Shapes of Organic Molecules and Isomerism

AS · 15 min

Isomers are different compounds with the same molecular formula. The formula $\ce{C4H10O}$ belongs to four alcohols and three ethers; $\ce{C4H8}$ belongs to six different hydrocarbons. Isomers can differ in which atoms are bonded to which (structural isomerism) or only in how the same bonded atoms are arranged in space (stereoisomerism). To understand the second kind you first need the three-dimensional shapes of organic molecules, which come from hybridisation. Isomer questions appear in every Paper 1 and Paper 2: "how many isomers", "draw the isomers", "identify the chiral centres", "explain why this compound shows cis/trans isomerism".

The shapes of organic molecules

Every carbon atom in an organic molecule is hybridised in one of three ways, depending on how many other atoms it is bonded to. You met the theory in Sigma and pi bonds and hybridisation; here is how it applies to organic molecules.

Key result

carbon forms	hybridisation	bonds	shape around that carbon	bond angle	example
four single bonds	sp³	four $\sigma$	tetrahedral	$109.5^\circ$	every C in ethane, $\ce{CH3CH3}$
one double bond, two single	sp²	three $\sigma$ , one $\pi$	trigonal planar	$120^\circ$	both C in ethene; the C of C=O
one triple bond, one single (or two double)	sp	two $\sigma$ , two $\pi$	linear	$180^\circ$	the C of $\ce{C#N}$ ; both C in ethyne

Every single bond is a $\sigma$ bond. A double bond is one $\sigma$ plus one $\pi$ . A triple bond is one $\sigma$ plus two $\pi$ .
A $\pi$ bond forms by sideways overlap of p orbitals, one on each carbon, above and below the line between the nuclei. This only works if the p orbitals are parallel, which locks the two carbons and all four atoms attached to them into one plane.
A molecule or part of a molecule is planar when all its atoms lie in one plane. Ethene is planar: both carbons are sp², and all six atoms lie in the same plane with angles close to $120^\circ$ .

The same rules apply to oxygen and nitrogen. In an alcohol the oxygen has two bonding pairs and two lone pairs (C–O–H angle about $104.5^\circ$ ); in an amine the nitrogen has three bonding pairs and one lone pair (pyramidal, about $107^\circ$ ).

Method

Counting $\sigma$ and $\pi$ bonds

Draw the displayed formula, with every bond shown.
Count every line: each single, double or triple bond contributes exactly one $\sigma$ bond.
Count the extra lines: each double bond adds one $\pi$ bond, each triple bond adds two.

Structural isomerism

Definition

Structural isomers are compounds with the same molecular formula but different structural formulae.

There are three kinds.

Chain isomerism

The carbon skeleton is different: a straight chain in one isomer, branched in another. $\ce{C5H12}$ has three chain isomers:

isomer	structural formula	boiling point
pentane	$\ce{CH3CH2CH2CH2CH3}$	$36\ ^\circ\text{C}$
2-methylbutane	$\ce{CH3CH(CH3)CH2CH3}$	$28\ ^\circ\text{C}$
2,2-dimethylpropane	$\ce{C(CH3)4}$	$10\ ^\circ\text{C}$

Chain isomers have the same functional group and similar chemistry, but different physical properties. The more branched the molecule, the lower its boiling point: branching makes the molecule more compact, so there is less surface contact between neighbouring molecules and the instantaneous dipole–induced dipole forces are weaker.

Positional isomerism

The carbon skeleton is the same, but the functional group is in a different position. Propan-1-ol, $\ce{CH3CH2CH2OH}$ , and propan-2-ol, $\ce{CH3CH(OH)CH3}$ , are positional isomers. So are but-1-ene and but-2-ene, and 1-bromobutane and 2-bromobutane. Positional isomers can react differently: propan-1-ol is a primary alcohol and is oxidised to an aldehyde and acid, while propan-2-ol is secondary and gives a ketone.

