This chapter covers Advanced Higher Chemistry content, extending beyond Higher level.

Analytical Techniques

Spectroscopy Overview

Analytical chemistry uses physical methods to determine the composition and structure of substances. The main spectroscopic techniques are:

• Infrared (IR) spectroscopy
• Mass spectrometry (MS)
• Nuclear magnetic resonance (NMR) spectroscopy
• UV-Visible spectroscopy

Comparison of Techniques

Technique	Information provided	Sample type	Destructive?
IR	Functional groups	Solid/liquid/gas	No
MS	Molecular mass, fragments	Solid/liquid/gas	Yes
NMR	Molecular structure, proton environments	Solution	No
UV-Vis	Concentration, conjugation	Solution	No
TLC	Number of components, purity	Solution	No
GC	Separation, identification (with MS)	Volatile compounds	No (GC) / Yes (GC-MS)

Infrared Spectroscopy

IR spectroscopy identifies functional groups by measuring the absorption of infrared radiation. Covalent bonds absorb IR radiation at characteristic frequencies, causing them to vibrate.

Key absorptions:

Bond	Wavenumber (cm$^{-1}$)	Type
O-H (alcohol)	3200-3600	Broad
O-H (carboxylic acid)	2500-3300	Very broad
N-H	3300-3500	Medium
C-H	2850-3100	Weak-medium
C=O	1680-1750	Strong
C=C	1620-1680	Medium
C-O	1000-1300	Strong
C-Cl	600-800	Strong
C≡N	2210-2260	Medium

Fingerprint region: Below 1500 \mathrm{ cm^{-1}The pattern is unique to each molecule and can Be compared with reference spectra.

Worked Example 1: An organic compound with molecular formula \mathrm{C_3\mathrm{H_6\mathrm{O Shows a strong absorption at 1715 \mathrm{ cm^{-1} and a broad absorption at 2500-3300 \mathrm{ cm^{-1}. Identify the compound.

The strong absorption at 1715 \mathrm{ cm^{-1} indicates a \mathrm{C=\mathrm{O group. The very Broad absorption at 2500-3300 \mathrm{ cm^{-1} is characteristic of the \mathrm{O-\mathrm{H in a Carboxylic acid. The compound is propanoic acid (\mathrm{CH_3\mathrm{CH_2\mathrm{COOH).

Worked Example 2: An IR spectrum shows strong absorptions at 3300 \mathrm{ cm^{-1} (broad) and 1050 \mathrm{ cm^{-1}. Identify the functional groups present.

The broad absorption at 3300 \mathrm{ cm^{-1} indicates an O-H group (alcohol). The absorption at 1050 \mathrm{ cm^{-1} is in the C-O stretching region, consistent with a primary alcohol (\mathrm{C-O stretch at 1050 \mathrm{ cm^{-1} is characteristic of primary alcohols).

Mass Spectrometry

Mass spectrometry determines the molecular mass and structure by ionising molecules and separating Them by their mass-to-charge ratio ($m/z$).

Process:

1. Ionisation: Molecules are bombarded with electrons, ejecting an electron to form a molecular ion (\mathrm{M^+^\bullet)
2. Acceleration: Ions are accelerated by an electric field
3. Deflection: Ions are deflected by a magnetic field (lighter ions deflected more)
4. Detection: Ions hit a detector, producing a mass spectrum

Molecular ion peak: The peak at the highest $m/z$ value (for the most abundant isotope) gives The molecular mass.

Isotopic patterns:

• Chlorine: \mathrm{Cl^{35}:\mathrm{Cl^{37} \approx 3:1 ratio
• Bromine: \mathrm{Br^{79}:\mathrm{Br^{81} \approx 1:1 ratio
• Carbon: \mathrm{C^{12}:\mathrm{C^{13} \approx 99:1 ratio

Fragmentation:

Common fragmentation patterns:

Fragment	$m/z$	Origin
\mathrm{CH_3^+	15	Methyl group
\mathrm{OH^+	17	Hydroxyl
\mathrm{C_2\mathrm{H_5^+	29	Ethyl group
\mathrm{CH_3\mathrm{CO^+	43	Acetyl
\mathrm{C_4\mathrm{H_9^+	57	Butyl
\mathrm{C_6\mathrm{H_5^+	77	Phenyl

Worked Example 3: The mass spectrum of a chlorinated compound shows peaks at $m/z = 78$ and 80 In a 3:1 ratio. Identify the molecular ion.

The 3:1 ratio is characteristic of one chlorine atom. If the compound contains one chlorine, then:

• Peak at 78: contains \mathrm{Cl^{35}So the rest of the molecule has mass $78 - 35 = 43$.
• Peak at 80: contains \mathrm{Cl^{37}So the rest has mass $80 - 37 = 43$.

The molecular formula is \mathrm{C_2\mathrm{H_3\mathrm{Cl (mass: $24 + 3 + 35 = 62$… That does not Match). Let me recalculate.

$M$ with \mathrm{Cl^{35}: $78 - 35 = 43$. $M$ with \mathrm{Cl^{37}: $80 - 37 = 43$. The remainder Is 43, which could be \mathrm{C_2\mathrm{H_3\mathrm{O (mass $24 + 3 + 16 = 43$) or \mathrm{C_3\mathrm{H_7 (mass $36 + 7 = 43$). The molecular formula is \mathrm{C_2\mathrm{H_3\mathrm{ClO (chloroethanal, $M_r = 78$) or \mathrm{C_3\mathrm{H_7\mathrm{Cl (1-chloropropane or 2-chloropropane, $M_r = 78.5$… Not quite). Given the exact match, the answer Is chloroethanal (\mathrm{CH_2\mathrm{ClCHO).

Worked Example 4: The mass spectrum of a compound shows a molecular ion peak at $m/z = 58$ and a Prominent peak at $m/z = 43$. Suggest a structure.

$M = 58$. Possible formula: \mathrm{C_3\mathrm{H_6\mathrm{O (propanone, $M_r = 58$).

The peak at $m/z = 43$ corresponds to \mathrm{CH_3\mathrm{CO^+Formed by alpha cleavage of Propanone.

\mathrm{CH_3\mathrm{COCH_3 \to \mathrm{CH_3\mathrm{CO^+ + \mathrm{CH_3^\bullet

The compound is likely propanone.

Nuclear Magnetic Resonance (NMR) Spectroscopy

^1\mathrm{H NMR

Proton NMR provides information about the hydrogen environments in a molecule.

Chemical shift ($\delta$): Measured in ppm relative to TMS (tetramethylsilane) as reference ($\delta = 0$).

Proton environment	$\delta$ (ppm)
\mathrm{R-\mathrm{CH_3	0.9-1.2
\mathrm{R_2-\mathrm{CH_2	1.2-1.5
\mathrm{R-\mathrm{OH	1.0-5.5 (variable)
\mathrm{R-\mathrm{CH_2-\mathrm{X	2.0-4.5
\mathrm{R-\mathrm{CHO	9.0-10.0
\mathrm{R-\mathrm{COOH	10.0-13.0
Aromatic	6.5-8.5

Key features of an NMR spectrum:

1. Number of signals: Equals the number of different proton environments
2. Integration (area under peak): Proportional to the number of protons in that environment
3. Splitting (multiplicity): Given by the $n + 1$ rule (a signal is split into $n + 1$ peaks by $n$ neighbouring protons)

Derivation of the $n + 1$ rule:

Neighbouring protons on adjacent carbons interact (spin-spin coupling) because their nuclear spins Can be aligned with or against the external magnetic field. For $n$ equivalent neighbouring protons, There are $n + 1$ possible spin arrangements (and therefore $n + 1$ possible local magnetic fields Experienced by the observed proton). This gives $n + 1$ peaks with intensities following Pascal’s Triangle.

Worked Example 5: The ^1\mathrm{H NMR spectrum of a compound with formula \mathrm{C_3\mathrm{H_8\mathrm{O shows:

• A triplet at $\delta = 1.2$ (integration 3)
• A quartet at $\delta = 3.7$ (integration 2)
• A singlet at $\delta = 2.5$ (integration 1, exchanges with \mathrm{D_2\mathrm{O)

Identify the compound.

The singlet that exchanges with \mathrm{D_2\mathrm{O is an \mathrm{O-\mathrm{H proton. The triplet (3H) and quartet (2H) indicate a \mathrm{CH_3\mathrm{CH_2 group. The quartet is shifted to $\delta = 3.7$Suggesting the \mathrm{CH_2 is adjacent to an electronegative oxygen. The compound Is ethanol (\mathrm{CH_3\mathrm{CH_2\mathrm{OH).

Worked Example 6: A compound with molecular formula \mathrm{C_4\mathrm{H_{10}\mathrm{O has two ^1\mathrm{H NMR signals: a septet at $\delta = 3.9$ (1H) and a doublet at $\delta = 1.2$ (6H). Identify the compound.

The septet (7 peaks) indicates 6 neighbouring protons. The doublet (2 peaks) indicates 1 Neighbouring proton. The chemical shift of $\delta = 3.9$ suggests the proton is on a carbon bonded To oxygen. The compound is 2-methylpropan-2-ol: (\mathrm{CH_3)_3\mathrm{COH has three equivalent \mathrm{CH_3 groups (6H, doublet split by 1H) and one \mathrm{CH proton (1H, septet split by 6H).

Wait, (\mathrm{CH_3)_3\mathrm{COH has a quaternary carbon with OH, so the H on oxygen would be a Singlet. The compound is better described as (\mathrm{CH_3)_2\mathrm{CHCH_2\mathrm{OH… No. Let us Re-analyse.

A septet from 1 proton split by 6 equivalent protons, and a doublet from 6 protons split by 1 Proton. This pattern is characteristic of an isopropyl group: (\mathrm{CH_3)_2\mathrm{CH-. The Chemical shift of the CH proton at $\delta = 3.9$ suggests it is attached to oxygen. The compound is Propan-2-ol (\mathrm{CH_3\mathrm{CH(OH)CH_3).

^{13}\mathrm{C NMR

Carbon-13 NMR identifies the number of different carbon environments.

• No splitting (proton-decoupled)
• Each signal represents a unique carbon environment
• Chemical shifts range from 0-220 ppm

Carbon environment	$\delta$ (ppm)
\mathrm{C-\mathrm{C	0-50
\mathrm{C-\mathrm{O	50-90
\mathrm{C=\mathrm{C (alkene)	100-150
Aromatic	100-150
\mathrm{C=\mathrm{O	160-220

UV-Visible Spectroscopy

UV-Vis spectroscopy measures the absorption of ultraviolet and visible light by molecules.

Beer-Lambert Law:

$A = \varepsilon c l$

Where $A$ is absorbance, $\varepsilon$ is the molar absorptivity (L mol$^{-1}$ cm$^{-1}$), $c$ is Concentration (mol/L), and $l$ is the path length (cm).

Derivation:

The absorbance is proportional to the number of absorbing molecules in the light path. If a solution Of concentration $c$ and path length $l$ contains $c \times l$ moles per unit area, then doubling $c$ or $l$ doubles the number of absorbers and hence the absorbance. The proportionality constant is The molar absorptivity $\varepsilon$Which depends on the substance and wavelength.

Applications: Determining concentrations, following reaction kinetics, identifying conjugated Systems.

Worked Example 7: A solution of a compound with \varepsilon = 12500 \mathrm{ L mol^{-1}\mathrm{ cm^{-1} has an absorbance of 0.625 in a 1 \mathrm{ cm cuvette. Find the concentration.

c = \frac{A}{\varepsilon l} = \frac{0.625}{12500 \times 1} = 5.0 \times 10^{-5} \mathrm{ mol/L

Worked Example 8: A solution has absorbance 0.450 at 520 nm in a 2 cm cuvette. If \varepsilon = 15000 \mathrm{ L mol^{-1}\mathrm{ cm^{-1}Find the concentration.

c = \frac{A}{\varepsilon l} = \frac{0.450}{15000 \times 2} = 1.5 \times 10^{-5} \mathrm{ mol/L

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Chromatography

Principles

Chromatography separates mixtures based on differential distribution between a stationary phase and A mobile phase.

