Mass spectrometry is an analytical technique used to determine the relative molecular mass (\( M_r \)) of an organic compound and gain structural information. By examining how a molecule breaks apart in the spectrometer, we can deduce its carbon skeleton and functional groups.
🔑 Key Principle 1
When an organic sample is introduced into a mass spectrometer, it is bombarded with high-energy electrons. This causes the molecule to lose an electron, forming a positive molecular ion (\( \text{M}^+ \)). The molecular ion is unstable and can fragment into a smaller positive fragment ion and a neutral radical. Only positively charged species are accelerated and detected.
🔑 Key Principle 2
High-resolution mass spectrometers measure mass-to-charge ratios (\( m/z \)) to four decimal places. Since different atoms have unique, non-integer isotopic masses, this precision allows us to distinguish between compounds that have the same integer relative molecular mass, confirming the exact molecular formula.
The Mass Spectrum
A mass spectrum shows the relative abundance of each detected ion on the y-axis against its mass-to-charge ratio (\( m/z \)) on the x-axis. Because the charge (\( z \)) on the detected ions is almost always \( +1 \), the \( m/z \) value corresponds directly to the mass of the ion.
The positive ion formed when a whole molecule loses a single electron in a mass spectrometer. Its \( m/z \) value corresponds to the relative molecular mass (\( M_r \)) of the compound.
The process by which an unstable molecular ion breaks apart into smaller positive fragment ions and neutral radicals inside a mass spectrometer.
The tallest peak in a mass spectrum, representing the most abundant and stable positive ion formed. It is assigned a relative abundance of 100%.
A highly precise form of mass spectrometry that measures the \( m/z \) ratio of ions to four decimal places, allowing the exact molecular formula to be calculated.
1. The Molecular Ion Peak (M⁺)
The molecular ion peak is the peak furthest to the right in the mass spectrum, excluding the very small \( \text{M}+1 \) peak. It is formed by the reaction:
\[ \text{M(g)} + \text{e}^- \rightarrow \text{M}^+\text{(g)} + 2\text{e}^- \]The \( m/z \) value of the \( \text{M}^+ \) peak gives the integer relative molecular mass (\( M_r \)) of the organic compound. For example, in the spectrum of propanone above, the \( \text{M}^+ \) peak is at \( m/z = 58 \), confirming that \( M_r = 58 \).
You may see a tiny peak at \( \text{M}+1 \). This peak is caused by the presence of carbon-13 (\( ^{13}\text{C} \)) atoms. Carbon-13 is a naturally occurring stable isotope that makes up approximately 1.1% of all carbon atoms. It increases the mass of a small fraction of the molecular ions by 1 unit. You should ignore this peak when determining the molecular mass of the compound.
2. Fragmentation
Because the molecular ion has high internal energy, it is unstable. It frequently breaks apart at covalent bonds in a process called fragmentation. Fragmentation produces a positive fragment ion and a neutral radical:
\[ \text{M}^+\text{(g)} \rightarrow \text{Fragment}^+\text{(g)} + \text{Radical}^\bullet\text{(g)} \]Only the positive fragment ion is accelerated and detected by the spectrometer. The neutral radical has no charge and is swept away by the vacuum pumps. Common fragmentations provide clues about structural features. For example, a peak at \( m/z = 15 \) suggests the presence of a \( \text{CH}_3^+ \) ion.
Common Fragment Losses and Ions
You can identify functional groups by calculating the mass lost from the molecular ion peak (\( M_r - m/z \)) or by identifying the mass of the fragment ion itself:
| m/z Value of Ion | Possible Fragment Ion | Mass Lost from M⁺ | Suggested Group Lost |
|---|---|---|---|
| \( m/z = 15 \) | \( \text{CH}_3^+ \) | 15 | \( \text{CH}_3^\bullet \) (methyl radical) |
| \( m/z = 17 \) | \( \text{OH}^+ \) | 17 | \( \text{OH}^\bullet \) (hydroxyl radical) |
| \( m/z = 29 \) | \( \text{C}_2\text{H}_5^+ \) or \( \text{CHO}^+ \) | 29 | \( \text{C}_2\text{H}_5^\bullet \) (ethyl) or \( \text{CHO}^\bullet \) (formyl) |
| \( m/z = 31 \) | \( \text{CH}_2\text{OH}^+ \) or \( \text{OCH}_3^+ \) | 18 | \( \text{H}_2\text{O} \) (water molecule) |
| \( m/z = 43 \) | \( \text{C}_3\text{H}_7^+ \) or \( \text{CH}_3\text{CO}^+ \) | 45 | \( \text{COOH}^\bullet \) or \( \text{OC}_2\text{H}_5^\bullet \) |
| \( m/z = 45 \) | \( \text{COOH}^+ \) or \( \text{OC}_2\text{H}_5^+ \) | 44 | \( \text{CO}_2 \) (carbon dioxide molecule) |
When identifying fragments responsible for peaks in a mass spectrum, you MUST include the positive charge on the formula (e.g. \( \text{CH}_3^+ \), \( \text{CH}_3\text{CO}^+ \)). Writing the formulas without a positive charge is a common error and will result in zero marks for that part of the question.