Functional group isomerism

The isomers have different functional groups, so they belong to different homologous series and have very different chemistry. Common pairs at AS:

molecular formula	isomers
$\ce{C_{n}H_{2n}}$	alkene and cycloalkane: propene, $\ce{CH2=CHCH3}$ , and cyclopropane
$\ce{C_{n}H_{2n}O}$	aldehyde and ketone: propanal, $\ce{CH3CH2CHO}$ , and propanone, $\ce{CH3COCH3}$
$\ce{C_{n}H_{2n}O2}$	carboxylic acid and ester: propanoic acid, $\ce{CH3CH2COOH}$ , and methyl ethanoate, $\ce{CH3COOCH3}$
$\ce{C_{n}H_{2n+2}O}$	alcohol and ether: ethanol, $\ce{CH3CH2OH}$ , and methoxymethane, $\ce{CH3OCH3}$

Stereoisomerism

Definition

Stereoisomers are compounds with the same structural formula but a different arrangement of their atoms in space.

Stereoisomers have the same atoms bonded to the same atoms. They differ only in three-dimensional arrangement. There are two kinds: geometrical (cis/trans) isomerism and optical isomerism.

Geometrical (cis/trans) isomerism

Around a C–C single bond the two ends of a molecule rotate freely, so $\ce{CH2ClCH2Cl}$ has only one form, whichever way you draw it. Around a C=C double bond they cannot: rotating one carbon relative to the other would twist the p orbitals out of line and break the $\pi$ bond. This restricted rotation means that groups fixed on the same side of a double bond stay on the same side.

Geometrical isomers of but-2-ene. In the cis isomer the two methyl groups are on the same side of the C=C bond; in the trans isomer they are on opposite sides. The π bond prevents rotation, so one cannot turn into the other at room temperature.

Key result

Geometrical (cis/trans) isomerism needs both:

a C=C bond (or a ring), which prevents rotation, and
two different groups on each carbon of the double bond.

In the cis isomer, the two like groups (or the two non-hydrogen groups) are on the same side of the double bond. In the trans isomer, they are on opposite sides.

The second condition is the one students forget. But-1-ene, $\ce{CH2=CHCH2CH3}$ , has no geometrical isomers: carbon 1 carries two hydrogen atoms, so swapping them makes no difference. Methylpropene, $\ce{CH2=C(CH3)2}$ , has none either: carbon 2 carries two identical methyl groups.

Cis and trans isomers are different compounds with different physical properties. Cis-but-2-ene boils at about $4\ ^\circ\text{C}$ and trans-but-2-ene at about $1\ ^\circ\text{C}$ .

Tip

The E/Z system is the general IUPAC method. For simple cases like but-2-ene, cis is Z and trans is E. The syllabus says E/Z naming is acceptable but not required; cis/trans is enough.

Cis/trans isomerism in rings. A ring also prevents rotation about its C–C bonds. In 1,2-dimethylcyclohexane, the two methyl groups can be on the same face of the ring (cis) or on opposite faces (trans). The test is the same: each of the two ring carbons must carry two different groups (here, $\ce{H}$ and $\ce{CH3}$ ).

Optical isomerism

Definition

A chiral centre is a carbon atom bonded to four different atoms or groups. A molecule with one chiral centre exists as two optical isomers (enantiomers): non-superimposable mirror images of each other.

Your hands are a familiar chiral object: each is the mirror image of the other, but you cannot place one exactly on top of the other. A carbon with four different groups is the same. Butan-2-ol, $\ce{CH3CH(OH)CH2CH3}$ , has a chiral centre at carbon 2: it carries H, OH, $\ce{CH3}$ and $\ce{C2H5}$ .

The two enantiomers of butan-2-ol. The central carbon is a chiral centre (bonded to H, CH₃, OH and C₂H₅). Each molecule is the mirror image of the other, and no rotation makes them superimposable.

