Organic chemistry

Class 12 Chemistry, unit 7, book pages 128–141

Organic chemistry is the chemistry of carbon compounds, and carbon forms millions of them because of catenation, tetravalency and multiple bonding. This unit recaps classification, functional groups, homologous series, IUPAC names and structural isomerism, then studies stereoisomerism in depth: geometric (cis-trans and E/Z) isomerism caused by the rigid bond, and optical isomerism caused by chiral carbons. It ends with why chirality matters in medicines, using thalidomide, and how racemic mixtures are separated or avoided with chiral catalysts.

Learning outcomes (8)
  • Explain stereoisomerism and its division into geometric (cis/trans) and optical isomerism.
  • Describe geometrical (cis/trans) isomerism in alkenes and explain its origin in terms of restricted rotation due to the presence of a pi () bond.
  • Describe the shape of benzene and other aromatic molecules, including hybridisation, in terms of sigma () bonds and a delocalized pi () system.
  • Explain what is meant by a chiral centre and that such a centre gives rise to two optical isomers (enantiomers).
  • Describe that enantiomers have identical physical and chemical properties except for their ability to rotate the plane of polarized light and their potential biological activity.
  • Apply the terms optically active, racemic mixture and meso compound to a given structure.
  • Describe the effect of the two optical isomers of a single substance on plane-polarized light.
  • Explain the significance of chirality in the synthetic preparation of drug molecules, including the different biological activity of enantiomers, the need to separate racemic mixtures, and the use of chiral catalysts to produce a single pure optical isomer, using thalidomide as an example.

Unit 7 introductionWhat organic chemistry is

Organic chemistry is the study of the preparation, properties, identification and modification of organic compounds. Every organic compound contains carbon. Almost all of them also contain hydrogen, and many contain oxygen, nitrogen, halogens, sulphur or phosphorus. A short definition used in the key points of the book is: organic chemistry is the chemistry of hydrocarbons and their derivatives.

Organic compounds are everywhere in daily life: carbohydrates, proteins and fats in food, medicines, cloth fibres, plastics (polymers), insecticides, cosmetics and dyes. Many industries exist only to make them.

methane the simplest organic compound
ethanol in hand sanitisers
ethanoic acid in vinegar
urea first made in a lab

A few simple carbon compounds are still treated as inorganic: carbon monoxide, carbon dioxide, carbonates such as , hydrogencarbonates, cyanides and carbides. So a compound containing carbon is not automatically organic; organic compounds are built on a framework of carbon atoms bonded to hydrogen and to each other.

Typical differences (there are exceptions to every row).
PropertyOrganic compoundsInorganic compounds
Bondingcovalentoften ionic
Effect of heatingburn (combustion) or charusually melt; most do not burn
Melting and boiling pointsgenerally lowgenerally high
Solubilitymostly soluble in organic solventsmany are soluble in water
Speed of reactionsslow, between moleculesoften fast, between ions
Isomerismvery commonrare
Number knownabout 20 millionfar fewer

Nearly 19 to 20 million organic compounds are known. One big reason for this huge number is isomerism: the same set of atoms can be joined in different ways, or arranged differently in space, to give different compounds. Isomers exist because covalent bonds are directional (they point in fixed directions), so the shape of a molecule is fixed.

Recap (grade 11)Why carbon forms so many compoundsOften examined

This is a recap from grade 11, and a favourite short question. Carbon forms more compounds than all other elements put together because of a few special properties of the carbon atom.

1. Catenation

Catenation is the ability of atoms of an element to bond to each other to form long chains, branched chains and rings. Carbon shows it more than any other element because the bond is strong (about ). Silicon, the element below carbon, also catenates, but the bond is weaker (about ), so long silicon chains are not stable.

hexane straight chain
2-methylpentane branched chain
cyclohexane ring

2. Tetravalency

A carbon atom has four valence electrons (), so it forms four covalent bonds. In an carbon the four bonds point to the corners of a tetrahedron ( apart). With four bonds per atom, chains can branch in many ways, and each carbon can carry different atoms or groups.

3. Multiple bonds

Carbon forms stable double and triple bonds, both with itself (, ) and with oxygen and nitrogen (, ). This multiplies the number of possible structures. Carbon's small size lets the p orbitals of two carbon atoms come close enough for good sideways overlap, which is why its bonds are strong.

ethene C=C double bond
ethyne triple bond
ethanal C=O double bond

4. Bonding with other elements

Carbon bonds strongly with H, O, N, S, P and the halogens. These atoms form functional groups (such as , , ), and each functional group gives a new family of compounds.

5. Isomerism

The same molecular formula can have many structures. The number of possible isomers grows very fast as the number of carbon atoms increases:

Structural isomers of alkanes (stereoisomers not counted).
Alkane formula
Number of structural isomers2351875366 319

Recap (grade 11)Classification of organic compounds

Organic compounds are classified by the way their carbon atoms are joined: in open chains, or in closed rings.

Classification of organic compounds, with one example of each type.
Open-chain (acyclic or aliphatic) compounds
The carbon atoms form straight or branched chains, not rings. Saturated ones have only single bonds (alkanes). Unsaturated ones have at least one or (alkenes, alkynes).
Closed-chain (cyclic) compounds
The atoms form one or more rings.
Homocyclic (carbocyclic) compounds
The ring contains only carbon atoms. Alicyclic compounds (such as cyclohexane) behave like open-chain compounds. Aromatic compounds contain a benzene-type ring with delocalised electrons (such as benzene and naphthalene).
Heterocyclic compounds
The ring contains at least one atom other than carbon (a heteroatom), usually N, O or S. Examples: pyridine, furan, thiophene.
butane open chain, saturated
but-1-ene open chain, unsaturated
cyclohexane alicyclic
benzene aromatic
naphthalene aromatic, two rings
pyridine heterocyclic (N)
furan heterocyclic (O)
thiophene heterocyclic (S)

Recap (grade 11)Functional groupsOften examined

A functional group is an atom or group of atoms in a molecule that decides its characteristic chemical reactions. The rest of the molecule, the hydrocarbon part, is written (an alkyl group such as or ). All compounds with the same functional group react in a similar way, so learning one member teaches you the whole family.

Common functional groups (, = alkyl groups; = aryl group such as ).
FamilyFunctional groupGeneral formulaExample
Alkene (double bond)ethene,
Alkyne (triple bond)ethyne,
Halogenoalkane ( = F, Cl, Br, I)chloroethane,
Alcohol (hydroxyl)ethanol,
Phenol on a benzene ringphenol,
Ethermethoxymethane,
Aldehydeethanal,
Ketone (carbonyl, inside the chain)propanone,
Carboxylic acid (carboxyl)ethanoic acid,
Esterethyl ethanoate,
Acyl chlorideethanoyl chloride,
Amideethanamide,
Amine (primary) (amino)ethanamine,
Nitrileethanenitrile,
ethanol alcohol
ethanal aldehyde
propanone ketone
ethanoic acid carboxylic acid
ethyl ethanoate ester
ethanamide amide
ethanamine amine
ethanenitrile nitrile

Recap (grade 11)Homologous seriesOften examined

A homologous series is a family of organic compounds that have the same functional group and the same general formula, in which each member differs from the next by one unit. Each member is a homologue of the others.

  • All members have the same general formula, for example for alkanes.
  • Successive members differ by , which is units of relative molecular mass.
  • All members have the same functional group, so they have similar chemical properties.
  • Physical properties (boiling point, melting point, density) change gradually as the chain gets longer.
  • Members can be made by similar general methods.
The first eight straight-chain alkanes: rises by 14 each time.
AlkaneFormulaBoiling point /
methane16−162
ethane30−89
propane44−42
butane58−1
pentane7236
hexane8669
heptane10098
octane114126
Boiling points of straight-chain alkanes
1
2
3
4
5
6
7
8
9
−200
−150
−100
−50
50
100
150
O
Boiling point rises steadily with chain length because longer molecules have stronger London (dispersion) forces. The alkanes with 1 to 4 carbon atoms are gases at room temperature.
SeriesGeneral formulaFirst three members
Alkanes, ,
Alkenes (), ,
Alkynes (), ,
Alcohols, ,
Carboxylic acids (), ,
methanol first member
ethanol + CH2
propan-1-ol + CH2 again
Spot the homologues
Which pair belongs to the same homologous series: (A) and , (B) and , (C) and , (D) and ?
  1. (A) Both fit (alkanes, and ) and differ by exactly . Same series.
  2. (B) is an alkane, is an alkene; they differ by , not .
  3. (C) An alcohol and a carboxylic acid: different functional groups.
  4. (D) An alkyne and an aldehyde: different functional groups.

Recap (grade 11)IUPAC names: a quick recap

You need names to talk about isomers, so here is a short reminder of the IUPAC system. A name has three parts: prefix (substituents such as methyl or chloro, with their positions), word root (number of carbons in the main chain) and suffix (the family: -ane, -ene, -ol and so on).

Carbons12345678
Word rootmethethpropbutpenthexheptoct
FamilySuffix or prefixExample
alkane-anepropane
alkene-enebut-2-ene
alkyne-ynepropyne
alcohol-olpropan-2-ol
aldehyde-alethanal
ketone-onepropanone
carboxylic acid-oic acidethanoic acid
amine-amineethanamine
halogen, alkyl side chainchloro-, bromo-, methyl-, ethyl-2-chlorobutane
  1. Find the longest chain that contains the functional group or the multiple bond.
  2. Number it from the end that gives the functional group (or the \ce{C=C}) the lowest number; then give the side chains the lowest numbers.
  3. Name side chains as prefixes in alphabetical order, each with its position number; use di-, tri-, tetra- for repeats (2,3-dimethyl).
  4. Use commas between numbers and hyphens between numbers and words (2,3,4-trimethylhept-3-ene).
  5. For stereoisomers put cis/trans or (E)/(Z) in front of the name: trans-but-2-ene, (E)-but-2-ene.
Name these compounds
Name (a) \ce{CH3CH2CH(CH3)CH2CH2CH3}, (b) \ce{(CH3)2C(OH)CH2CH3}, (c) \ce{CH3CH2CH2C(CH3)=C(CH3)CH(CH3)2}.
(a) 3-methylhexane
(b) 2-methylbutan-2-ol
(c) 2,3,4-trimethylhept-3-ene
  1. (a) The longest chain has 6 carbons (hexane). Numbering from either end puts the methyl group on C3, so the name is 3-methylhexane.
  2. (b) The longest chain containing the \ce{C-OH} carbon has 4 carbons. Numbering from the end nearer the \ce{-OH} puts both the \ce{-OH} and the extra methyl on C2: 2-methylbutan-2-ol.
  3. (c) The longest chain through the \ce{C=C} has 7 carbons. Numbering from the isopropyl end gives the double bond the number 3 (not 4), and methyl groups on C2, C3 and C4: 2,3,4-trimethylhept-3-ene.

7.1IsomerismOften examined

Isomerism is the existence of two or more compounds with the same molecular formula but different structures. The compounds are called isomers. Isomers are different substances: they have different physical properties, and often different chemical properties too.

Isomerism is divided into two broad classes:

Structural (constitutional) isomerism
The atoms are joined in a different order (different connectivity), so the isomers have different structural formulae. Example: butane and 2-methylpropane, both \ce{C4H10}.
Stereoisomerism (configurational isomerism in the book)
The atoms are joined in the same order (same structural formula) but are arranged differently in space. Example: cis- and trans-but-2-ene.
butane C4H10
2-methylpropane C4H10, different chain
cis-but-2-ene C4H8
trans-but-2-ene C4H8, same connectivity
Structural isomersStereoisomers
Molecular formulasamesame
Order of bonding (structural formula)differentsame
Arrangement in spacenot the pointdifferent
Examplepropan-1-ol and propan-2-olcis- and trans-but-2-ene

Recap (grade 11)Structural isomerismOften examined

The book revises structural isomerism only briefly, but it is examined (for example 'draw three structural isomers of \ce{C4H10O}'). There are five types.

1. Chain isomerism

The isomers have different carbon skeletons (straight or branched chains). Example: the three isomers of \ce{C5H12}.

pentane b.p. 36 °C
2-methylbutane b.p. 28 °C
2,2-dimethylpropane b.p. 10 °C

2. Position isomerism

Same carbon skeleton and same functional group, but the functional group (or multiple bond) is at a different position.

propan-1-ol
propan-2-ol
but-1-ene
but-2-ene

3. Functional group isomerism

Same molecular formula, but different functional groups, so the isomers belong to different families.

ethanol C2H6O, alcohol
methoxymethane C2H6O, ether
propanal C3H6O, aldehyde
propanone C3H6O, ketone

4. Metamerism

Different alkyl groups on the two sides of the same functional group, such as \ce{-O-}, \ce{-CO-} or \ce{-NH-}.

ethoxyethane C2H5–O–C2H5
1-methoxypropane CH3–O–C3H7

5. Tautomerism

Two structural isomers that change into each other rapidly by the movement of a hydrogen atom and a double bond, and exist together in equilibrium. The common case is keto-enol tautomerism:

\ce{CH3-CO-CH3 <=> CH2=C(OH)-CH3}

propanone (keto form, over 99.9%) \rightleftharpoons prop-1-en-2-ol (enol form)

propanone (keto form)
prop-1-en-2-ol (enol form)
All the structural isomers of \ce{C4H10O}
Draw the structural isomers of \ce{C4H10O} and classify them.

\ce{C4H10O} fits \ce{C_{n}H_{2n+2}O}, so it is saturated: it can be an alcohol (\ce{R-OH}) or an ether (\ce{R-O-R'}). There are 7 isomers.

butan-1-ol primary alcohol
butan-2-ol secondary alcohol
2-methylpropan-1-ol primary alcohol
2-methylpropan-2-ol tertiary alcohol
ethoxyethane ether
1-methoxypropane ether
2-methoxypropane ether
  • Butan-1-ol and butan-2-ol: position isomers.
  • Butan-1-ol and 2-methylpropan-1-ol: chain isomers.
  • Any alcohol and any ether: functional group isomers.
  • Ethoxyethane and 1-methoxypropane: metamers.

7.1.1StereoisomerismOften examined

Stereoisomers are molecules that have the same molecular formula and the same structural formula (the same atoms joined in the same order) but a different arrangement of atoms or groups in space. The phenomenon is called stereoisomerism.

Why does this matter? Most reactions in our body are controlled by enzymes. An enzyme's active site has a fixed 3D shape, so it accepts only the stereoisomer that fits it. A medicine made as the wrong stereoisomer may not work, or may cause side effects, because it does not fit the target site.

The book's classification of stereoisomerism, with an example of each type.
Conformational isomers (conformers)
Different arrangements of the same molecule produced by rotation about a single bond. No bond is broken, the energy needed is small, and conformers change into each other millions of times a second at room temperature, so they cannot be separated.
Configurational isomers
Stereoisomers that can be changed into each other only by breaking and re-forming bonds. They are stable, separable compounds. They are of two kinds: geometrical (cis-trans) and optical isomers.
Enantiomers
Configurational isomers that are non-superimposable mirror images of each other.
Diastereomers
Stereoisomers that are not mirror images of each other, for example meso-tartaric acid and (+)-tartaric acid.
Conformations of ethane seen along the \ce{C-C} bond (Newman projections). Only about 12\,\mathrm{kJ\,mol^{-1}} separates them, so rotation is free at room temperature.

7.1.1 (1)Geometric isomerism and restricted rotationOften examined

Geometric isomers have the same molecular formula and the same structural formula (same connectivity) but a different arrangement of atoms or groups around a double bond. This type of stereoisomerism is called geometric isomerism (or cis-trans isomerism).

Bonding in an alkene

Each carbon atom of the \ce{C=C} bond is \mathrm{sp^2} hybridised. Its three \mathrm{sp^2} orbitals lie in one plane at 120^\circ to each other, and one unhybridised \mathrm{2p_z} orbital stands at right angles to that plane.

  • The **sigma (\sigma) bond forms by head-on overlap** of one \mathrm{sp^2} orbital from each carbon, along the line joining the two nuclei.
  • The **pi (\pi) bond forms by sideways (parallel) overlap** of the two \mathrm{2p_z} orbitals. Its electron cloud lies above and below the plane of the molecule.
The \pi bond exists only while the two p orbitals stay parallel. Twisting one carbon destroys the overlap.

Why rotation is restricted

To rotate one carbon of a double bond relative to the other, the two p orbitals must stop being parallel, so the \pi bond must break. Breaking the \pi bond of ethene needs about 264\,\mathrm{kJ\,mol^{-1}}. Molecules at room temperature do not have this much energy, so there is no free rotation about a \ce{C=C} bond. The groups attached to it are locked in place.

In an alkane, rotating about the \ce{C-C} single bond does not break anything, because the \sigma bond is symmetrical around the bond axis. It needs only about 12\,\mathrm{kJ\,mol^{-1}} (the staggered to eclipsed barrier), which collisions easily supply. So a major difference between alkanes and alkenes is the freedom of rotation: free about \ce{C-C}, restricted about \ce{C=C}.

Conditions for geometric isomerism

  1. Restricted rotation: a \ce{C=C} bond (or a ring).
  2. Two different groups on each doubly bonded carbon: the alkene must be \ce{abC=Cab} or \ce{abC=Cde} with a\ne b and d\ne e.

If one carbon of the double bond carries two identical atoms or groups, swapping the positions of the groups on the other carbon gives the same molecule, so no geometric isomers exist. The book shows this with but-1-ene: C1 carries two hydrogen atoms, and 'rotating' the double bond only turns the molecule over.

but-1-ene C1 has two H: no isomers
2-methylpropene two CH3 on C2: no isomers
but-2-ene (a) cis form
but-2-ene (b) trans form
Apply condition 2 to each carbon of the double bond separately.
AlkeneGroups on first C=C carbonGroups on second C=C carbonGeometric isomers?
propene \ce{CH2=CHCH3}H, H\ce{CH3}, Hno
but-1-ene \ce{CH2=CHCH2CH3}H, H\ce{C2H5}, Hno
but-2-ene \ce{CH3CH=CHCH3}\ce{CH3}, H\ce{CH3}, Hyes
pent-2-ene \ce{CH3CH=CHC2H5}\ce{CH3}, H\ce{C2H5}, Hyes
2-methylbut-2-ene \ce{(CH3)2C=CHCH3}\ce{CH3}, \ce{CH3}\ce{CH3}, Hno
1,1-dichloroethene \ce{CCl2=CH2}Cl, ClH, Hno
1,2-dichloroethene \ce{ClCH=CHCl}Cl, HCl, Hyes
cis-1,2-dimethylcyclopropane both CH3 on one face
trans-1,2-dimethylcyclopropane CH3 on opposite faces

7.1.1 (i)The cis-trans system of namingOften examined

IUPAC recommends two ways of naming geometric isomers: the cis-trans system and the E/Z system. The cis-trans system was first used for disubstituted alkenes of the type \ce{RCH=CHR'}.

cis isomer
The two similar groups (or the two main-chain carbons) are on the same side of the double bond. 'Cis' is Latin for 'on the same side'.
trans isomer
The two similar groups (or the two main-chain carbons) are on opposite sides of the double bond. 'Trans' is Latin for 'across'.

In but-2-ene the two methyl groups (C1 and C4 of the main chain) decide the name. They are on the same side in cis-but-2-ene and on opposite sides in trans-but-2-ene.

cis-but-2-ene b.p. 3.7 °C
trans-but-2-ene b.p. 0.9 °C

Pent-2-ene has three different kinds of substituent (methyl, ethyl and hydrogen), which looks harder. Treat every alkyl group as a 'main-chain' group: the cis isomer has the two alkyl groups on the same side and the two hydrogen atoms on the same side. Equivalently (book's key information), you can decide cis or trans by looking only at the two hydrogen atoms on the double-bond carbons.

cis-pent-2-ene alkyl groups same side
trans-pent-2-ene alkyl groups opposite
cis-1,2-dichloroethene
trans-1,2-dichloroethene
Concept Assessment Exercise 7.1
Which of the following compounds can and cannot show cis-trans isomerism, and why? (i) 1-chloroprop-1-ene (ii) 3-chloroprop-1-ene (iii) hex-2-ene
  1. (i) 1-chloroprop-1-ene, \ce{ClCH=CHCH3}. C1 carries Cl and H (different); C2 carries \ce{CH3} and H (different). Both conditions hold, so it shows cis-trans isomerism.
  2. (ii) 3-chloroprop-1-ene, \ce{CH2=CHCH2Cl}. C1 carries two hydrogen atoms (identical), so it cannot show cis-trans isomerism. The Cl is on C3, which is not part of the double bond.
  3. (iii) hex-2-ene, \ce{CH3CH=CHCH2CH2CH3}. C2 carries \ce{CH3} and H; C3 carries \ce{C3H7} and H. Both carbons have two different groups, so it shows cis-trans isomerism.
cis-1-chloroprop-1-ene
trans-1-chloroprop-1-ene
3-chloroprop-1-ene no isomers
cis-hex-2-ene
trans-hex-2-ene
\text{(i) yes,\quad (ii) no,\quad (iii) yes}

7.1.1Properties of cis and trans isomersOften examined

Cis and trans isomers are different compounds, so their physical properties (melting point, boiling point, density, solubility, dipole moment) are different. Their chemical properties are similar, because they have the same functional groups, but not identical.

Polarity and boiling point

In cis-1,2-dichloroethene both polar \ce{C-Cl} bonds point towards the same side, so their bond dipoles add up to a resultant dipole moment of about 1.9\,\mathrm{D}. The molecule is polar. In the trans isomer the two equal bond dipoles point in opposite directions and cancel, so the resultant dipole moment is zero and the molecule is non-polar.

Dipole moments of the two isomers of 1,2-dichloroethene. The arrows point from the positive carbon to the negative chlorine.

Polar cis molecules attract each other by dipole-dipole forces as well as London forces, so the cis isomer usually has the higher boiling point. The more symmetrical trans molecules pack more neatly in a crystal, so the trans isomer usually has the higher melting point.

Physical data for three cis/trans pairs (rounded).
PairDipole moment / DBoiling point / ^\circ\mathrm{C}Melting point / ^\circ\mathrm{C}
cis-1,2-dichloroethene1.960−80
trans-1,2-dichloroethene048−50
cis-but-2-ene0.33.7−139
trans-but-2-ene00.9−106
maleic acid (cis-butenedioic acid)polar(decomposes)135
fumaric acid (trans-butenedioic acid)0(sublimes)287

Stability and reactivity

In the cis isomer the two bulky groups sit close together on the same side of the double bond. Their electron clouds repel each other (steric hindrance or steric strain), which raises the energy of the molecule. So the cis isomer is less stable than the trans isomer. Because it has more energy stored, the cis isomer is also more reactive. The book adds that the electron cloud of the cis isomer is pushed towards one side, which leaves the \pi bond more exposed to attack by electrophiles.

Steric strain makes cis-but-2-ene about 4\,\mathrm{kJ\,mol^{-1}} higher in energy than trans-but-2-ene: less heat is released when the trans isomer is hydrogenated.
\begin{aligned}\ce{CH3CH=CHCH3 + H2 &->[\text{Ni}] CH3CH2CH2CH3}\\\Delta H_{\text{cis}}&=-120\ \mathrm{kJ\,mol^{-1}}\\\Delta H_{\text{trans}}&=-116\ \mathrm{kJ\,mol^{-1}}\end{aligned}

Both isomers give the same product (butane), so the difference in heat released is the difference in their energies.

A chemical difference: maleic and fumaric acid

Maleic acid (cis) has both \ce{-COOH} groups on the same side, close enough to lose a molecule of water on heating to about 140^\circ\mathrm{C} and form a cyclic anhydride. Fumaric acid (trans) cannot do this, because its \ce{-COOH} groups are too far apart.

maleic acid (cis) m.p. 135 °C
fumaric acid (trans) m.p. 287 °C
maleic anhydride formed from the cis acid only

Different behaviour in living things

Cisplatin, \ce{[Pt(NH3)2Cl2]}, is a cancer medicine; its trans isomer is not.

7.1.1 (key points)The E/Z system

The book's key points and exercise use the E/Z system, but the text does not explain it, so here it is. The cis-trans names work well only when each double-bond carbon carries one hydrogen atom. For tri- and tetra-substituted alkenes they become unclear. For example, in 1-bromo-1-chloroprop-1-ene the carbon C1 carries Br and Cl and no hydrogen, so 'cis' has no clear meaning. IUPAC therefore recommends the E/Z system, which works for every alkene.

Cahn-Ingold-Prelog (CIP) priority rules

Look at each carbon of the double bond separately and decide which of its two groups has the higher priority:

  1. Rule 1. Compare the atoms directly attached to the double-bond carbon. The atom with the higher atomic number has higher priority: \ce{I}>\ce{Br}>\ce{Cl}>\ce{S}>\ce{F}>\ce{O}>\ce{N}>\ce{C}>\ce{H}.
  2. Rule 2. If these atoms are the same, move one atom further out and compare the sets of atoms attached, at the first point of difference. For example \ce{-CH2CH3} (C, H, H) beats \ce{-CH3} (H, H, H), and \ce{-CH2OH} (O, H, H) beats \ce{-C(CH3)3} (C, C, C) because O beats C at the first point of difference.
  3. Rule 3. A double or triple bond counts its atoms twice or three times: \ce{-CH=CH2} is treated as (C, C, H).
  4. Rule 4. For isotopes, the higher mass number wins: D beats H.
Z isomer
The two higher-priority groups are on the same side of the double bond (German zusammen, together).
E isomer
The two higher-priority groups are on opposite sides (German entgegen, opposite).
Assigning E and Z to 1-bromo-1-chloroprop-1-ene. The circled groups have the higher priority on their carbon.
(Z)-1-bromo-1-chloroprop-1-ene
(E)-1-bromo-1-chloroprop-1-ene
Assign E or Z
Name the isomer of 3-methylpent-2-ene, \ce{CH3CH=C(CH3)CH2CH3}, in which the \ce{CH3} on C2 and the ethyl group on C3 are on opposite sides.
  1. C2 carries \ce{CH3} and H. Rule 1: C beats H, so \ce{CH3} is higher.
  2. C3 carries \ce{CH3} and \ce{CH2CH3}. Both attach through C, so use Rule 2: ethyl has (C, H, H), methyl has (H, H, H). Ethyl is higher.
  3. The higher groups (\ce{CH3} on C2, ethyl on C3) are on opposite sides, so the isomer is E.
(E)-3-methylpent-2-ene
(E)\text{-3-methylpent-2-ene}
(E)-2-chlorobut-2-ene methyl groups cis, yet E

SLO (see also unit 8)Shape of benzene and other aromatic molecules

The SLO box asks for the shape of benzene, which is treated in detail in unit 8. Benzene, \ce{C6H6}, is a flat regular hexagon of six carbon atoms, each carrying one hydrogen atom.

  • Each carbon atom is **\mathrm{sp^2} hybridised**. Its three \mathrm{sp^2} orbitals form three \sigma bonds: two \ce{C-C} bonds (by \mathrm{sp^2}-\mathrm{sp^2} overlap) and one \ce{C-H} bond (by \mathrm{sp^2}-\mathrm{1s} overlap).
  • All 12 atoms lie in one plane and every bond angle is 120^\circ. This planar set of \sigma bonds is the **\sigma framework**.
  • Each carbon keeps one unhybridised \mathrm{2p} orbital (holding one electron) at right angles to the ring. Each p orbital overlaps sideways with the p orbitals on both neighbours, not just one.
  • The result is a continuous ring of \pi electron density above and below the plane of the ring: a **delocalised \pi system** holding 6 electrons that belong to all six carbons.
Benzene's \sigma framework (left) and delocalised \pi cloud (right).

Evidence for delocalisation: all six \ce{C-C} bonds have the **same length, 139\,\mathrm{pm}**, between a single bond (154\,\mathrm{pm}) and a double bond (134\,\mathrm{pm}). Benzene is also much more stable than a ring of three separate double bonds would be: its enthalpy of hydrogenation is about -208\,\mathrm{kJ\,mol^{-1}}, not the 3\times(-120)=-360\,\mathrm{kJ\,mol^{-1}} expected, so it is about 150\,\mathrm{kJ\,mol^{-1}} more stable.

benzene (Kekulé structure)
methylbenzene (toluene)
naphthalene 10 delocalised π electrons

Other aromatic molecules have the same features: a flat ring of \mathrm{sp^2} atoms with a delocalised \pi cloud. Naphthalene (two fused rings) and pyridine (one N in the ring) are examples.

Benzene and isomerism

Disubstituted benzenes show position isomerism: the two groups can be at positions 1,2 (ortho), 1,3 (meta) or 1,4 (para). Benzene does not show cis-trans isomerism, because all the substituents lie in the flat plane of the ring, so there are no two 'sides' to put them on.

1,2-dimethylbenzene ortho
1,3-dimethylbenzene meta
1,4-dimethylbenzene para

7.1.2Optical isomerism and chiralityOften examined

The easiest way to understand optical isomerism is to look at your hands. Your right hand is the mirror image of your left hand, yet the two cannot be placed one on top of the other so that every part matches, however you turn them. They are non-superimposable mirror images. A right-handed glove does not fit your left hand for the same reason.

Chiral
An object or molecule that is non-superimposable on its mirror image (Greek 'cheir', meaning hand). Examples of chiral objects: hands, gloves, shoes, scissors, screws.
Achiral
An object or molecule that is superimposable on its mirror image. It usually has a plane of symmetry. Examples: a ball, a cup, a pencil.
Chirality
The property of 'handedness': having a left-handed and a right-handed form.
Chiral (asymmetric) carbon
A carbon atom bonded to four different atoms or groups. It is also called a chiral centre or stereocentre and is marked with an asterisk (\ce{C}^*).

Many organic molecules exist as a pair of non-superimposable mirror images. Such molecules usually contain a chiral carbon. For example, bromochlorofluoromethane, \ce{CHBrClF}, has a carbon bonded to H, F, Cl and Br: four different atoms. Its mirror image cannot be superimposed on it. But in \ce{CHBrF2} the carbon carries two identical F atoms; the molecule has a plane of symmetry, and its mirror image is the same molecule.

A plain line is a bond in the plane of the paper, a solid wedge points towards you, a hashed wedge points away from you.
Enantiomers
A pair of stereoisomers that are non-superimposable mirror images of each other, like a right and a left hand (Greek 'enantion', opposite). The phenomenon is enantiomerism.

The book's example is 3-methylhexane. Its C3 carries four different groups, H, \ce{CH3}, \ce{C2H5} and \ce{C3H7}, so it exists as two enantiomers. Even a simple alkane can be chiral.

3-methylhexane, enantiomer I
3-methylhexane, enantiomer II
butan-2-ol, enantiomer I
butan-2-ol, enantiomer II
lactic acid, enantiomer I
lactic acid, enantiomer II

In each pair above the two drawings are identical except that the wedge and the hashed bond are swapped. Swapping any two groups on a chiral carbon turns one enantiomer into the other.

How to find a chiral carbon
  1. Ignore every carbon that cannot be chiral: \ce{CH3} and \ce{CH2} carbons (two or three identical H atoms), and carbons in \ce{C=C}, \ce{C#C} or \ce{C=O} (only three or two groups).
  2. For each remaining \mathrm{sp^3} carbon, write down its four groups.
  3. Compare the groups as whole groups, not just the first atom: \ce{-CH2CH3} and \ce{-CH2CH2CH3} are different; two \ce{-CH3} groups are the same.
  4. If all four are different, the carbon is chiral; mark it with *. A molecule with one chiral carbon is always chiral.
Concept Assessment 7.2
Separate chiral and achiral molecules: (a) \ce{CF3CH2CCl3} (b) \ce{CF2HCHFCCl3} (c) \ce{CHCl2(CH2)3CH(OH)CH3} (d) \ce{CH(OH)ClCH2CH3} (e) \ce{CH2ClCH(OH)CH3} (f) \ce{CF3CH2CCl3}
Carbon examinedIts four groupsResult
(a)\ce{CF3}, \ce{CH2}, \ce{CCl3}each has at least two identical atomsachiral
(b)middle \ce{CHF}H, F, \ce{CHF2}, \ce{CCl3}chiral (\ce{C}^*)
(c)\ce{CH(OH)} carbonH, OH, \ce{CH3}, \ce{(CH2)3CHCl2}chiral
(d)\ce{CH(OH)Cl} carbonH, OH, Cl, \ce{CH2CH3}chiral
(e)\ce{CH(OH)} carbonH, OH, \ce{CH3}, \ce{CH2Cl}chiral
(f)same as (a)no carbon with four different groupsachiral
(a) and (f): achiral
(b): chiral, one C*
(e): chiral, one C*

In (c) the \ce{CHCl2} carbon is not chiral (two Cl atoms). Part (f) repeats part (a), probably a misprint in the book.

\text{chiral: (b), (c), (d), (e);\quad achiral: (a), (f)}

7.1.2Plane-polarized light and optical activityOften examined

Ordinary light is a wave whose vibrations occur in all planes at right angles to its direction of travel. When it passes through a polarizing filter, such as a calcite (Nicol) prism or a Polaroid sheet, only the vibrations in one plane get through. Light vibrating in only one plane is called plane-polarized light.

A polarimeter measures the angle \alpha through which a sample rotates the plane of polarized light.
Optically active substance
A substance that rotates the plane of plane-polarized light. Chiral compounds (optical isomers) are optically active.
Optical isomerism
Isomerism shown by compounds that rotate the plane of polarized light, either clockwise or anticlockwise. The isomers are optical isomers.
Dextrorotatory, (+)
Rotates the plane clockwise (to the right, as seen by the observer). Latin 'dexter' means right.
Levorotatory, (-)
Rotates the plane anticlockwise (to the left). Latin 'laevus' means left.
Polarimeter
The instrument that measures the rotation and so detects optical activity.

The two enantiomers of a compound rotate the plane by the same angle but in opposite directions. For example, a standard solution of one enantiomer of butan-2-ol rotates the plane by +13.5^\circ and the other by -13.5^\circ. The older symbols 'd' (dextro) and 'l' (levo) are now obsolete; IUPAC recommends (+) and (-).

(+)-butan-2-ol rotates +13.5°
(−)-butan-2-ol rotates −13.5°

7.1.2Racemic mixtures and meso compoundsOften examined

Racemic mixture

A racemic mixture (racemate) contains equal amounts (50:50) of the two enantiomers: 50% dextrorotatory and 50% levorotatory. It is optically inactive. Each enantiomer rotates the plane by the same angle in opposite directions, so the two effects cancel. This cancelling between molecules of two different kinds is called external compensation. A racemic mixture is written with the prefix (\pm).

\alpha_{\text{net}}=\alpha_{(+)}+\alpha_{(-)}=(+\alpha)+(-\alpha)=0

A reaction that makes a chiral centre from achiral starting materials gives a racemic mixture, because the reagent attacks the two faces of the flat reactant equally often. For example, lactic acid made in the laboratory is (\pm), while the lactic acid formed in our muscles is a single enantiomer.

Meso compounds

A meso compound has two (or more) chiral carbon atoms but also an internal plane of symmetry. One half of the molecule is the mirror image of the other half, so the rotation caused by one half is cancelled by the other half inside the same molecule. This is internal compensation. A meso compound is superimposable on its mirror image, so it is optically inactive (the book: 'optically inactive although it has chiral carbon atoms').

Tartaric acid, \ce{HOOC-CH(OH)-CH(OH)-COOH}, has two chiral carbons. It exists as three stereoisomers: a pair of enantiomers, (+)- and (-)-tartaric acid, and meso-tartaric acid. Meso-tartaric acid is a diastereomer of each of the other two.

(+)-tartaric acid optically active
(−)-tartaric acid its enantiomer
meso-tartaric acid inactive: internal compensation

Molecules with more than one chiral centre

A molecule with n chiral carbons can have at most 2^n stereoisomers. If the molecule can be divided into two identical halves, meso forms appear and the number is smaller. Tartaric acid has n=2, so up to 2^2=4 are possible, but only 3 exist because the 'fourth' isomer is the meso form again.

N_{\max}=2^{n}

The book's second example is 1,2-dimethylcyclobutane. Its trans isomer is chiral and exists as a pair of enantiomers. Its cis isomer has a plane of symmetry through the ring, so it is a meso compound.

trans-1,2-dimethylcyclobutane (I)
trans-1,2-dimethylcyclobutane (II)
cis-1,2-dimethylcyclobutane meso

Natural products are often chiral. Nicotine, made by the tobacco plant, has one chiral carbon (the ring carbon joined to the pyridine ring), and the plant makes only one enantiomer, (-)-nicotine.

(−)-nicotine the C bonded to both rings is chiral
SampleContains chiral carbon?Optically active?Reason
one pure enantiomeryesyesall molecules rotate the plane the same way
racemic mixtureyesnoequal (+) and (-) molecules: external compensation
meso compoundyes (two or more)noplane of symmetry: internal compensation
achiral compound such as propan-2-olnonosuperimposable on its mirror image

7.1.3Properties of optical isomersOften examined

Enantiomers have identical physical and chemical properties in an ordinary (achiral) environment: the same melting point, boiling point, density and solubility, and the same reactions with achiral reagents. For example, both enantiomers of a compound have the same density and the same melting point.

They differ in only two ways: (1) they rotate plane-polarized light by the same angle in opposite directions, and (2) they behave differently towards other chiral things, such as enzymes, receptors in the nose and tongue, and drug targets in the body.

PropertyTwo enantiomersWhy
Melting point, boiling point, density, solubilityidenticalsame bonds and same intermolecular forces
Reactions with achiral reagentsidenticalsame functional groups; mirror-image paths have equal energy
Rotation of plane-polarized lightequal angle, opposite directionmirror-image shapes
Reactions with chiral molecules (enzymes, receptors)differentonly one shape fits the chiral site
Taste, smell, effect as a drugoften differentdetected by chiral receptors
Why the body can tell enantiomers apart: a chiral receptor needs three matching contacts, and only one enantiomer can make all three.
Enantiomers the body can tell apart
Give examples from the book of enantiomers with different biological effects.

Asparagine (an amino acid): one enantiomer tastes bitter, the other tastes sweet.

L-asparagine bitter
D-asparagine sweet

Limonene: (+)-limonene is found in lemons and oranges, while (-)-limonene is found in pine needles, peppermint and spearmint. Our nose detects them as different smells.

(+)-limonene citrus smell
(−)-limonene pine-like smell

Ibuprofen (a common painkiller): the (+) enantiomer is the active painkiller; the (-) enantiomer is inactive.

(+)-ibuprofen active painkiller, star carbon next to COOH
(−)-ibuprofen inactive enantiomer of the drug
Pharmacokinetics
The study of how the body absorbs, distributes, metabolises and excretes a drug.
Pharmacodynamics
The study of what the drug does to the body and how it binds to its target site.

7.2Chirality and preparation of drugsOften examined

Designing, making and selling medicines is a huge industry, and every new drug is expensive to develop. Because the targets of drugs in the body (enzymes and receptors) are chiral, the two enantiomers of a chiral drug can act very differently, so chemists must know which enantiomer does what.

Propoxyphene: one enantiomer is an analgesic (pain reliever) and its mirror image relieves coughing. The two were even sold under mirror-image trade names, Darvon and Novrad.

dextropropoxyphene (Darvon) pain reliever
levopropoxyphene (Novrad) cough suppressant

The thalidomide tragedy

Thalidomide was sold from 1957 as a racemic mixture and prescribed as a sedative and for morning sickness (nausea) in pregnant women. It was then found to cause serious birth defects (for example, missing or shortened limbs) in thousands of babies whose mothers took it, and it was withdrawn in 1961. Research showed that the (+)-enantiomer is the useful sedative, while the (-)-enantiomer is teratogenic (causes birth defects).

(+)-thalidomide (sedative) the chiral carbon carries the H drawn in the book
(−)-thalidomide (teratogenic) mirror image: causes birth defects

Unfortunately, the two enantiomers of thalidomide interconvert in the body. The hydrogen on the chiral carbon sits next to a \ce{C=O} group, so it can be removed and put back from either side under body conditions. A dose of pure (+)-thalidomide therefore turns partly into the harmful (-) form within hours, and giving one enantiomer does not prevent the harm.

7.2.1Separation of racemic thalidomideOften examined

Most chiral drugs are first made as racemic mixtures (50% (+) and 50% (-)), because ordinary reactions cannot tell the two faces of a flat reactant apart. Health professionals do not advise racemic drugs, because one enantiomer may be active while the other is inactive, less effective or even harmful, as thalidomide showed. There are two ways to obtain a single enantiomer.

Method 1: optical resolution

Optical resolution is the separation of a racemic mixture into its two enantiomers, after which the unwanted (-)-thalidomide is removed. Enantiomers have identical physical properties, so ordinary distillation or crystallisation cannot separate them; a chiral 'helper' is needed:

  • High-performance liquid chromatography (HPLC) on a column packed with a chiral stationary phase: one enantiomer is held more strongly and comes out later.
  • Crystallisation: the racemate is reacted with a pure chiral acid or base (a resolving agent). This gives two diastereomeric salts, which have different solubilities and can be separated by fractional crystallisation; each pure enantiomer is then set free again.
  • Enzymes: an enzyme is chiral, so it reacts with (or is made to act on) only one enantiomer, leaving the other untouched.

Method 2: chiral (asymmetric) synthesis

Asymmetric synthesis makes mainly the desired enantiomer, such as (+)-thalidomide, and suppresses the formation of the other. It uses a chiral catalyst, which creates a chiral environment around the reacting molecule so that the reaction on one face is much faster than on the other. The book mentions BINOL (1,1'-bi-2-naphthol) catalysts combined with palladium or rhodium.

In the route described by the book, a \ce{C=C} double bond in the starting molecule (at the position that will become the chiral carbon) is selectively hydrogenated in the presence of the chiral catalyst. The catalyst lets \ce{H2} add from one face only, so one enantiomer forms in preference to the other.

\ce{\text{precursor with C=C} + H2 ->[\text{chiral Rh or Pd catalyst}] \text{mainly one enantiomer}}
BINOL (1,1'-bi-2-naphthol) chiral although it has no chiral carbon
Two ways to obtain single-enantiomer thalidomide, and the problem that remains.

Advantages of chiral catalysts: a small amount of catalyst makes a large amount of product; no half of the product is thrown away as the unwanted isomer; fewer separation steps are needed, so the process is cheaper and produces less waste. The 2001 Nobel Prize in Chemistry (Knowles, Noyori and Sharpless) was awarded for chiral catalysts of this kind.

Formula sheet

General formula of alkanes
\ce{C_{n}H_{2n+2}}
n: number of carbon atoms (n \ge 1)
General formula of alkenes
\ce{C_{n}H_{2n}}
n \ge 2; one \ce{C=C} double bond
General formula of alkynes
\ce{C_{n}H_{2n-2}}
n \ge 2; one \ce{C#C} triple bond
General formula of alcohols
\ce{C_{n}H_{2n+1}OH}
n \ge 1; functional group \ce{-OH}
General formula of carboxylic acids
\ce{C_{n}H_{2n+1}COOH}
n \ge 0 (n=0 is methanoic acid); functional group \ce{-COOH}
Maximum number of optical isomers
N_{\max}=2^{n}
n: number of chiral carbon atoms; the number is smaller when a meso form exists
Rotation by a racemic mixture
\alpha_{\text{net}}=(+\alpha)+(-\alpha)=0
\alpha: angle through which one pure enantiomer turns the plane of polarized light; equal amounts cancel (external compensation)
Unit: degrees (^\circ)
Hydrogenation of but-2-ene (stability of cis and trans)
\ce{CH3CH=CHCH3 + H2 ->[\text{Ni}] CH3CH2CH2CH3}
\Delta H_{\text{cis}}=-120\,\mathrm{kJ\,mol^{-1}}, \Delta H_{\text{trans}}=-116\,\mathrm{kJ\,mol^{-1}}: the cis isomer releases more energy, so it had more energy (is less stable)
Unit: \mathrm{kJ\,mol^{-1}}

Key terms

Organic chemistry
The chemistry of hydrocarbons and their derivatives: their preparation, properties, identification and modification.
Catenation
The ability of atoms of an element to bond to each other to form chains and rings; carbon shows it most because the \ce{C-C} bond is strong.
Tetravalency
The ability of a carbon atom to form four covalent bonds, because it has four valence electrons.
Functional group
An atom or group of atoms in a molecule that decides its characteristic chemical properties, for example \ce{-OH}, \ce{-CHO}, \ce{-COOH}.
Homologous series
A family of compounds with the same functional group and the same general formula, in which each member differs from the next by a \ce{CH2} unit; members show similar chemical properties and gradually changing physical properties.
Isomers
Compounds with the same molecular formula but different structures or different arrangements of atoms in space.
Structural isomerism
Isomerism in which the atoms are joined in a different order; types are chain, position, functional group, metamerism and tautomerism.
Chain isomerism
Structural isomers that differ in the carbon skeleton (straight or branched chain).
Position isomerism
Structural isomers that have the same carbon skeleton and functional group but the group is at a different position.
Functional group isomerism
Structural isomers that have different functional groups, for example an alcohol and an ether of the same formula.
Metamerism
Isomerism in which different alkyl groups are attached on either side of the same polyvalent functional group, for example in ethers.
Tautomerism
A special structural isomerism in which two isomers change into each other by the movement of a hydrogen atom, for example keto-enol tautomerism in propanone.
Stereoisomerism
Isomerism in which compounds have the same structural formula but a different arrangement of atoms or groups in space; its types are geometric and optical isomerism.
Geometric (cis-trans) isomerism
Stereoisomerism caused by restricted rotation about a double bond (or in a ring), when each doubly bonded carbon carries two different groups.
Cis isomer
The geometric isomer in which similar groups are on the same side of the double bond.
Trans isomer
The geometric isomer in which similar groups are on opposite sides of the double bond.
E/Z system
A naming system for geometric isomers based on priority (atomic number) rules: Z (zusammen) when the higher-priority groups are on the same side, E (entgegen) when they are on opposite sides.
Chiral molecule
A molecule that is non-superimposable on its mirror image (it has handedness).
Chiral carbon (chiral centre)
A carbon atom bonded to four different atoms or groups; it is also called an asymmetric carbon.
Enantiomers
A pair of stereoisomers that are non-superimposable mirror images of each other.
Optical isomerism
Stereoisomerism in which isomers differ in their effect on plane-polarized light.
Plane-polarized light
Light whose vibrations occur in one plane only; it is produced by passing ordinary light through a Nicol prism or polaroid.
Optically active
Able to rotate the plane of plane-polarized light.
Dextrorotatory (+)
An optical isomer that rotates the plane of polarized light clockwise (to the right).
Laevorotatory (−)
An optical isomer that rotates the plane of polarized light anticlockwise (to the left).
Polarimeter
The instrument that measures the angle and direction through which an optically active substance rotates plane-polarized light.
Racemic mixture
An equimolar (50:50) mixture of the dextro and laevo forms of a compound; it is optically inactive by external compensation.
Meso compound
A compound that has chiral carbons but is optically inactive because it has an internal plane of symmetry (internal compensation).
Resolution
The separation of a racemic mixture into its two pure enantiomers.
Chiral catalyst
A catalyst that is itself chiral and makes one enantiomer much faster than the other, so a single pure optical isomer is produced.
Thalidomide
A drug sold in the 1950s as a sedative; one enantiomer relieves morning sickness while the other causes birth defects (teratogenic).
Delocalised pi system
In benzene, the six unhybridised p orbitals overlap sideways all round the ring, so six \pi electrons are spread over all six carbon atoms.

Multiple choice questionsp. 139–140

i
Which one of the following molecules will have the ability to rotate the plane of polarized light?
molecule I
molecule II
molecule III
molecule IV
  1. AI and II only
  2. BI and III only
  3. CI, II and IV onlyAnswer
  4. DII and IV only
Why

A molecule rotates plane-polarized light if it is chiral. I (3-ethylcyclohexene): the ring carbon carrying the ethyl group has H, \ce{C2H5}, a \ce{CH=} ring carbon and a \ce{CH2} ring carbon, four different groups. II has two chiral carbons (the \ce{CH(OH)} and the \ce{CH(CH3)} ring carbons), and IV (2-methylcyclohexanamine) has two (the \ce{CH(NH2)} and \ce{CH(CH3)} carbons).

III (isopropylcyclohexane) is achiral: its ring carbon has two identical ring paths, and the isopropyl CH carries two \ce{CH3} groups. So I, II and IV only.

Note. The book labels the fourth structure 'VI'; it is the structure called IV in the options.

ii
Which one of the following molecules shows optical isomerism?
  1. A\ce{CH2=CHCH2CH3}
  2. B\ce{(CH3)2CHCl}
  3. C\ce{CH2=CHCH(Cl)CH3}Answer
  4. D\ce{CH3CH2CH2CH=CH2}
Why

In 3-chlorobut-1-ene, \ce{CH2=CH-\overset{*}{C}H(Cl)-CH3}, the starred carbon carries H, Cl, \ce{CH3} and \ce{CH=CH2}: four different groups, so it is chiral. In (B) the carbon carries two \ce{CH3} groups, and (A) and (D) have no carbon with four different groups.

iii
How many chiral centers are present in this molecule?
the molecule (as printed)
  1. A1Answer
  2. B2
  3. C3
  4. D5
Why

The molecule is 3-methylcyclohexene. Only the ring carbon carrying the methyl group is chiral: its groups are H, \ce{CH3}, the \ce{CH=} side of the ring and the \ce{CH2} side of the ring. The two double-bond carbons are \mathrm{sp^2} and all other ring carbons are \ce{CH2} groups.

Note. The book prints the third option as 'b) 3'; it should be 'c) 3'.

iv
Geometric isomerism is shown by
  1. AalkenesAnswer
  2. Balkynes
  3. Calkanes
  4. Dbenzene
Why

Geometric isomerism needs restricted rotation plus two different groups on each carbon of the \ce{C=C} bond, which alkenes such as but-2-ene have. Alkynes are linear, open-chain alkanes rotate freely, and benzene's substituents all lie in the plane of the ring.

v
Which one is correct statement about cis-trans isomerism?
  1. AIt is shown by alkenes due to restricted rotation of double bondAnswer
  2. BIt is shown by alkenes due to free rotation of sigma bond
  3. CIt is not shown by alkynes because it cannot have different atoms/groups
  4. Dalkanes cannot show geometric isomerism due to free rotation of sigma bonds
Why

The \pi bond locks the two carbons of \ce{C=C}, so the groups cannot rotate: this restricted rotation is the cause of cis-trans isomerism (A). (B) gives the wrong reason; (C) gives the wrong reason for alkynes, which fail because they are linear.

Note. Statement D is also true (open-chain alkanes cannot show it because rotation about \ce{C-C} is free). A is the intended answer because it states the cause of cis-trans isomerism directly.

vi
Which compounds need to be designated by E/Z system?
  1. Alow substituted alkanes
  2. Bhigh substituted alkenesAnswer
  3. Chighly substituted alkanes
  4. Dlow substituted alkenes
Why

Cis-trans names become unclear when the double-bond carbons carry three or four non-hydrogen groups (tri- and tetra-substituted alkenes), so these highly substituted alkenes need the E/Z system. Alkanes have no geometric isomers at all.

vii
Which one is the least stable compound?
  1. Acis-but-2-eneAnswer
  2. Btrans-but-2-ene
  3. Ctrans-hex-3-ene
  4. Dtrans-pent-2-ene
Why

Only cis-but-2-ene has its two alkyl groups on the same side of the double bond. They crowd each other (steric strain), so it has the highest energy and is the least stable. The other three are trans isomers with the alkyl groups far apart.

viii
Which molecule is chiral and can rotate the plane of polarized light?
  1. Acyclohexane
  2. Bbenzene
  3. C2-methylbutan-2-ol
  4. Dpentan-2-olAnswer
Why

In pentan-2-ol, \ce{CH3-\overset{*}{C}H(OH)-CH2CH2CH3}, C2 carries H, OH, \ce{CH3} and \ce{C3H7}, so it is chiral. In 2-methylbutan-2-ol, C2 carries two \ce{CH3} groups; cyclohexane and benzene have planes of symmetry.

ix
Which one of the following is optically active?
  1. Aracemic mixture
  2. Bmeso-compounds
  3. Call chiral molecules
  4. Dall asymmetric moleculesAnswer
Why

A molecule with no symmetry (no plane or centre of symmetry) cannot be superimposed on its mirror image, so it rotates plane-polarized light. A racemic mixture (external compensation) and a meso compound (internal compensation) are optically inactive.

The book rejects (C) because it calls any molecule with chiral carbons 'chiral', and its key points say 'all chiral molecules are not optically active' (meso compounds have chiral carbons but are inactive).

Note. Under the strict modern definition (chiral = not superimposable on its mirror image), every chiral molecule is optically active and C would also be correct. The book's answer is D.

x
Which one is incorrect statement about optical isomers?
  1. Athey cause different rotation of plane polarized lightAnswer
  2. Bthey have no functional groups
  3. Calkanes cannot show optical isomerism
  4. Dthe double bond carbon in alkenes acts chiral center
Why

Only statement A is true: the two optical isomers rotate the plane by equal angles in opposite directions. B is false (lactic acid has \ce{-OH} and \ce{-COOH}), C is false (3-methylhexane, an alkane, is chiral) and D is false (a double-bond carbon has only three groups and cannot be a chiral centre).

Note. Misprint: as printed ('incorrect'), three options (B, C and D) are incorrect and only A is correct, so the question works only as 'Which one is the correct statement?', with answer A. If an examiner keeps the word 'incorrect', D is the statement usually intended.

Short answer questionsp. 141

i3 marks
Designate each of the following alkenes as E or Z, and name it.
Solution

For each carbon of the double bond, find the higher-priority group using the Cahn-Ingold-Prelog rules (higher atomic number first; if tied, compare the next atoms out). If the two higher-priority groups are on the same side the isomer is Z, if on opposite sides it is E.

(a) Structure a
structure a (as printed)
  1. The compound is \ce{ClCH2-C(CH3)=C(CH3)-CH2CH3}: the longest chain through the \ce{C=C} has 5 carbons, with Cl on C1 and methyl groups on C2 and C3. Name: 1-chloro-2,3-dimethylpent-2-ene.
  2. C2 carries \ce{CH2Cl} and \ce{CH3}. Both attach through C, so look further out: (Cl, H, H) beats (H, H, H). \ce{CH2Cl} has the higher priority.
  3. C3 carries \ce{CH3} and \ce{CH2CH3}: (C, H, H) beats (H, H, H), so ethyl has the higher priority.
  4. In the structure the \ce{CH2Cl} group and the ethyl group lie on opposite sides of the double bond, so the isomer is E.
(E)\text{-1-chloro-2,3-dimethylpent-2-ene}
(b) Structure b
structure b (as printed)
  1. The compound is \ce{BrClC=CH-CH3}: 1-bromo-1-chloroprop-1-ene. C1 has no hydrogen, so cis/trans cannot be used; E/Z is needed.
  2. C1 carries Br and Cl. Atomic number Br (35) > Cl (17), so Br has the higher priority.
  3. C2 carries \ce{CH3} and H. C (6) > H (1), so \ce{CH3} has the higher priority.
  4. Br and \ce{CH3} are on the same side of the double bond, so the isomer is Z.
(Z)\text{-1-bromo-1-chloroprop-1-ene}
(c) Structure c
structure c (as printed)
  1. The compound is \ce{CH3CH2CH2-C(CH3)=C(CH3)-CH(CH3)2}. The longest chain through the \ce{C=C} has 7 carbons. Numbering from the isopropyl end gives the double bond locant 3: 2,3,4-trimethylhept-3-ene.
  2. C3 carries \ce{CH3} and \ce{CH(CH3)2}: (C, C, H) beats (H, H, H), so the isopropyl group is higher.
  3. C4 carries \ce{CH3} and \ce{CH2CH2CH3}: (C, H, H) beats (H, H, H), so the propyl group is higher.
  4. The isopropyl and propyl groups lie on opposite sides of the double bond (the zig-zag main chain passes straight through it), so the isomer is E.
(E)\text{-2,3,4-trimethylhept-3-ene}

Note. The book prints only the three structures, with no instruction. Since the unit's key points and MCQ (vi) deal with the E/Z system, the task is taken to be: name each alkene and designate it E or Z.

ii3 marks
Draw the cis-trans isomers of hex-3-ene.
Solution

Hex-3-ene is \ce{CH3CH2-CH=CH-CH2CH3}. Each carbon of the double bond carries two different groups (an ethyl group and a hydrogen atom), and the \pi bond prevents rotation, so hex-3-ene has two geometric isomers:

cis-hex-3-ene (Z): ethyl groups on the same side
trans-hex-3-ene (E): ethyl groups on opposite sides

In cis-hex-3-ene the two ethyl groups (and the two H atoms) are on the same side of the double bond; in trans-hex-3-ene they are on opposite sides. The trans isomer is the more stable one, because its bulky ethyl groups are far apart and do not crowd each other.

iii3 marks
Highlight the importance of chirality in drugs. Why do we need to separate optical isomers in the preparation of drugs?
Solution

Importance. The targets of drugs in the body (enzymes and receptors) are themselves chiral, so they fit only one enantiomer of a chiral drug, like a right hand fits only a right glove. The two enantiomers therefore have different biological effects: one may be active while the other is inactive, has a different action, or is harmful. For example, (+)-thalidomide is a sedative but (-)-thalidomide causes birth defects; one enantiomer of propoxyphene relieves pain, the other relieves coughing; only (+)-ibuprofen is an active painkiller.

Why separate. A racemic drug gives the patient 50% of an isomer that is at best useless and at worst dangerous. Separating the isomers (or making only one) gives the full effect at a smaller dose, avoids side effects from the unwanted isomer and makes the drug safer, which is why regulators ask for each enantiomer to be tested.

iv3 marks
What is meant by levo-rotatory and dextro-rotatory isomers? Give one example.
Solution

Optical isomers rotate the plane of plane-polarized light. A dextro-rotatory isomer rotates the plane clockwise (to the right, Latin 'dexter'); it is given the sign (+), formerly 'd'. A levo-rotatory isomer rotates the plane anticlockwise (to the left, Latin 'laevus'); it is given the sign (-), formerly 'l'. The two enantiomers of a compound rotate the plane by the same angle in opposite directions.

Example: butan-2-ol. (+)-Butan-2-ol rotates the plane by +13.5^\circ (dextro-rotatory) and (-)-butan-2-ol by -13.5^\circ (levo-rotatory). A 50:50 mixture of the two is a racemic mixture and shows no rotation.

(+)-butan-2-ol dextro-rotatory
(−)-butan-2-ol levo-rotatory
v3 marks
Why can't geometric isomerism occur in alkenes where identical atoms or groups are attached to the same carbon atom of the double bond?
Solution

Geometric isomers differ only in which side of the double bond a group is on. If one carbon of the \ce{C=C} carries two identical groups (such as two H atoms), then putting a group from the other carbon 'on the same side' as one of them or 'on the opposite side' gives exactly the same arrangement: the two drawings are superimposable after turning the molecule over. So only one compound exists.

For example, in propene, \ce{CH2=CH-CH3}, C1 carries two hydrogen atoms. The methyl group on C2 is always cis to one H and trans to the other, so there is only one propene. The same is true for but-1-ene and 2-methylpropene. Geometric isomerism needs two different groups on each carbon of the double bond (\ce{abC=Cab} or \ce{abC=Cde}).

vi3 marks
How does the presence of a π (pi) bond in alkenes restrict rotation, and what energy implication does this have for isomerism?
Solution

In an alkene each double-bond carbon is \mathrm{sp^2} hybridised and keeps one unhybridised p orbital. The \pi bond forms by the sideways overlap of these two parallel p orbitals, above and below the plane of the molecule. If one carbon were rotated, the p orbitals would no longer be parallel and the overlap would be lost, so rotation means **breaking the \pi bond**.

Energy implication: breaking the \pi bond of ethene needs about 264\,\mathrm{kJ\,mol^{-1}}, far more than molecules have at room temperature (compare only about 12\,\mathrm{kJ\,mol^{-1}} for rotation about the \ce{C-C} bond in ethane). So cis and trans isomers do not interconvert at room temperature; they are separate, stable compounds that can be isolated. They change into each other only when a lot of energy is supplied, for example by strong heating or by light, as in the conversion of cis-retinal to trans-retinal in the eye.

vii3 marks
Why are cis-isomers generally more reactive but less stable than their trans-isomer counterparts?
Solution

In a cis isomer the two bulky groups are on the same side of the double bond, close to each other. Their electron clouds repel (steric hindrance or steric strain), which raises the energy of the molecule. A higher-energy molecule is less stable. In the trans isomer the bulky groups are on opposite sides, far apart, so there is little strain.

The evidence is the heat of hydrogenation: cis-but-2-ene releases about 120\,\mathrm{kJ\,mol^{-1}} and trans-but-2-ene about 116\,\mathrm{kJ\,mol^{-1}} when both form butane, so the cis isomer starts about 4\,\mathrm{kJ\,mol^{-1}} higher in energy. Because it already has more energy, the cis isomer needs less extra energy to react, so it is more reactive. Also its electron cloud is pushed towards one side, which leaves the \pi bond more exposed to attack by electrophiles.

viii3 marks
How can a molecule be chiral yet remain optically inactive? Give the term used for such a molecule.
Solution

Term: meso compound. A meso compound contains two (or more) chiral carbon atoms, but the molecule also has an internal plane of symmetry: one half is the mirror image of the other half. The first half rotates the plane of polarized light one way and the second half rotates it by the same amount the other way, so the rotations cancel inside the molecule (internal compensation). The compound is optically inactive even though it has chiral centres. Example: meso-tartaric acid.

meso-tartaric acid two chiral carbons, plane of symmetry

Strictly, the plane of symmetry makes a meso molecule achiral as a whole, although its carbons are chiral. A sample of truly chiral molecules is also inactive when it is a racemic mixture (50:50 of the enantiomers), because the two kinds of molecule cancel each other (external compensation).

ix3 marks
Why is it not effective to administer only the (+)-thalidomide enantiomer to patients?
Solution

Because the two enantiomers of thalidomide interconvert in the body (racemisation). The hydrogen atom on thalidomide's chiral carbon sits next to a \ce{C=O} group, which makes it slightly acidic. Under body conditions this hydrogen can be removed and then put back from either side of the molecule, which turns some (+)-thalidomide into (-)-thalidomide within hours.

So even if a patient takes pure (+)-thalidomide (the sedative), a mixture containing the teratogenic (-)-isomer soon forms in the blood, and the risk of birth defects remains. Separating the enantiomers before giving the drug cannot prevent the harmful effect, which is why thalidomide is never given in pregnancy.

x3 marks
Why is the use of chiral catalysts preferred in the pharmaceutical industry for synthesizing optical isomers of drugs?
Solution

A chiral catalyst creates a chiral environment in which the reaction leading to one enantiomer is much faster than the reaction leading to the other, so the desired isomer (for example (+)-thalidomide, using a BINOL catalyst with Rh or Pd) is made directly. This is preferred because:

  • no half of the product is wasted, whereas resolving a racemic mixture throws away up to 50%;
  • a small amount of catalyst, used again and again, makes a large amount of product, so it is cheaper;
  • fewer steps and separations are needed, giving less waste and a purer, safer drug that meets the rules of drug regulators.

Long answer questionsp. 141

i2026 Annual I, Set D, Q.2 (vi)4 marks
Define geometric isomerism. Draw cis-trans isomers of any alkene.
Solution

Definition

Geometric isomerism is a type of stereoisomerism in which compounds have the same molecular formula and the same structural formula, but a different arrangement of atoms or groups around a carbon-carbon double bond (or a ring). The isomers are called geometric or cis-trans isomers.

Cause: restricted rotation

Each carbon of \ce{C=C} is \mathrm{sp^2} hybridised. The \sigma bond forms by head-on overlap of \mathrm{sp^2} orbitals and the \pi bond by sideways overlap of parallel p orbitals. Rotation about the double bond would break the \pi bond (about 264\,\mathrm{kJ\,mol^{-1}}), so the groups are fixed in position on one side or the other.

Condition

Each carbon of the double bond must carry two different atoms or groups (\ce{abC=Cab} or \ce{abC=Cde}).

Diagram: cis and trans but-2-ene

But-2-ene, \ce{CH3CH=CHCH3}, has \ce{CH3} and H on each double-bond carbon, so it has two isomers:

cis-but-2-ene CH3 groups on the same side
trans-but-2-ene CH3 groups on opposite sides

In cis-but-2-ene the two methyl groups are on the same side of the double bond; in trans-but-2-ene they are on opposite sides.

How the isomers differ

  • Physical properties differ: cis-but-2-ene boils at 3.7^\circ\mathrm{C}, trans-but-2-ene at 0.9^\circ\mathrm{C}; the cis isomer is slightly polar, the trans isomer non-polar.
  • The trans isomer is more stable: the methyl groups are far apart, so there is less steric strain.
  • Chemical properties are similar because both have the same functional group (\ce{C=C}).
ii4 marks
Why propene cannot show cis-trans isomerism?
Solution

Structure of propene

Propene is \ce{CH2=CH-CH3}. The first carbon of the double bond (C1) carries two hydrogen atoms. The second carbon (C2) carries a hydrogen atom and a methyl group.

Condition for cis-trans isomerism

Rotation about \ce{C=C} is restricted, but that alone is not enough. Cis-trans isomers exist only if each carbon of the double bond carries two different atoms or groups (\ce{abC=Cab} or \ce{abC=Cde}).

Applying the condition

On C1 of propene both groups are H, so the condition fails. If we try to draw a 'cis' form (methyl on the same side as an H on C1) and a 'trans' form (methyl opposite an H on C1), the methyl group is always on the same side as one H and on the opposite side from the other H. Turning the molecule over makes the two drawings match exactly: they are superimposable, so they are the same compound.

propene only one form exists

This is why propene has only one structure and no cis-trans isomers. The same reasoning applies to but-1-ene (\ce{CH2=CHCH2CH3}) and 2-methylpropene (\ce{(CH3)2C=CH2}).

Contrast with but-2-ene

In but-2-ene, \ce{CH3CH=CHCH3}, each double-bond carbon carries \ce{CH3} and H, so the two methyl groups can be on the same side (cis) or on opposite sides (trans). These two arrangements are not superimposable, so but-2-ene shows cis-trans isomerism while propene does not.

cis-but-2-ene
trans-but-2-ene
iii4 marks
Explain the way of separating a racemic mixture of thalidomide.
Solution

Background

Thalidomide has one chiral carbon, so it has two enantiomers. It is normally made as a racemic mixture (50% (+) and 50% (-)). The (+)-enantiomer is the useful sedative, but the (-)-enantiomer is teratogenic (it causes birth defects), so a single enantiomer is wanted.

(+)-thalidomide (sedative)
(−)-thalidomide (teratogenic)

Why ordinary methods fail

Enantiomers have identical melting points, boiling points and solubilities, so simple distillation or crystallisation cannot separate them. A chiral agent is needed, which interacts differently with the two enantiomers. There are two approaches.

Method 1: optical resolution

  • Chiral HPLC. The racemate is passed through a high-performance liquid chromatography column containing a chiral stationary phase. One enantiomer binds more strongly and comes out later, so the two are collected separately and the (-)-isomer is removed.
  • Crystallisation. The racemate is combined with a pure chiral resolving agent to form two diastereomers. Diastereomers have different solubilities, so they separate by fractional crystallisation; the pure enantiomer is then recovered.
  • Enzymes. An enzyme (a chiral biological catalyst) acts on one enantiomer only, so the other can be separated.

Method 2: chiral (asymmetric) synthesis

Instead of separating, only the desired (+)-isomer is made. A double bond in the starting molecule is hydrogenated in the presence of a chiral catalyst (for example a BINOL-based palladium or rhodium catalyst), which allows hydrogen to add from one face only. The formation of the (-)-isomer is suppressed.

Summary of both methods.

Limitation

Even pure (+)-thalidomide changes partly into the (-)-form inside the body, because the two enantiomers interconvert. So separation makes a purer drug but cannot make thalidomide safe in pregnancy.

iv4 marks
How a catalyst can play its role in this separation.
Solution

Idea

An ordinary (achiral) catalyst speeds up the formation of both enantiomers equally, so it always gives a racemic mixture. A chiral catalyst is itself one pure enantiomer. It creates a chiral environment around the reacting molecule, so it can tell the two faces of a flat reactant apart. This is used in asymmetric synthesis to make one enantiomer of thalidomide, (+)-thalidomide, instead of a racemic mixture that would need separating.

How it works

  1. The reactant (a thalidomide precursor containing a \ce{C=C} bond) binds to the metal of the catalyst, for example rhodium or palladium carrying a chiral BINOL-type ligand.
  2. Because the ligand is chiral, the reactant fits well only with one of its faces towards the metal.
  3. Hydrogen is delivered from the metal to that face only. The route to the (+)-enantiomer has a lower activation energy than the route to the (-)-enantiomer, so it is much faster.
  4. The product leaves, and the catalyst is free to repeat the cycle many times.
\begin{aligned}&\ce{\text{precursor (C=C)} + H2}\\&\qquad\ce{->[\text{chiral Rh or Pd--BINOL}][\text{catalyst}] \text{mainly }(+)\text{-thalidomide}}\end{aligned}
Chiral catalyst: two routes of different activation energy
With a chiral catalyst the two enantiomers form through transition states of different energy. The route with the lower activation energy is much faster, so mainly one enantiomer forms. Both products have the same energy (they are mirror images).

Enzymes as natural chiral catalysts

In optical resolution, an enzyme can act as the chiral catalyst: it speeds up a reaction of one enantiomer only (for example, converting it into a different compound), leaving the other enantiomer unchanged so that the two can be separated easily.

Advantages

  • Mainly one enantiomer forms, so up to half of the product is not wasted.
  • A tiny amount of catalyst makes a large amount of product.
  • Fewer separation steps: cheaper, less waste and a purer, safer drug.
v2026 Annual I, Set D, Q.2 (vi)2026 Annual I, Set D, Q.2 (vi)8 marks
Explain why the trans-isomer of but-2-ene is more stable than the cis-isomer. What role do steric interactions play? Given the compound 1,2-dichloroethene, identify and draw all possible geometrical isomers.
Solution

Part (a) explains stability using steric strain; part (b) applies the conditions for geometric isomerism to \ce{ClCH=CHCl}.

(a) Explain why the trans-isomer of but-2-ene is more stable than the cis-isomer. What role do steric interactions play?

The two isomers

cis-but-2-ene
trans-but-2-ene

Both isomers are \ce{CH3CH=CHCH3}. In the cis isomer both methyl groups are on the same side of the double bond; in the trans isomer they are on opposite sides.

Role of steric interactions

The bond angles around each double-bond carbon are about 120^\circ, so in the cis isomer the two methyl groups are held close together on the same side. Their electron clouds (the C-H bonds of the two methyl groups) repel each other. This repulsion between groups that are forced too close is called steric interaction (steric hindrance or steric strain). It raises the energy of the cis molecule and slightly distorts its bond angles.

In the trans isomer the methyl groups are on opposite sides, far apart, so there is almost no steric strain. The trans isomer therefore has lower energy, which means it is more stable.

Steric repulsion in the cis isomer.

Evidence from energy

When the two isomers are hydrogenated they give the same product, butane. The cis isomer releases about 120\,\mathrm{kJ\,mol^{-1}} and the trans isomer about 116\,\mathrm{kJ\,mol^{-1}}, so the cis isomer started about 4\,\mathrm{kJ\,mol^{-1}} higher in energy. The higher-energy cis isomer is also the more reactive one.

(b) Given the compound 1,2-dichloroethene, identify and draw all possible geometrical isomers.

1,2-Dichloroethene is \ce{ClCH=CHCl}. Each double-bond carbon carries two different atoms, Cl and H, so it has two geometrical isomers:

cis-1,2-dichloroethene (Z); both Cl on the same side
trans-1,2-dichloroethene (E); Cl atoms on opposite sides
  • cis-1,2-dichloroethene, (Z): both Cl atoms on the same side; polar (\mu\approx1.9\,\mathrm{D}); b.p. 60^\circ\mathrm{C}.
  • trans-1,2-dichloroethene, (E): Cl atoms on opposite sides; non-polar (\mu=0); b.p. 48^\circ\mathrm{C}.
vi4 marks
Indicate which isomer would likely have a higher boiling point and why.
Solution

This question continues question (v): the isomers are cis- and trans-1,2-dichloroethene.

Answer

cis-1,2-Dichloroethene has the higher boiling point: about 60^\circ\mathrm{C}, compared with about 48^\circ\mathrm{C} for the trans isomer.

Reason: polarity

Each \ce{C-Cl} bond is polar (\ce{C^{\delta+}-Cl^{\delta-}}) because chlorine is more electronegative than carbon. In the cis isomer both \ce{C-Cl} bond dipoles point towards the same side of the molecule, so they add up to a resultant dipole moment of about 1.9\,\mathrm{D}: the molecule is polar. In the trans isomer the two equal bond dipoles point in exactly opposite directions and cancel, so the resultant dipole moment is zero: the molecule is non-polar.

Dipole moments of the two isomers.

Reason: intermolecular forces

Both isomers have the same mass and the same number of electrons, so their London dispersion forces are similar. But polar cis molecules also attract each other by dipole-dipole forces. More energy is needed to separate the cis molecules from each other in the liquid, so the cis isomer boils at a higher temperature.

vii4 marks
What structural features must a molecule have to exhibit optical isomerism? Give an example and explain why it is optically active.
Solution

Structural features

  • The molecule must be chiral: it must be non-superimposable on its mirror image, like a right hand and a left hand.
  • Usually this is because it contains a chiral (asymmetric) carbon atom: an \mathrm{sp^3} carbon bonded to four different atoms or groups. The four groups point to the corners of a tetrahedron, so they can be arranged in two mirror-image ways.
  • The molecule must have no plane of symmetry (and no centre of symmetry). A molecule with chiral carbons but an internal plane of symmetry (a meso compound) is not optically active.

Example: butan-2-ol

In butan-2-ol, \ce{CH3-\overset{*}{C}H(OH)-CH2CH3}, carbon 2 is bonded to H, OH, \ce{CH3} and \ce{C2H5}: four different groups. So butan-2-ol exists as two enantiomers:

(+)-butan-2-ol C2 is the chiral carbon
(−)-butan-2-ol its mirror image

Why it is optically active

Because the molecule has no symmetry, it interacts with plane-polarized light in an unsymmetrical way and turns the plane of vibration through an angle. A pure sample of one enantiomer turns the plane in one direction, (+)-butan-2-ol clockwise by 13.5^\circ, while its mirror image, (-)-butan-2-ol, turns it by the same angle anticlockwise. Each pure enantiomer is therefore optically active.

FBISE 2026 questions

Questions from the FBISE HSSC-II 2026 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

62026 Annual I, Set B, MCQ 61 mark
Which of the following is a type of stereoisomerism?
  1. AChain isomerism
  2. BPosition isomerism
  3. CGeometrical isomerismAnswer
  4. DFunctional group isomerism
Why

Stereoisomers have the same structural formula (same atoms joined in the same order) but a different arrangement of groups in space. Its two types are geometrical (cis-trans) and optical isomerism. Chain, position and functional group isomerism change the order in which atoms are joined, so they are types of structural isomerism.

vi2026 Annual I, Set B, Q.2 (vi)1+2 = 3 marks
Determine whether the following compound is optically active: \ce{CH3-CH(OH)-Br}. Justify your answer.
Solution

Answer (1 mark)

Yes. \ce{CH3-CH(OH)-Br} (1-bromoethan-1-ol) is chiral, so each of its pure enantiomers is optically active.

Justification (2 marks)

Look at the middle carbon, \ce{CH3-\overset{*}{C}H(OH)-Br}. It is bonded to four different groups: \ce{-H}, \ce{-OH}, \ce{-Br} and \ce{-CH3}. So it is a chiral carbon (asymmetric centre). The molecule has no plane of symmetry, and its mirror image cannot be superimposed on it. The two mirror images are a pair of enantiomers: one turns plane-polarized light clockwise (+), the other turns it anticlockwise (−) by the same angle.

The two enantiomers of \ce{CH3CH(OH)Br} are non-superimposable mirror images.
72026 Annual I, Set D, MCQ 71 mark
An optically active compound is one that:
  1. AHas double bonds
  2. BIs colored
  3. CHas fruity smell
  4. DRotates plane-polarized lightAnswer
Why

Optical activity is defined by what the compound does to light: it rotates the plane of plane-polarized light, clockwise (dextrorotatory, +) or anticlockwise (laevorotatory, −). It needs a chiral molecule; double bonds, colour or smell have nothing to do with it.

vi2026 Annual I, Set D, Q.2 (vi)2+1 = 3 marks
Draw the Cis and Trans isomers of 2-butene and identify the stable isomer between them.
Solution

Structures (2 marks)

But-2-ene is \ce{CH3-CH=CH-CH3}. Each carbon of the double bond carries two different groups (\ce{H} and \ce{CH3}), and the \pi bond stops rotation, so two geometric isomers exist:

cis-but-2-ene both CH3 groups on the same side
trans-but-2-ene CH3 groups on opposite sides

More stable isomer (1 mark)

trans-But-2-ene is more stable. In the cis isomer the two bulky \ce{CH3} groups are on the same side and close together, so their electron clouds repel (steric strain). In the trans isomer they are far apart, so there is no such strain. Evidence: the enthalpy of hydrogenation of cis-but-2-ene (-120\,\mathrm{kJ\,mol^{-1}}) is more negative than that of trans-but-2-ene (-116\,\mathrm{kJ\,mol^{-1}}), so the cis isomer has more energy to start with.

vii2026 Annual I, Set D, Q.2 (vii)1x3 = 3 marks
Separate the chiral and achiral molecules in the following: a. \ce{CF3CH2CCl3} b. \ce{CF2HCHFCCl3} c. \ce{CH2ClCH(OH)CH3}
Solution

Rule: a molecule with one carbon bonded to four different atoms or groups is chiral. Check every carbon of each molecule.

MoleculeCarbon checkedFour groups on itResult
a. \ce{CF3-CH2-CCl3}C1 (\ce{CF3}), C2 (\ce{CH2}), C3 (\ce{CCl3})C1 has three F, C2 has two H, C3 has three Cl: no carbon has four different groupsachiral
b. \ce{CHF2-\overset{*}{C}HF-CCl3}C2\ce{-H}, \ce{-F}, \ce{-CHF2}, \ce{-CCl3}: all differentchiral
c. \ce{CH2Cl-\overset{*}{C}H(OH)-CH3}C2\ce{-H}, \ce{-OH}, \ce{-CH2Cl}, \ce{-CH3}: all differentchiral
a. achiral 1,1,1-trichloro-3,3,3-trifluoropropane
b. chiral C2 is the chiral centre
c. chiral C2 is the chiral centre
viii2026 Annual I, Set D, Q.2 (viii)2+1 = 3 marks
Draw a pair of enantiomers for 2-butanol and show the chiral center.
Solution

Chiral centre (1 mark)

2-Butanol (butan-2-ol) is \ce{CH3-\overset{*}{C}H(OH)-CH2-CH3}. Carbon 2 (marked *) is the chiral centre: it is bonded to four different groups, \ce{-H}, \ce{-OH}, \ce{-CH3} and \ce{-C2H5}.

Pair of enantiomers (2 marks)

Draw the chiral carbon as a tetrahedron (plain lines in the page, a wedge coming towards you, a dashed bond going away), then draw its mirror image. The two cannot be superimposed, so they are enantiomers.

The enantiomers of butan-2-ol: (+)-butan-2-ol and (−)-butan-2-ol.
(R)-butan-2-ol one enantiomer
(S)-butan-2-ol the other enantiomer

The two enantiomers have the same boiling point and density, but they rotate plane-polarized light by equal angles in opposite directions (+13.5^\circ and -13.5^\circ).

62026 Annual I, Old syllabus (Curriculum 2006), MCQ 6, old syllabus1 mark
Which of the following compounds is organic?
  1. A\ce{NaCl}
  2. B\ce{CH4}Answer
  3. C\ce{H2SO4}
  4. D\ce{CaCO3}
Why

Methane, \ce{CH4}, is a hydrocarbon, the simplest organic compound. \ce{CaCO3} contains carbon, but carbonates (like \ce{CO}, \ce{CO2} and cyanides) are treated as inorganic. \ce{NaCl} and \ce{H2SO4} contain no carbon.

72026 Annual I, Old syllabus (Curriculum 2006), MCQ 7, old syllabus1 mark
Which of the following pairs belong to the same homologous series?
  1. A\ce{C2H6} and \ce{C3H8}Answer
  2. B\ce{CH4} and \ce{C2H4}
  3. C\ce{CH3OH} and \ce{C2H5COOH}
  4. D\ce{C2H2} and \ce{C2H4O}
Why

Ethane and propane both fit the alkane formula \ce{C_{n}H_{2n+2}} and differ by one \ce{CH2} unit. In B, \ce{C2H4} is an alkene. In C, one is an alcohol and the other an acid (different functional groups). In D, an alkyne and an aldehyde.

102026 Annual I, Old syllabus (Curriculum 2006), MCQ 10, old syllabus1 mark
The compound \ce{CH3-CH(OH)-CH3}:
  1. AHas one chiral center
  2. BHas two chiral centers
  3. CExhibits optical isomerism
  4. DHas no chiral centerAnswer
Why

The middle carbon of propan-2-ol carries \ce{-H}, \ce{-OH} and two identical \ce{-CH3} groups, so it is not chiral. The end carbons carry three H atoms each. With no chiral centre the molecule is superimposable on its mirror image and shows no optical isomerism.

iv2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (iv), old syllabus1.5+1.5 = 3 marks
Why carbon forms more compounds than any other element? Give at least two reasons related to its properties.
Solution

Carbon forms more compounds (about 20 million) than any other element because of the following properties of the carbon atom.

1. Catenation (1.5 marks)

Carbon atoms join to one another to form long straight chains, branched chains and rings. This is possible because the \ce{C-C} bond is very strong (about 348\,\mathrm{kJ\,mol^{-1}}), much stronger than the \ce{Si-Si} bond (226\,\mathrm{kJ\,mol^{-1}}). Example: hexane, 2-methylpentane and cyclohexane all have six carbons.

2. Tetravalency (1.5 marks)

Carbon has four valence electrons (\mathrm{2s^2\,2p^2}), so it forms four covalent bonds. Four bonds per atom allow branching and let each carbon carry different atoms or groups, which gives a huge number of structures and isomers.

Other reasons (any can be used)

  • It forms stable multiple bonds: \ce{C=C}, \ce{C#C}, \ce{C=O}, \ce{C#N}.
  • It bonds strongly with H, O, N, S and halogens, giving many functional groups.
  • Isomerism: one molecular formula can have many structures, for example \ce{C10H22} has 75 structural isomers.
ix OR2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (ix) OR, old syllabus1x3 = 3 marks
Draw any three possible structural isomers of \ce{C4H10O}.
Solution

\ce{C4H10O} fits \ce{C_{n}H_{2n+2}O}, so it is saturated and can be an alcohol (\ce{R-OH}) or an ether (\ce{R-O-R'}). Any three of the seven isomers earn full marks (1 mark each). Three alcohols:

butan-1-ol \ce{CH3CH2CH2CH2OH}
butan-2-ol \ce{CH3CH2CH(OH)CH3}
2-methylpropan-2-ol \ce{(CH3)3COH}

The other four isomers are listed below. Butan-1-ol and butan-2-ol are position isomers; butan-1-ol and 2-methylpropan-1-ol are chain isomers; an alcohol and an ether are functional group isomers.

IsomerCondensed formulaType
2-methylpropan-1-ol\ce{(CH3)2CHCH2OH}alcohol (1°)
1-methoxypropane\ce{CH3OCH2CH2CH3}ether
2-methoxypropane\ce{CH3OCH(CH3)2}ether
ethoxyethane\ce{CH3CH2OCH2CH3}ether

Exam focus

  1. Why does carbon form more compounds than any other element? (short)
    Asked in the 2026 old-syllabus paper (1.5 + 1.5 marks); catenation and tetravalency are the two expected reasons.
    Go to answer
  2. Draw cis and trans isomers of but-2-ene and say which is more stable. (short)
    Asked in 2026 Set D Q.2 (vi); also a book long question. Steric strain and hydrogenation enthalpies give the reason.
    Go to answer
  3. Why can propene not show cis-trans isomerism? (long)
    The condition 'two different groups on each doubly bonded carbon' is tested every year.
    Go to answer
  4. Identify chiral carbons and decide whether a molecule is optically active. (short)
    Three 2026 questions (Set B Q.2 (vi), Set D Q.2 (vii), old-syllabus MCQ 10) test only this skill.
    Go to answer
  5. Draw a pair of enantiomers of butan-2-ol and mark the chiral centre. (short)
    Asked in 2026 Set D Q.2 (viii); wedge-dash drawing on both sides of a mirror earns the marks.
    Go to answer
  6. Racemic mixture and meso compound: why are both optically inactive? (short)
    External versus internal compensation is a standard short question and MCQ.
    Go to answer
  7. Designate alkenes as E or Z using priority rules. (short)
    Book short question (i); needs the atomic-number priority rule applied to each carbon.
    Go to answer
  8. Importance of chirality in drugs and the thalidomide example; separation of a racemic mixture. (long)
    Book long questions (iii) and (iv) and short questions (iii), (ix), (x) all come from this SLO.
    Go to answer
  9. Why is the trans isomer more stable while the cis isomer has the higher boiling point? (long)
    Book long questions (v) and (vi): stability from steric strain, boiling point from polarity.
    Go to answer
  10. Structural isomers of \ce{C4H10O} and types of structural isomerism. (short)
    Asked in the 2026 old-syllabus paper (Q.2 (ix) OR); also common in MCQs.
    Go to answer

Mark this unit as done to fill the Class 12 Chemistry ring.

Next, unit 8
Hydrocarbons

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