$R_f$ values:

R_f = \frac{\mathrm{distance travelled by substance}{\mathrm{distance travelled by solvent front}

Thin Layer Chromatography (TLC)

• Stationary phase: silica gel on a glass plate
• Mobile phase: solvent
• Visualisation: UV light or iodine vapour
• Used for quick, qualitative analysis

Gas Chromatography (GC)

• Stationary phase: high-boiling liquid coating inside a capillary column
• Mobile phase: inert carrier gas (He, N$_2$)
• Separates volatile compounds
• Coupled with mass spectrometry (GC-MS) for identification

Retention time: The time a compound takes to pass through the column. Used for identification by Comparison with standards.

High Performance Liquid Chromatography (HPLC)

• Stationary phase: packed column with small particles
• Mobile phase: liquid solvent (under high pressure)
• Used for non-volatile or thermally unstable compounds

Comparison of Chromatographic Techniques

Feature	TLC	GC	HPLC
Mobile phase	Liquid	Gas	Liquid
Stationary phase	Silica gel	Liquid on solid support	Solid particles
Sample type	Non-volatile, thermally stable	Volatile	Any (especially non-volatile)
Detection	Visual, UV	FID, MS	UV, MS
Resolution	Low	High	Very high
Speed	Minutes	Minutes	Minutes

Worked Example 9: In TLC, substance A travels 2.5 cm and substance B travels 4.0 cm while the Solvent front travels 8.0 cm. Calculate the $R_f$ values and identify which substance is more polar.

R_f(\mathrm{A) = \frac{2.5}{8.0} = 0.31 R_f(\mathrm{B) = \frac{4.0}{8.0} = 0.50

Substance A has a lower $R_f$ value, meaning it travels less far. On silica gel (a polar stationary Phase), more polar substances are more strongly attracted to the stationary phase and travel less Far. Therefore, substance A is more polar than substance B.

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Combined Spectroscopic Analysis

Systematic Approach to Structure Determination

1. Molecular formula: From mass spectrometry (molecular ion peak) or given data.
2. Degree of unsaturation (DBE): \mathrm{DBE = \frac{2C + 2 + N - H - X}{2} where C, N, H, X are the numbers of carbon, nitrogen, hydrogen, and halogen atoms.
3. Functional groups: From IR spectroscopy.
4. Carbon framework: From ^{13}\mathrm{C NMR (number of signals = number of different C environments).
5. Hydrogen environments: From ^1\mathrm{H NMR (chemical shift, integration, splitting).
6. Confirm: Check that the proposed structure is consistent with all data.

Worked Example 10: Combined IR, mass spec, and NMR data for an unknown compound \mathrm{C_3\mathrm{H_6\mathrm{O_2 shows: IR absorption at 1740 \mathrm{ cm^{-1}Molecular ion at $m/z = 74$And ^1\mathrm{H NMR singlet at $\delta = 3.7$ (3H), singlet at $\delta = 2.1$ (3H).

Step 1: Molecular formula \mathrm{C_3\mathrm{H_6\mathrm{O_2, $M_r = 74$.

Step 2: DBE $= (2 \times 3 + 2 - 6)/2 = (6 + 2 - 6)/2 = 1$. One double bond or ring.

Step 3: IR absorption at 1740 \mathrm{ cm^{-1} indicates a C=O group. This accounts for the DBE Of 1.

Step 4: ^1\mathrm{H NMR shows only two signals (singlets), indicating two types of protons that Are not coupled to each other. Integration 3:3, so 3H each.

Step 5: Singlet at $\delta = 3.7$ (3H) is characteristic of a \mathrm{OCH_3 group (methyl Ester). Singlet at $\delta = 2.1$ (3H) is a \mathrm{CH_3\mathrm{CO group (acetyl).

Conclusion: The compound is methyl ethanoate (\mathrm{CH_3\mathrm{COOCH_3).

Worked Example 11: A compound \mathrm{C_4\mathrm{H_8\mathrm{O_2 has ^1\mathrm{H NMR signals at $\delta = 1.2$ (triplet, 3H), $\delta = 2.3$ (quartet, 2H), and $\delta = 11.5$ (singlet, 1H). Identify the compound.

Step 1: Molecular formula \mathrm{C_4\mathrm{H_8\mathrm{O_2.

