Source: Carey, Organic Chemistry, 11th ed., Chapters 4 and 14 (skip 14.14-14.19 on 13C NMR)
Tags: chirality, stereochemistry, enantiomers, diastereomers, R and S configuration, optical activity, meso compounds, spectroscopy, IR spectroscopy, mass spectrometry, 1H NMR, proton NMR, chemical shift, splitting, integration, CHEM 2301 UMN
Difficulty: Intermediate | Prerequisites: Chapters 1-3 study notes (Lewis structures, hybridisation, conformational analysis)
Big picture: Chapter 4 introduces stereochemistry: the study of molecules that have the same connectivity but differ in the three-dimensional arrangement of their atoms. This is where organic chemistry moves from flat drawings to genuine 3D thinking, and it matters enormously in biology and medicine (your left hand and right hand are mirror images, and so are many drug molecules). Chapter 14 teaches you how chemists figure out what a molecule looks like without seeing it directly, using infrared spectroscopy, mass spectrometry, and proton NMR. These two chapters together give you the tools to describe molecular structure precisely and to identify unknown compounds from spectral data.
Chirality is the property of a molecule that makes it non-superposable on its mirror image, usually because it has a carbon bonded to four different groups (a chirality centre). Enantiomers are mirror-image pairs; diastereomers are stereoisomers that are not mirror images. You assign R/S configuration using Cahn-Ingold-Prelog priority rules. Spectroscopy (Ch. 14) provides three complementary tools: IR tells you which functional groups are present, mass spectrometry gives you the molecular weight and fragmentation clues, and 1H NMR reveals the carbon-hydrogen framework through chemical shifts, integration, and splitting patterns.
Chirality
The geometric property of a molecule (or any object) that makes it non-superposable on its mirror image. In simple terms, a chiral molecule and its mirror image are related the way your left and right hands are: similar but not identical.
Chirality centre (stereogenic centre, stereocentre)
A carbon atom bonded to four different substituents. This is the most common source of chirality in organic molecules. Sometimes called an "asymmetric carbon" or "chiral carbon."
Enantiomers
Stereoisomers that are non-superposable mirror images of each other. They have identical physical properties (boiling point, melting point, solubility) except for the direction in which they rotate plane-polarised light.
Diastereomers
Stereoisomers that are not mirror images of each other. Unlike enantiomers, diastereomers have different physical properties (different melting points, boiling points, solubilities, etc.).
Optical activity
The ability of a chiral substance to rotate the plane of plane-polarised light. Measured with a polarimeter. A sample that rotates light clockwise is dextrorotatory (+); counterclockwise is levorotatory (-).
Specific rotation
The standardised measurement of optical activity, corrected for path length and concentration. Symbol: [alpha]. You cannot predict the sign of rotation from the R/S designation; it must be measured experimentally.
Racemic mixture (racemate)
A 50:50 mixture of two enantiomers. It shows zero net optical rotation because the rotations cancel.
Meso compound
A molecule that contains chirality centres but is achiral overall because it possesses an internal plane of symmetry. Think of it as a molecule where the chiral centres cancel each other out.
R and S configuration (Cahn-Ingold-Prelog rules)
The system for assigning absolute configuration at a chirality centre. Substituents are ranked by atomic number (higher = higher priority). With the lowest-priority group pointing away from you, if the remaining three groups decrease in priority going clockwise, the centre is R (Latin: rectus, right); counterclockwise is S (Latin: sinister, left).
Fischer projection
A two-dimensional representation of a three-dimensional molecule at a chirality centre. Horizontal lines represent bonds coming out of the page; vertical lines represent bonds going into the page.
Look for sp3 carbons bonded to four different groups.
A carbon in a CH2 or CH3 group is never a chirality centre (it has at least two identical substituents).
Trace each of the four substituent "branches" outward from the carbon. If all four are different, the carbon is a chirality centre.
Rings can contain chirality centres. Treat the two paths around the ring from the candidate carbon as two separate substituents and compare them.
Step 1: Assign priorities 1 (highest) through 4 (lowest) to the four substituents using atomic number at the first point of difference.
Higher atomic number = higher priority.
If two atoms are identical at the first position, move outward along each chain until a difference is found.
Atoms in double or triple bonds are "expanded": a C=O is treated as if C is bonded to O twice and O is bonded to C twice.
Step 2: Orient the molecule so that priority 4 points away from you (into the page).
Step 3: Trace a path from priority 1 to 2 to 3. Clockwise = R. Counterclockwise = S.
Shortcut: if the lowest-priority group is on a dash (going into the page), you can assign directly. If it is on a wedge (coming toward you), trace 1-2-3 and then reverse the answer.
One chirality centre: two possible stereoisomers (one pair of enantiomers).
Two chirality centres: up to four stereoisomers (two pairs of enantiomers), unless a meso form exists.
General maximum: 2^n stereoisomers for n chirality centres.
Enantiomers: mirror images, identical physical properties except optical rotation.
Diastereomers: not mirror images, different physical properties. Cis/trans isomers on rings are a type of diastereomer.
A meso compound has chirality centres but an internal plane of symmetry that makes the overall molecule achiral.
Classic example: (2R,3S)-tartaric acid, where the top half of the molecule is the mirror image of the bottom half.
A meso compound is optically inactive (no rotation of plane-polarised light) despite having chirality centres.
To spot a meso compound: look for a molecule with two or more chirality centres whose substituent pattern creates an internal mirror plane.
A pure enantiomer will rotate plane-polarised light by a characteristic amount.
A racemic mixture has zero net rotation.
Enantiomeric excess (ee) = |% of one enantiomer - % of the other|.
100% ee = pure enantiomer.
0% ee = racemic.
Resolution: the separation of a racemic mixture into its component enantiomers, typically using a chiral resolving agent.
Electromagnetic spectrum
The full range of electromagnetic radiation, from radio waves to gamma rays. Organic spectroscopy uses infrared (IR), radio-frequency (NMR), and high-energy electron beams (mass spectrometry).
Infrared (IR) spectroscopy
A technique that measures the absorption of infrared light by molecular bonds. Different bond types absorb at characteristic frequencies (reported in wavenumbers, cm-1). Think of it as a fingerprint for functional groups.
Wavenumber
The unit used in IR spectroscopy, measured in cm-1 (reciprocal centimetres). Higher wavenumber = higher energy = higher frequency.
Mass spectrometry (MS)
A technique that ionises molecules, fragments them, and measures the mass-to-charge ratio (m/z) of the resulting ions. It gives you the molecular weight and, from the fragmentation pattern, structural clues.
Molecular ion (M+)
The peak in a mass spectrum corresponding to the intact molecule after losing one electron. Its m/z value equals the molecular weight.
Base peak
The tallest peak in a mass spectrum (assigned 100% relative abundance). It corresponds to the most stable fragment ion.
1H NMR spectroscopy (proton NMR)
A technique that detects hydrogen atoms in a molecule by placing the sample in a strong magnetic field and irradiating with radio-frequency energy. It reveals: how many distinct types of hydrogen are present (number of signals), their electronic environment (chemical shift), how many of each type (integration), and which hydrogens are near each other (splitting pattern).
Chemical shift (delta)
The position of an NMR signal on the spectrum, measured in parts per million (ppm) relative to a reference compound (TMS, tetramethylsilane, defined as 0 ppm). Electronegative groups and pi bonds shift signals downfield (higher ppm).
Shielding and deshielding
A hydrogen surrounded by high electron density is shielded (absorbs at lower ppm, upfield). A hydrogen near electronegative atoms or pi systems is deshielded (absorbs at higher ppm, downfield).
Integration
The relative area under each NMR signal, which is proportional to the number of hydrogens producing that signal. If one signal integrates for 3 and another for 2, the ratio of hydrogens is 3:2.
Spin-spin splitting (coupling)
The splitting of an NMR signal into multiple lines caused by neighbouring non-equivalent hydrogens. Follows the n+1 rule: a hydrogen with n equivalent neighbours appears as n+1 lines.
Coupling constant (J)
The distance (in Hz) between the lines of a split NMR signal. Hydrogens that are coupled to each other share the same J value.
O-H stretch: broad absorption around 3200-3600 cm-1. Very distinctive.
N-H stretch: medium absorption around 3300-3500 cm-1 (can be sharp or broad).
C-H stretch: around 2850-3000 cm-1 for sp3 C-H; above 3000 cm-1 for sp2 C-H.
C=O stretch (carbonyl): strong, sharp absorption around 1700-1750 cm-1. One of the most recognisable IR absorptions.
C=C stretch: around 1600-1680 cm-1, usually weaker than C=O.
C-O stretch: around 1000-1260 cm-1.
The "fingerprint region" (below ~1500 cm-1) is complex and unique to each molecule, but individual peaks are harder to assign.
The molecular ion (M+) peak gives the molecular weight. Not always the tallest peak; sometimes it is very small or absent.
The nitrogen rule: a compound with an odd molecular weight contains an odd number of nitrogen atoms.
Common fragmentations to recognise:
Loss of 15 = loss of CH3
Loss of 18 = loss of H2O (from alcohols)
Loss of 29 = loss of CHO or C2H5
Loss of 31 = loss of OCH3
m/z = 57 = tert-butyl cation, C4H9+
m/z = 91 = tropylium cation (from benzyl/toluene fragmentation)
0-1 ppm: alkyl H (CH3, CH2 far from electronegative groups)
1-2 ppm: allylic H, H alpha to carbonyl
2-4.5 ppm: H on carbon attached to electronegative atom (O, N, halogen)
4.5-6.5 ppm: vinylic H (H on C=C)
6.5-8.5 ppm: aromatic H
9-10 ppm: aldehyde H
10-13 ppm: carboxylic acid H (variable, broad)
A hydrogen with n equivalent neighbouring hydrogens on adjacent carbons appears as a multiplet with n+1 lines.
0 neighbours = singlet (s)
1 neighbour = doublet (d)
2 neighbours = triplet (t)
3 neighbours = quartet (q)
And so on.
Equivalent hydrogens do not split each other.
The coupling constant J is the same for both partners in a coupled pair. This helps you match which signals are coupled.
The integral curve (or reported ratio) tells you the relative number of hydrogens for each signal.
Combine integration with the molecular formula to get absolute numbers. If the formula has 10 hydrogens and the integration ratio is 3:2:2:3, those signals correspond to 3H, 2H, 2H, and 3H.
Start with the molecular formula. Calculate degrees of unsaturation.
Check the IR for functional groups (OH, NH, C=O, C=C).
Use the mass spectrum for molecular weight confirmation and key fragments.
In the NMR: count the number of distinct signals. Use chemical shifts to propose environments. Use integration for hydrogen counts. Use splitting to identify neighbours.
Propose a structure. Verify it accounts for all data.
Maximum number of stereoisomers
2^n, where n = number of chirality centres (reduced if meso forms exist)
Enantiomeric excess
ee = [(major enantiomer - minor enantiomer) / total] x 100%
Specific rotation
[alpha] = observed rotation / (path length in dm x concentration in g/mL)
n+1 rule (NMR splitting)
A signal is split into n+1 lines, where n = number of equivalent H on adjacent carbons
Degree of unsaturation (review)
DoU = (2C + 2 + N - H - X) / 2
Chirality is central to pharmacology. Thalidomide is the most cited example: one enantiomer treated morning sickness, while the other caused severe birth defects. Modern drug development routinely produces single enantiomers rather than racemates, because the two mirror-image forms often have different biological activities.
NMR spectroscopy is the same physics behind MRI scans in hospitals. In MRI, the "nuclei" being detected are the hydrogen atoms in water molecules in your body. The chemical shift differences between hydrogen in different tissue environments produce the contrast in the image.
Students often assume that R configuration means the compound rotates light to the right (+). There is no connection between R/S labels and the sign of optical rotation. R/S is assigned from a set of priority rules; the direction of rotation must be measured experimentally.
A common mistake is confusing enantiomers with diastereomers. If two stereoisomers are mirror images, they are enantiomers. If they are not mirror images, they are diastereomers. The distinction matters because diastereomers have different physical properties and can be separated by ordinary techniques, while enantiomers cannot.
Students frequently forget to orient the lowest-priority group away from them before assigning R or S. If the lowest-priority group faces toward you, you must reverse your initial assignment.
In NMR, students often forget that equivalent hydrogens do not split each other. The two H atoms in a CH2 group that are equivalent by symmetry produce one signal, not a doublet.
⚠️ R/S assignment is a guaranteed exam topic. Practise until you can assign configuration in under 30 seconds per centre, including cases where the lowest-priority group is on a wedge.
⚠️ Be able to identify meso compounds. The key clue is an internal mirror plane in a molecule with two or more chirality centres.
⚠️ Know the relationship chart: same compound, enantiomers, diastereomers, or constitutional isomers. Draw structures and compare systematically.
⚠️ For spectroscopy, memorise the key IR absorptions (O-H, C=O, N-H, C-H). You do not need to memorise the entire fingerprint region.
⚠️ Be fluent with NMR interpretation: given a molecular formula and a spectrum, propose a structure. Practise working backwards from spectra to structures.
⚠️ Note: Sections 14.14-14.19 on 13C NMR are excluded from this course.
True or False: A molecule with two chirality centres must be chiral.
Fill in the blank: The maximum number of stereoisomers for a compound with three chirality centres is ______.
True or False: A broad absorption at 3300 cm-1 in an IR spectrum suggests the presence of an O-H or N-H bond.
Fill in the blank: A proton NMR signal that appears as a quartet indicates the hydrogen has ______ equivalent neighbours on adjacent carbons.
True or False: Enantiomers can be separated by distillation.
Answers: 1. False (it could be a meso compound). 2. 8 (2^3 = 8, though meso forms may reduce this). 3. True. 4. Three (n+1 = 4, so n = 3). 5. False (enantiomers have identical boiling points; separation requires a chiral environment or resolving agent).
Q: A compound has the molecular formula C4H8O. Its IR spectrum shows a strong absorption at 1715 cm-1 but no broad O-H stretch. Its 1H NMR shows two signals: a singlet at 2.1 ppm (integrating for 3H) and a triplet at 2.4 ppm (integrating for 2H). Wait, that does not add up to 8H. Propose a structure consistent with the available data.
A: The strong IR absorption at 1715 cm-1 indicates a carbonyl (C=O). No broad O-H means it is not a carboxylic acid or alcohol. The molecular formula C4H8O with one degree of unsaturation (the C=O) points to a ketone or aldehyde. Given the partial NMR data suggesting a methyl group near a carbonyl (singlet at 2.1 ppm), a reasonable candidate is butanone (methyl ethyl ketone, CH3COCH2CH3), which would show a singlet for the CH3 next to the carbonyl, a quartet for the CH2, and a triplet for the terminal CH3.
Q: Draw (R)-2-bromobutane. Assign priorities and confirm your configuration.
A: At C2, the four substituents are Br (priority 1, highest atomic number), CH2CH3 (priority 2), CH3 (priority 3), and H (priority 4). Orient H (priority 4) on a dash (going away). Trace 1 (Br) to 2 (CH2CH3) to 3 (CH3). If the path is clockwise, the configuration is R.
Q: Identify the relationship between (2R,3R)-tartaric acid and (2R,3S)-tartaric acid.
A: They are diastereomers. They differ at one chirality centre (C3) but not the other (C2), so they are not mirror images. Note: (2R,3S)-tartaric acid is the meso form.
Q: A mass spectrum shows M+ at m/z = 86 and a base peak at m/z = 57. What fragment was lost?
A: 86 - 57 = 29. Loss of 29 suggests loss of CHO or C2H5. The m/z = 57 peak is consistent with a C4H9+ cation (tert-butyl). A reasonable structure is 3,3-dimethylbutanal or 2,2-dimethylbutanal, where loss of the formyl group gives the tert-butyl cation.
Chirality connects directly to Chapters 5-7 (substitution and elimination reactions), where the stereochemical outcome of a reaction (inversion, retention, or racemisation) depends on the mechanism (SN1, SN2, E1, E2). Spectroscopy is used throughout the rest of the course whenever you need to confirm what product was formed in a reaction. The ability to read NMR and IR spectra is cumulative: later chapters will assume you can do it.
chirality, chiral centre, stereocentre, asymmetric carbon, enantiomers, diastereomers, meso compound, optical activity, specific rotation, polarimeter, racemic mixture, racemate, enantiomeric excess, R configuration, S configuration, Cahn-Ingold-Prelog priority rules, Fischer projection, stereoisomers, infrared spectroscopy, IR spectrum, wavenumber, functional group identification, mass spectrometry, molecular ion, base peak, fragmentation, nitrogen rule, 1H NMR, proton NMR, chemical shift, ppm, TMS, shielding, deshielding, integration, spin-spin splitting, coupling, n+1 rule, coupling constant, singlet doublet triplet quartet, Carey organic chemistry chapter 4, chapter 14, CHEM 2301 UMN, organic chemistry I stereochemistry