1. Multi-Step Organic Synthesis Design
In advanced organic synthesis, target molecules are built from simpler starting materials by chaining functional group transformations. Synthetic design requires mastery of reaction types, reagents, reaction conditions, and hazards:
Key Synthetic Strategies for A2 Organic Chemistry
- Carbon Chain Extension:
- Haloalkane + KCN(ethanolic) -> Nitrile (adds 1 carbon).
- Carbonyl + KCN/H+ -> 2-Hydroxynitrile (adds 1 carbon).
- Benzene + Acyl chloride + AlCl3 -> Aromatic ketone (Friedel-Crafts acylation). - Functional Group Interconversions:
- Nitrile (R-CN) + 2H2O + HCl -> Carboxylic acid (R-COOH) + NH4Cl [Acid Hydrolysis].
- Nitrile (R-CN) + 4[H] (H2/Ni or LiAlH4) -> Primary amine (R-CH2-NH2) [Reduction].
- Primary alcohol -> Aldehyde (distillation with K2Cr2O7/H2SO4) -> Carboxylic acid (reflux).
- Nitrobenzene + Sn/HCl + NaOH -> Phenylamine.
2. Carbon-13 (13C) NMR Spectroscopy
Carbon-13 is a stable isotope of carbon with nuclear spin (unlike the abundant carbon-12, which has zero spin). 13C NMR spectroscopy provides direct structural information about the carbon skeleton:
- Number of peaks = Number of non-equivalent carbon environments. If a molecule has 6 carbons but exhibits only 3 peaks, molecular symmetry is present.
- Chemical shift (delta in ppm): Indicates the electronic environment of the carbon atom:
- Aliphatic alkyl carbons (C-C): 0 to 50 ppm.
- Carbons bonded to electronegative atoms (C-O, C-Cl, C-N): 50 to 90 ppm.
- Alkene and aromatic carbons (C=C, aromatic ring): 100 to 160 ppm.
- Carbonyl carbons (C=O in esters, acids, amides): 160 to 185 ppm.
- Carbonyl carbons (C=O in aldehydes, ketones): 190 to 220 ppm.
3. High-Resolution 1H NMR Spectrum Diagram
The spectrum below displays the high-resolution 1H NMR spectrum of ethyl ethanoate (CH3COOCH2CH3), illustrating chemical shift positions, integration values, and splitting patterns:
4. Proton (1H) NMR Spectroscopy Principles
High-resolution proton NMR is the most powerful spectroscopic tool for organic structure determination. Every spectrum provides four crucial pieces of information:
| Spectral Feature | Physical Information Revealed | Example Deduction |
|---|---|---|
| 1. Number of Signals | Number of non-equivalent proton (hydrogen) environments. | Propan-1-ol has 4 signals (-CH3, -CH2-, -CH2-O, -OH). Propan-2-ol has 3 signals (two equivalent -CH3 groups). |
| 2. Chemical Shift (delta / ppm) | The chemical environment and shielding/deshielding by nearby electronegative atoms relative to TMS at 0 ppm. | Protons on -CH3 appear at ~0.9 ppm; protons on -CH2-O- appear deshielded at ~4.0 ppm. |
| 3. Integration Trace | The relative ratio of protons responsible for each signal. | Integration ratio of 3 : 2 : 1 indicates 3 protons in one environment, 2 in another, and 1 in the third. |
| 4. Splitting Pattern | The number of non-equivalent protons on immediately adjacent carbon atoms. | A triplet indicates 2 adjacent protons (2 + 1 = 3). A quartet indicates 3 adjacent protons (3 + 1 = 4). |
Why TMS and Deuterated Solvents (CDCl3) are Used
- Tetramethylsilane (TMS, (CH3)4Si): Used as the internal standard assigned delta = 0.00 ppm. All 12 protons are in identical environments, giving a single sharp, intense signal well away from almost all organic signals. It is non-toxic, chemically unreactive, and volatile (b.p. 27 degrees C), so it evaporates easily to recover the sample.
