Revision sheet: nomenclature, degree of unsaturation, Z/E isomerism and R/S chirality (202-SNU-RE)
This chapter is decided in two places: the name you write and the spatial arrangement you read. Nomenclature is lost on a ranking done too fast (the longest chain, the principal functional group, alphabetical order), the degree of unsaturation on a forgotten atom, Z/E isomerism on a group judged too big or too small, and chirality on an asymmetric carbon mistaken for the whole molecule.
This sheet does not copy the course, which you already have. It gathers the moves that decide, from IUPAC nomenclature of alkanes, alkenes, rings and functional groups to the meso compound, with the cost of each mistake as it appears on papers in the CEGEP organic chemistry course 202-SNU-RE, an elective of the Natural Science program that follows General Chemistry 202-SN1-RE.
Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.
The thread of the chapter
A name, a degree of unsaturation, a Z or R descriptor: three answers read off a DRAWING, and lost in the same way, by a ranking done too fast. The longest chain is not the one you see, the priority group is not the biggest, and an asymmetric carbon is not the whole molecule.
•1. Main chain: the longest one, but it must contain the principal functional group, or failing that the multiple bond. Look for it in ALL directions of the drawing, branches included.
•2. Numbering: the lowest locant to the principal functional group, then to the multiple bond, then to the set of substituents, series compared locant by locant at the FIRST difference.
•3. Substituents: an alkane that loses an H becomes a group ending in -yl (methyl, ethyl, propyl); a halogen becomes a prefix (chloro-, bromo-). Identical groups take di-, tri-, tetra- and one locant each.
•4. Alphabetical order: it works on the name of the group, WITHOUT counting di-, tri-, tetra-. Ethyl comes before methyl, and “dimethyl” is filed under m.
•5. Assembly: locants and substituents, name of the chain, suffix. Commas between two numbers, hyphens between a number and a letter: 4-ethyl-2,3-dimethylhexane.
The drawing does not lay the main chain out horizontally for you. You find it by counting the carbons of every possible path, you do not simply read it.
Several functional groups: only one gives the suffix
•Order of priority, from highest to lowest: carboxylic acid, ester, amide, nitrile, aldehyde, ketone, alcohol, amine, then the multiple bonds (alkene, alkyne), finally the alkane.
•The principal functional group gives the SUFFIX (-oic acid, -oate, -amide, -nitrile, -al, -one, -ol, -amine) and gets the lowest locant. Acid, ester, amide, nitrile and aldehyde sit at the end of the chain: their carbon is C1.
•The other groups become PREFIXES: hydroxy- (alcohol), oxo- (ketone or aldehyde), amino- (amine), along with halogens and alkyl groups. The alkene stays an infix: but-3-en-1-ol.
•The same group changes form with its rank: -ol or hydroxy-, -one or oxo-, -amine or amino-.
Two rankings carry the word “priority”: the one in nomenclature, above, which reads functional groups, and the Cahn-Ingold-Prelog one, below, which reads atomic numbers. They do not give the same order.
Alkenes, alkynes, rings and benzene
•Alkene or alkyne: the chain contains the multiple bond, which gets the lowest locant BEFORE the substituents. The locant is that of the first carbon of the bond (pent-2-ene). With equal locants, the double bond goes before the triple bond: pent-1-en-4-yne.
•Ring: prefix cyclo-, numbering that starts at a substituted carbon and runs in the direction giving the lowest locants. A single substituent takes no locant: methylcyclohexane.
•Ring or chain: the structure with more carbons is the parent, the other becomes a substituent. Propylcyclohexane (six carbons against three), but cyclopropylheptane (seven against three).
•Benzene: the group C6H5− is phenyl; ortho, meta and para mean positions 1,2, 1,3 and 1,4; phenol, toluene and aniline are retained common names.
A double bond inside a ring gets locants 1 and 2 automatically, so we write cyclohexene and 3-methylcyclohexene, never “cyclohex-1-ene”.
Degree of unsaturation: count before drawing
•For CcHhNnOoXx: DU=22c+2+n−h−x. Oxygen does not count, each nitrogen adds, each halogen X counts as a hydrogen.
•Each ring is worth 1 and each π bond is worth 1: a double bond C=C or C=O is worth 1, a triple bond is worth 2.
•A DU of 4 or more suggests a benzene ring: 1 ring and 3 π bonds.
•The DU COUNTS the unsaturations, it does not say which: C5H8 has DU=2 and can be an alkyne, a diene or a ring carrying a double bond.
