Difficulty: Intermediate | Prerequisites: General Chemistry, introductory organic chemistry (functional groups, bonding, molecular formula notation)
This material sits at the core of any second-semester organic chemistry course: working backwards from spectral data to determine an unknown compound's structure. If you can read a mass spectrum, interpret an IR spectrum, and calculate degrees of unsaturation, you can tackle most structure-determination problems on an exam. You should already be comfortable with Lewis structures, electronegativity, and the basics of covalent bonding. If terms like "molecular ion" or "base peak" are unfamiliar, start here before moving to the NMR companion notes.
Compound 55 has the molecular formula C₄H₈OCl₂ (determined by combining elemental analysis percentages with the molecular ion peak at m/e 142). Its degree of unsaturation is zero, meaning no rings or pi bonds. The mass spectrum, IR spectrum, and structural reasoning all point to a dichlorinated ether with only single bonds.
Molecular ion (parent ion, M⁺)
The peak in the mass spectrum that corresponds to the intact molecule after it has lost one electron. Its m/e value equals the compound's relative molecular mass. In simple terms, this is the heaviest meaningful peak and tells you the molecular weight.
Base peak
The tallest (most intense) peak in a mass spectrum, set to 100% relative abundance. Think of it as the fragment the molecule most readily breaks into.
Fragment ion
Any peak in the mass spectrum lighter than the molecular ion, produced when the molecule breaks apart in the ionisation chamber. Each fragment ion's m/e value gives clues about which bonds broke and which pieces formed.
Degree of unsaturation (index of hydrogen deficiency, IHD)
A calculation that tells you how many rings or pi bonds a molecular formula implies. A value of zero means the molecule is fully saturated with no rings, double bonds, or triple bonds.
Elemental analysis
An experimental technique that reports the percentage by mass of each element in a compound. Combined with the molecular ion mass, it lets you work out the molecular formula.
IR absorption band
A region in an infrared spectrum where a particular bond type absorbs energy and vibrates. Each functional group has characteristic absorption ranges (measured in cm⁻¹), so the presence or absence of a band confirms or rules out that group.
C–Cl stretch
An IR absorption appearing in the 600–800 cm⁻¹ region. In simple terms, if you see a strong peak down there, your molecule likely contains a carbon-chlorine bond.
C–O stretch (ether)
An IR absorption for sp³ C–O bonds appearing around 1000–1150 cm⁻¹. This is the signature band for an ether linkage.
C–H stretch
An IR absorption for sp³ C–H bonds appearing around 2850–2960 cm⁻¹. Nearly every organic compound shows this, but its position tells you whether the hydrogens are on saturated or unsaturated carbons.
Elemental analysis of Compound 55 gives: C 33.83%, H 4.97%, O 11.27%. The remainder (49.93%) accounts for halogen, confirmed as chlorine.
The molecular ion cluster appears at m/e 141–146, with the key peak at m/e 142. This is the relative molecular mass.
When you compute the empirical formula from the percentages and then scale it up to match 142 g/mol, the empirical and molecular formulae turn out to be the same: C₄H₈OCl₂.
The isotope pattern at m/e 141–146 (a cluster of peaks rather than a single line) is characteristic of a compound containing two chlorine atoms, because ³⁵Cl and ³⁷Cl produce multiple M, M+2, and M+4 peaks.
Molecular ion (parent ion), m/e 142: confirms the molecular weight and, together with the elemental analysis, locks down the molecular formula.
Base peak, m/e 31: the most intense peak. This fragment corresponds to CHO⁺ (or CH₃O⁺), consistent with cleavage next to the oxygen in an ether.
Fragment at m/e 45: likely CH₃CH₂O⁺ or OC₂H₅⁺, pointing to an ethoxy fragment.
Fragment at m/e 78: loss of one HCl unit or a larger rearrangement; consistent with partial loss of chlorine-containing fragments.
Fragments at m/e 106–109: these correspond to M–35 and M–37 losses (loss of ³⁵Cl and ³⁷Cl, respectively), directly confirming chlorine is present in the molecule.
The pattern of M–35 and M–37 is a classic test for chlorine. If you see the molecular ion drop by exactly 35 or 37 mass units, you can be confident a C–Cl bond is breaking.
For C₄H₈OCl₂, the formula is:
Degree of unsaturation = (2C + 2 – H – X) / 2
Where C = number of carbons, H = number of hydrogens, and X = number of halogens. Oxygen does not appear in the formula because it does not change the hydrogen count.
Plugging in: (2 × 4 + 2 – 8 – 2) / 2 = (10 – 10) / 2 = 0
A result of zero means no double bonds, no triple bonds, and no rings. The molecule is fully saturated and acyclic.
This immediately rules out carbonyl groups (C=O), alkenes (C=C), alkynes, and aromatic rings. Every bond in this molecule is a single bond.
C–H stretch, ~2900 cm⁻¹: present, as expected for any organic compound with sp³ C–H bonds. Confirms saturated (alkane-type) hydrogens.
No absorption above 3000 cm⁻¹: rules out alkene (=C–H) and alkyne (≡C–H) stretches, consistent with zero degrees of unsaturation.
No strong absorption at ~1700 cm⁻¹: rules out a carbonyl (C=O) group. This is important because C₄H₈OCl₂ could, in theory, contain a carbonyl, but the IR confirms it does not.
C–O stretch, ~1100 cm⁻¹ (sp³ C–O, 1000–1150 cm⁻¹ range): a strong peak here confirms an ether linkage. For an sp² C–O the absorption would shift to 1200–1300 cm⁻¹, but that is not the case here.
C–Cl stretch, 650–700 cm⁻¹ (broad range 600–800 cm⁻¹): strong absorption in this region confirms carbon-chlorine bonds, consistent with two Cl atoms in the formula.
Alkane bending, 1365–1450 cm⁻¹: CH₂ and CH₃ bending modes appear in this region, consistent with the saturated carbon chain.
All expected peaks are observed. No unexpected peaks appear. The IR data fully support a saturated, acyclic, chlorinated ether.
Degree of unsaturation
DoU = (2C + 2 + N – H – X) / 2
C = carbons, H = hydrogens, N = nitrogens, X = halogens. Oxygen and sulfur are not counted.
Key IR ranges to memorise
Bond / Group | Wavenumber (cm⁻¹) | Notes |
|---|---|---|
O–H (alcohol) | 3200–3550 (broad) | Absent in this compound |
N–H | 3300–3500 | Absent in this compound |
sp³ C–H | 2850–2960 | Present |
sp² C–H | 3010–3100 | Absent (no unsaturation) |
C=O (carbonyl) | 1690–1760 | Absent (no carbonyl) |
C=C (alkene) | 1620–1680 | Absent |
sp³ C–O (ether) | 1000–1150 | Present, strong |
C–Cl | 600–800 | Present, confirms chlorine |
Students often think oxygen must appear in the degree-of-unsaturation formula. It does not. Oxygen (and sulfur) do not change the hydrogen count, so they drop out of the calculation entirely.
Students sometimes confuse the base peak with the molecular ion. The base peak is the tallest peak (most abundant fragment), not necessarily the heaviest. In this compound, the base peak (m/e 31) is far lighter than the molecular ion (m/e 142).
Seeing a cluster of peaks near the molecular ion (m/e 141–146) does not mean the molecular weight is uncertain. The cluster arises from isotope patterns (especially ³⁵Cl vs ³⁷Cl), and the molecular weight is the lowest peak in that cluster (the all-³⁵Cl version).
Students often assume that because a molecular formula contains oxygen, a carbonyl (C=O) must be present. It need not be. The oxygen can sit in an ether linkage (C–O–C), which shows up at a completely different IR frequency.
