Difficulty: Intermediate | Prerequisites: Mass spectrometry and IR companion notes, basic understanding of molecular formula determination and functional group identification
This is the second half of the Compound 55 spectral analysis. Where mass spectrometry and IR told you the molecular formula (C₄H₈OCl₂) and confirmed the functional groups present, NMR tells you exactly how the atoms are connected. ¹H NMR reveals how many distinct hydrogen environments exist and which hydrogens are neighbours; ¹³C NMR counts the distinct carbon types and, through DEPT data, tells you how many hydrogens each carbon carries. Together, these let you draw the full structural formula. If you are not yet comfortable with the n+1 rule or the concept of chemical shift, review those before diving in.
The ¹H NMR of Compound 55 shows three distinct proton environments (CH, two CH₂ groups, and CH₃) with integration ratios of 1:4:3. The ¹³C NMR confirms four unique carbon types. Splitting patterns and chemical shifts point to the structure CH₃CH₂–O–CHCl₂–CH₂– (an ethoxy group connected to a dichloromethine), consistent with the molecular formula C₄H₈OCl₂.
Chemical shift (δ, ppm)
The position of an NMR signal on the horizontal axis, measured in parts per million relative to a reference compound (TMS at 0 ppm). In simple terms, the further downfield (higher ppm) a signal appears, the more deshielded that hydrogen or carbon is, usually because an electronegative atom is nearby.
Splitting pattern (multiplicity)
The shape of an NMR peak, determined by how many neighbouring hydrogens are coupled to the observed nucleus. A singlet has no coupled neighbours; a doublet has one; a triplet has two; a quartet has three. Think of it as: count the neighbours, add one, and that is the number of lines you see.
n+1 rule
The number of lines in a ¹H NMR signal equals n + 1, where n is the number of equivalent hydrogens on adjacent carbons. This is the workhorse rule for interpreting splitting patterns in first-order spectra.
Integration
The area under an NMR peak, proportional to the number of hydrogens producing that signal. The numbers printed on top of each peak in the ¹H NMR (1, 4, 3 in this compound) give the relative hydrogen count.
DEPT (Distortionless Enhancement by Polarisation Transfer)
A ¹³C NMR technique that distinguishes CH₃, CH₂, CH, and quaternary carbons by the multiplicity of each ¹³C signal. Quartets correspond to CH₃, triplets to CH₂, doublets to CH, and absent signals to quaternary carbons.
Deshielding
The effect of nearby electronegative atoms (O, Cl, N) pulling electron density away from a nucleus, shifting its NMR signal downfield (to higher ppm). In simple terms, the more electronegative groups attached to or near a hydrogen or carbon, the higher its chemical shift.
Inequivalent protons
Hydrogens in different chemical environments, which appear as separate signals in the ¹H NMR spectrum. Two CH₂ groups can be inequivalent if one is next to oxygen and the other is not.
The ¹H NMR spectrum of Compound 55 shows three distinct signal regions, corresponding to three types of proton environment:
~5.5 ppm, integration = 1: one hydrogen (CH). This is the most downfield proton signal.
~3–4 ppm, integration = 4: four hydrogens total. Two overlapping CH₂ groups produce this signal.
~1–1.5 ppm, integration = 3: three hydrogens (CH₃). This is the most upfield signal.
Three signal regions with an integration ratio of 1:4:3 account for all eight hydrogens in C₄H₈OCl₂. Because the two CH₂ groups overlap in the 3–4 ppm region, you see what looks like one broad signal with an integration of 4 rather than two separate signals of 2 each.
CH at 5.4–5.8 ppm (triplet, 1H): this hydrogen sits on the carbon bearing two chlorines (CHCl₂). It appears as a triplet because it has two neighbouring hydrogens on the adjacent CH₂. The unusually high chemical shift (5.5 ppm) is explained by the strong deshielding effect of two electronegative chlorine atoms on the same carbon. Ordinarily a C–H next to a halogen would appear lower, around 3–4 ppm, but two chlorines push it well downfield.
CH₂ groups at 3.2–4.2 ppm (complex multiplet, 4H total): these four hydrogens belong to two different CH₂ groups that happen to overlap. One CH₂ is adjacent to the oxygen (part of the ether linkage), and one CH₂ is adjacent to both the CHCl₂ carbon and the oxygen. Their 3–4 ppm chemical shifts are consistent with C–O and C–X bond environments. The multiple lines in this region reflect coupling to neighbours on both sides.
CH₃ at 1.0–1.5 ppm (triplet, 3H): the methyl group appears as a triplet because it has two neighbouring hydrogens on the adjacent CH₂ (n+1 = 2+1 = 3 lines). Its low chemical shift places it in the typical alkane range, consistent with a methyl group at the end of the chain, away from electronegative atoms.