Same formula C5H8, same DU=2, three unrelated structures: the number says HOW MANY unsaturations, never which ones.
The DU is computed before drawing, and recomputed on the drawing: if it changes, the structure or the name is wrong.
The four possible relationships between two structures
•Different molecular formulas: they are not isomers, and the discussion stops there.
•Same molecular formula, different connectivity of the atoms: CONSTITUTIONAL isomers, of chain (different skeletons), of position (same functional group, different carbon) or of functional group (different functional groups).
•Same connectivity, different spatial arrangement: STEREOISOMERS. Non-superimposable mirror images: enantiomers. Everything else, including Z/E pairs: diastereomers.
•Interconvertible by simple rotation about a single bond: conformers, therefore the SAME molecule, not isomers.
Same formula C4H10O or C4H8: the connectivity changes (constitutional), the space changes (Z/E, mirror), or a simple rotation is enough, and it is the same molecule.
The word “isomer” used alone earns nothing: the marker expects the exact family, and that is what carries the marks.
Z or E: two conditions, one rule
•Condition 1: a double bond. Condition 2: EACH carbon of the double bond carries two DIFFERENT groups. A single carbon with identical groups removes the pair (but-1-ene carries two H on its first carbon).
•Rule: on each carbon, rank the two groups by the CIP priorities. The two priority groups on the same side: Z (zusammen). On opposite sides: E (entgegen).
•Cis and trans compare two identical groups or two hydrogens; Z and E compare the PRIORITY groups, which are not always the ones you look at. Cis is not a synonym of Z.
•The two isomers are separable diastereomers: going from one to the other requires breaking the π bond, which room temperature does not provide.
The descriptor is written in parentheses at the start of the name, with the locant of each double bond when there are several: (2Z,4E).
Chirality: the definition that decides, and the one that misleads
•The definition that decides: a molecule is chiral if it cannot be superimposed on its mirror image. It is a property of the WHOLE molecule.
•An asymmetric carbon, written C∗, carries four groups that are all different. It is a property of an ATOM, and it does not imply that the molecule is chiral.
•An internal plane of symmetry makes the molecule achiral whatever the number of C∗: this is the meso case.
•A chiral molecule and its image form a pair of enantiomers: same melting and boiling points, same solubility in an achiral solvent, opposite optical rotations.
Two drawings identical up to the mirror, the hydrogen on the hashed wedge toward the back: 1→2→3 turns clockwise on the left, the other way on the right.
On an exam, the question “is this molecule chiral?” is always handled in two steps: count the C∗, then look for internal symmetry. Skipping the second step is the costliest mistake of the chapter.
The CIP rules in four moves
•1. Rank the four atoms directly bonded to the C∗ by DECREASING atomic number: I>Br>Cl>S>P>O>N>C>H.
•2. Ties: move away from one atom and compare the triplets of atoms carried, ordered by decreasing atomic number, at the FIRST point of difference. Ethyl gives (C,H,H), methyl (H,H,H), so ethyl wins.
•3. A multiple bond counts as many times as it has bonds: the carbon of −CHO is seen as (O,O,H), that of −COOH as (O,O,O).
•4. Place group 4 toward the back, then read 1→2→3: clockwise gives R, counterclockwise gives S.
We do not count the atoms of a group: we compare the first atom, then the next triplet, a double bond counting twice.
The ranking is settled at the first point of difference, never by adding up atomic numbers: −CH2Cl beats −C(CH3)3 at the very first comparison, with a single chlorine against three carbons.
What is counted and what is measured
•n stereogenic centres give AT MOST 2n stereoisomers. Each meso compound removes one.
•Biot's law: α=[α]lc, with l the cell length in decimetres and c the concentration in g/mL.
•Enantiomeric excess: ee=∣αpure∣∣αobs∣×100%, and the percentage of the major enantiomer is 2100+ee.
•Racemic mixture: 50/50, so α=0 and no optical activity, even though every molecule in it remains chiral.
The problem always gives [α] in degrees for a 1 dm cell and a concentration of 1 g/mL: the units of l and c are not negotiable.
The mistakes that cost marks
These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.
1.Choosing the main chain you can see instead of the longest one
the whole nomenclature question, often 2 marks
What not to write
“The horizontal chain has five carbons: it is a pentane carrying a propyl on carbon 2, 2-propylpentane.”
What to write
“The main chain is the longest one in the whole drawing. Going through the branch, it has seven carbons: it is a heptane carrying a methyl at 4, 4-methylheptane.”
The same drawing read two ways: in red the horizontal chain, five carbons; in green the longest chain, seven carbons, which goes through the branch. The bond left blue is the methyl.
Why: A name built on a chain shorter than the longest is wrong even when the molecule is the right one. The test is mechanical: count the carbons of every path between two ends, the longest wins.
2.Alphabetizing substituents by counting “di” or “tri”
1 mark
What not to write
“2,3-dimethyl-4-ethylhexane: the d of dimethyl comes before the e of ethyl.”
What to write
“The multiplying prefixes di-, tri-, tetra- do not count in alphabetical order. We rank ethyl and methyl, so ethyl is written first: 4-ethyl-2,3-dimethylhexane.”
Why: Alphabetical order chooses the numbering only when the locants tie, but it always decides the ORDER of writing, and a name written in the wrong order is marked wrong.
3.Giving the suffix to the alcohol when a higher-priority group is present
the whole question, often 3 marks
What not to write
“The molecule has an OH group, so it is an alcohol: propan-1-ol carrying an aldehyde group as a substituent.”
What to write
“The CHO group is an aldehyde, higher priority than the alcohol: it gives the suffix -al and fixes carbon 1. The alcohol becomes the prefix hydroxy-: 3-hydroxypropanal.”
Blue box: the principal functional group, which gives the suffix and carbon 1. Orange box: the group that becomes a prefix.
Why: Only one functional group is principal in a name. The others change form: -ol becomes hydroxy-, -one becomes oxo-. Read the priority list from top to bottom and stop at the first group present.
4.Numbering an alkene for the substituents before the multiple bond
1 mark, and the name is wrong
What not to write
“The methyl must get the lowest locant: 2-methylbut-3-ene.”
What to write
“The double bond gets the lowest locant before the substituents. We number from the end where it starts: 3-methylbut-1-ene.”
Why: The criteria are tested in a fixed order: principal functional group, multiple bond, substituents. We move to the next one only on a tie at the previous one. “But-3-ene” does not exist: a double bond at the end of a chain is always but-1-ene.
5.Forgetting the nitrogen in the degree of unsaturation
2 marks, and the structure drawn from this number is wrong
What not to write
“Caffeine C8H10N4O2: DU=22×8+2−10=4.”
What to write
“Each nitrogen adds: DU=22×8+2+4−10=6. Oxygen changes nothing, and a halogen counts as a hydrogen.”
Why: A saturated amine carries one more hydrogen than the alkane with the same number of carbons, while an alcohol or an ether changes none: that is the reason for the +n and for the silence of oxygen.
6.Concluding Z because two identical groups are on the same side
2 marks
What not to write
“The two methyls are on the same side of the double bond: it is the Z.”
What to write
“On carbon 3, CH3 beats H; on carbon 2, Cl beats CH3. The priority groups, CH3 at the top and Cl at the bottom, are on opposite sides: (E)-2-chlorobut-2-ene.”
Orange box: the priority group on each carbon. On the left they are on the same side, Z; on the right they are opposite, E, even though the two methyls end up on the same side.
Why: Cis and trans compare identical groups, and are unambiguous only if each carbon carries a hydrogen. Z and E compare the priority groups: here chlorine (Z=17) beats carbon (Z=6) and reverses the answer.
7.Looking for a Z/E pair on a double bond where one carbon carries two identical groups
1 mark
What not to write
“But-1-ene exists as Z and as E.”
What to write
“The first carbon of but-1-ene carries two hydrogens, identical: swapping them changes nothing, so there is no Z/E pair. But-2-ene carries H and CH3 on each carbon: (Z) and (E) exist.”
Why: The condition applies to EACH carbon of the double bond. A single carbon with identical groups is enough to void it, as in 2-methylprop-1-ene, where one carbon carries two H and the other two methyls.
8.Counting 2n stereoisomers on a symmetric diene
1 mark
What not to write
“Hexa-2,4-diene has two Z/E double bonds, so 22=4 stereoisomers.”
What to write
“The molecule is symmetric: (2Z,4E) and (2E,4Z) are the same molecule read from the other end. There are three: (2Z,4Z), (2E,4E) and (2Z,4E). With an ethyl end, hepta-2,4-diene, all four exist.”
Why: 2n is a maximum, never a result: a symmetry reduces it, exactly like the meso compound for asymmetric centres.
9.Concluding “chiral” as soon as an asymmetric carbon is seen
the whole question, often 4 marks
What not to write
“The molecule has two asymmetric carbons, so it is chiral and has an enantiomer.”
What to write
“The molecule has two asymmetric carbons, but an internal plane of symmetry exchanges its two halves: it is superimposable on its image, so it is achiral. It is the meso compound, and its optical rotation is zero.”
Two asymmetric carbons on each side, but on the left the dashed line is a true internal plane of symmetry, the upper half being the image of the lower: this molecule is achiral.
Why: The asymmetric carbon is a local property, chirality a global property of the molecule. Tartaric acid is the example every exam uses, precisely because it separates the two.
10.Reading the direction 1→2→3 without looking at where group 4 points
2 marks, and the error carries into every following question
What not to write
“The direction 1→2→3 is clockwise on the drawing, so the configuration is R.”
What to write
“Group 4 points toward the viewer: the direction read is clockwise, so the actual configuration is the opposite, that is S.”
The same clockwise direction read on both drawings: on the left 4 is behind and it is R, on the right it is in front and the answer reverses to S.
Why: The clockwise rule holds only when group 4 is behind the plane of the sheet. Exams deliberately draw the hydrogen in front, because that is exactly what they want to test.
11.Ranking −COOH before −OH in the CIP priorities
1 mark, and the configuration announced ends up reversed
What not to write
“The carboxylic acid is the highest-priority functional group, so −COOH is group 1.”
What to write
“We compare the ATOMS attached to the asymmetric carbon: the oxygen of −OH (Z=8) beats the carbon of −COOH (Z=6). The ranking is therefore −OH>−COOH.”
We look at the atom attached to the asymmetric carbon, not at the functional group: the oxygen of the OH beats the carbon of the COOH.
Why: Two rankings carrying the same word “priority” coexist in the chapter: the one in nomenclature, which chooses the suffix, and the CIP one, which looks only at atomic numbers. They do not give the same order.
12.Ranking two carbon groups by counting atoms
1 mark, and the requested justification is not given
What not to write
“Ethyl contains more atoms than methyl, so it has priority.”
What to write
“The first atoms of the two groups are carbons, tied. We compare the triplets: ethyl gives (C,H,H), methyl (H,H,H); at the first point of difference C beats H, so ethyl wins.”
The bigger group loses: at the first rank, the chlorine of −CH2Cl beats the three carbons of tert-butyl.
Why: The conclusion is right, the justification is wrong: a bigger group can perfectly well lose, like −C(CH3)3 against −CH2Cl. It is the justification that is marked.
13.Rotating a Fischer projection by 90∘
the whole question
What not to write
“I turn the cross by a quarter turn to see better, the molecule is the same.”
What to write
“A rotation by 90∘ in the plane of the sheet swaps the front and back bonds: it turns the drawing into that of the ENANTIOMER. Only a rotation by 180∘ keeps the configuration.”
The same cross turned by a quarter turn: the front and back bonds swap, and the drawing becomes that of the enantiomer.
Why: The Fischer projection is not an ordinary drawing: it encodes space by the direction of the lines, horizontal toward the front and vertical toward the back. A quarter turn swaps these two directions.
14.Concluding “no chiral molecule” because α=0
2 marks
What not to write
“The measured optical rotation is zero, so the product obtained is not chiral.”
What to write
“The optical rotation is zero because the two enantiomers are present in equal amounts and exactly cancel their rotations: it is a racemic mixture, and every molecule in it is indeed chiral.”
Why: A synthesis starting from achiral reagents attacks both faces of the plane with the same probability: it can only give a racemic mixture, however carefully it is carried out. The measured zero is a cancellation, not an absence.
15.Announcing enantiomers when only one centre has been inverted
1.5 marks, and the next question on separation becomes wrong
What not to write
“The two compounds differ in the configuration of carbon 2, so they are enantiomers.”
What to write
“Only carbon 2 is inverted, carbon 3 is unchanged: the two molecules are not mirror images of each other, they are diastereomers.”
Enantiomers: ALL the centres inverted; one centre inverted out of two gives a diastereomer.
Why: Two enantiomers have ALL their centres inverted, without exception. The distinction governs what follows: diastereomers can be separated by distillation or crystallization, enantiomers never by these simple methods.
16.Counting conformers as stereoisomers
1 mark
What not to write
“The staggered and eclipsed conformations of butane are two stereoisomers of the molecule.”
