Difficulty: Advanced | Prerequisites: Nomenclature, structure/bonding, reactions roadmap, mechanisms
Multistep synthesis asks you to string known reactions together to build a target molecule from a simple starting material, typically in four or fewer steps. NMR spectroscopy gives you the reverse problem: deduce an unknown structure from a molecular formula, an IR spectrum and a proton NMR spectrum. Together these sections were worth 41 of 150 points (27%). Synthesis rewards methodical backwards planning; NMR rewards pattern recognition.
Retrosynthetic analysis
Working backwards from the target molecule to identify simpler precursors. You mentally "undo" each reaction to find a viable starting material and reagent sequence.
In simple terms, start at the finish line and ask "what reaction could have made this?" repeatedly until you reach something simple.
Degree of unsaturation (index of hydrogen deficiency, IHD)
Calculated from the molecular formula: IHD = (2C + 2 + N - H - X) / 2, where C = carbons, N = nitrogens, H = hydrogens, X = halogens. Each degree represents one ring or one double bond; two degrees can represent one triple bond.
Think of it as a count of "missing hydrogens" that tells you how many rings and/or pi bonds the molecule contains.
Chemical shift (delta, ppm)
The position of an NMR signal on the horizontal axis, measured in parts per million relative to a reference (TMS). It tells you about the electronic environment of the hydrogen(s) producing that signal.
In simple terms, chemical shift is the "address" of a proton signal on the NMR spectrum: higher ppm means the proton is more deshielded (near electronegative atoms or pi systems).
Splitting pattern (multiplicity)
The number of peaks in an NMR signal, determined by the n + 1 rule: a proton with n equivalent neighbouring protons appears as an (n + 1)-line pattern. A singlet (s) = 0 neighbours, doublet (d) = 1 neighbour, triplet (t) = 2, quartet (q) = 3.
In simple terms, splitting tells you how many hydrogens are on the carbons next door.
Integration
The area under an NMR signal, proportional to the number of equivalent protons producing it. Given as a ratio (e.g. 2H, 3H).
Think of integration as a headcount: how many hydrogens are in that particular environment.
Hydroboration-oxidation
BH3-THF, 2. NaOH/H2O2. Anti-Markovnikov, syn addition of OH to an alkene. The boron initially adds to the less hindered carbon.
Catalytic hydrogenation
H2 with a metal catalyst (Pd, Pt or Ni). Reduces alkenes to alkanes (and alkynes to alkanes) via syn addition. Lindlar's catalyst stops alkynes at the cis alkene stage.
The exam gave four possible reagent sequences to convert 2-methylpropene (isobutylene) to an ether product. The correct answer was B: i) NBS, heat; ii) CH3ONa; iii) H2/Pt.
The logic:
NBS with heat brominates at the allylic position (radical mechanism), installing Br on the carbon adjacent to the double bond.
CH3ONa (sodium methoxide) performs an SN2 substitution, replacing the allylic bromide with an OCH3 group (Williamson ether synthesis).
H2/Pt reduces the remaining double bond to give the saturated ether product.
Why the other options fail:
Option A starts with H2/Pt (reduces the double bond first), then Br2/light on a fully saturated compound gives a mixture of regioisomers, not selective allylic bromination.
Option C does NBS first (correct), then reduces the double bond before substitution, but now SN2 on a saturated secondary bromide gives poor selectivity.
Option D uses HBr with peroxide (anti-Markovnikov radical addition), which puts Br on the terminal carbon and changes the regiochemistry entirely.
Target: convert a secondary alcohol to a primary alcohol at a different position.
Approach:
Dehydrate with H2SO4 to form an alkene (elimination).
Use hydroboration-oxidation (1. BH3-THF, 2. NaOH/H2O2) to place the OH at the anti-Markovnikov position.
This two-step sequence effectively moves the OH from a secondary to a primary position.
Target: convert a methylcyclohexene (or toluene derivative with a methyl group) into a ketone.
Approach:
NBS, heat to brominate at the benzylic position.
Substitution to install the desired functional group.
Oxidation or hydration to reach the ketone.
Use Markovnikov hydration (HgSO4, H2SO4, H2O) to convert a terminal alkyne to a methyl ketone, if an alkyne intermediate is involved.
Work backwards from the target: what functional group is present, and what reaction creates it?