3. High-Resolution Mass Spectrometry
Low-resolution mass spectrometry measures \( m/z \) values to the nearest whole number. High-resolution mass spectrometry measures \( m/z \) values to four decimal places. This is useful because the precise atomic masses of isotopes are not whole numbers:
- \( ^1\text{H} = 1.0078 \)
- \( ^{12}\text{C} = 12.0000 \) (by definition)
- \( ^{16}\text{O} = 15.9949 \)
- \( ^{14}\text{N} = 14.0031 \)
Using these precise values, compounds with the same integer molecular mass will have different high-resolution molecular masses. For example, both propene (\( \text{C}_3\text{H}_6 \)) and diazene (\( \text{N}_2\text{H}_2 \)) have an integer mass of 42. However, their precise masses are:
- \( \text{C}_3\text{H}_6 \): \( 3(12.0000) + 6(1.0078) = 42.0468 \)
- \( \text{N}_2\text{H}_2 \): \( 2(14.0031) + 2(1.0078) = 30.0218 \) (if compared to different molecules) or for diazene: \( 2(14.0031) + 2(1.0078) = 28.0062 + 2.0156 = 30.0218 \)
Let's look at another example: carbon monoxide (\( \text{CO} \)) vs nitrogen gas (\( \text{N}_2 \)) both have integer mass 28. Their precise masses are \( 27.9949 \) and \( 28.0062 \), which are easily distinguished using high-resolution mass spectrometry.
Worked Examples
Candidate A: \( \text{C}_3\text{H}_8\text{O} \)
Candidate B: \( \text{C}_2\text{H}_4\text{O}_2 \)
(Use precise masses: \( ^1\text{H} = 1.0078 \), \( ^{12}\text{C} = 12.0000 \), \( ^{16}\text{O} = 15.9949 \)).
Solution:
Step 1: Calculate the precise mass for Candidate A (\( \text{C}_3\text{H}_8\text{O} \)):
\[ \text{Mass} = 3(12.0000) + 8(1.0078) + 1(15.9949) \] \[ \text{Mass} = 36.0000 + 8.0624 + 15.9949 = 60.0573 \]Step 2: Calculate the precise mass for Candidate B (\( \text{C}_2\text{H}_4\text{O}_2 \)):
\[ \text{Mass} = 2(12.0000) + 4(1.0078) + 2(15.9949) \] \[ \text{Mass} = 24.0000 + 4.0312 + 31.9898 = 60.0210 \]Conclusion: The calculated precise mass for Candidate B matches the experimental value of 60.0210 exactly. Therefore, the molecular formula of the compound is \( \text{C}_2\text{H}_4\text{O}_2 \) (which could be ethanoic acid or methyl methanoate).
Solution:
1. Identify the compound: A ketone contains a carbonyl group (\( \text{C}=\text{O} \)). The general formula of a saturated ketone is \( \text{C}_n\text{H}_{2n}\text{O} \). Let's solve for \( n \):
\[ 12n + 2n + 16 = 86 \implies 14n = 70 \implies n = 5 \]The ketone contains 5 carbon atoms, so its molecular formula is \( \text{C}_5\text{H}_{10}\text{O} \). The possible structural isomers are pentan-2-one or pentan-3-one.
2. Analyze the fragmentation: The base peak is at \( m/z = 43 \). This corresponds to a fragment ion of mass 43, which is the acylium ion \( \text{CH}_3\text{CO}^+ \) (\( 12 \times 2 + 3 \times 1.0 + 16 = 43 \)).
3. Deduce the correct isomer:
- If the compound is pentan-2-one (\( \text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_3 \)), cleavage of the \( \text{C-C} \) bond adjacent to the carbonyl group yields the \( \text{CH}_3\text{CO}^+ \) ion (\( m/z = 43 \)) and a neutral propyl radical (\( {}^\bullet\text{CH}_2\text{CH}_2\text{CH}_3 \)). This is a highly stable cleavage.
- If the compound is pentan-3-one (\( \text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3 \)), cleavage adjacent to the carbonyl group yields the propionyl ion \( \text{CH}_3\text{CH}_2\text{CO}^+ \) (\( m/z = 57 \)) and an ethyl radical. It cannot produce a fragment of \( m/z = 43 \) by a single bond cleavage.
Therefore, the compound must be pentan-2-one.
4. Write the fragmentation equation:
\[ [\text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_3]^+ \rightarrow \text{CH}_3\text{CO}^+ + {}^\bullet\text{CH}_2\text{CH}_2\text{CH}_3 \]The detected ion is \( \text{CH}_3\text{CO}^+ \).
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