Organic Chemistry II Problem Solving

The ability to solve multi-step organic chemistry problems is the ultimate test of your understanding at A-Level. This lesson brings together everything from Organic Chemistry II -- carbonyl compounds, carboxylic acids, esters, amines, amino acids, polymers, synthesis, and mechanisms -- and applies it to the kinds of extended problems you will encounter in the exam.

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Multi-Step Synthesis Design

Synthesis problems at A-Level typically provide a starting material and a target molecule, and ask you to plan the route between them. The approach is always the same:

1. Compare the starting material and the target -- what has changed? (functional group, chain length, or both)
2. Work backwards from the target (retrosynthesis) to identify the immediate precursor
3. Choose reagents and conditions for each step
4. Check that each step is chemically reasonable and does not introduce unwanted side reactions
5. Verify the carbon count at each stage

Worked Example 1

Starting material: Propan-1-ol Target: Propanoyl chloride

Analysis: The functional group changes from alcohol (-OH) to acyl chloride (-COCl). The carbon chain length stays the same (3 carbons). The route must go through the carboxylic acid, because there is no direct route from alcohol to acyl chloride.

Route:

1. Propan-1-ol --> Propanoic acid: Oxidise with K2Cr2O7/H2SO4 under reflux
2. Propanoic acid --> Propanoyl chloride: React with SOCl2 (producing SO2 and HCl as gaseous by-products)

Worked Example 2

Starting material: 1-bromobutane (4C) Target: Pentanoic acid (5C)

Analysis: Chain must extend by 1 carbon. The nitrile route is needed.

Route:

1. 1-bromobutane + KCN --> Pentanenitrile (5C) -- KCN in ethanol/water, reflux
2. Pentanenitrile --> Pentanoic acid -- Dilute H2SO4, heat under reflux (hydrolysis)

Worked Example 3

Starting material: Ethanol Target: N-ethylethanamide (CH3CONHCH2CH3)

Analysis: The target is an amide. It requires an acyl component (ethanoyl, from ethanoic acid) and an amine component (ethylamine). Both must come from ethanol.

Route:

1. Ethanol --> Ethanal: K2Cr2O7/H2SO4, distil
2. Ethanal --> Ethanoic acid: K2Cr2O7/H2SO4, reflux
3. Ethanoic acid --> Ethanoyl chloride: SOCl2
4. Separately: Ethanol --> Bromoethane: NaBr/conc. H2SO4
5. Bromoethane --> Ethylamine: Excess NH3, sealed tube, heat
6. Ethanoyl chloride + Ethylamine --> N-ethylethanamide + HCl

This is a 6-step synthesis from a single starting material. The key insight is that the ethanol must be split into two streams -- one for the acid pathway and one for the amine pathway.

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Identifying Unknown Compounds from Reactions

A common exam format provides you with an unknown compound and the results of several chemical tests. You must deduce the identity of the compound.

Systematic Approach

1. Note the molecular formula if given -- calculate the degree of unsaturation (index of hydrogen deficiency, IHD)

   • IHD = (2C + 2 + N - H - X) / 2 (where C, N, H, X are the number of carbon, nitrogen, hydrogen, and halogen atoms)
   • IHD = 1 means one ring or one double bond
   • IHD = 4 usually means a benzene ring

2. Interpret each test result to identify or rule out functional groups:

Test	Positive Result	Functional Group Indicated
Na2CO3 (effervescence)	CO2 gas produced	Carboxylic acid (-COOH)
Tollens' reagent (warm)	Silver mirror	Aldehyde (-CHO)
Fehling's solution (warm)	Brick-red precipitate	Aldehyde (-CHO)
2,4-DNP	Yellow/orange precipitate	Carbonyl (C=O) -- aldehyde or ketone
Br2(aq) (decolourises)	Orange to colourless	Alkene (C=C)
PCl5 (white fumes of HCl)	HCl gas produced	-OH group (alcohol or acid)
NaBH4 reduction	Gives secondary alcohol	Ketone
NaBH4 reduction	Gives primary alcohol	Aldehyde