The ¹³C NMR reports four signals, each with a DEPT multiplicity that identifies the hydrogen count on that carbon:
Doublet at 97.3 ppm → CH (one hydrogen): this is the CHCl₂ carbon. Its high chemical shift reflects the heavy deshielding from two chlorine atoms. In a standard ¹³C, carbons bonded to chlorine typically appear in the 25–50 ppm range, but two chlorines on the same carbon push it much further downfield.
Triplet at 61.0 ppm → CH₂ (two hydrogens): this CH₂ is adjacent to the CHCl₂ on one side and the oxygen on the other. Its 61 ppm shift is consistent with a carbon bonded to oxygen (the 50–90 ppm range is typical for carbons attached to oxygen).
Triplet at 54.3 ppm → CH₂ (two hydrogens): this is the other CH₂, also attached to the oxygen but on the ethyl side of the ether. Its shift at 54.3 ppm also falls in the C–O range.
Quartet at 14.7 ppm → CH₃ (three hydrogens): the terminal methyl carbon. Its low shift (below 20 ppm) is characteristic of an alkane-type methyl group with no nearby electronegative atoms.
One of the trickiest aspects of this spectrum is the chemical shift of the CHCl₂ group. Students expect C–X bonds to place a proton or carbon in a moderate range (roughly 3–4 ppm for ¹H, 25–50 ppm for ¹³C). But here, the CH appears at 5.5 ppm in the ¹H NMR and 97.3 ppm in the ¹³C.
The reason is cumulative deshielding. Each chlorine atom withdraws electron density from the carbon and its attached hydrogen through induction. With two chlorines on the same carbon, the effect roughly doubles, pushing the signals well beyond the single-chlorine range.
This is a pattern worth memorising: geminal dihalides (two halogens on the same carbon) produce unusually downfield shifts in both ¹H and ¹³C NMR. Dichloromethane (CH₂Cl₂), for comparison, shows its ¹H signal at 5.3 ppm and its ¹³C at 53.5 ppm, right in this neighbourhood.
Combining all the spectral evidence, Compound 55 is chloroacetal (1,1-dichlorodiethyl ether), with the connectivity:
CH₃–CH₂–O–CHCl₂
However, accounting for all four carbons and the integration of 4H in the CH₂ region, the full structure is:
CH₃–CH₂–O–CH(Cl₂)–? or, more precisely, the data supports a structure where the ethyl group (CH₃CH₂–) connects through an oxygen to a CHCl₂ unit, with a second CH₂ also present.
The ¹H integration of 1 + 4 + 3 = 8 hydrogens and four distinct ¹³C signals confirm four unique carbon environments. The structure is consistent with bis(2-chloroethyl) ether or a chloral-derived acetal. The exact connectivity is: the oxygen bridges two carbon chains, each carrying chlorine substituents, with no unsaturation.
The key diagnostic evidence, in order:
Zero degrees of unsaturation: no rings or pi bonds
IR confirms ether (1100 cm⁻¹) and C–Cl (650–700 cm⁻¹), no carbonyl
¹H NMR shows three proton environments with splitting consistent with a CH₃–CH₂–O– fragment and a CHCl₂ group
¹³C DEPT confirms CH₃, two CH₂, and one CH, with shifts matching oxygen-bearing and chlorine-bearing carbons
¹H NMR typical ranges
Environment | δ (ppm) | Example in this compound |
|---|---|---|
Alkyl (R–CH₃, R–CH₂–R) | 0.5–1.5 | CH₃ at ~1.2 ppm |
Adjacent to O (R–O–CH₂–) | 3.2–3.8 | CH₂ groups at ~3.5 ppm |
Adjacent to Cl (R–CHCl–) | 3.5–4.5 | Single Cl on C |
Geminal dihalide (–CHCl₂) | 5.0–6.0 | CH at ~5.5 ppm |
¹³C NMR typical ranges
Environment | δ (ppm) | Example in this compound |
|---|---|---|
Alkyl CH₃ | 5–25 | 14.7 ppm |
C adjacent to O | 50–90 | 54.3 and 61.0 ppm |
C with two Cl (geminal) | 60–100 | 97.3 ppm |
C=O (carbonyl) | 170–220 | Absent |
Students often assume that a triplet in ¹³C DEPT means the same thing as a triplet in ¹H NMR. In DEPT, a triplet identifies a CH₂ group (two attached hydrogens). In ¹H NMR, a triplet means the observed hydrogen has two neighbouring hydrogens (n+1 rule). These are different pieces of information.
Students sometimes think that two CH₂ groups must always appear as two separate ¹H NMR signals. They will overlap if their chemical shifts are similar, as they are here (both near 3.5 ppm). The ¹³C data, where they appear at 54.3 and 61.0 ppm, confirms they are distinct.
A common error is to assume a ¹H signal above 5 ppm must indicate an alkene. In this compound, the 5.5 ppm signal comes from a CHCl₂ group, not an unsaturated carbon. Always cross-check with the degree of unsaturation and the IR before concluding an alkene is present.
Students sometimes forget that the n+1 rule applies only to first-order spectra with equivalent coupling constants. In more complex molecules, the rule still works as an approximation but the patterns may not be perfectly clean.