- Deuterated Trichloromethane (CDCl3): Deuterium (2H or D) has an even mass number and nuclear spin 1, so it produces no signal in proton (1H) NMR. It dissolves the organic sample without obscuring the analyte peaks.
5. Spin-Spin Splitting & The (n + 1) Rule
The magnetic field experienced by a proton is perturbed by the nuclear spin magnetic fields of non-equivalent protons on adjacent carbon atoms. This spin-spin coupling causes peaks to split according to the (n + 1) rule, where n is the number of non-equivalent protons on adjacent carbons:
| Adjacent Protons (n) | Splitting Pattern (n + 1) | Relative Peak Area Ratios | Typical Structural Fragment |
|---|---|---|---|
| 0 | Singlet | 1 | Isolated methyl group (e.g. -CO-CH3, -O-CH3) or quaternary carbon. |
| 1 | Doublet | 1 : 1 | Proton adjacent to a -CH- group. |
| 2 | Triplet | 1 : 2 : 1 | Proton adjacent to a -CH2- group (e.g. the -CH3 of an ethyl group: -CH2-CH3). |
| 3 | Quartet | 1 : 3 : 3 : 1 | Proton adjacent to a -CH3 group (e.g. the -CH2- of an ethyl group: -CH2-CH3). |
Whenever you see a triplet (integrating to 3H) paired with a quartet (integrating to 2H) in a 1H NMR spectrum, you have identified an ethyl group (-CH2-CH3)!
6. D2O Exchange for Identifying -OH and -NH Protons
Protons bonded to electronegative oxygen (-OH in alcohols and carboxylic acids) or nitrogen (-NH in amines and amides) undergo rapid proton exchange in solution. Consequently, they usually appear as broad singlets that do NOT participate in spin-spin splitting.
- Run the initial 1H NMR spectrum and record the peak position.
- Add a few drops of deuterium oxide (D2O) and shake the NMR tube.
- The acidic labile proton rapidly exchanges with deuterium:
R-OH + D2O <=> R-OD + HOD - Because deuterium produces no signal in 1H NMR, the original -OH (or -NH) peak completely disappears from the spectrum!
7. Worked Spectral Deductions
IR: Sharp strong absorption at 1740 cm-1.
1H NMR:
- delta 1.25 ppm: Triplet, integration = 3H
- delta 2.05 ppm: Singlet, integration = 3H
- delta 4.10 ppm: Quartet, integration = 2H
Deduce the structure of the ester and justify your answer.
Step 1: Functional group identification:
IR absorption at 1740 cm-1 corresponds to a carbonyl group (C=O) of an ester.
Step 2: Analyse 1H NMR signals:
- Triplet at 1.25 ppm (3H) and Quartet at 4.10 ppm (2H): Proves an ethyl group (-CH2-CH3). Because the -CH2- quartet is deshielded at 4.10 ppm, it must be bonded directly to the ester oxygen: -O-CH2-CH3.
- Singlet at 2.05 ppm (3H): Represents a methyl group with zero adjacent protons, bonded to the carbonyl carbon: CH3-C=O.
Step 3: Assemble the molecule:
CH3-CO-O-CH2-CH3 (Ethyl ethanoate)
(a) Pentan-2-one: CH3-CO-CH2-CH2-CH3
(b) Pentan-3-one: CH3-CH2-CO-CH2-CH3
Solution:
- (a) Pentan-2-one: Asymmetrical molecule. All 5 carbons are in different electronic environments (C1 methyl, C2 carbonyl, C3 methylene, C4 methylene, C5 methyl). Expected: 5 peaks.
- (b) Pentan-3-one: Symmetrical molecule with a plane of symmetry through the C3 carbonyl group. Carbons 1 and 5 are identical; Carbons 2 and 4 are identical; Carbon 3 is unique. Expected: 3 peaks.