Step 2: DBE $= (2 \times 4 + 2 - 8)/2 = 1$. One double bond or ring.

Step 3: Singlet at $\delta = 11.5$ (1H) is characteristic of a carboxylic acid proton.

Step 4: Triplet (3H) at $\delta = 1.2$ and quartet (2H) at $\delta = 2.3$ indicate a \mathrm{CH_3\mathrm{CH_2 group. The quartet at $\delta = 2.3$ is shifted downfield, suggesting the \mathrm{CH_2 is adjacent to a carbonyl group.

Conclusion: The compound is propanoic acid (\mathrm{CH_3\mathrm{CH_2\mathrm{COOH).

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Common Pitfalls

1. IR spectroscopy: The O-H stretch in carboxylic acids is very broad (2500-3300 cm$^{-1}$), not a sharp peak.

2. Mass spectrometry: The molecular ion peak may be very weak or absent for fragmented compounds.

3. NMR splitting: The $n + 1$ rule applies only to chemically equivalent neighbouring protons. Protons on adjacent carbons that are chemically different give complex splitting patterns.

4. $R_f$ values: These are always between 0 and 1. They depend on the solvent system and must be compared with a standard run on the same plate.

5. Beer-Lambert Law: Only valid for dilute solutions ( $A < 2$).

6. Isotopic patterns: Chlorine gives 3:1 ratio; bromine gives 1:1. Do not confuse these.

7. NMR integration: Integration gives the ratio of protons, not the absolute number. A 3:2 ratio could mean 3H:2H, 6H:4H, etc.

8. TLC: Substances with similar $R_f$ values may not be resolved. A different solvent system should be tried.

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Practice Questions

1. An IR spectrum shows strong absorptions at 3300 cm$^{-1}$ (broad) and 1050 cm$^{-1}$. Identify the functional groups present.

2. The mass spectrum of a chlorinated compound shows peaks at $m/z = 78$ and 80 in a 3:1 ratio. Identify the molecular ion.

3. A compound \mathrm{C_4\mathrm{H_8\mathrm{O_2 has ^1\mathrm{H NMR signals at $\delta = 1.2$ (triplet, 3H), $\delta = 2.3$ (quartet, 2H), and $\delta = 11.5$ (singlet, 1H). Identify the compound and explain the spectrum.

4. A solution has absorbance 0.450 at 520 nm in a 2 cm cuvette. If \varepsilon = 15000 \mathrm{ L mol^{-1}\mathrm{ cm^{-1}Find the concentration.

5. Explain how GC-MS can be used to identify an unknown compound in a mixture.

6. A compound with molecular formula \mathrm{C_4\mathrm{H_{10}\mathrm{O has two ^1\mathrm{H NMR signals: a septet at $\delta = 3.9$ (1H) and a doublet at $\delta = 1.2$ (6H). Identify the compound.

7. In TLC, substance A travels 2.5 cm and substance B travels 4.0 cm while the solvent front travels 8.0 cm. Calculate the $R_f$ values and identify which substance is more polar.

8. Combined IR, mass spec, and NMR data for an unknown compound \mathrm{C_3\mathrm{H_6\mathrm{O_2 shows: IR absorption at 1740 \mathrm{ cm^{-1}Molecular ion at $m/z = 74$And ^1\mathrm{H NMR singlet at $\delta = 3.7$ (3H), singlet at $\delta = 2.1$ (3H). Identify the compound.

9. Calculate the degree of unsaturation for each of the following molecular formulae: (a) \mathrm{C_6\mathrm{H_{12}(b) \mathrm{C_4\mathrm{H_{10}\mathrm{O(c) \mathrm{C_7\mathrm{H_{12}\mathrm{O_2(d) \mathrm{C_8\mathrm{H_8.

10. A compound \mathrm{C_4\mathrm{H_{10}\mathrm{O has ^1\mathrm{H NMR: triplet at $\delta = 0.9$ (3H), multiplet at $\delta = 1.4$ (2H), multiplet at $\delta = 1.6$ (2H), triplet at $\delta = 3.6$ (2H), singlet at $\delta = 2.5$ (1H, exchanges with \mathrm{D_2\mathrm{O). Identify the compound and explain the splitting pattern.