⚠️ Predicting and interpreting splitting patterns using the n+1 rule is one of the most commonly tested NMR skills. Expect to be asked: "How many lines would the CH₃ signal show, and why?"
⚠️ The relationship between ¹H integration ratios and the number of hydrogens is tested frequently. If the total integration ratio is 1:4:3 and the formula has 8 hydrogens, you need to recognise that the ratio maps directly (1H, 4H, 3H).
⚠️ Unusual chemical shifts, like the 5.5 ppm CHCl₂ signal, are favourite exam questions. You will be expected to explain why a particular shift is higher or lower than expected.
⚠️ Matching ¹³C DEPT multiplicities to hydrogen counts is a standard exam skill. Know that doublet = CH, triplet = CH₂, quartet = CH₃ in DEPT.
NMR spectroscopy is the primary tool used in research labs, pharmaceutical development, and forensic chemistry to confirm molecular structures. Whenever a new compound is synthesised, its ¹H and ¹³C NMR spectra are recorded and compared against predicted values, exactly as done in this exercise.
A ¹H NMR triplet indicates the observed hydrogen has ___ neighbouring hydrogens on adjacent carbons.
True or false: in ¹³C DEPT, a quartet signal corresponds to a CH₂ group. ___
The ¹³C signal at 97.3 ppm is unusually downfield because the carbon bears ___ chlorine atoms.
True or false: if two CH₂ groups have similar chemical shifts, they will appear as one broad signal in the ¹H NMR. ___
The integration ratio 1:4:3 for a compound with 8 total hydrogens means the signals correspond to ___, ___, and ___ hydrogens.
Answers: 1. Two. 2. False (a quartet = CH₃). 3. Two. 4. True. 5. 1H, 4H, and 3H.
Q: The ¹H NMR of Compound 55 shows a triplet at 5.5 ppm with an integration of 1. What structural feature does this signal represent, and why is the chemical shift so high?
A: The triplet at 5.5 ppm represents a CH group. It appears as a triplet because it is coupled to two adjacent CH₂ hydrogens (n+1 = 3 lines). The shift is unusually high because two chlorine atoms on the same carbon (a geminal dihalide) strongly deshield the hydrogen through inductive electron withdrawal.
Q: How do you distinguish between a CH₂ and a CH₃ group using ¹³C DEPT data?
A: In DEPT, CH₂ groups appear as triplets and CH₃ groups appear as quartets. For Compound 55, the triplets at 61.0 and 54.3 ppm identify two CH₂ carbons, while the quartet at 14.7 ppm identifies the CH₃ carbon.
Q: The ¹H NMR shows a signal at 3–4 ppm with an integration of 4. Could this be a single CH₂ group? Explain.
A: No. An integration of 4 means four hydrogens, which requires two CH₂ groups (2H each), not one. The two CH₂ groups have similar chemical environments (both near oxygen), so their signals overlap in the same region of the spectrum.
Q: Why does the ¹³C signal for the CHCl₂ carbon (97.3 ppm) appear so far from the typical C–Cl range of 25–50 ppm?
A: The typical 25–50 ppm range applies to carbons bearing a single halogen. With two chlorines on the same carbon, the cumulative inductive effect of both electronegative atoms pulls far more electron density away, deshielding the carbon and pushing its signal to 97.3 ppm.
Q: Based on the ¹H NMR splitting patterns alone, reconstruct the connectivity of the carbon backbone.
A: The CH₃ triplet says the methyl is next to a CH₂. The CH triplet says the CH is next to a CH₂. The 4H multiplet at 3–4 ppm sits between these two groups and connects to both. The connectivity is therefore CH₃–CH₂–(link)–CH, with the IR and ¹³C data placing an oxygen in the chain: CH₃–CH₂–O–CH(Cl₂).
This connects to mass spectrometry and IR spectroscopy (the companion notes), which provide the molecular formula and functional group confirmation before NMR narrows down the connectivity. In practice, you use all three techniques together, and exam problems will often require you to do the same.
The concept of deshielding and chemical shift ties directly to electronegativity and inductive effects from general chemistry. The more electronegative substituents on or near a carbon, the further downfield its NMR signal. This principle extends to every NMR problem you will encounter.
DEPT analysis connects to the broader topic of carbon-hydrogen bonding environments, which is revisited in spectroscopic identification of alcohols, amines, carbonyls, and aromatic compounds later in the course.
NMR spectroscopy, proton NMR, 1H NMR, carbon-13 NMR, 13C NMR, DEPT, chemical shift, ppm, splitting pattern, multiplicity, n+1 rule, doublet, triplet, quartet, integration, deshielding, shielding, electronegativity, inductive effect, geminal dihalide, CHCl2, dichloromethine, ether NMR, C-O chemical shift, alkyl NMR, spectral unknown, structure determination, Compound 55, C4H8OCl2, organic chemistry II, Penn State CHEM, spectral analysis, NMR peak assignment, inequivalent protons, carbon environments