Identify the carbon skeleton changes: if carbons were added, a C-C bond-forming reaction was used (e.g. cyanide substitution, Grignard in Organic Chemistry II).
If no carbon-carbon bonds were made, you are doing functional-group interconversions on the same skeleton.
Keep syntheses to four or fewer steps for this exam.
Show all reagents for partial credit, even if you are unsure of the order.
The exam gave molecular formulae, an IR clue and two 1H NMR spectra. The workflow for solving these problems:
Calculate the degree of unsaturation (IHD) from the molecular formula.
Check the IR spectrum for key functional groups (broad O-H around 3200-3600 cm-1, sharp C=O around 1700 cm-1, C-H stretches above 3000 cm-1 for sp2 C-H).
Count the number of distinct signals in the 1H NMR: this tells you how many types of non-equivalent protons the molecule has.
Read the integration (number of H per signal) and the splitting pattern.
Use chemical shift ranges to assign each signal to a functional-group environment.
0-1 ppm: alkyl CH3, CH2 (far from electronegative groups)
1-2 ppm: allylic, alkyl CH, CH2 near unsaturation
2-3 ppm: benzylic CH, propargylic, CH next to C=O
3-4 ppm: CH next to O or N (ethers, alcohols, amines)
4-5 ppm: O-H (variable, often broad), allylic/benzylic alcohols
5-6.5 ppm: vinyl (alkene) C-H
6.5-8.5 ppm: aromatic C-H
9-10 ppm: aldehyde C-H
Molecule A (C9H10)
IHD = (2 x 9 + 2 - 10) / 2 = 5. Four degrees from a benzene ring, one more from an additional double bond or ring.
A is treated with BH3/NaOH/H2O2 to give B (C9H12O), which is an anti-Markovnikov alcohol addition, meaning A contains a double bond.
A is also reduced with H2/Ni to give C (C9H12), a fully saturated product.
A is therefore a styrene derivative: a benzene ring with a propenyl side chain. The structure is para-methylstyrene (or a related C9H10 isomer with a vinyl group on a substituted benzene).
Molecule B (C9H12O) from hydroboration-oxidation of A
1H NMR of B: 2H doublet + 2H doublet near 7 ppm (para-substituted benzene ring, 4 aromatic H appearing as two doublets), 1H broad singlet around 5 ppm (O-H), 2H triplet and 2H triplet around 2.5-3.5 ppm (CH2-CH2 adjacent to each other), 3H singlet around 2 ppm (aromatic methyl, CH3).
IR stretch at 3100 cm-1 suggests sp2 C-H or O-H.
The structure: a para-substituted benzene ring with a CH3 group on one side and a -CH2CH2OH chain on the other.
Molecule C (C9H12) from catalytic hydrogenation of A
1H NMR of C: 2H doublet + 2H doublet near 7 ppm (para-disubstituted benzene), 2H quartet around 2.5 ppm, 3H singlet around 2.3 ppm, 3H triplet around 1.2 ppm.
This pattern (quartet + triplet) indicates an ethyl group (CH2CH3). Combined with the methyl singlet and para-substituted ring, C is 4-ethyltoluene (1-ethyl-4-methylbenzene).
NMR spectroscopy is the primary tool for confirming the identity and purity of synthesised compounds in pharmaceutical research, chemical manufacturing and forensic chemistry. Every new drug candidate is characterised by 1H and 13C NMR before it advances to biological testing.
Students often forget to calculate the degree of unsaturation before attempting NMR interpretation. The IHD immediately narrows the structural possibilities. Four or more degrees almost always means a benzene ring is present.
Students often confuse the n + 1 rule: the splitting pattern of a signal is determined by the number of non-equivalent neighbouring protons, not the protons on the same carbon. A CH3 group next to a CH2 appears as a triplet (2 + 1 = 3 peaks), and the CH2 appears as a quartet (3 + 1 = 4 peaks).
Students often assume a broad singlet must be an N-H. In 1H NMR at this level, a broad singlet between 1 and 5 ppm is more commonly an O-H (alcohol), especially if the IR confirms an O-H stretch.
Students often treat multistep synthesis as a forward problem. Working backwards from the target (retrosynthetic analysis) is almost always more efficient: identify the last reaction that could form the target's key functional group or bond, then repeat for the precursor.
