Combined Techniques Revision Notes for OCR A-Level Chemistry A

Combined Techniques

Introduction to structure determination

In organic chemistry, identifying the structure of an unknown compound requires combining information from multiple analytical techniques. No single method provides enough information on its own, so chemists use a systematic approach that brings together data from different sources.

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The key to successful structure determination is using complementary techniques - each method provides different information that, when combined, reveals the complete molecular structure.

The main analytical techniques used for structure determination are:

Elemental analysis - determines the empirical formula
Mass spectrometry - identifies the molecular mass and molecular formula
Infrared (IR) spectroscopy - identifies functional groups present
Nuclear magnetic resonance (NMR) spectroscopy - reveals the arrangement of carbon and hydrogen atoms

The systematic approach to structure determination

Structure determination follows a logical sequence where each technique provides specific information that builds toward the complete structure. Working through these techniques in order ensures you gather all necessary information before proposing a final structure.

Step 1: Elemental analysis

Elemental analysis provides the percentage composition by mass of each element in the compound. This data allows you to calculate the empirical formula - the simplest whole number ratio of atoms in the compound.

How to calculate the empirical formula:

Convert each percentage to moles by dividing by the relative atomic mass
Divide all mole values by the smallest number to find the simplest ratio
If necessary, multiply all ratios by a small whole number to remove decimals

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Worked Example: Calculating Empirical Formula

If elemental analysis gives C: 73.17%, H: 7.32%, O: 19.51%:

Step 1: Convert percentages to moles

$n(\text{C}) = \frac{73.17}{12.0} = 6.10 \text{ mol}$

$n(\text{H}) = \frac{7.32}{1.0} = 7.32 \text{ mol}$

$n(\text{O}) = \frac{19.51}{16.0} = 1.22 \text{ mol}$

Step 2: Divide by the smallest value (1.22)

$\text{C} : \text{H} : \text{O} = \frac{6.10}{1.22} : \frac{7.32}{1.22} : \frac{1.22}{1.22} = 5 : 6 : 1$

Step 3: Write the empirical formula

Therefore, the empirical formula is C₅H₆O

Step 2: Mass spectrometry

The mass spectrum shows the molecular ion peak ( $M^+$ peak), which appears at the highest m/z value. This peak corresponds to the molecular mass of the compound and is essential for determining the molecular formula.

Finding the molecular formula:

Calculate the relative mass of the empirical formula
Divide the molecular mass by the empirical formula mass
Multiply the empirical formula by this number

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Worked Example: Finding Molecular Formula

Continuing the example above, if the molecular ion peak is at m/z = 164:

Step 1: Calculate empirical formula mass

$\text{Empirical formula mass} = (12.0 \times 5) + (1.0 \times 6) + (16.0 \times 1) = 82.0$

Step 2: Find the multiplier

$\text{Number of empirical units} = \frac{164}{82.0} = 2$

Step 3: Calculate molecular formula

$\text{Molecular formula} = \text{C}_5\text{H}_6\text{O} \times 2 = \text{C}_{10}\text{H}_{12}\text{O}_2$

chatImportant

Exam Tip: The molecular ion peak is usually the peak with the highest m/z value, though it may not always be the tallest peak in the spectrum. Don't confuse the tallest peak with the molecular ion peak!

Step 3: Infrared (IR) spectroscopy

The IR spectrum identifies functional groups by showing characteristic absorption peaks at specific wavenumbers. You should use the data sheet provided in the exam to interpret IR spectra.

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Always refer to your data sheet for IR wavenumbers. You don't need to memorize exact values, but you should recognize the key functional groups and their approximate positions.

Key functional group absorptions:

$\text{C=O}$ (carbonyl) group: strong absorption around 1710 cm⁻¹
$\text{O-H}$ (alcohol): broad absorption 3200-3600 cm⁻¹
$\text{O-H}$ (carboxylic acid): very broad absorption 2500-3300 cm⁻¹
$\text{C-H}$ (aromatic): absorption around 3000-3100 cm⁻¹

In the example, a peak at 1710 cm⁻¹ indicates a $\text{C=O}$ group in an aldehyde, ketone, or ester. The absence of a broad peak at 2500-3300 cm⁻¹ rules out a carboxylic acid.

Step 4: Carbon-13 NMR spectroscopy

The $^{13}\text{C}$ NMR spectrum shows the number of different carbon environments in the molecule. Each peak represents carbons in a unique chemical environment.

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Important Points:

The number of peaks = number of different carbon environments
Symmetry in the molecule reduces the number of peaks
Chemical shift values (in ppm) indicate the type of carbon environment

In the worked example, eight peaks indicate eight different carbon environments, matching the eight carbons in the molecular formula $\text{C}_{10}\text{H}_{12}\text{O}_2$ .

Step 5: Proton NMR spectroscopy

The $^1\text{H}$ NMR (proton NMR) spectrum provides three key pieces of information that help determine the hydrogen arrangement in your molecule.