To draw a pair of enantiomers, draw the chiral carbon in three dimensions using two bonds in the plane of the paper, one wedge (towards the viewer) and one hashed bond (away from the viewer). Then draw its reflection in a mirror line. The four groups must be attached in the reflected positions.

Enantiomers have identical physical properties (same boiling point, melting point, solubility) and identical chemical reactions with non-chiral reagents. They differ in one way that can be measured: each rotates the plane of plane-polarised light by the same angle in opposite directions. For this reason they are described as optically active.

Tip

A mixture of equal amounts of the two enantiomers is called a racemic mixture; it does not rotate plane-polarised light because the two effects cancel. Racemic mixtures, and why they matter in drug synthesis, are covered at A Level.

Finding chiral centres

Method

Identifying chiral centres

Ignore every carbon with two or more hydrogen atoms ( $\ce{CH3}$ , $\ce{CH2}$ ): it cannot have four different groups.
Ignore every carbon in a double or triple bond: it has only three or two attached groups.
For each remaining carbon, write down the whole group on each of its four bonds, not just the next atom. Two groups that both start with $\ce{CH2}$ may still be different further along.
If all four groups are different, mark the carbon with an asterisk (*).
In a ring, compare the two ways round the ring from that carbon. If they are identical, the carbon is not chiral.

A molecule can contain more than one chiral centre. Each one doubles the number of possible stereoisomers, so a molecule with $n$ chiral centres has at most $2^n$ stereoisomers. A molecule with a chiral centre and a cis/trans double bond has up to $2 \times 2 = 4$ stereoisomers.

Deducing all the isomers of a formula

Method

Finding every isomer of a molecular formula

Work out which functional groups are possible. Use the general formulae: $\ce{C_{n}H_{2n}}$ means one C=C or one ring; $\ce{C_{n}H_{2n}O}$ suggests an aldehyde or ketone; and so on.
For each functional group, draw the longest chain first, then shorten the chain by one carbon and add branches.
For each skeleton, move the functional group to every distinct position.
Name every structure. If two structures have the same name, they are the same compound: delete one.
Check each structure for cis/trans isomerism (C=C with two different groups on each carbon) and optical isomerism (a chiral centre).

Worked examples

Routine: hybridisation, shape and bond counting

Propenenitrile, $\ce{CH2=CHCN}$ , is used to make acrylic fibres.

(a) State the hybridisation of each carbon atom. (b) Give the number of $\sigma$ bonds and $\pi$ bonds in one molecule. (c) State the H–C–H bond angle at carbon 1 and the C–C–N bond angle at carbon 3. (d) Is the molecule planar?

Solution

Number the carbons from the left: $\ce{CH2=CH-C#N}$ has C1 ( $\ce{CH2}$ ), C2 (CH) and C3 (the nitrile carbon).

(a) C1: one double bond, two single bonds: sp². C2: sp². C3: one triple bond, one single bond: sp.

(b) Bonds: two C1–H, one C2–H, C1=C2, C2–C3, C3≡N. Every bond has one $\sigma$ : $2 + 1 + 1 + 1 + 1 = 6$ $\sigma$ bonds. $\pi$ : one in C=C and two in C≡N: 3 $\pi$ bonds.

(d) Yes. C1 and C2 are sp², so C1, C2 and the atoms attached to them (two H on C1, one H on C2, and C3) all lie in one plane. C3 and N lie on a straight line through C2 and C3, so they are in the same plane too.

Routine: cis/trans or not?

State which of these show geometrical isomerism, explaining each answer: (a) pent-1-ene, (b) pent-2-ene, (c) 2-methylbut-2-ene, (d) 1,2-dichloroethene.

Solution

(a) $\ce{CH2=CHCH2CH2CH3}$ : carbon 1 carries two H atoms. No.

(b) $\ce{CH3CH=CHCH2CH3}$ : C2 carries H and $\ce{CH3}$ ; C3 carries H and $\ce{C2H5}$ . Two different groups on each carbon. Yes: cis- and trans-pent-2-ene.

