-OH$CH2​=CH−CH2​−CH2​−OH).

(c) Two-step synthesis: but-3-en-1-ol → butanoic acid.

Step 1: hydrogenation of the C=C. Reagent: H2 with Ni catalyst, heat.

$CH_2=CH-CH_2-CH_2-OH \xrightarrow{H_2, \text{Ni}} CH_3-CH_2-CH_2-CH_2-OH \text{ (butan-1-ol)}$CH2​=CH−CH2​−CH2​−OHH2​,Ni​CH3​−CH2​−CH2​−CH2​−OH (butan-1-ol)

M1 — H2/Ni catalyst correctly identified.

A1 — product butan-1-ol.

Step 2: oxidation of primary alcohol to carboxylic acid. Reagent: acidified K2Cr2O7, heat under reflux.

$CH_3-CH_2-CH_2-CH_2-OH + 2[O] \xrightarrow{K_2Cr_2O_7/H^+, \text{reflux}} CH_3-CH_2-CH_2-COOH + H_2O$CH3​−CH2​−CH2​−CH2​−OH+2[O]K2​Cr2​O7​/H+,reflux​CH3​−CH2​−CH2​−COOH+H2​O

M1 — K2Cr2O7/H+ under reflux identified.

M1 — product butanoic acid identified.

A1 — overall scheme complete.

Total: 12 marks (M6 A6).

---

Specimen question modelled on the Edexcel 9CH0 Paper 3 format

Question (8 marks): A pharmaceutical intermediate Y is identified by:

• Mass spectrometry: M+ = 88 (with M+2 peak at 90, ratio 3:1, indicating one Cl).
• IR: sharp absorption 3300 cm$^{-1}$−1 (narrow); 1620 cm$^{-1}$−1 (C=C).
• $^{1}H$1H NMR: δ 5.8 (1H, m), 5.2 (2H, m), 3.3 (1H, broad), 1.4 (3H, d), no signals between 0.8 and 1.2 ppm.

(a) Determine the molecular formula of Y, given that it contains C, H, O and Cl. (2)

(b) Propose a structure for Y, justifying the assignment. (4)

(c) State whether Y is a primary, secondary or tertiary halogenoalkane and predict the product of hydrolysis with aqueous NaOH. (2)

Mark scheme decomposition by AO: