Isomerism and chirality in organic chemistry: exercises

In Year 5 of the technical course in chemistry, materials and biotechnology, isomerism is covered in class between October and November, right after the hybridisation of carbon. The test on organic chemistry usually falls between mid-November and December, that is, six to ten weeks from today. The topic is the same in the science-oriented high schools (liceo scientifico and applied sciences): at most the depth of the reaction mechanisms and the demands on nomenclature change, but the reasoning about isomers is the same.

In the post on hydrocarbons in Year 5 isomerism appears in a single section. Here I take it in full, because it is the topic where I see the most marks lost: half of the tests ask you to count, draw and name isomers, and anyone without a method goes by trial and error.

The tree of isomers

Two compounds are isomers if they have the same molecular formula but different formulas in space or in the connection of the atoms. It helps to draw a tree, because every question is solved by choosing the right branch.

Constitutional (or structural) isomers: the atoms are bonded in a different order.

• chain: different carbon skeleton (butane and 2-methylpropane);
• positional: same skeleton, same functional group in a different position (propan-1-ol and propan-2-ol);
• functional group: different functional groups (ethanol ${\text{CH}}_3{\text{CH}}_2\text{OH}$ and dimethyl ether ${\text{CH}}_3{\text{OCH}}_3$, both ${\text{C}}_2{\text{H}}_6\text{O}$). The post on functional groups serves as a review for this branch.

Stereoisomers: same connection of the atoms, but a different arrangement in space.

• geometric (cis/trans, E/Z), due to the double bond that prevents rotation (or to a ring);
• optical, due to chirality: enantiomers (non-superimposable mirror images) and diastereoisomers (stereoisomers that are not mirror images).

The second level, that of stereoisomers, is where marks are lost. Constitutional isomers have different physical and chemical properties. Enantiomers have identical physical properties (melting point, solubility) and are distinguished only by the way they rotate polarised light and by how they react with other chiral molecules.

Geometric isomerism: cis/trans and E/Z

On a $\text{C=C}$ double bond rotation is blocked. If each carbon of the double bond has two different substituents, there are two isomers. In but-2-ene ${\text{CH}}_3{\text{CH=CHCH}}_3$ the two methyls can be on the same side (cis) or on opposite sides (trans). If one carbon of the double bond has two identical substituents (as in but-1-ene, ${\text{CH}}_2{\text{=CHCH}}_2{\text{CH}}_3$, where one end is ${\text{CH}}_2$), geometric isomerism does not exist.

The E/Z nomenclature is more general because it also works with three or four different substituents. On each carbon of the double bond you give priority to the two substituents using the CIP rules (I describe them below). If the two of higher priority are on the same side the compound is Z (from German zusammen, together), if they are on opposite sides it is E (from entgegen, opposite).

The classic mistake is to believe that cis is always Z and trans always E. It is not so: cis and trans refer to two substituents chosen for similarity, E and Z to the two of higher priority. Example: in 2-bromobut-2-ene ${\text{CH}}_3{\text{CBr=CHCH}}_3$, on carbon 2 the priority is Br > ${\text{CH}}_3$, on carbon 3 it is ${\text{CH}}_3$ > H. The isomer that has the two methyls on the same side (cis) therefore has the bromine on the side opposite the methyl of carbon 3: it is the E isomer.

Chirality and the asymmetric carbon

A molecule is chiral when it is not superimposable on its mirror image, like the right hand and the left. In most of the cases you meet at school chirality arises from an asymmetric carbon (also called a stereocentre): an $s{p}^3$-hybridised carbon bonded to four substituents that are all different.

To recognise it, proceed like this: a ${\text{CH}}_2$ carbon is never asymmetric (it has two identical H), neither is a ${\text{CH}}_3$, neither is a carbon with a double bond or in $s{p}^2$ hybridisation. Then, for each candidate carbon, look at the four groups as far as needed to tell them apart: two chains that look the same at the first atom may differ further along. If the groups are all different, you have a stereogenic centre.

Example: in butan-2-ol ${\text{CH}}_3{\text{CH(OH)CH}}_2{\text{CH}}_3$ carbon 2 has OH, H, ${\text{CH}}_3$ and ${\text{CH}}_2{\text{CH}}_3$: four different, so asymmetric. Carbon 3 (${\text{CH}}_2$) and the ends are not.

How to draw the two enantiomers

With wedges and dashes: the central carbon lies in the plane of the page, two bonds stay in the plane, one goes forward (solid wedge) and one goes back (dash). To obtain the enantiomer you just swap two substituents, for example the one on the wedge and the one on the dash. Swapping two converts a molecule into its mirror image, swapping four (two swaps) brings it back identical.

The Fischer projection is a five-line shortcut: the carbon sits at the centre of a cross, the vertical bonds go behind the page, the horizontal ones come in front. Two rules are enough: rotating the projection by 180 degrees in the plane does not change the molecule, while swapping two substituents turns it into the enantiomer.

The CIP rules for assigning R and S

To tell the two enantiomers apart the R/S nomenclature is used. The working method has three moves.

1. Assign priority to the four substituents according to the atomic number of the atom directly bonded to the centre: higher atomic number, higher priority. Example: Br > Cl > F > C > H.
2. If two atoms are the same, move on to the next atoms. You compare the atoms bonded to them, one level at a time: ${\text{CH}}_2{\text{CH}}_3$ is a carbon bonded to (C, H, H), while ${\text{CH}}_3$ is a carbon bonded to (H, H, H), so ethyl takes priority over methyl. A double bond counts as two single bonds (the atom is duplicated).
3. Look at the molecule with the lowest-priority substituent (usually H) at the back, that is, on the side away from your eye. Follow 1 → 2 → 3: if the turn is clockwise the centre is R (rectus), if it is anticlockwise it is S (sinister).

If the lowest-priority substituent is not at the back, you can also look at it from wherever and then reverse the answer, because looking from the opposite side reverses the direction of rotation. In the Fischer projection, if the lowest-priority substituent is on a horizontal bond (in front of the page) you reverse the result you read.

Optical activity, racemates and meso compounds

Enantiomers rotate the plane of polarised light by an equal angle but in opposite directions. The rotatory power is indicated by $\alpha$ and an enantiomer is called dextrorotatory (+) if it rotates clockwise, levorotatory (−) if it rotates anticlockwise. A 50% mixture of the two enantiomers is a racemate (or racemic mixture) and shows no optical activity, because the two rotations cancel out.

A confusion I often see: R/S has no relation to +/−. R/S is a geometric convention on the priority of the substituents, (+)/(−) is an experimental datum measured with the polarimeter. An R molecule can be dextrorotatory or levorotatory, and you find out only by measuring.

With $n$ stereogenic centres the maximum number of stereoisomers is

The maximum becomes an exact equality only if there are no symmetries. If the molecule has an internal plane of symmetry despite having stereogenic centres, it is a meso compound: it is achiral, not optically active, and brings the number of stereoisomers below $2^n$ because two of the “theoretical” structures coincide.

The mistakes I see most often

• Counting a ${\text{CH}}_2$ as asymmetric, or a carbon with two identical substituents. If two groups are the same, the centre is not stereogenic.
• Confusing E/Z with cis/trans: they are not synonyms (see the example of 2-bromobut-2-ene).
• Saying that R means dextrorotatory. There is no connection.
• Forgetting to push the comparison beyond the first atom when two substituents begin with the same atom.
• Applying $2^n$ without checking whether there is a plane of symmetry (meso compounds).
• Drawing the mirror image and not checking that it really is non-superimposable.

Worked exercises

Exercise 1: how many isomers do ${\text{C}}_5{\text{H}}_{12}$ and ${\text{C}}_4{\text{H}}_8$ have?

${\text{C}}_5{\text{H}}_{12}$ is an alkane: the formula ${\text{C}}_n{\text{H}}_{2n+2}$ with $n=5$ gives 12, so there are no double bonds or rings, and the isomers are chain isomers only. You start from the longest chain and shorten it:

1. pentane: ${\text{CH}}_3{\text{CH}}_2{\text{CH}}_2{\text{CH}}_2{\text{CH}}_3$;
2. 2-methylbutane: chain of 4 with a methyl on carbon 2;
3. 2,2-dimethylpropane: chain of 3 with two methyls on the central carbon.

The isomers number 3. There is no “3-methylbutane”: numbering from the other end makes it 2-methylbutane, because you number so as to give the lowest number to the substituents.

${\text{C}}_4{\text{H}}_8$ has one degree of unsaturation (two hydrogens fewer than the alkane ${\text{C}}_4{\text{H}}_{10}$): it can be an alkene or a cycloalkane. The isomers are:

1. but-1-ene: ${\text{CH}}_2{\text{=CHCH}}_2{\text{CH}}_3$;
2. cis-but-2-ene;
3. trans-but-2-ene;
4. 2-methylpropene: ${\text{CH}}_2{\text{=C(CH}}_3{\text{)}}_2$;
5. cyclobutane;
6. methylcyclopropane.

The isomers number 6 in all: 5 constitutional, and one of these (but-2-ene) exists in two geometric forms. The mistake to avoid is stopping at the alkenes and forgetting the cycloalkanes, which have the same formula ${\text{C}}_n{\text{H}}_{2n}$.

Exercise 2: asymmetric carbons and number of stereoisomers

Find the stereogenic centres and the number of stereoisomers of a) butan-2-ol, b) 2,3-dichlorobutane.

a) Butan-2-ol, ${\text{CH}}_3{\text{CH(OH)CH}}_2{\text{CH}}_3$. Carbon 2 is bonded to OH, H, ${\text{CH}}_3$, ${\text{CH}}_2{\text{CH}}_3$: four different substituents. The other carbons have repeated groups (two H or three H). A single centre: $2^1=2$ stereoisomers, which are a pair of enantiomers, (R)-butan-2-ol and (S)-butan-2-ol.

b) 2,3-dichlorobutane, ${\text{CH}}_3{\text{CHClCHClCH}}_3$. Carbons 2 and 3 are both bonded to Cl, H, ${\text{CH}}_3$ and the other asymmetric carbon: two centres. The maximum is $2^2=4$, but the molecule is symmetric: the two halves are equal. The combinations are (R,R), (S,S), (R,S) and (S,R). The first two are enantiomers of each other. The last two are the same molecule, because there is a plane of symmetry passing through the middle of the C2–C3 bond: it is the meso form, achiral.

The real stereoisomers are therefore 3: the pair (R,R) and (S,S) and the meso (R,S). The meso is a diastereoisomer of each of the two enantiomers. If you applied $2^n$ without thinking, you got 4: it is the wrong answer the question expects.

Exercise 3: assigning R/S

Bromochlorofluoromethane $\text{CHFClBr}$ has a carbon with four different substituents. Drawn with H on a dash (behind the page), Br at the top, Cl at the bottom right and F at the bottom left, assign the configuration.

Priorities. Compare the atomic numbers: Br (35) > Cl (17) > F (9) > H (1). So 1 = Br, 2 = Cl, 3 = F, 4 = H.

Direction. The H is already at the back, so you read it directly. From Br (top) to Cl (bottom right) to F (bottom left): it is a clockwise turn.

Result. R configuration. The enantiomer, obtained by swapping for example F and Cl, has an anticlockwise turn and is S. We cannot say whether it is dextrorotatory or levorotatory without measuring it.

Quick check, the butan-2-ol case: the priorities are OH > ${\text{CH}}_2{\text{CH}}_3$ > ${\text{CH}}_3$ > H. Ethyl beats methyl because its first carbon is bonded to (C, H, H) against (H, H, H) for methyl.

Where to practise

There is no specific simulator for organic chemistry yet: to train on science reasoning you can use the high-school practice tests, and for the overall picture of the subject there is the sciences page. The practical advice is to do one or two isomer exercises by hand every day: start from a molecular formula, draw all the chains, then look for the stereogenic centres. For what comes next, the typical reactions of alkenes and alkanes are in addition and substitution, while anyone with doubts about electronic structure can revisit electron configuration and quantum numbers.

If a step in the tree of isomers or the CIP rules does not click, bring one of your own worked exercises and we will look at it together: from the contact page you can ask for a free introductory call.