What to write
“We go from one to the other by simple rotation about a single bond, without breaking any bond: they are two conformers of the SAME molecule, not isolable at room temperature.”
A rotation by 60∘ about C2−C3 goes from staggered to eclipsed without breaking anything: they are conformers, not isomers.
Why: A stereoisomer can be isolated in a bottle, a conformer cannot: the rotation barrier of butane is 16 to 19 kJ/mol, crossed billions of times per second at 25∘C.
17.Claiming that two enantiomers have exactly the same properties
1 mark on a recall question, more on an open question
What not to write
“Two enantiomers have the same properties, so the body reacts in the same way to both.”
What to write
“Two enantiomers have the same properties only in an achiral environment. Facing a biological receptor, itself chiral, they behave differently: the two enantiomers of limonene smell of orange and of lemon.”
Why: This is why a drug is often sold as a single enantiomer: the other is at best inactive. The formula learned by heart, “same properties”, is true under a condition that must be quoted.
Which method to choose
Which name, which main chain?
Look first for what the name must contain (functional group, multiple bond), then the chain, and only at the end the substituents.
If a principal functional group is present (acid, ester, amide, nitrile, aldehyde, ketone, alcohol, amine) → the main chain is the longest one that CONTAINS it; it gives the suffix and its carbon gets the lowest locant
Example: HOCH2CH2CHO: 3-hydroxypropanal
the other groups become prefixes: hydroxy-, oxo-, amino-
If no functional group, but a double or triple bond → the longest chain that contains the multiple bond; it gets the lowest locant, before the substituents
Example: CH2=CH−CH(CH3)−CH3: 3-methylbut-1-ene
with equal locants, the double bond takes the lowest: pent-1-en-4-yne
If a ring and a chain → the structure with more carbons is the parent, the other becomes a substituent
Example: propylcyclohexane, but cyclopropylheptane
If a branched alkane, no functional group → the longest chain of the whole drawing; with equal length, the one carrying more substituents
Example: 4-methylheptane, never 2-propylpentane
If the chain is chosen → number for the lowest locants (functional group, multiple bond, substituents at the first difference), then rank the substituents alphabetically without counting di- or tri-
Example: 4-ethyl-2,3-dimethylhexane
The final test is to redraw the molecule from the name: if the redrawn picture does not look like the statement, the name is wrong. The most frequent mistake remains a chain chosen without going through the branches.
What does the degree of unsaturation say?
Compute the DU before drawing, then read it as a budget: the number of rings plus the number of π bonds.
If DU=0 → saturated molecule with no ring: alkane, alcohol, ether, saturated amine
Example: C4H10O: DU=0
If DU=1 → one π bond (C=C or C=O) OR one ring
Example: C3H6O: DU=1, propanal, propanone or an alcohol with a double bond
If DU=2 → two π bonds, a triple bond, or one ring and one π bond
Example: C5H8: DU=2, pent-1-yne, isoprene or cyclopentene
If DU≥4 → think first of a benzene ring, which uses up 4 unsaturations
Example: C7H8O: DU=4, benzyl alcohol, cresol or anisole
If the formula contains nitrogen or a halogen → add the number of nitrogens, subtract the number of halogens, do nothing for oxygen
Example: C8H10N4O2: DU=6
A DU is a budget, not a plan: it bounds the possible structures, it does not choose between them.
Assigning R or S: the reading depends on the drawing
Find the priority 4 group, almost always the hydrogen, then look WHERE it points before reading anything.
On the left, 4 is on the hashed wedge, so toward the back, and the direction is read as is; on the right, it is on the horizontal branch of a Fischer projection, so toward you, and the reading must be reversed.
If Wedge-and-dash, group 4 on a hashed bond, toward the back → read 1→2→3 and conclude directly
Example: clockwise, so R
it is the only case that needs no correction
If Wedge-and-dash, group 4 on a solid wedge, toward the viewer → read the direction, then REVERSE the answer
Example: clockwise read, so S
the most frequent case in problems, because it is the most discriminating
If Wedge-and-dash, group 4 in the plane of the sheet → swap group 4 with the one on the hashed bond, read, then reverse
one swap reverses the configuration, two swaps restore it
If Fischer projection, group 4 on a VERTICAL branch → read directly
Example: H at the bottom, counterclockwise, so S
vertical means toward the back, the comfortable position
If Fischer projection, group 4 on a HORIZONTAL branch → read the direction, then reverse
horizontal means toward the viewer, like the solid wedge of the wedge-and-dash drawing
No branch applies if the carbon carries two identical groups: it is then not asymmetric, and the question simply has no answer. It is the first thing to check.
Which relationship between two given structures?
First compare the molecular formulas, then the connectivity of the atoms, and only then the space.
If different molecular formulas → they are not isomers, go no further
Example: C4H10 and C4H8
If same molecular formula, different connectivity of the atoms → constitutional isomers
Example: butan-1-ol and butan-2-ol (position), butan-1-ol and ethoxyethane (functional group)
If same connectivity, superimposable by rotation about a single bond → same molecule, simple conformers
Example: staggered butane and eclipsed butane
If same connectivity, non-superimposable mirror images → enantiomers
Example: (2R,3R) and (2S,3S) tartaric acids
check that ALL the centres are inverted
If same connectivity, stereoisomers that are not mirror images → diastereomers
Example: the Z/E pair of an alkene, or meso tartaric acid against the (2R,3R) acid
The deciding test is superposition in space, never the resemblance of the drawings: two very different drawings can show the same molecule, and two almost identical drawings two enantiomers.
How the answer is expected to be written
A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.
Naming a molecule on an exam paper
When to use it: The statement gives a drawing or a condensed formula and asks for the IUPAC name.
1Spot the functional groups and keep only ONE, the highest in the priority list: it gives the suffix.
2Find the longest chain that contains this group, or failing that the multiple bond, going through all the branches, and write its number of carbons.
3Number in both directions and write the two series of locants: functional group first, then multiple bond, then substituents, at the first difference.
4Name each substituent, group the identical ones (di-, tri-) with their locants, then rank them alphabetically ignoring di- and tri-.
5Assemble the name, then redraw the molecule from it to check.
Concluding sentence
“The principal functional group is the carboxylic acid: the main chain must contain its carbon, which gets locant 1. The longest chain has seven carbons, numbered from the carboxyl, and carries a methyl at position 4: 4-methylheptanoic acid.”
The trap: Choosing the chain without going through the branches: a name built on the wrong chain is wrong as a whole, even if all its locants are correct.
Marking: 1 mark the functional group and suffix, 1 mark the main chain, 1 mark the numbering, 1 mark the substituents and their order.
Assigning Z or E on an exam paper
When to use it: The statement asks for the configuration of an alkene, or whether a Z/E pair exists.
1Check the condition: each carbon of the double bond carries two different groups. Otherwise, conclude that there is no Z/E stereoisomerism and stop.
2On the first carbon, rank the two groups by the CIP rules, writing the atoms compared, and underline the priority group.
3Do the same on the second carbon, never comparing a group of the first carbon with a group of the second.
4See whether the two underlined groups are on the same side of the double bond: same side, Z; opposite sides, E.
5Write the descriptor in parentheses at the start of the name, with the locant of the double bond if there are several.
Concluding sentence
“On carbon 3, CH3 beats H; on carbon 2, Cl (Z=17) beats CH3 (Z=6). The two priority groups are on opposite sides of the double bond: the configuration is E, that is (E)-2-chlorobut-2-ene.”
The trap: Comparing crosswise: the ranking is done carbon by carbon, never between the two carbons of the double bond.
Marking: 1 mark the condition checked, 1 mark each carbon ranked with its atoms, 1 mark the conclusion and the descriptor.
Assigning a configuration on an exam paper
When to use it: The question says “assign the configuration”, “specify R or S”, or gives a name that lacks the descriptor.
1Circle the carbon concerned and write that it carries four different groups, naming them.
2Write out the CIP ranking in words, with the reason for each tie-break.
3State explicitly where group 4 points on the drawing provided: hashed, solid, or horizontal in Fischer.
4Read the direction 1→2→3 and say whether it must be reversed, before giving the answer.
5Conclude with the descriptor in parentheses in front of the compound name.
Concluding sentence
“Carbon 2 carries −OH, −COOH, −CH3 and −H: it is asymmetric. The CIP ranking gives −OH>−COOH>−CH3>−H. The hydrogen pointing toward the viewer, the clockwise direction read corresponds to the opposite configuration: the compound is (S)-lactic acid.”
The trap: Skipping the step “where group 4 points”: it is the one that is marked, and reading the direction is worth nothing without it.
Marking: Usually 1 mark for the justified CIP ranking, 1 mark for the orientation of group 4, 1 mark for the final descriptor.
Justifying that a compound is meso
When to use it: The question asks for the ACTUAL number of stereoisomers, or has you explain why there are only three.
1Count the stereogenic centres and announce 2n as a MAXIMUM, never as a result.
2Draw the Fischer projection and look for an internal plane of symmetry, horizontal, at mid-height.
3Assign the configurations of the two centres and show that they are opposite.
4Conclude that the molecule is superimposable on its image, therefore achiral and optically inactive.
5Remove this compound from the count and give the actual number.
Concluding sentence
“Tartaric acid has two asymmetric carbons, so at most 22=4 stereoisomers. The (2R,3S) isomer has an internal plane of symmetry: it is superimposable on its mirror image, therefore achiral, and its optical rotation is zero. It is the meso compound, so only three stereoisomers exist.”
The trap: Writing “there are 2n stereoisomers” as a conclusion: the rule gives an upper bound.
Marking: 1 mark for the announced count, 1 mark for the plane of symmetry identified, 1 mark for the conclusion of achirality.
Check before you hand in
Five minutes of checking recover more marks than one more problem started in a hurry.
Folding the Fischer cross in half
Before concluding a pair of enantiomers, mentally fold the projection at mid-height. If the two halves superimpose, the molecule is meso: achiral, zero optical rotation, and it has no enantiomer.
A failed check: you have just announced two enantiomers and the fold superimposes. The right answer was “same compound, achiral”.
Recounting the four groups
Before reading any direction, check that the four substituents are really different, going out one more atom when two look alike.
On pentan-3-ol, the central carbon carries two identical ethyl groups: it is not asymmetric, and any R or S answer is wrong from the start.
The consistency test of the announced pair
If you conclude “enantiomers”, all the centres must be inverted. If one remains identical, the answer is “diastereomers”, and the physical properties differ.
(2R,3R) and (2S,3S): enantiomers. (2R,3R) and (2R,3S): diastereomers.
Bounding the enantiomeric excess
ee falls between 0 and 100%, and the percentage of the major enantiomer between 50% and 100%. A value under 50% signals that you took the wrong enantiomer as the reference.
ee=80% gives 90% and 10%, never 80% and 20%: it is the most frequent misreading on the polarimetry question.
Redraw the molecule from the name
Before handing in a name, start again from it: draw the chain, place the substituents at the given locants, and compare with the statement. This is the reverse substitution of nomenclature.
A “2-propylpentane” is redrawn as a pentane carrying a propyl at 2: you then see a chain of seven carbons, so the name is not that of a properly named alkane.
Recompute the degree of unsaturation on the drawing
Once the name is written, count the carbons and hydrogens of the drawing, recompute the DU and compare it with what the name announces: an -ene, a ring, a C=O are worth 1 each.
A name ending in -ene whose formula gives DU=0: the drawing is an alkane, so the name or the formula is wrong.
The typical problem, taken apart
Lactic acid: configuration, then enantiomeric excess
Lactic acid CH3−CHOH−COOH is shown below, the hydrogen of the central carbon pointing toward the viewer.
1. Show that this carbon is asymmetric and rank its four groups by CIP. 2. Assign the configuration of the compound drawn. 3. A sample gives an apparent specific rotation of +2.00∘, whereas the pure (S) enantiomer has +3.82∘. Determine the enantiomeric excess and the composition of the mixture.
The four bonds of the central carbon, the hydrogen turned toward the viewer and not toward the back: the direction read directly on this drawing will be the opposite of the actual configuration.
Step 1
The central carbon carries −OH, −COOH, −CH3 and −H, four groups that are all different: it is an asymmetric carbon.
Why
This sentence is worth a mark on its own and conditions everything else: without an asymmetric carbon, the question would have no answer.
Step 2
CIP ranking: −OH is 1, its oxygen having the highest atomic number. Between −COOH and −CH3, the first atoms are two tied carbons: we compare (O,O,O) with (H,H,H), so −COOH is 2 and −CH3 is 3. Hydrogen is 4.
Why
We compare the atoms directly bonded to the centre, not the priority of functional groups in nomenclature, which would put the carboxylic acid first and reverse the answer.
Step 3
On the drawing, going from −OH to −COOH then to −CH3, the direction is clockwise.
Why
The raw reading is written separately from the correction: the marker wants to see both moves, and the second makes no sense without the first.
Step 4
Group 4 points toward the viewer: the actual configuration is the opposite of the direction read. The compound drawn is therefore (S)-lactic acid.
Why
This is the step that half the papers skip, and the only one that tells the two possible answers apart. Without it, the right conclusion is only a stroke of luck.
Step 5
ee=3.822.00=0.524, that is 52.4%. The observed rotation is positive like that of pure (S), so (S) is the major enantiomer: 2100+52.4=76.2% of (S) and 23.8% of (R).
Why
The sign of the observed rotation names the major enantiomer; the excess alone does not, and forgetting this swaps the two percentages.
Step 6
Check: 0.762×(+3.82)+0.238×(−3.82)=+2.00∘.
Why
The reverse substitution closes the calculation: if the weighted sum does not give back the measurement in the problem, the percentages have been swapped.
The conclusion, written out
“The compound shown is (S)-lactic acid. The sample has an enantiomeric excess of 52.4% and contains 76.2% of (S) for 23.8% of (R).”
The classic mistake on this problem: Announcing R because the direction read is clockwise, without taking into account that the hydrogen is drawn toward the front.
Learn by heart
•Name: principal functional group for the suffix, longest chain that contains it, lowest locants at the first difference, substituents in alphabetical order without di- or tri-.
•Priority of functional groups: acid, ester, amide, nitrile, aldehyde, ketone, alcohol, amine, then multiple bonds; the others become prefixes (oxo-, hydroxy-, amino-).
•DU=22c+2+n−h−x: oxygen does not count, nitrogen adds, a halogen counts as an H; a ring or a π bond is worth 1, a triple bond is worth 2.
•Z/E: EACH carbon of the double bond carries two different groups; priority groups on the same side, Z; on opposite sides, E.
•Asymmetric carbon: four DIFFERENT groups on the same carbon. Chirality: a molecule not superimposable on its image.
•CIP priorities: decreasing atomic number, ties settled at the FIRST point of difference, multiple bonds duplicated.
•Group 4 toward the back: clockwise gives R, counterclockwise gives S. Group 4 toward the front: we read, then REVERSE.
•Fischer: horizontal toward the front, vertical toward the back. Rotation by 180∘ allowed, by 90∘ forbidden.
•n centres: at most 2n stereoisomers, minus one for each meso compound.
•Enantiomers: same physical properties in an achiral medium, opposite optical rotations. Diastereomers: all properties differ.
•Racemic: 50/50, α=0, and yet every molecule is chiral.
Frequently asked questions
How do I name a branched organic molecule?
Look first for the highest-priority functional group (acid, ester, amide, aldehyde, ketone, alcohol, amine): it gives the suffix. Take the longest chain that contains it, going through the branches, then number for the lowest locant for the functional group, the multiple bond and the substituents. Rank the substituents alphabetically without counting di or tri, and redraw the molecule from the name to check.
How can I tell whether a carbon is asymmetric?
A carbon is asymmetric when it carries four groups that are all different. Compare the substituents two by two, going out atom by atom to the first point of difference. On pentan-3-ol, the central carbon carries two identical ethyl groups: it is not asymmetric, despite appearances.
How do I assign R or S to an asymmetric carbon?
Rank the four groups by the CIP rules, from the highest atomic number to the lowest. Place the fourth-ranked group toward the back, then read the direction going from the first to the third: clockwise gives R, counterclockwise gives S. If this fourth group points toward you, read the direction normally, then reverse the answer.
What is the difference between enantiomers and diastereomers?
Two enantiomers are mirror images of each other and non-superimposable: all their centres are inverted, and they share the same physical properties except optical rotation, which is opposite. Two diastereomers are stereoisomers that are not mirror images, because only some of their centres have changed, or the configuration of a double bond: their melting points and solubilities differ, so they can be separated.
Why is a meso compound not chiral?
Because it has an internal plane of symmetry: one half of the molecule is the image of the other, so the whole molecule is superimposable on its mirror image. Tartaric acid is the classic example: its two centres have opposite configurations, its optical rotation is zero, and there are only three stereoisomers instead of four.
How do I read R or S on a Fischer projection?
In a Fischer projection, the horizontal bonds point toward you and the vertical bonds point away from you. If the priority four group is at the top or bottom, read the direction directly. If it is on the right or left, read the direction then reverse the answer. Never turn the cross by a quarter turn: that gives the enantiomer.
Practise it
Corrected exercises: Nomenclature and isomerism, 202-SNU-RE
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Get in touch for a first session. We work through nomenclature and stereochemistry on real drawings, those of CEGEP tests, until the reflexes “longest chain” and “where does group 4 point” are automatic.
