Difficulty: Intermediate-Advanced | Prerequisites: Functional groups, bonding, degree of unsaturation
Spectroscopy lets you work backwards from physical data to a molecular structure. Mass spectrometry tells you the molecular weight and can reveal the presence of halogens (from isotope patterns). IR spectroscopy identifies functional groups by their characteristic bond vibrations. ¹H NMR tells you how many distinct types of hydrogen are present, how many of each there are (integration) and what their chemical environment looks like (chemical shift and splitting). Combining all three is how you solve "identify the structure" problems on the exam.
Molecular ion peak (M⁺)
The peak in a mass spectrum corresponding to the intact molecule after it loses one electron. Its m/z value equals the molecular weight.
In simple terms, this is the "weight" of your molecule as measured by the mass spectrometer.
Isotope pattern
The pattern of peaks near the molecular ion caused by naturally occurring isotopes. Bromine (⁷⁹Br and ⁸¹Br in roughly 1:1 ratio) gives a distinctive pair of peaks 2 mass units apart with nearly equal intensity. Chlorine (³⁵Cl and ³⁷Cl in roughly 3:1 ratio) gives a pair 2 mass units apart with the lighter peak about three times taller.
Degree of unsaturation (index of hydrogen deficiency)
Formula: DoU = (2C + 2 + N - H - X) / 2, where C = carbons, N = nitrogens, H = hydrogens, X = halogens. Each degree corresponds to one ring or one double bond; a triple bond counts as two.
IR spectroscopy
Measures the absorption of infrared light by molecular bonds. Different functional groups absorb at characteristic wavenumbers (cm⁻¹). The key region for Organic Chemistry I:
Broad O-H stretch: ~3200-3600 cm⁻¹ (alcohols, carboxylic acids)
Sharp C=O stretch: ~1700-1750 cm⁻¹ (ketones, aldehydes, carboxylic acids, esters)
C-H stretch: ~2850-3000 cm⁻¹
O-H of carboxylic acid: very broad, ~2500-3300 cm⁻¹, overlapping with C-H
¹H NMR (proton nuclear magnetic resonance)
Reveals the number of chemically distinct hydrogen environments (number of signals), the relative number of hydrogens in each environment (integration), the electronic environment of each hydrogen (chemical shift, in ppm) and the number of neighbouring hydrogens (splitting pattern, via the n+1 rule).
Chemical shift (delta, ppm)
The position of an NMR signal on the x-axis. Hydrogens near electron-withdrawing groups appear further downfield (higher ppm). Typical ranges: alkyl C-H (0.5-2.0 ppm), C-H next to C=C or halogen (2.0-4.5 ppm), vinyl C-H (4.5-6.5 ppm), aromatic C-H (6.5-8.0 ppm), aldehyde C-H (9-10 ppm).
Integration
The area under each NMR signal, proportional to the number of hydrogens producing that signal. If one signal integrates for 6H and another for 2H, the ratio is 3:1.
Step 1: Molecular formula from MS. The molecular ion peak gives you the molecular weight. Use the isotope pattern to spot halogens: twin peaks 2 units apart in ~1:1 ratio means bromine; ~3:1 ratio means chlorine.
Step 2: Degree of unsaturation. Calculate DoU. Each ring or double bond adds 1; a triple bond adds 2. This narrows the structural possibilities immediately.
Step 3: IR for functional groups. Look for the big diagnostic peaks. Broad absorption around 3300 cm⁻¹ = O-H (alcohol or acid). Sharp peak at ~1710 cm⁻¹ = C=O. If both are present, it may be a carboxylic acid. No broad O-H but a C=O = ketone, aldehyde or ester.
Step 4: ¹H NMR for hydrogen environments. Count the number of signals (= number of distinct hydrogen types). Note integration (how many H per signal). Note chemical shifts (what environment). Note splitting (how many neighbouring H). Piece together fragments that fit all the data.
From the exam: Compound A has a molecular peak at 56. Its ¹H NMR shows two signals: 2H at ~4.5 ppm and 6H at ~1.7 ppm.
MW 56 with only C and H: C₄H₈ (DoU = 1, one double bond).
The 6H singlet around 1.7 ppm suggests two equivalent methyl groups.
The 2H signal around 4.5 ppm suggests vinyl (C=C-H) protons.
Structure: 2-methylpropene (isobutylene), CH₂=C(CH₃)₂. The two methyl groups are equivalent (6H), and the two vinyl H atoms are equivalent (2H).
Compound A is treated with HBr/peroxides (anti-Markovnikov radical addition), giving Compound B.
Twin molecular peaks at 136 and 138 in 1:1 ratio confirm one bromine atom.
¹H NMR of B shows three signals: 2H (~3.3 ppm), 1H (~1.9 ppm) and 6H (~1.0 ppm).
Anti-Markovnikov addition puts Br on the less substituted carbon (CH₂ end).
Structure of B: 1-bromo-2-methylpropane, BrCH₂CH(CH₃)₂. The -CH₂Br gives 2H downfield, the -CH- gives 1H (septet region), and the two equivalent methyls give 6H (doublet).
When given an IR spectrum and five possible structures, look for the diagnostic peaks:
If the spectrum shows both a broad O-H stretch (~3300 cm⁻¹) and a strong C=O stretch (~1710 cm⁻¹), the compound is a carboxylic acid.
If it shows a broad O-H but no C=O, it is an alcohol.
If it shows a C=O but no broad O-H, it is a ketone, aldehyde or ester.
An ester will show C=O near 1735-1750 cm⁻¹ and C-O stretches but no broad O-H.
From the exam, a compound with both a broad O-H and a C=O peak, where one of the options is a hydroxy-acid (compound E with both -OH and -COOH features), is the match.
To predict the number of ¹H NMR signals a compound will show:
Identify all hydrogen atoms in the structure.
Group them by chemical equivalence: hydrogens related by a plane of symmetry, a rotation axis, or rapid conformational interconversion are equivalent.
Each unique group produces one signal.
For a chloro-substituted cyclopentene, carefully check for symmetry elements. The exam answer gives 5 signals for the specific structure shown, meaning five distinct hydrogen environments exist after accounting for any molecular symmetry.
SN2 rate depends on steric hindrance at the carbon bearing the leaving group:
Methyl halides: fastest.
Primary halides: fast.
Secondary halides: slow.
Tertiary halides: SN2 does not occur.
Neopentyl-type (primary but with a quaternary carbon next door): extremely slow, effectively no SN2.
From the exam, three chlorides ranked slow to fast: the neopentyl-type (III) is slowest, the secondary (I) is intermediate and the primary (II) is fastest: III < I < II.
Students often confuse the number of NMR signals with the number of hydrogen atoms. The number of signals equals the number of chemically distinct hydrogen environments, not the total hydrogen count. Integration tells you how many hydrogens are in each signal.
Students sometimes assume that every C-H in a molecule gives a separate NMR signal. Equivalent hydrogens (related by symmetry or free rotation) produce the same signal.
Students often forget that bromine's isotope pattern is nearly 1:1 (⁷⁹Br:⁸¹Br), while chlorine's is roughly 3:1 (³⁵Cl:³⁷Cl). Getting these backwards leads to wrong halogen identification.
Students sometimes look at an IR spectrum and try to assign every peak. Focus on the diagnostic region: O-H, C=O, N-H and the presence or absence of these features. The fingerprint region below ~1500 cm⁻¹ is useful only for matching, not for exam-style identification.
⚠️ Structure determination problems combine MS, IR and NMR data. You must be comfortable using all three together. Start with MS for molecular weight and halogen presence, move to IR for functional groups, then use NMR for the full structure.
⚠️ ¹H NMR signal counting is a stand-alone question type. Draw the structure, find symmetry elements, and count distinct H environments.
⚠️ IR identification is typically multiple-choice: given a spectrum, pick the matching compound. Focus on the 1700-1750 and 3200-3600 cm⁻¹ regions.
⚠️ SN2 rate ranking appears as a multiple-choice ordering question. The key variable is steric bulk around the electrophilic carbon.
True or false: Twin molecular ion peaks at m/z 136 and 138 in a 1:1 ratio indicate the presence of chlorine. (False: 1:1 ratio indicates bromine. Chlorine gives roughly 3:1.)
Fill in the blank: A broad absorption at 3300 cm⁻¹ in an IR spectrum indicates an ________ bond. (O-H, i.e., alcohol or carboxylic acid.)
True or false: In ¹H NMR, integration tells you the chemical environment of the hydrogens. (False: integration tells you the relative number of hydrogens. Chemical shift tells you the environment.)
Q: A compound with molecular formula C₄H₈ (MW 56) shows two ¹H NMR signals: 6H near 1.7 ppm and 2H near 4.5 ppm. What is the structure?
A: 2-Methylpropene (isobutylene), CH₂=C(CH₃)₂. The two equivalent methyl groups give the 6H signal and the two vinyl protons give the 2H signal.
Q: That compound is treated with HBr/peroxides. The product shows twin MS peaks at 136/138 (1:1). What is the product?
A: 1-Bromo-2-methylpropane, BrCH₂CH(CH₃)₂. Anti-Markovnikov radical addition places Br on the terminal carbon. The bromine isotope pattern confirms one Br atom.
Q: An IR spectrum shows a broad O-H stretch and a strong C=O peak. Which functional group class does this indicate?
A: Carboxylic acid. The combination of broad O-H and C=O is the signature of -COOH.
Q: A compound with a chlorine substituent on a cyclopentane ring is predicted to show 5 signals in its ¹H NMR. How do you determine this?
A: Draw the structure, identify the plane of symmetry (if any), and count the distinct hydrogen environments. Hydrogens related by the symmetry plane are equivalent and produce one shared signal.
Q: Rank these substrates for SN2 rate (slow to fast): neopentyl chloride, isopropyl chloride, n-propyl chloride.
A: Neopentyl (slowest, massive beta-branching blocks nucleophile approach) < isopropyl (secondary, moderate steric hindrance) < n-propyl (primary, least hindered, fastest).
Spectroscopy is used throughout Organic Chemistry II for identifying products of new reaction types (aromatic substitution, carbonyl chemistry, amines). The ability to interpret NMR and IR is cumulative and does not reset between semesters. Structure determination also connects to synthesis: after running a reaction, spectroscopy is how you confirm the product is what you expected.
Mass spectrometry, molecular ion, isotope pattern, bromine isotope ratio, chlorine isotope ratio, degree of unsaturation, index of hydrogen deficiency, IR spectroscopy, infrared, O-H stretch, C=O stretch, carbonyl, carboxylic acid IR, alcohol IR, ¹H NMR, proton NMR, chemical shift, ppm, integration, splitting, n+1 rule, signal count, equivalent hydrogens, symmetry, SN2 rate, steric hindrance, neopentyl, primary secondary tertiary, structure determination, spectroscopy problem, Organic Chemistry I, CHEM 2301, UMN