(d) $\ce{CHCl=CHCl}$ : each carbon carries H and Cl. Yes: cis-1,2-dichloroethene (both Cl on the same side) and trans-1,2-dichloroethene.

Standard: all the isomers of C4H8

Draw and name all the isomers with molecular formula $\ce{C4H8}$ , including stereoisomers.

Solution

$\ce{C4H8}$ fits $\ce{C_{n}H_{2n}}$ : either one C=C (alkene) or one ring (cycloalkane).

Alkenes, four-carbon chain: but-1-ene $\ce{CH2=CHCH2CH3}$ ; but-2-ene $\ce{CH3CH=CHCH3}$ , which has cis and trans forms.

Alkenes, three-carbon chain with a methyl branch: methylpropene $\ce{CH2=C(CH3)2}$ (no number needed; no cis/trans, since C1 has two H).

Cycloalkanes: cyclobutane (a four-membered ring) and methylcyclopropane (a three-membered ring with a methyl group).

Total: six isomers (but-1-ene, cis-but-2-ene, trans-but-2-ene, methylpropene, cyclobutane, methylcyclopropane). None has a chiral centre.

Standard: isomers of C4H9Br including optical

(a) Give the structural formulae and names of the four structural isomers of $\ce{C4H9Br}$ . (b) Which one shows optical isomerism? Identify its chiral centre and explain why it is chiral. (c) How many isomers of $\ce{C4H9Br}$ are there in total?

Solution

(a) Four-carbon chain: $\ce{CH3CH2CH2CH2Br}$ 1-bromobutane; $\ce{CH3CH2CHBrCH3}$ 2-bromobutane. Three-carbon chain with a methyl branch: $\ce{(CH3)2CHCH2Br}$ 1-bromo-2-methylpropane; $\ce{(CH3)3CBr}$ 2-bromo-2-methylpropane.

(b) 2-bromobutane. Carbon 2 is bonded to H, Br, $\ce{CH3}$ and $\ce{CH2CH3}$ : four different groups, so it is a chiral centre and the molecule exists as two non-superimposable mirror images. (In 1-bromo-2-methylpropane, C2 carries two identical $\ce{CH3}$ groups, so it is not chiral.)

Exam-hard: counting stereoisomers

(a) Pent-3-en-2-ol has the structural formula $\ce{CH3CH(OH)CH=CHCH3}$ . Explain why it has four stereoisomers.

(b) 3-methylpentan-2-ol, $\ce{CH3CH(OH)CH(CH3)CH2CH3}$ . Identify each chiral centre by listing the four groups on it, and state the maximum number of stereoisomers.

Solution

(a) Carbon 2 is bonded to H, OH, $\ce{CH3}$ and $\ce{CH=CHCH3}$ : four different groups, so it is a chiral centre (two optical forms). The C3=C4 double bond has H and $\ce{CH(OH)CH3}$ on C3 and H and $\ce{CH3}$ on C4: two different groups on each carbon and restricted rotation, so cis/trans isomerism (two forms). The two features are independent: each optical form can be cis or trans, giving $2 \times 2 = 4$ stereoisomers.

(b) C2: H, OH, $\ce{CH3}$ , $\ce{CH(CH3)CH2CH3}$ . All different: chiral.

C3: H, $\ce{CH3}$ , $\ce{CH2CH3}$ , $\ce{CH(OH)CH3}$ . All different: chiral.

Two chiral centres, so at most $2^2 = 4$ stereoisomers.

Watch out

"Two different groups" means on each carbon of the C=C. A compound such as but-1-ene has a double bond but no cis/trans isomers. State both conditions in any explanation.
Free rotation about single bonds. $\ce{CH2BrCH2Br}$ drawn with the two Br atoms on "the same side" and on "opposite sides" is the same molecule. Only a double bond or a ring prevents rotation.
Chiral centre groups. Look at the whole group, not just the first atom. In 3-methylhexane, C3 carries $\ce{CH2CH3}$ and $\ce{CH2CH2CH3}$ : both begin with $\ce{CH2}$ , but they are different, so C3 is chiral.
Drawing enantiomers. The two drawings must be genuine mirror images with 3D bonds (wedge and hash). Two flat drawings with groups swapped do not earn the mark.
Saying the isomers "react differently". Enantiomers have identical chemical properties (with non-chiral reagents) and identical physical properties apart from the direction in which they rotate plane-polarised light.

Exam tip

"Explain why compound X shows geometrical isomerism" needs two points: restricted rotation about the C=C (because of the $\pi$ bond) and two different groups on each carbon of the C=C.
"Explain what is meant by a chiral centre": a carbon atom attached to four different atoms or groups. Optical isomers: non-superimposable mirror images.
Draw optical isomers as 3D tetrahedral structures with a mirror line. Use wedge and hashed bonds and make sure each bond goes to the correct atom (for example, the bond to OH must end at the O).
Marks for "how many isomers" questions are lost by drawing the same structure twice. Name every structure you draw; a repeated name reveals a duplicate.
Mark chiral centres with an asterisk on the carbon itself, as the question instructs.

Summary

sp³ carbon: four $\sigma$ bonds, tetrahedral, $109.5^\circ$ . sp² carbon: three $\sigma$ + one $\pi$ , trigonal planar, $120^\circ$ . sp carbon: two $\sigma$ + two $\pi$ , linear, $180^\circ$ .
Every bond contains one $\sigma$ bond; double bonds add one $\pi$ , triple bonds add two.
Structural isomers: same molecular formula, different structural formula. Three kinds: chain, positional, functional group.
Stereoisomers: same structural formula, different arrangement in space. Two kinds: geometrical (cis/trans) and optical.
Cis/trans needs restricted rotation (C=C or ring) and two different groups on each carbon.
A chiral centre is a carbon with four different groups; it gives two enantiomers, non-superimposable mirror images that rotate plane-polarised light in opposite directions.
$n$ chiral centres give at most $2^n$ stereoisomers.

Practice

Question

State the hybridisation of each carbon atom in propanal, $\ce{CH3CH2CHO}$ , and give the bond angle around each carbon.
How many $\sigma$ bonds and how many $\pi$ bonds are there in a molecule of but-2-ynoic acid, $\ce{CH3C#CCOOH}$ ?
Name the type of structural isomerism shown by each pair: (a) butan-1-ol and butan-2-ol; (b) butane and methylpropane; (c) propanal and propanone.
Explain why but-2-ene shows geometrical isomerism but but-1-ene does not.
Which of these molecules contains a chiral centre? Identify it. (a) propan-2-ol (b) 2-chlorobutane (c) 2-hydroxypropanoic acid (d) 3-chloropentane
Draw the two optical isomers of 2-hydroxypropanoic acid, $\ce{CH3CH(OH)COOH}$ , showing the 3D arrangement.
Give the structural formulae and names of all the alkenes with molecular formula $\ce{C5H10}$ , and state which show cis/trans isomerism. (There are six, counting cis and trans forms separately.)
A compound has the molecular formula $\ce{C3H6O}$ . Draw and name three structural isomers of it that contain a C=O group or an OH group, and name the type of isomerism between the aldehyde and the ketone.
Compound T, $\ce{C5H10O}$ , is an aldehyde. Its molecule contains a chiral centre. Deduce the structure of T and name it, showing that no other aldehyde of this formula is chiral.
Citral contains the fragment $\ce{(CH3)2C=CHCH2CH2C(CH3)=CHCHO}$ . (a) Identify which of the two C=C bonds can show cis/trans isomerism, explaining your answer. (b) How many stereoisomers of this fragment are possible? (c) State the hybridisation of the carbonyl carbon and the C=C–C bond angle at that alkene carbon.