11. Explain why GC is not suitable for analysing ionic compounds, and suggest an alternative technique.

12. The mass spectrum of a compound shows the molecular ion peak at $m/z = 120$ with a smaller peak at $m/z = 122$ of approximately half the intensity. Explain what this indicates about the composition of the compound.

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Advanced NMR Concepts

Spin-Spin Coupling Constants

The separation between the peaks in a multiplet is called the coupling constant $J$Measured in Hz. Coupling constants provide information about the spatial relationship between coupled protons.

Coupling type	Typical $J$ (Hz)	Relationship
Vicinal ($^3J$)	5-8	Protons on adjacent carbons
Geminal ($^2J$)	0-20	Protons on the same carbon
Long-range ($^4J$)	0-3	Protons separated by three bonds

Application: In aromatic systems, ortho coupling ($J \approx 7$ Hz) is larger than meta coupling ($J \approx 2$ Hz) or para coupling ($J \approx 0$ Hz).

DEPT ^{13}\mathrm{C NMR

DEPT (Distortionless Enhancement by Polarization Transfer) is a technique that distinguishes between Different types of carbon atoms:

Experiment	CH$_3$	CH$_2$	CH	C (quaternary)
DEPT-90	—	—	Up	—
DEPT-135	Up	Down	Up	—

Quaternary carbons appear only in the standard broadband-decoupled ^{13}\mathrm{C NMR spectrum.

Worked Example 13: A compound \mathrm{C_5\mathrm{H_{10}\mathrm{O_2} has the following ^{13}\mathrm{C NMR data: $\delta = 14.1$ (CH$_3$), 22.6 (CH$_2$), 31.8 (CH$_2$), 60.5 (CH$_2$), 174.4 (C=O). Identify the compound.

The signal at $\delta = 174.4$ is a carbonyl carbon (carboxylic acid or ester). The signal at $\delta = 60.5$ is a CH$_2$ bonded to oxygen. The three signals in the 14-32 ppm range are alkyl Carbons. The pattern is consistent with butanoic acid (\mathrm{CH_3\mathrm{CH_2\mathrm{CH_2\mathrm{COOH).

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Elemental Analysis

Combustion Analysis

When an organic compound containing C, H, and possibly other elements is burned completely, the Products are \mathrm{CO_2, \mathrm{H_2\mathrm{OAnd other oxides.

Worked Example 14: A compound contains 40.0% C, 6.7% H, and 53.3% O by mass. Find its empirical Formula.

Assume 100 g:

• n(\mathrm{C) = 40.0/12.0 = 3.33 \mathrm{ mol
• n(\mathrm{H) = 6.7/1.0 = 6.7 \mathrm{ mol
• n(\mathrm{O) = 53.3/16.0 = 3.33 \mathrm{ mol

Ratio: $1 : 2 : 1$. Empirical formula: \mathrm{CH_2\mathrm{O.

If the molecular mass is 180, then molecular formula is \mathrm{C_6\mathrm{H_{12}\mathrm{O_6.

Combustion Analysis with Other Elements

If nitrogen is present, it is collected as \mathrm{N_2. If sulfur is present, it forms \mathrm{SO_2. If halogens are present, they are collected as silver halides.

Worked Example 15: 0.200 g of a compound containing C, H, and N produced 0.440 g of \mathrm{CO_2 and 0.180 g of \mathrm{H_2\mathrm{O on combustion. Find the empirical formula.

n(\mathrm{C) = \frac{0.440}{44.0} = 0.0100 \mathrm{ mol, \quad m(\mathrm{C) = 0.120 \mathrm{ g

n(\mathrm{H) = \frac{2 \times 0.180}{18.0} = 0.0200 \mathrm{ mol, \quad m(\mathrm{H) = 0.0200 \mathrm{ g

m(\mathrm{N) = 0.200 - 0.120 - 0.020 = 0.060 \mathrm{ g

n(\mathrm{N) = \frac{0.060}{14.0} = 0.00429 \mathrm{ mol

Ratio: \mathrm{C : \mathrm{H : \mathrm{N = 0.0100 : 0.0200 : 0.00429 = 2.33 : 4.67 : 1 = 7 : 14 : 3.