3. Combine the evidence to narrow down the structure
4. Propose a structure consistent with ALL observations

Worked Deduction

An unknown compound C3H6O2:

• Gives effervescence with Na2CO3
• Reacts with ethanol + H2SO4 to give a sweet-smelling product
• Has a pH of approximately 2.8 in solution

Deduction: Effervescence with Na2CO3 indicates a carboxylic acid. The molecular formula C3H6O2 with -COOH gives CH3CH2COOH --> propanoic acid. Confirmed by ester formation with ethanol and acidic pH.

Another Worked Deduction

An unknown compound C4H8O:

• No effervescence with Na2CO3 (not a carboxylic acid)
• Yellow/orange precipitate with 2,4-DNP (contains C=O)
• No silver mirror with Tollens' (not an aldehyde)
• Reduction with NaBH4 gives butan-2-ol (a secondary alcohol)

Deduction: Contains C=O (2,4-DNP positive) but is not an aldehyde (Tollens' negative) -- so it is a ketone. Reduction gives butan-2-ol, meaning the C=O is at position 2. The compound is butanone (CH3COCH2CH3).

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Predicting Products and Mechanisms

When given a reaction, you should be able to:

1. Identify the functional group in the starting material
2. Identify the reagent type (nucleophile, electrophile, oxidising agent, reducing agent, base)
3. Predict the product based on the functional group interconversion rules
4. State or draw the mechanism if required

Quick Reference for Reagent Identification

Reagent	Role	What It Does
NaBH4	Reducing agent (mild)	Reduces aldehydes to 1 degree alcohols, ketones to 2 degree alcohols
LiAlH4	Reducing agent (strong)	Reduces carboxylic acids, aldehydes, ketones, nitriles
K2Cr2O7/H2SO4	Oxidising agent	Oxidises alcohols and aldehydes
HCN/KCN	Nucleophile	Nucleophilic addition to C=O (extends chain by 1C)
NaOH(aq)	Nucleophile	Substitution on halogenoalkanes; hydrolysis of esters
NaOH(ethanol)	Base	Elimination from halogenoalkanes
NH3 (excess)	Nucleophile	Substitution on halogenoalkanes (amine formation)
KCN	Nucleophile	Substitution on halogenoalkanes (nitrile formation, +1C)
SOCl2	Chlorinating agent	Converts -COOH to -COCl; converts -OH to -Cl
Conc. H2SO4	Catalyst / dehydrating agent	Esterification catalyst; elimination of alcohols
Br2(aq)	Electrophile / test reagent	Tests for C=C; electrophilic addition

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Combining Analytical Data with Organic Knowledge

At A-Level, you may be given spectroscopic data (infrared spectra, mass spectra) alongside chemical test results.

Infrared (IR) Absorption

Bond	Wavenumber Range (cm-1)	Appearance	Functional Group
O-H (acid)	2500-3300	Very broad	Carboxylic acid
O-H (alcohol)	3200-3600	Broad	Alcohol
N-H	3300-3500	Medium, may be sharp	Amine or amide
C-H	2850-3100	Medium	Present in most organic compounds
C=O	1680-1750	Strong, sharp	Carbonyl (aldehyde, ketone, acid, ester, amide)
C-O	1000-1300	Medium	Alcohol, ester

Key diagnostic patterns:

• Broad O-H at 2500-3300 cm-1 + C=O at ~1710 cm-1 = carboxylic acid
• Sharp C=O at ~1740 cm-1 without broad O-H = ester
• C=O at ~1720 cm-1 without broad O-H = aldehyde or ketone
• Broad O-H at 3200-3600 cm-1 without C=O = alcohol

Mass Spectrometry

The molecular ion peak (M+) gives the molecular mass. Fragmentation patterns can indicate the loss of specific groups:

Loss from M+	Fragment Lost	Possible Group
-15	CH3	Methyl group
-17	OH	Hydroxyl group
-18	H2O	Alcohol or carboxylic acid
-28	CO	Aldehyde or ketone
-29	CHO or C2H5	Aldehyde or ethyl group
-45	OC2H5 or COOH	Ethoxy group or carboxyl group

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Decision Framework for Multi-Step Problems

flowchart TD
    A[Compare starting material to target] --> B{Same carbon count?}
    B -->|Yes| C[Functional group interconversion only]
    B -->|No: target has more C| D[Need chain extension -- use KCN route]
    B -->|No: target has fewer C| E[Unusual at A-Level -- check question]
    C --> F{What functional group changes?}
    F -->|"Alcohol to acid"| G["Oxidise: K2Cr2O7/H2SO4, reflux"]
    F -->|"Alcohol to amine"| H["Via halogenoalkane: 1. NaBr/H2SO4  2. Excess NH3"]
    F -->|"Acid to ester"| I["Via acyl chloride: 1. SOCl2  2. ROH"]
    F -->|"Acid to amide"| J["Via acyl chloride: 1. SOCl2  2. RNH2"]
    D --> K["1. Convert to halogenoalkane<br>2. React with KCN<br>3. Hydrolyse or reduce the nitrile"]

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Common Mistakes to Avoid

1. Confusing oxidation conditions: Distillation gives the aldehyde; reflux gives the carboxylic acid. Many students get this backwards. Remember: distil = "d" for "don't over-oxidise."

2. Forgetting that Fischer esterification is reversible: The yield is limited by equilibrium. Acyl chloride routes are irreversible and go to completion.

3. Using NaBH4 to reduce carboxylic acids: NaBH4 is too mild. You need LiAlH4 in dry ether.

4. Mixing up nucleophilic addition and addition-elimination: Addition has no leaving group (aldehydes/ketones). Addition-elimination has a leaving group that departs (acyl chlorides).

5. Ignoring the solvent for NaOH reactions: Aqueous NaOH --> substitution. Ethanolic NaOH --> elimination. This single word changes the entire reaction pathway.

6. Drawing curly arrows from the wrong place: Always from the electron-rich source to the electron-poor sink. Examiners are strict on this.

7. Not checking the carbon count: If the target has more carbons than the starting material, you need a chain extension step (KCN). If the count is the same, you do not.

8. Forgetting to activate the carboxylic acid: To form an amide efficiently, you cannot just mix a carboxylic acid with an amine (they form a salt). Convert the acid to an acyl chloride first, then react with the amine.

9. Proposing reactions that do not exist at A-Level: Stick to the reactions in the specification. Do not invent steps or reagents.

10. Not giving full reagent details: "Dichromate" is not enough -- write K2Cr2O7/H2SO4 and specify distil or reflux. "Ammonia" is not enough -- write "excess ammonia, sealed tube, heat."

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Summary

Problem solving in organic chemistry requires you to integrate functional group knowledge, reagent recognition, mechanism drawing, and analytical interpretation. The exam rewards students who can think systematically: compare structures, work backwards, and apply the correct reagent and mechanism for each transformation. Practice is essential -- the more synthesis routes and deduction problems you work through, the more instinctive the reasoning becomes.

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A-Level Deep Dive: Organic Chemistry II Problem Solving

Spec mapping

Edexcel 9CH0 specification — synoptic across Topics 16, 17 and 18 plus Topic 19 (Modern analytical techniques) covers multi-stage organic problem-solving on Paper 3 of the A-Level: identification of unknown compounds from spectroscopic data (IR, ¹H NMR, ¹³C NMR, mass spectrometry), proposal of synthesis routes from given starting materials to identified targets, mechanism interpretation, and design of analytical confirmation procedures (refer to the official specification document for exact wording). Examined predominantly on Paper 3 — General and Practical Principles in Chemistry, often as a single 10–14 mark question that integrates spectroscopy + mechanism + synthesis + practical method. Connects to all CP6, CP7, CP14, CP15 and CP16 because problem-solving questions typically require students to outline experimental procedures alongside theoretical interpretation.