Answers

$\ce{CH3}$ carbon: sp³, $109.5^\circ$ . $\ce{CH2}$ carbon: sp³, $109.5^\circ$ . CHO carbon: sp², $120^\circ$ (trigonal planar).
Displayed: three C–H, C–C, C≡C, C–C, C=O, C–O, O–H. $\sigma$ : $3 + 1 + 1 + 1 + 1 + 1 + 1 = 9$ . $\pi$ : two in C≡C, one in C=O, so 3.
(a) Positional. (b) Chain. (c) Functional group.
In but-2-ene, $\ce{CH3CH=CHCH3}$ , each carbon of the double bond carries two different groups (H and $\ce{CH3}$ ), and rotation about the C=C is restricted because rotation would break the $\pi$ bond, so the methyl groups can be fixed on the same side (cis) or opposite sides (trans). In but-1-ene, $\ce{CH2=CHCH2CH3}$ , carbon 1 carries two identical H atoms, so swapping sides gives the same molecule.
(a) No: C2 carries two $\ce{CH3}$ groups. (b) Yes: C2 carries H, Cl, $\ce{CH3}$ , $\ce{C2H5}$ . (c) Yes: C2 carries H, OH, $\ce{CH3}$ , COOH. (d) No: C3 carries H, Cl and two identical $\ce{C2H5}$ groups.
Draw the central carbon with H and $\ce{CH3}$ on in-plane bonds, OH on a wedge and COOH on a hashed bond; then draw its exact reflection in a vertical mirror line (H and $\ce{CH3}$ in the reflected in-plane positions, OH on the reflected wedge, COOH on the reflected hashed bond). The two are non-superimposable.
Pent-1-ene $\ce{CH2=CHCH2CH2CH3}$ (no cis/trans). Pent-2-ene $\ce{CH3CH=CHCH2CH3}$ (cis and trans). 2-methylbut-1-ene $\ce{CH2=C(CH3)CH2CH3}$ (no: C1 has two H). 3-methylbut-1-ene $\ce{CH2=CHCH(CH3)2}$ (no). 2-methylbut-2-ene $\ce{(CH3)2C=CHCH3}$ (no: C2 has two $\ce{CH3}$ ). Total six including cis- and trans-pent-2-ene.
Propanal $\ce{CH3CH2CHO}$ ; propanone $\ce{CH3COCH3}$ ; prop-2-en-1-ol $\ce{CH2=CHCH2OH}$ (also accepted: cyclopropanol). Propanal and propanone are functional group isomers.
The four aldehydes with formula $\ce{C5H10O}$ are: pentanal $\ce{CH3CH2CH2CH2CHO}$ (no carbon has four different groups); 3-methylbutanal $\ce{(CH3)2CHCH2CHO}$ (C3 carries two identical $\ce{CH3}$ groups, so not chiral); 2,2-dimethylpropanal $\ce{(CH3)3CCHO}$ (C2 carries three $\ce{CH3}$ groups); and 2-methylbutanal $\ce{CH3CH2CH(CH3)CHO}$ . In 2-methylbutanal, C2 is bonded to H, $\ce{CH3}$ , $\ce{C2H5}$ and CHO: four different groups. T is 2-methylbutanal.
(a) $\ce{(CH3)2C=CH-}$ : the first carbon carries two identical $\ce{CH3}$ groups, so no cis/trans. $\ce{-C(CH3)=CHCHO}$ : one carbon carries $\ce{CH3}$ and $\ce{CH2CH2-}$ (different), the other carries H and CHO (different), and rotation is restricted, so this C=C does show cis/trans isomerism. (b) One cis/trans double bond and no chiral centre (no carbon carries four different groups): 2 stereoisomers. (c) The carbonyl carbon is sp²; the alkene carbons are sp² too, so the C=C–C angle is about $120^\circ$ .

The shapes of organic molecules

Structural isomerism

Chain isomerism

Positional isomerism

Functional group isomerism

Stereoisomerism

Geometrical (cis/trans) isomerism

Optical isomerism

Finding chiral centres

Deducing all the isomers of a formula

Worked examples

Practice

Console