1. Number of peaks = number of different proton environments

2. Integration (relative peak areas) = ratio of protons in each environment

In the example, the peak areas give a ratio of 4 : 3 : 2 : 3, showing:

4 protons in one environment
3 protons in another environment
2 protons in a third environment
3 protons in a fourth environment

3. Splitting patterns = information about adjacent protons

The n+1 rule states that a proton with n adjacent protons will split into (n+1) peaks:

n = 0 → singlet (1 peak)
n = 1 → doublet (2 peaks)
n = 2 → triplet (3 peaks)
n = 3 → quartet (4 peaks)

Interpreting chemical shifts:

Different proton environments appear at characteristic chemical shift values:

$\delta$ = 6.9-8.0 ppm: aromatic protons ( $\text{Ar-H}$ )
$\delta$ = 3.8 ppm: protons in $\text{CH}_3$ next to oxygen ( $\text{O-CH}_3$ )
$\delta$ = 2.6 ppm: protons in $\text{CH}_2$ next to carbonyl ( $\text{CH}_2\text{-C=O}$ ) or aromatic ring ( $\text{CH}_2\text{-Ar}$ )
$\delta$ = 1.2 ppm: protons in $\text{CH}_3$ next to carbon ( $\text{C-CH}_3$ )

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Worked Example: Analysing Splitting Patterns

From the spectrum:

6.9-8.0 ppm: 4 aromatic protons (appears as two doublets in 1,4-disubstituted benzene)
3.8 ppm: $\text{CH}_3$ group (singlet, no adjacent H) → O-CH₃
2.6 ppm: $\text{CH}_2$ group (quartet, 3 adjacent H) → CH₂ next to CH₃
1.2 ppm: $\text{CH}_3$ group (triplet, 2 adjacent H) → CH₃ next to CH₂

The triplet at 1.2 ppm and quartet at 2.6 ppm indicate a $\text{CH}_3\text{CH}_2$ group.

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Study Tip: The splitting pattern of aromatic protons depends on the substitution pattern. In 1,4-disubstituted benzene rings, aromatic protons often appear as two doublets because protons at positions 2 and 6 are equivalent, as are protons at positions 3 and 5.

Step 6: Combining all information

The final step is to piece together all the information to propose a structure that fits all the data from every technique. This requires careful analysis and logical deduction.

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Worked Example: Determining the Final Structure

For the worked example:

Molecular formula: $\text{C}_{10}\text{H}_{12}\text{O}_2$
Functional group from IR: $\text{C=O}$ (ester, ketone, or aldehyde)
Eight carbon environments from $^{13}\text{C}$ NMR
Proton NMR shows: aromatic ring, $\text{CH}_3\text{CH}_2$ group, $\text{O-CH}_3$ group

The $^{13}\text{C}$ NMR suggests a 1,4-disubstituted aromatic ring (due to symmetry). Combining a benzene ring, a $\text{C=O}$ group, $\text{CH}_3\text{CH}_2$ , and $\text{O-CH}_3$ leads to two possible structures:

$\text{CH}_3\text{CH}_2\text{-C}_6\text{H}_4\text{-CO-O-CH}_3$ (methyl ester with ethyl substituent)
$\text{H}_3\text{C-O-C}_6\text{H}_4\text{-CO-CH}_2\text{CH}_3$ (ethyl ketone with methoxy substituent)

Distinguishing between structures:

To determine which structure is correct, you could use additional proton NMR evidence. The chemical shifts and splitting patterns of the aromatic protons would differ depending on whether the ring is attached to an ethyl group or a methoxy group.

Practice questions

When tackling exam questions on combined techniques:

Work systematically - follow the same order: empirical formula, molecular formula, functional groups, carbon environments, proton environments
Show your working - especially for empirical and molecular formula calculations
Use the data sheet - for IR wavenumbers and NMR chemical shifts
Consider symmetry - this explains why the number of NMR peaks may be less than the total number of atoms
Check your answer - does your proposed structure match ALL the data?

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Common Mistakes to Avoid:

Forgetting to divide by the smallest number when finding the empirical formula
Misidentifying the molecular ion peak (it's usually the highest m/z, not the tallest peak)
Ignoring splitting patterns in proton NMR
Proposing a structure that doesn't match the molecular formula

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Key Points to Remember:

Combined techniques work together - elemental analysis → mass spec → IR → NMR follows a logical sequence where each technique provides complementary information
Empirical formula first - convert percentages to moles, divide by the smallest, and find whole number ratios
Molecular ion peak = molecular mass - use this with the empirical formula mass to find the molecular formula
IR identifies functional groups - look for characteristic peaks like C=O at ~1710 cm⁻¹
¹³C NMR counts carbon environments - remember that symmetry reduces the number of peaks
¹H NMR gives three clues - number of peaks (environments), integration (ratio), and splitting (n+1 rule for adjacent protons)
Always combine ALL the evidence - your proposed structure must fit every piece of data from all four techniques

Combined Techniques (OCR A-Level Chemistry A): Revision Notes