Empirical formula: \mathrm{C_7\mathrm{H_{14}\mathrm{N_3 (or more practically, this would need Checking with molecular mass information).

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Electroanalytical Methods

Flame Emission Spectroscopy

Used for determination of metal ions. A sample is atomised in a flame, and the emitted light at Characteristic wavelengths is measured.

• Sodium: intense yellow (589 nm)
• Potassium: lilac (766 nm)
• Calcium: brick red (622 nm)
• Copper: blue-green (521 nm)

Atomic Absorption Spectroscopy (AAS)

Measures the absorption of light by ground-state atoms. A hollow cathode lamp provides light at the Specific wavelength of the element being analysed.

Applications: Determining trace metal concentrations in water, soil, biological samples.

Worked Example 16: A series of standard solutions of \mathrm{Cu^{2+} gave the following Absorbance readings:

Concentration (ppm)	Absorbance
0	0.00
2	0.15
4	0.30
6	0.45
8	0.60

An unknown sample gave an absorbance of 0.40. Find the concentration.

From the calibration curve, absorbance is proportional to concentration: c = \frac{0.40}{0.60} \times 8 = 5.3 \mathrm{ ppm

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Summary Table: Analytical Techniques

Technique	Information obtained	Sensitivity	Sample requirement
IR spectroscopy	Functional groups	Low	mg
Mass spectrometry	Molecular mass, fragments	Very high	ng
^1\mathrm{H NMR	H environments, structure	Moderate	mg
^{13}\mathrm{C NMR	C environments	Low-moderate	mg
UV-Vis	Concentration, conjugation	High	mL of solution
TLC	Number of components, $R_f$	Moderate	ug
GC	Separation, retention time	High	uL (volatile)
HPLC	Separation, retention time	High	uL
AAS	Metal ion concentration	Very high	mL of solution

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Quality of Analytical Data

Accuracy and Precision

• Accuracy: How close the result is to the true value.
• Precision: How close repeated measurements are to each other.

Uncertainty

Every measurement has uncertainty. Combined uncertainty for multiplication/division:

$\frac{\Delta y}{y} = \sqrt{\left(\frac{\Delta a}{a}\right)^2 + \left(\frac{\Delta b}{b}\right)^2}$

Significant Figures

Results should be quoted to an appropriate number of significant figures, consistent with the Precision of the measurements used.

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Practice Questions (Extended)

13. A compound \mathrm{C_5\mathrm{H_{12}\mathrm{O has ^1\mathrm{H NMR signals: $\delta = 0.9$ (t, 6H), $\delta = 1.5$ (m, 4H), $\delta = 3.4$ (t, 1H, exchanges with \mathrm{D_2\mathrm{O), $\delta = 1.6$ (m, 1H). Identify the compound and explain your reasoning.

14. Explain how you would use TLC to determine whether a reaction has gone to completion.

15. The molecular ion peak in the mass spectrum of a compound has $m/z = 92$. The base peak is at $m/z = 91$. Suggest what the compound might be and explain the fragmentation.

16. Describe the principles of HPLC and explain why it is preferred over GC for the analysis of pharmaceutical compounds.

17. A compound with molecular formula \mathrm{C_8\mathrm{H_8\mathrm{O has the following spectroscopic data:

• IR: 1700 cm$^{-1}$ (strong), 1600, 1500 cm$^{-1}$ (medium)
• ^1\mathrm{H NMR: $\delta = 2.6$ (s, 3H), $\delta = 7.5-8.0$ (m, 5H)
• MS: molecular ion at $m/z = 120$

Identify the compound and assign all spectral features.

18. Explain the difference between flame emission spectroscopy and atomic absorption spectroscopy, including the principle behind each technique.

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Advanced Mass Spectrometry

High-Resolution Mass Spectrometry (HRMS)

Low-resolution mass spectrometry gives $m/z$ values to the nearest whole number. High-resolution MS Can distinguish between compounds with the same nominal mass but different exact masses.

Example: \mathrm{CO has exact mass 27.9949 and \mathrm{N_2 has exact mass 28.0061. These Cannot be distinguished at low resolution (both appear at $m/z = 28$) but are separated at High resolution.

Tandem Mass Spectrometry (MS-MS)

In MS-MS, ions of a selected $m/z$ are fragmented further (collision-induced dissociation), Providing additional structural information. This is particularly useful for large biomolecules like Proteins.

Fragmentation Mechanisms

Alpha cleavage: Breaks the bond adjacent to a carbonyl group, common in ketones and aldehydes.

\mathrm{R-CO-R' \to \mathrm{R-CO^+ + \mathrm{R'^\bullet

McLafferty rearrangement: A hydrogen atom from the gamma carbon (three bonds away) transfers to The carbonyl oxygen, followed by cleavage of the beta-gamma bond. Produces an enol radical cation And a neutral alkene.

This rearrangement requires a gamma hydrogen and is common in carbonyl compounds with at least 3 Carbons in the chain.

Beta cleavage: Common in amines and halogenoalkanes. The charge stays on the fragment with the Heteroatom.

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Advanced Chromatography Concepts

Theoretical Plates

Column efficiency is measured in theoretical plates ($N$). A higher number of plates means better Separation.

$N = 16\left(\frac{t_R}{W}\right)^2$

Where $t_R$ is retention time and $W$ is peak width at the base.

Resolution

The ability to separate two adjacent peaks:

$R_s = \frac{2(t_{R2} - t_{R1})}{W_1 + W_2}$

Where $R_s > 1.5$ indicates baseline separation.

Factors Affecting Chromatographic Separation

Factor	Effect on separation
Column length	Longer column = more plates = better separation
Particle size (HPLC)	Smaller particles = better separation but higher pressure
Temperature (GC)	Optimal temperature gives best resolution
Flow rate	Too fast or too slow reduces efficiency
Mobile phase composition (HPLC)	Optimised gradient improves separation

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Interpretation Guide for Combined Spectroscopic Data

Step-by- Strategy

1. Mass spec: Determine molecular mass and molecular formula. Check for characteristic isotope patterns (Cl, Br, S).
2. Calculate DBE: Determine the number of rings and pi bonds.
3. IR: Identify functional groups (C=O, O-H, N-H, C=C, C≡N).
4. ^{13}\mathrm{C NMR: Count signals to determine the number of unique carbon environments. Use DEPT to classify carbons.
5. ^1\mathrm{H NMR: Use chemical shifts, integration, and splitting to piece together the carbon-hydrogen framework.
6. Cross-check: Verify that the proposed structure is consistent with ALL data.

Common Functional Group Signatures

Functional group	IR (cm$^{-1}$)	^1\mathrm{H NMR $\delta$ (ppm)	^{13}\mathrm{C NMR $\delta$ (ppm)
Alcohol (-OH)	3200-3600 (broad)	1-5 (variable, exchanges with \mathrm{D_2\mathrm{O)	50-90 (C-O)
Aldehyde (-CHO)	1720-1740	9-10	190-210
Ketone (C=O)	1705-1725	No H on C=O	200-220
Carboxylic acid	1700-1725, 2500-3300	10-13 (broad)	170-185
Ester	1735-1750	3.5-4.5 (O-CH$_2$)	160-180
Aromatic ring	1600-1580, 1500-1400	6.5-8.5	120-150
Alkene (C=C)	1620-1680	4.5-6.5	100-150

Worked Examples

Example 1: Rate equation

The reaction $\text{A} + \text{B} \rightarrow \text{C}$ has rate equation $\text{rate} = k[\text{A}][\text{B}]^2$. When $[\text{A}] = 0.10\,\text{mol\,dm}^{-3}$ and $[\text{B}] = 0.20\,\text{mol\,dm}^{-3}$, the rate is $2.0 \times 10^{-3}\,\text{mol\,dm}^{-3}\text{s}^{-1}$. Calculate $k$.

Solution:

$k = \frac{\text{rate}}{[\text{A}][\text{B}]^2} = \frac{2.0 \times 10^{-3}}{(0.10)(0.20)^2} = \frac{2.0 \times 10^{-3}}{4.0 \times 10^{-3}} = 0.50\,\text{mol}^{-2}\,\text{dm}^6\text{s}^{-1}$