Fill in the blank: The degree of unsaturation for C6H6 is ____.
Answer: 4 (consistent with benzene: three double bonds + one ring, or more precisely, benzene's delocalised structure)
True or false: A quartet in 1H NMR means the proton has four equivalent neighbours.
Answer: False. A quartet means it has three equivalent neighbours (n + 1 = 4).
Fill in the blank: Hydroboration-oxidation gives ____ (Markovnikov / anti-Markovnikov) addition of OH with ____ (syn / anti) stereochemistry.
Answer: anti-Markovnikov, syn
True or false: Aromatic protons typically appear between 6.5 and 8.5 ppm in a 1H NMR spectrum.
Answer: True.
True or false: In a para-disubstituted benzene ring, the aromatic protons appear as two doublets (each integrating for 2H).
Answer: True. The two pairs of equivalent protons on either side of the substituents couple to each other, giving two doublets.
Q: A compound has the molecular formula C8H8. Calculate its degree of unsaturation and suggest what structural features might be present.
A: IHD = (2 x 8 + 2 - 8) / 2 = 5. Five degrees of unsaturation suggests a benzene ring (4 IHD) plus one additional double bond or ring. A likely structure is styrene (vinylbenzene).
Q: You have an unknown with formula C4H8O. Its 1H NMR shows a triplet (3H) at 1.1 ppm, a singlet (3H) at 2.1 ppm and a quartet (2H) at 2.5 ppm. Propose a structure.
A: IHD = 1, so one double bond or ring. The singlet at 2.1 ppm suggests a CH3 next to a carbonyl. The quartet + triplet pattern indicates an ethyl group (CH2CH3). The structure is butanone (methyl ethyl ketone, CH3COCH2CH3).
Q: Propose a two-step synthesis to convert cyclohexanol to cyclohexene oxide (1,2-epoxycyclohexane). Assume any common reagents are available.
A: Step 1 is not needed since we already have cyclohexanol. We would first dehydrate to cyclohexene (H2SO4, heat), then epoxidise with mCPBA (or similar peracid). Alternatively, if the starting material is already cyclohexene, one step with mCPBA suffices.
Q: In 1H NMR, a signal at 7.2 ppm appears as two doublets (2H each). What does this splitting pattern suggest about the substitution on a benzene ring?
A: Two doublets of 2H each in the aromatic region indicate a para-disubstituted benzene ring. Each pair of equivalent aromatic protons couples with the adjacent pair.
Q: Why does NBS with heat selectively brominate at the benzylic position rather than elsewhere on the molecule?
A: The benzylic radical intermediate is stabilised by resonance with the aromatic ring. NBS maintains a low, steady concentration of Br2, which favours the radical chain mechanism through the most stable radical intermediate.
Multistep synthesis was worth 19 points (13%). NMR was worth 22 points (15%).
The synthesis multiple-choice question (Q5.i) tests whether you understand the order in which reactions must be performed. Think about what functional group each reagent needs as its starting point.
Open-ended synthesis problems (Q5.ii) give partial credit for showing correct reagents even if the overall sequence is wrong. Always write something.
NMR structural elucidation (Q6) is heavily weighted. Practise the workflow: formula to IHD to IR to NMR signals to structure. The exam provided chemical shift and IR reference tables on the last page.
Multistep synthesis is the capstone: it requires you to recall every reaction from the roadmap and every selectivity rule from the mechanisms section, then sequence them correctly. NMR connects to structure and bonding because chemical shift depends on hybridisation and electronegativity, and splitting patterns depend on molecular connectivity. Both topics prepare you for Organic Chemistry II, where longer syntheses and 13C NMR interpretation become standard.
multistep synthesis, retrosynthetic analysis, retrosynthesis, target molecule, starting material, reagent sequence, NMR spectroscopy, proton NMR, 1H NMR, chemical shift, ppm, splitting pattern, multiplicity, singlet doublet triplet quartet, integration, degree of unsaturation, index of hydrogen deficiency, IHD, molecular formula, IR spectroscopy, hydroboration oxidation, catalytic hydrogenation, Lindlar catalyst, NBS radical bromination, Williamson ether synthesis, para-disubstituted benzene, structural elucidation, CHEM 2301, organic chemistry I, Pomerantz, University of Minnesota