8HLCis–trans isomerism

SL students can skip to section 14.

Stereoisomers have the same constitution, meaning the same atoms, joined in the same order by the same kinds of bond, but a different arrangement of those atoms in space. HL covers two kinds: cis–trans isomers here, and enantiomers in section 9.

A C=C double bond cannot rotate: rotating one end would break the π bond. So if each carbon of the double bond carries two different groups, there are two distinct arrangements. In cis isomers the matching groups are on the same side of the double bond; in trans isomers they are on opposite sides. Figure 6(a) shows cis- and trans-but-2-ene. But-1-ene has no such isomers, because its C1 carries two identical H atoms, so swapping them changes nothing.

A small ring locks rotation the same way. In a 1,2-disubstituted cyclopropane or cyclobutane, the two groups can be on the same face of the ring (cis) or opposite faces (trans), as Figure 6(b) shows. The E–Z naming system is not assessed.

Cis–trans isomers are different compounds with different physical properties. Cis-but-2-ene boils at about 4 °C and trans at about 1 °C. The difference can be larger when the groups are polar: in cis-1,2-dichloroethene both C–Cl dipoles point the same way and the molecule is polar, while in the trans isomer they cancel.

The guide's HL link to S2.2 belongs here too: there are only three dibromobenzenes (1,2-, 1,3- and 1,4-). If benzene had alternating single and double bonds, 1,2-dibromobenzene would come in two forms, bromines across a single or across a double bond. Only one exists, which supports the model of six identical, delocalised C–C bonds.

9HLChirality and enantiomers

A carbon atom bonded to four different atoms or groups is a chiral carbon. A molecule with one has two forms that are mirror images of each other and cannot be superimposed, however you rotate them, the way a left hand cannot fit a right glove. The two forms are enantiomers. Figure 7 shows the enantiomers of butan-2-ol, whose C2 carries H, OH, CH₃ and CH₂CH₃.

Draw them the way the guide asks: the chiral carbon with two bonds in the plane of the paper, one wedge coming towards you and one hashed bond going away, then the mirror image beside it. Check with a model: if swapping any two groups on one drawing turns it into the other, you have drawn a genuine pair.

Enantiomers have identical boiling points, melting points and densities, and react identically with non-chiral reagents. They differ in two ways:

• Optical activity. Each enantiomer rotates the plane of plane-polarised light, by the same angle but in opposite directions. A compound that does this is optically active.
• Behaviour in a chiral environment. Enzymes and receptors in the body are themselves chiral, so they can respond to one enantiomer and not the other. The two enantiomers of carvone, for example, smell different, one of spearmint and one of caraway.

A racemic mixture contains equal amounts of the two enantiomers. Their rotations cancel, so it is not optically active. Reactions that make a chiral centre from non-chiral starting materials usually produce a racemic mixture.

10HLMass spectrometry of organic compounds

In a mass spectrometer, molecules are ionised and many break into pieces. Only positive ions are detected, and the spectrum plots their relative abundance against mass-to-charge ratio, m/z (for a 1+ ion, simply its mass). Two kinds of peak matter.

The molecular ion, M⁺, is the whole molecule minus one electron. It is usually the peak with the highest m/z (ignore a tiny peak one unit above it, caused by ¹³C), and it gives the relative molecular mass.

Fragment ions form when bonds in the molecular ion break. The gaps between M⁺ and the fragments tell you what was lost. The data booklet lists common fragments; the ones you meet most:

Figure 8 shows why this matters. Pentan-2-one and pentan-3-one both give M⁺ at 86. Pentan-2-one breaks beside its C=O to give CH₃CO⁺ at 43, the tallest peak (the base peak), and loses CH₃ to give 71. Pentan-3-one has an ethyl group on each side of the C=O, so it gives C₂H₅CO⁺ at 57 and C₂H₅⁺ at 29, and no peak at 43. The fragments place the carbonyl.

11HLInfrared spectroscopy

Covalent bonds vibrate, stretching and bending, at frequencies that depend on the masses of the atoms and the strength of the bond. A bond absorbs infrared radiation whose frequency matches its vibration. Chemists quote this as a wavenumber, in cm⁻¹. Because each type of bond absorbs in a characteristic range, an IR spectrum tells you which bonds, and so which functional groups, a molecule contains.

A spectrum plots transmittance against wavenumber, with wavenumber decreasing from left to right. Each dip is an absorption. The region above about 1500 cm⁻¹ is the functional group region; below it lies the fingerprint region, a pattern unique to each compound but too complex to read bond by bond. The data booklet gives the full table. The ranges you will use most:

Figure 9 compares propan-1-ol and propanoic acid. Both have C–H and C–O. Only the acid has the sharp C=O near 1715 cm⁻¹, and its O–H is broader and lower than the alcohol's.

A vibration absorbs IR only if it changes the dipole moment of the molecule. That is the link to climate. N₂ and O₂ are symmetrical diatomic molecules: stretching them changes no dipole, so they are not IR active. H₂O, CH₄ and CO₂ have vibrations that do change the dipole. CO₂ is non-polar at rest, but its asymmetric stretch and its bend create a temporary dipole. So these greenhouse gases absorb the infrared radiation emitted by the Earth's warm surface and re-emit it in all directions, some back towards the surface, which warms the lower atmosphere. Reactivity 1.3 takes this further.

12HL¹H NMR spectroscopy

In a strong magnetic field, hydrogen nuclei absorb radio waves. The exact frequency depends on the chemical environment of each hydrogen atom: which atoms are nearby, and how much electron density surrounds it. Hydrogen atoms in the same environment, such as the three H of a CH₃ group, absorb together. A ¹H NMR spectrum gives four pieces of evidence.

Number of signals = the number of different hydrogen environments. Look for symmetry: propanone, CH₃COCH₃, has one environment, because both CH₃ groups are equivalent.

Chemical shift, δ, in ppm, measured from the signal of tetramethylsilane, TMS, Si(CH₃)₄, which is set at 0. TMS is used because its twelve equivalent H give one strong signal away from almost everything else, and it is inert and easily removed. Hydrogen near an electronegative atom or a C=O absorbs at higher δ. The data booklet lists the ranges; approximately:

Integration: the area under each signal is proportional to the number of hydrogen atoms in that environment. The spectrum shows it as an integration trace, whose step heights give the ratio.

Splitting: a signal is split by the hydrogen atoms on adjacent carbon atoms. By the n + 1 rule, n neighbouring H split a signal into n + 1 peaks: 0 neighbours gives a singlet, 1 a doublet, 2 a triplet (intensities 1:2:1), 3 a quartet (1:3:3:1). Equivalent H do not split each other, and the H of an O–H usually appears as a singlet.

Figure 10 reads all four for methyl propanoate, CH₃CH₂COOCH₃.

Three signals: three environments. Areas 3 : 2 : 3. The CH₃ at δ ≈ 1.1 has two H next door, so it is a triplet. The CH₂ at δ ≈ 2.3, next to the C=O, has three H next door, so it is a quartet. The O–CH₃ at δ ≈ 3.7 is next to the O, and its neighbour is an oxygen with no H, so it is a singlet. A triplet (3H) with a quartet (2H) is the signature of an ethyl group.

13HLPutting the evidence together

Each technique answers a different question, and exam problems give you several at once. Mass spectrometry gives the molecular mass and pieces; IR gives the functional groups; NMR gives the hydrogen skeleton. Use each for what it is good at, and check that the final structure explains every piece of data.

Worked example. An unknown has molecular formula C₃H₆O. Its mass spectrum shows M⁺ at 58 and a base peak at 43. Its IR spectrum has a strong absorption at 1715 cm⁻¹ and nothing between 3200 and 3600 cm⁻¹. Its ¹H NMR spectrum has a single signal at δ = 2.2. Deduce the structure.

Propanal would have shown three signals, including one near δ 9.7 for the CHO hydrogen. Say which evidence rules out each alternative; that is where the "deduce" marks are.