Structural isomerism and stereoisomerism
The gas in a patio-heater cylinder is mostly butane and methylpropane. Both have the molecular formula $\mathrm{C_4H_{10}}$, yet they are different substances with different boiling points and different shapes. Two compounds with the same molecular formula but different structures are called isomers. Isomers differ either in how their atoms are connected (structural isomerism) or in how the same connections are arranged in space (stereoisomerism). By the end of this page you will be able to explain why butane and methylpropane are different.
Structural isomerism: chain, position and functional group
Two bottles hold liquids with the molecular formula $\mathrm{C_2H_6O}$. One is ethanol, which boils at $78\ ^\circ\text{C}$ and mixes freely with water. The other is methoxymethane (dimethyl ether), a flammable gas that boils at $-24\ ^\circ\text{C}$. How can two compounds made of exactly the same atoms behave so differently?
Different connections mean different bonding and intermolecular forces, and so different physical and chemical properties. Ethanol has an $\mathrm{-OH}$ group and forms hydrogen bonds. Methoxymethane has no O–H bond and cannot. That is why ethanol is a liquid and methoxymethane is a gas at room temperature.
Aldehydes and ketones are also functional group isomers of each other. $\mathrm{C_3H_6O}$ can be propanal, $\mathrm{CH_3CH_2CHO}$ (an aldehyde), or propanone, $\mathrm{CH_3COCH_3}$ (a ketone). It can also be the unsaturated alcohol prop-2-en-1-ol, $\mathrm{CH_2=CHCH_2OH}$.
Finding all the isomers
Work systematically. Start with the longest straight chain, then shorten the main chain by one carbon and place the leftover carbon as a branch in every different position. Keep going until you run out of possibilities. Remember the limit for this course: up to C6.
Organic compounds of molecular formula $\mathrm{C_3H_6O}$ show different types of structural isomerism depending on the connectivity of their atoms and the functional…
Organic compounds of molecular formula $\mathrm{C_3H_6O}$ show different types of structural isomerism depending on the connectivity of their atoms and the functional groups present.
- a) Write the semi-structural formula and the systematic name of an aldehyde and of a ketone that share this molecular formula, stating the type of isomerism between them.
- b) Write the semi-structural formula and the systematic name of an unsaturated acyclic alcohol with the same molecular formula $\mathrm{C_3H_6O}$.
The saturated molecular formula $\mathrm{C_4H_{10}O}$ includes various constitutional isomers belonging to different oxygen-containing functional groups. a) Write the…
The saturated molecular formula $\mathrm{C_4H_{10}O}$ includes various constitutional isomers belonging to different oxygen-containing functional groups.
- a) Write the semi-structural formula and the systematic name of the four possible isomeric alcohols with this formula, and state what type of structural isomerism (chain or position) exists between butan-1-ol and butan-2-ol, and between butan-1-ol and 2-methylpropan-1-ol.
- b) Write the semi-structural formula and the systematic name of three isomeric ethers with this molecular formula. What type of isomerism exists between butan-1-ol and diethyl ether?
Geometric (cis–trans) isomerism
Pure but-2-ene can be obtained as two different liquids. One boils at $3.7\ ^\circ\text{C}$ and the other at $0.9\ ^\circ\text{C}$, even though both are $\mathrm{CH_3CH=CHCH_3}$. What stops one form from turning into the other simply by twisting?
The answer lies in the double bond. A $\mathrm{C=C}$ bond is made of a $\sigma$ bond and a $\pi$ bond. The $\pi$ bond holds the two carbons in a flat arrangement, so there is no free rotation around the double bond.
Different shapes, different properties
Geometric isomers are different substances. They differ in melting point, boiling point and polarity. In the cis isomer the bond dipoles of the two substituents point to the same side and partly add up, so the molecule has a net dipole moment. In the trans isomer they point in opposite directions and cancel (completely, if the two substituents are identical). Take 1,2-dichloroethene as an example. The cis isomer is polar (boiling point about $60\ ^\circ\text{C}$) and the trans isomer is non-polar (boiling point about $48\ ^\circ\text{C}$). The stronger intermolecular forces in the polar cis form explain its higher boiling point.
The following illustration shows the spatial configurations of two unsaturated hydrocarbons labelled A and B. Observe carefully the arrangement of their atoms. a)…
The following illustration shows the spatial configurations of two unsaturated hydrocarbons labelled A and B. Observe carefully the arrangement of their atoms.
- a) Identify the type of isomerism between compounds A and B, and assign to each structure its complete systematic name, including the cis or trans descriptors.
- b) Justify which of the two isomers will have an appreciable net dipole moment (polar molecule) and which will have zero dipole moment (non-polar molecule).
The E/Z system
Cis and trans work well when each carbon has an obvious "matching pair" of groups, as in but-2-ene. What do you call an alkene like 2-chlorobut-2-ene, where the four groups are all different or none are obviously alike?
The Cahn–Ingold–Prelog (CIP) rules give a way to decide. On each carbon of the double bond, rank the two groups by the atomic number $Z$ of the atom directly attached (higher $Z$ means higher priority). If the two atoms tie, compare the atoms attached to them, working outwards until a difference appears.
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If the two higher-priority groups are on the same side, the isomer is (Z) (from German zusammen, together).
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If they are on opposite sides, the isomer is (E) (from entgegen, opposite).
The E/Z nomenclature system proposed by Cahn, Ingold and Prelog (CIP) assigns priorities to the substituents attached to each carbon of the double bond according to…
The E/Z nomenclature system proposed by Cahn, Ingold and Prelog (CIP) assigns priorities to the substituents attached to each carbon of the double bond according to decreasing atomic number ($Z$). If the two highest-priority groups on each carbon lie on the same side of the double bond, it is designated (Z), and if they are on opposite sides, (E).
Consider 2-chlorobut-2-ene, whose molecular formula is $\mathrm{C_4H_7Cl}$.
Atomic number data: $Z(\mathrm{Cl}) = 17$; $Z(\mathrm{C}) = 6$; $Z(\mathrm{H}) = 1$.
- a) Determine the order of priority of the two substituents on carbon 2 and on carbon 3 of the double bond.
- b) Draw the semi-structural spatial arrangement and name, with justification, the two possible geometric stereoisomers using the (E) and (Z) notation.
You can do this on paper and send a photo with your answer.
Chirality: asymmetric carbon and enantiomers
R-carvone smells of caraway seeds and S-carvone smells of spearmint. They have the same formula, the same connectivity and the same boiling point. What can possibly be different between two such molecules?
The difference is that the molecules are mirror images of each other that cannot be placed on top of one another, just as your left hand cannot be superimposed on your right.
In diagrams, a solid wedge means a bond pointing towards you, a dashed wedge means a bond pointing away, and plain lines lie in the plane of the paper.
Properties of enantiomers
Putting it together: from a chiral halogenoalkane to alkenes
Of the four isomers of $\mathrm{C_4H_9Cl}$ only one has a carbon with four different groups: 2-chlorobutane, $\mathrm{CH_3CHClCH_2CH_3}$. Heating a haloalkane with potassium hydroxide in concentrated ethanol removes H and Cl from neighbouring carbons (an elimination reaction) and forms a $\mathrm{C=C}$ bond. From 2-chlorobutane, losing H from carbon 1 gives but-1-ene. Losing H from carbon 3 gives but-2-ene, which exists as cis and trans isomers. Elimination is covered in more detail under organic reactions.
Butan-2-ol is a secondary alcohol of molecular formula $\mathrm{C_4H_{10}O}$. a) Write its semi-structural formula and mark with an asterisk () the carbon that acts as…
Butan-2-ol is a secondary alcohol of molecular formula $\mathrm{C_4H_{10}O}$.
- a) Write its semi-structural formula and mark with an asterisk (*) the carbon that acts as the chiral centre (asymmetric carbon), justifying why it meets this condition.
- b) Deduce whether an enantiomerically pure sample of butan-2-ol shows optical activity, and state what the two non-superimposable spatial forms that are mirror images of each other are called.
Lactic acid (2-hydroxypropanoic acid) is a fundamental metabolite in anaerobic muscle biological processes and in lactic fermentation. a) Write its semi-structural…
Lactic acid (2-hydroxypropanoic acid) is a fundamental metabolite in anaerobic muscle biological processes and in lactic fermentation.
- a) Write its semi-structural formula, identify its chiral centre and represent the pair of enantiomers in three dimensions using wedge formulae (bond in the plane, solid wedge towards the front and dashed wedge towards the back).
- b) Explain which physical and chemical properties both pure enantiomers share in an achiral environment, in which measurable physical property they differ quantitatively, and what the macroscopic behaviour of an equimolar mixture of both (racemic mixture) is towards polarised light in a polarimeter.
Vuelve al principio: butane (a straight chain of four carbons) and methylpropane (a three-carbon chain with a methyl branch) share the formula $\mathrm{C_4H_{10}}$ but have different carbon skeletons. They are structural isomers of the chain type, which is why they are different substances.