⚠️ You will almost certainly be asked to calculate the degree of unsaturation for an unknown compound. Forgetting that halogens count as hydrogen equivalents (they subtract from H) is the most common error.
⚠️ Exams frequently give you a molecular formula with oxygen and ask whether a carbonyl is present. The IR is how you decide, not the formula alone. No peak at ~1700 cm⁻¹ = no carbonyl.
⚠️ Recognising the M–35/M–37 loss pattern in a mass spectrum is a fast way to confirm chlorine on exam problems. Bromine gives M–79/M–81.
⚠️ The base peak is tested separately from the molecular ion. Know the difference: molecular ion = molecular weight; base peak = most abundant fragment.
This is the kind of problem where spectral analysis does what a formula alone cannot. An engineer designing a process to synthesise a chlorinated ether would use exactly these techniques to confirm the product's identity after a reaction.
True or false: oxygen must be included in the degree-of-unsaturation formula. ___
The tallest peak in a mass spectrum is called the ___.
True or false: a degree of unsaturation of zero means the compound can still contain a ring. ___
A loss of 35 or 37 mass units from the molecular ion suggests the presence of ___.
The IR absorption for an sp³ C–O ether bond falls in the range ___ to ___ cm⁻¹.
Answers: 1. False. 2. Base peak. 3. False. 4. Chlorine. 5. 1000 to 1150.
Q: Given the elemental analysis C 33.83%, H 4.97%, O 11.27%, and a molecular ion at m/e 142, determine the molecular formula.
A: The remaining percentage (49.93%) is chlorine. Converting to moles and finding the simplest ratio gives C₄H₈OCl₂, which already matches the molecular ion mass of ~143 (using ³⁵Cl: 4(12) + 8(1) + 16 + 2(35) = 142). The empirical and molecular formulae are the same.
Q: Why does the mass spectrum of Compound 55 show a cluster of peaks at m/e 141–146 rather than a single molecular ion peak?
A: Chlorine has two abundant isotopes, ³⁵Cl and ³⁷Cl. With two chlorine atoms, there are three isotopic combinations (both ³⁵Cl, one of each, both ³⁷Cl), producing peaks at M, M+2, and M+4. This isotope pattern is diagnostic for two chlorines.
Q: What does a degree of unsaturation of zero tell you about the structure of C₄H₈OCl₂?
A: The molecule has no double bonds, triple bonds, or rings. Every carbon-carbon and carbon-heteroatom bond is a single bond, and the carbon skeleton is acyclic.
Q: An unknown compound shows a strong IR absorption at 1100 cm⁻¹ but nothing at 1700 cm⁻¹. What does this suggest about the oxygen-containing functional group?
A: The 1100 cm⁻¹ band is characteristic of an sp³ C–O stretch (ether), while the absence of a 1700 cm⁻¹ band rules out a carbonyl. The oxygen is in an ether linkage, not a ketone, aldehyde, or ester.
Q: The base peak of Compound 55 appears at m/e 31. Propose a fragment that accounts for this peak.
A: m/e 31 corresponds to CHO⁺ (12 + 1 + 16 = 29) or, more likely, CH₃O⁺ (12 + 3 + 16 = 31), which is an oxocarbenium fragment produced by alpha-cleavage adjacent to the ether oxygen.
This material connects directly to NMR spectroscopy (covered in the companion study notes), where you use ¹H and ¹³C data to assign specific hydrogen and carbon environments and confirm the structure. Mass spectrometry and IR narrow the possibilities; NMR pins down the connectivity.
The degree-of-unsaturation concept reappears throughout organic chemistry whenever you encounter a new molecular formula, from simple alkyl halides to complex natural products. It is the first calculation you should do with any unknown.
The chlorine isotope pattern in mass spectrometry connects to the broader topic of isotope effects and how isotope ratios are used in analytical chemistry, environmental science, and forensic work.
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