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Worked example with full mark scheme

Question (12 marks — extended Paper 3 style):

An unknown organic acid X has the following analytical data:

• Molecular formula: C8H8O3
• IR spectrum: broad absorption 2500–3300 cm⁻¹, sharp 1660 cm⁻¹, broad 3200 cm⁻¹
• ¹H NMR: δ 11.2 (1H, broad s), δ 9.8 (1H, broad s), δ 7.9 (1H, dd), δ 7.4 (1H, t), δ 6.9 (2H, m), δ 0 ppm (0H — silent)
• Mass spectrum: M+ at m/z = 152, fragment at m/z = 135 (loss of OH), fragment at m/z = 120 (loss of H2O from 138)

(a) Deduce the structure of X with full justification. (5) (b) Suggest a 3-step synthesis of X starting from phenol. State the regioselectivity at each step. (4) (c) Suggest one analytical method to confirm the identity of X by comparison to a literature standard. (3)

Solution with mark scheme:

(a) Step 1 — molecular formula analysis.

C8H8O3, M+ = 152. Degree of unsaturation = (2×8 + 2 − 8) / 2 = 5 (likely 1 aromatic ring = 4, plus 1 additional double bond or ring).

M1 — degree of unsaturation calculated, aromatic ring inferred.

Step 2 — IR interpretation.

• Broad 2500–3300 cm⁻¹ + sharp 1660 cm⁻¹ → carboxylic acid (-COOH).
• Broad 3200 cm⁻¹ → phenolic -OH (separate from -COOH).

M1 — both functional groups identified from IR.

Step 3 — NMR interpretation.

• δ 11.2 (1H, broad s) → -COOH proton.
• δ 9.8 (1H, broad s) → phenolic -OH proton.
• δ 7.9, 7.4, 6.9 ppm (4H total in aromatic region) → 4 aromatic H's → ortho-substituted benzene (1,2-pattern in NMR).

M1 — 4-aromatic-H pattern consistent with a 1,2-disubstituted benzene.

Step 4 — assemble the structure.

C8H8O3 with -COOH and -OH on adjacent positions of a benzene ring gives 2-hydroxybenzoic acid (salicylic acid).

M1 — structure assembled.

A1 — final answer: salicylic acid (HO-C6H4-COOH, 2-isomer).

(b) Synthesis of salicylic acid from phenol — Kolbe-Schmitt-like sequence:

Step 1 — Kolbe synthesis (carbonation). Treat phenoxide (formed from phenol + NaOH) with CO2 under high pressure (≈ 100 atm) and high temperature (≈ 125 °C). The CO2 inserts into the ortho-position of the phenol ring, giving sodium salicylate (after acidification). Alternative: at A-level, students may instead suggest a 3-step route: (i) sulfonation of phenol with conc. H2SO4 → 2-hydroxybenzenesulfonic acid; (ii) replacement of -SO3H by -COOH via NaCN + heat — but this is not standard A-Level.

A simpler 3-step route: (i) react phenol with NaOH(aq) to form sodium phenoxide; (ii) react with CO2 under pressure (Kolbe-Schmitt) to give sodium salicylate; (iii) acidify with dilute HCl to liberate salicylic acid.

M1 M1 — three steps with reagents (acceptable variation in details).

Regioselectivity: the -OH group of phenol is 2,4-directing; the Kolbe-Schmitt reaction gives the 2-product preferentially because of intramolecular chelation of the -O⁻ with the inserting CO2.

M1 — regioselectivity to 2-position justified.

A1 — overall product correctly identified.

(c) Analytical confirmation methods: