Hydrocarbons

Class 11 Chemistry, unit 16, book pages 318–338

Practise 68 MCQs Take the unit quiz Review 46 cards

Hydrocarbons are compounds of carbon and hydrogen only. The unit classifies them as aliphatic and aromatic, explains why alkanes are so unreactive and how they react by free radical substitution, and describes the shapes of ethane, cyclopropane and ethene. It then covers how ethene is prepared and its addition reactions, including the electrophilic addition mechanism and Markovnikov\'s rule, followed by conjugation, redox reactions of organic compounds and stereoisomerism (optical and geometric).

Learning outcomes (18)

Classify hydrocarbons as aliphatic and aromatic compounds.
Describe nomenclature of alkanes and cycloalkanes.
Explain the shape of ethane and cyclopropane.
Explain unreactive nature of alkanes towards polar reagents.
Define the homolytic and heterolytic fission reactions.
Describe the mechanism of free radical substitution in alkanes exemplified by methane and ethane.
Identify redox reactions of organic compounds.
Explain the nomenclature of alkenes.
Explain shape of ethene molecule in terms of all bonds.
Describe the structure and reactivity of alkenes exemplified by ethene.
Explain stereoisomerism with suitable examples.
Explain dehydration of alcohols to prepare ethene.
Explain dehydrohalogenation of halogenoalkanes for preparation of ethene.
Explain hydrogenation, hydrohalogenation, hydration, halogenation, halohydration, epoxidation, ozonolysis and polymerization of ethene.
Explain the concept of conjugation of alkenes with alternate double bonds.
Use the IUPAC naming system for alkenes.
Describe the mechanism of electrophilic addition in alkenes, using bromine/ethene and hydrogen bromide/propene as example.
Explain the inductive effects of alkyl groups on the stability of primary, secondary and tertiary carbocations formed during electrophilic additions, following Markovnikov's rule.

16.1HydrocarbonsOften examined

Hydrocarbons are organic compounds that contain the elements hydrogen and carbon only. They are the simplest and most important class of organic compounds, and they are broadly divided into two classes: aliphatic hydrocarbons and aromatic hydrocarbons.

Aliphatic hydrocarbons: Hydrocarbons that are open chain, or closed chain (ring) compounds whose properties resemble those of open chain compounds. The word aliphatic comes from the Greek for "fatty".
Aromatic hydrocarbons: Benzene and the compounds whose properties resemble those of benzene. The word aromatic comes from the Greek for "fragrant".

Figure 16.1: Classification of hydrocarbons into open chain (acyclic) and closed chain (cyclic) classes, with an example under each.

Classification of hydrocarbons (Figure 16.1 in table form)
Class	Sub-class	Name for the family	Example
Open chain (acyclic)	Saturated	Alkanes (paraffins)	ethane, $C X_{2} H X_{6}$
Open chain (acyclic)	Unsaturated	Alkenes (olefins)	ethene, $H X_{2} C = CH X_{2}$
Open chain (acyclic)	Unsaturated	Alkynes (acetylenes)	ethyne, $HC \equiv CH$
Closed chain (cyclic)	Alicyclic	Cycloalkanes	cyclopentane
Closed chain (cyclic)	Aromatic	Compounds with a benzene ring	benzene

16.1.1Open chain hydrocarbonsOften examined

Open chain hydrocarbons have a carbon skeleton in which the carbon atoms form chains, either straight or branched. They are also called acyclic hydrocarbons, which means that they have no cycle (ring). There must be terminal carbon atoms in the chain. They are divided into two classes: saturated and unsaturated hydrocarbons.

Saturated hydrocarbons

Saturated hydrocarbons have only single covalent bonds among their carbon atoms. Each carbon is bonded to four atoms, so no more atoms can be added to it, and this is why they are called saturated. Every singly bonded carbon is $s p^{3}$ hybridized. They may be straight chain or branched chain. Alkanes are the example of saturated hydrocarbons.

butane straight chain alkane

2-methylpropane branched chain alkane

Unsaturated hydrocarbons

Unsaturated hydrocarbons have at least one multiple (double or triple) bond. A doubly bonded carbon is bonded to three atoms and a triply bonded carbon to two atoms, so more atoms can be added to them; this is why they are called unsaturated. They include alkenes and alkynes.

Alkenes

Alkenes are commonly known as olefins. They contain at least one double bond in their parent chain. Each carbon that has a double bond is $s p^{2}$ hybridized. If a molecule has more than one double bond it is called a polyene.

4-methylpent-2-ene

5-methylhexa-1,2-diene

Alkynes

The common name of the alkynes is acetylenes. They contain at least one triple bond between carbon atoms. Each triply bonded carbon is $s p$ hybridized. Alkynes with more than one triple bond are called polyynes.

3-methylpent-1-yne

5-methylhexa-1,3-diyne

Saturated and unsaturated open chain hydrocarbons
Type	Bonds between carbons	Atoms bonded to each C	Hybridization of that C
Alkane (saturated)	single only	4	$s p^{3}$
Alkene	at least one double	3 (for the C=C carbons)	$s p^{2}$
Alkyne	at least one triple	2 (for the C≡C carbons)	$s p$

16.1.2Closed chain hydrocarbonsOften examined

Closed chain hydrocarbons have no terminal carbon in the principal carbon skeleton. Every carbon atom is bonded to at least two other carbon atoms. They are divided into two classes: alicyclic and aromatic hydrocarbons.

Alicyclic hydrocarbons

The word alicyclic is a blend of the two words aliphatic and cyclic: they are aliphatic hydrocarbons with rings of carbon atoms and they resemble aliphatic hydrocarbons in their characteristics. An alicyclic hydrocarbon has two hydrogen atoms fewer than the corresponding open chain hydrocarbon, for example $C X_{4} H X_{10}$ (butane) and $C X_{4} H X_{8}$ (cyclobutane).

cyclobutane

cyclopentene

cyclohexane

Aromatic hydrocarbons

Aromatic hydrocarbons have at least one benzene ring in their molecules. The simplest aromatic hydrocarbon is benzene. The benzene molecule is hexagonal in shape with alternating single and double covalent bonds, and every carbon of the benzene ring is $s p^{2}$ hybridized.

benzene

ethylbenzene

1-methylethylbenzene

Concept assessment exercise 16.1

(i) Differentiate between saturated and unsaturated compounds. Give examples. (ii) Define aliphatic and aromatic compounds. Give two examples of each. (iii) What type of hybridization is shown by each carbon of hexa-1,2-diene?

(i) Saturated and unsaturated compounds

Point	Saturated compounds	Unsaturated compounds
Bonds between carbon atoms	only single bonds	at least one double or triple bond
Atoms bonded to each carbon	four (maximum)	three (C=C) or two (C≡C), so more atoms can be added
Hybridization	$s p^{3}$	$s p^{2}$ or $s p$ on the multiply bonded carbons
Examples	ethane $CH X_{3} CH X_{3}$ , propane, cyclohexane	ethene $CH X_{2} = CH X_{2}$ , ethyne $HC \equiv CH$

(ii) Aliphatic and aromatic compounds

Aliphatic compounds are open chain hydrocarbons, or closed chain hydrocarbons whose properties resemble the open chain ones. Examples: propane ( $CH X_{3} CH X_{2} CH X_{3}$ ) and cyclohexane. Aromatic compounds are benzene and the compounds whose properties resemble those of benzene (they contain a benzene ring). Examples: benzene and ethylbenzene.

(iii) Hybridization of the carbons of hexa-1,2-diene

Hexa-1,2-diene is $CH X_{2} = C = CH - CH X_{2} - CH X_{2} - CH X_{3}$ (double bonds from C-1 to C-2 and from C-2 to C-3). Count the atoms bonded to each carbon: a carbon with four single bonds is $s p^{3}$ , one with a double bond is $s p^{2}$ and one with two double bonds (bonded to two atoms) is $s p$ .

hexa-1,2-diene

Carbon	Bonds	Hybridization
C-1, $CH X_{2}$	one double bond	$s p^{2}$
C-2	two double bonds (one to C-1, one to C-3)	$s p$
C-3, $CH$	one double bond	$s p^{2}$
C-4, C-5, C-6	single bonds only	$s p^{3}$

C-1: s p^{2}, C-2: s p, C-3: s p^{2}, C-4, C-5, C-6: s p^{3}

16.2Reactivity of alkanesOften examined

Alkanes are non-polar hydrocarbons because the electronegativity difference between carbon and hydrogen is not enough to make the bond polar. Alkanes are also called paraffins. The word paraffin combines two Latin words: parum means "little" and affinis means "reactivity". The literal meaning is therefore "less reactive". The low reactivity of alkanes is attributed to two factors.

1. Non-polar bonds

The electronegativity values of carbon and hydrogen are $2.5$ and $2.1$ . The difference is negligible, so the bonding electrons are shared symmetrically and the C–H bonds (and the C–C bonds) in alkanes are non-polar.

Δ E N = 2.5 - 2.1 = 0.4

Because alkanes are non-polar, acids, bases, oxidizing agents and reducing agents (polar reagents) do not react with them. This is the unreactive nature of alkanes towards polar reagents.

2. Strength of the sigma bond

Alkanes have only sigma ( $σ$ ) bonds and no pi ( $π$ ) bonds. A sigma bond is strong because its electrons lie very close to the bonded nuclei, between them, and are strongly held. The $σ$ electrons in alkanes cannot easily be used for chemical reactions, which reflects their inertness.

Why alkanes are unreactive
Factor	What it means	Result
Non-polar C–H and C–C bonds	no positive or negative end for attack by ions or polar reagents	acids, bases, oxidizing and reducing agents do not react
Only strong $σ$ bonds	electrons held tightly between the nuclei	electrons cannot easily be used in a reaction

Alkanes do not react with other reagents under normal conditions, but they do react with some substances at high temperature. Combustion of alkanes (burning sui gas in the kitchen) is a common reaction in daily life. The two important reactions of alkanes are:

free radical substitution reaction, which needs ultraviolet light to form free radicals;
cracking, which needs heating to break large molecules and convert them into smaller, more useful molecules.

The free radical substitution of ethane is studied below with its mechanism.

16.2.1Free radical substitution of ethaneOften examined

Substitution reactions of ethane involve halogen free radicals, which are very reactive species that attack ethane molecules. Ultraviolet light is needed. The reaction takes place in three steps: initiation, propagation and termination.

Homolytic fission: Breaking of a covalent bond so that each of the two atoms keeps one electron of the shared pair. It gives two free radicals, each with an unpaired electron.
Heterolytic fission: Breaking of a covalent bond so that both electrons of the shared pair go to one atom. It gives a positive ion and a negative ion (for example a carbocation and an anion).
Free radical: An atom or group with an unpaired electron, written with a dot, for example $Cl \cdot$ . It is very reactive.

1. Initiation

The ultraviolet light breaks the chlorine molecule by homolytic fission and gives two chlorine free radicals.

Cl - Cl h ν 2 Cl \cdot

2. Propagation

This step is called propagation because the substitution of ethane progresses like a chain reaction: each step uses up one radical and produces a new one. As soon as the chlorine radical is formed it attacks an ethane molecule to give the ethyl free radical.

CH X_{3} - CH X_{3} + Cl \cdot CH X_{3} - CH X_{2} \cdot + HCl

The ethyl free radical reacts with a chlorine molecule to form chloroethane and another chlorine free radical, which starts the cycle again.

CH X_{3} - CH X_{2} \cdot + Cl - Cl CH X_{3} - CH X_{2} - Cl + Cl \cdot

C X_{2} H X_{6} + Cl X_{2} h ν C X_{2} H X_{5} Cl + HCl

overall reaction of the two propagation steps

If excess chlorine is used the propagation reactions continue, giving more chlorine free radicals and chloro-substituted alkyl free radicals, each step producing one free radical at a time. The result is a mixture of chloro-substituted alkanes such as 1,1-dichloroethane, 1,1,1-trichloroethane and 1,1,2,2-tetrachloroethane. In the presence of enough chlorine the final product is 1,1,1,2,2,2-hexachloroethane, $C X_{2} Cl X_{6}$ .

3. Termination

Termination is the end of the chain. It can take place at any stage of the propagation: two of the different free radicals present combine, which removes radicals from the mixture.

Cl \cdot + Cl \cdot Cl X_{2}

CH X_{3} - CH X_{2} \cdot + Cl \cdot CH X_{3} - CH X_{2} - Cl

chloroethane

CH X_{3} - CH X_{2} \cdot + \cdot CH X_{2} - CH X_{3} CH X_{3} - CH X_{2} - CH X_{2} - CH X_{3}

butane

The three steps of free radical chlorination of ethane
Stage	Reaction	What happens to radicals
Initiation	$Cl X_{2} h ν 2 Cl \cdot$	radicals are formed
Propagation	$C X_{2} H X_{6} + Cl \cdot C X_{2} H X_{5} \cdot + HCl$ and $C X_{2} H X_{5} \cdot + Cl X_{2} C X_{2} H X_{5} Cl + Cl \cdot$	a radical is used and a new one formed, so the chain goes on
Termination	$Cl \cdot + Cl \cdot Cl X_{2}$ , $C X_{2} H X_{5} \cdot + Cl \cdot C X_{2} H X_{5} Cl$ , $C X_{2} H X_{5} \cdot + C X_{2} H X_{5} \cdot C X_{4} H X_{10}$	two radicals combine and the chain stops

Worked example: one radical, many molecules in the propagation steps

In the chlorination of ethane, one chlorine radical made by ultraviolet light can lead to thousands of molecules of chloroethane. Explain this with the propagation steps, and say how the chain ends.

The two propagation steps form a cycle: the chlorine radical is regenerated, so one radical can convert many ethane molecules.

Initiation makes the first radicals. Ultraviolet light splits one chlorine molecule by homolytic fission.
$Cl - Cl h ν 2 Cl \cdot$
Propagation step 1: a chlorine radical takes a hydrogen atom from ethane. One radical is used up and one new radical (ethyl) is made.
$CH X_{3} CH X_{3} + Cl \cdot CH X_{3} CH X_{2} \cdot + HCl$
Propagation step 2: the ethyl radical takes a chlorine atom from a chlorine molecule. The product is chloroethane, and a chlorine radical is made again.
$CH X_{3} CH X_{2} \cdot + Cl - Cl CH X_{3} CH X_{2} Cl + Cl \cdot$
The chlorine radical from step 2 is the reactant of step 1, so the cycle repeats. Adding the two steps, the radicals cancel and only the overall equation is left.
$C X_{2} H X_{6} + Cl X_{2} C X_{2} H X_{5} Cl + HCl$
Termination ends the chain: two radicals meet and form a molecule, so no radical is left to continue.
$CH X_{3} CH X_{2} \cdot + Cl \cdot CH X_{3} CH X_{2} Cl$

each propagation step uses one radical and makes one radical, so the chain repeats until termination

The same mechanism for methane

The book refers to section 15.6.1 for the chlorination of methane. The three steps are the same, with $CH X_{3} \cdot$ in place of $C X_{2} H X_{5} \cdot$ :

Initiation: chlorine is split by ultraviolet light.
$Cl X_{2} h ν 2 Cl \cdot$
Propagation (a): a chlorine radical removes a hydrogen atom from methane.
$CH X_{4} + Cl \cdot CH X_{3} \cdot + HCl$
Propagation (b): the methyl radical removes a chlorine atom from a chlorine molecule.
$CH X_{3} \cdot + Cl X_{2} CH X_{3} Cl + Cl \cdot$
Termination: two radicals combine.
$CH X_{3} \cdot + Cl \cdot CH X_{3} Cl, CH X_{3} \cdot + CH X_{3} \cdot C X_{2} H X_{6}$

16.3Shapes of ethane and cyclopropaneOften examined

1. Shape of the ethane molecule, C₂H₆

The ethane molecule has two carbon atoms bonded to each other by a single covalent bond. Three hydrogen atoms are bonded to each carbon by single covalent bonds. A covalent bond formed by end-on overlap of atomic orbitals is called a **sigma (\sigma) bond**, and every single covalent bond is a \sigma bond.

Each carbon atom of ethane is sp^3 hybridized. Every orbital shared by the two carbons, either with hydrogen atoms or with each other, is an sp^3 hybrid orbital: the C–C bond is formed by sp^3–sp^3 overlap and each C–H bond by sp^3–s overlap.

Ethane is not planar, because all the covalent bonds around a carbon atom make angles of 109.5^\circ with one another. This arrangement of the atoms in space gives a tetrahedral shape around each carbon atom.

Figure 16.2: Shape of ethane. Each carbon is tetrahedral (all bond angles 109.5°) and the C–C bond is a σ bond; solid wedges come out of the page and hashed wedges go behind it.

2. Shape of the cyclopropane molecule, C₃H₆

The cyclopropane molecule has three carbon atoms bonded by single covalent bonds to one another in the form of a ring. Two hydrogen atoms are bonded by single covalent bonds to each carbon atom. Each carbon is sp^3 hybridized, and the orbitals shared between carbon atoms or with hydrogen atoms are sp^3 hybrid orbitals with a tetrahedral arrangement.

The bonds in cyclopropane are directed in three dimensions, so the molecule is not planar. The C–C–C angle is 60^\circ whereas the H–C–H and C–C–H angles are 109.5^\circ.

Figure 16.3: Shape of cyclopropane. The three carbons form a ring with C–C–C angles of 60°; the two hydrogens on each carbon lie above and below the plane of the ring with H–C–H angles of 109.5°.

Shapes of ethane and cyclopropane
Molecule	Hybridization of C	Shape around each C	Angles
ethane \ce{C2H6}	sp^3	tetrahedral, not planar	all 109.5^\circ
cyclopropane \ce{C3H6}	sp^3	three carbons in a ring; H atoms above and below it	C–C–C 60^\circ; H–C–H and C–C–H 109.5^\circ

16.4Preparation of alkenesOften examined

Alkenes can be prepared by several methods, but the book studies two: dehydration of alcohols and dehydrohalogenation of halogenoalkanes. Both are elimination reactions: a small molecule (\ce{H2O} or \ce{HX}) is removed from the reactant and a double bond is formed.

1. Dehydration of alcohols

Dehydration is an elimination reaction that involves the removal of water. It takes place when ethanol vapours are passed over hot aluminium oxide powder used as a catalyst. Aluminium oxide (alumina) also acts as a dehydrating agent. Concentrated sulphuric acid (\ce{H2SO4}) and phosphoric acid (\ce{H3PO4}) also act as catalysts and dehydrating agents in this reaction.

\ce{CH3-CH2-OH ->[Al2O3][heat] CH2=CH2 + H2O}

ethanol to ethene

2. Dehydrohalogenation of halogenoalkanes

Dehydrohalogenation is the removal of hydrogen and halogen in the form of a hydrogen halide molecule. It is also an elimination reaction. Ethene can be prepared from bromoethane with hot ethanolic sodium hydroxide.

\ce{CH3-CH2-Br + NaOH(ethanolic) ->[heat] CH2=CH2 + H2O + NaBr}

bromoethane to ethene

The two preparations of ethene
Point	Dehydration of alcohols	Dehydrohalogenation of halogenoalkanes
Starting material	alcohol (ethanol)	halogenoalkane (bromoethane)
Reagent and conditions	hot \ce{Al2O3} (or conc. \ce{H2SO4}, \ce{H3PO4})	hot ethanolic \ce{NaOH}
Small molecule removed	\ce{H2O}	\ce{HBr} (taken up by NaOH as \ce{NaBr} and \ce{H2O})
Product	ethene	ethene

Concept assessment exercise 16.2

Prepare but-1-ene by the following preparatory methods: (i) dehydration of alcohols; (ii) dehydrohalogenation of halogenoalkanes.

But-1-ene is \ce{CH2=CH-CH2-CH3}. To get the double bond between C-1 and C-2 the starting material must have its functional group on C-1.

(i) Dehydration. Start from butan-1-ol, \ce{CH3CH2CH2CH2OH}. Pass its vapour over hot aluminium oxide (or heat with conc. \ce{H2SO4} or \ce{H3PO4}). The \ce{OH} on C-1 and an \ce{H} from C-2 are removed as water.
\ce{CH3-CH2-CH2-CH2-OH ->[Al2O3][heat] CH3-CH2-CH=CH2 + H2O}
(ii) Dehydrohalogenation. Start from 1-bromobutane, \ce{CH3CH2CH2CH2Br}, and heat with ethanolic \ce{NaOH}. The \ce{Br} on C-1 and an \ce{H} from C-2 are removed as \ce{HBr}, which NaOH converts to \ce{NaBr} and \ce{H2O}.
\ce{CH3-CH2-CH2-CH2-Br + NaOH(ethanolic) ->[heat] CH3-CH2-CH=CH2 + NaBr + H2O}

16.5Structure and reactivity of ethene moleculeOften examined

The ethene molecule, \ce{C2H4}, has two carbon atoms bonded by a double covalent bond and two hydrogen atoms singly bonded to each carbon. The double bond consists of one sigma (\sigma) bond and one pi (\pi) bond. Each doubly bonded carbon is sp^2 hybridized with a trigonal planar arrangement.

The **\sigma bond** between the two carbons is formed by head-on overlap of two sp^2 hybrid orbitals, one from each carbon. The carbons hold the \sigma electrons strongly, so this bond is difficult to break.
The **\pi bond** is formed by parallel (sideways) overlap of the two unhybridized p orbitals, one on each carbon. The \pi electrons are weakly held, so the \pi bond is weaker than the \sigma bond.
Each carbon forms two \sigma bonds with two hydrogen atoms by sp^2–s overlap and one \sigma bond with the other carbon by sp^2–sp^2 overlap.

Figure 16.4: Orbitals of ethene. The two sp² hybrid orbitals overlap head-on to give the σ bond between the carbons, and the two unhybridized p orbitals overlap sideways, above and below the molecular plane, to give the π bond. The hydrogen atoms are left out.

Ethene is planar: all its atoms lie in the same plane and the bond angles are almost 120^\circ. A molecule in which all the atoms lie in the same plane is called a planar molecule.

Figure 16.5: Shape of the ethene molecule. All six atoms lie in one plane and the H–C–H and H–C=C angles are about 120°. The double bond is one σ bond and one π bond.

The two bonds of the carbon–carbon double bond
	\sigma bond in ethene	\pi bond in ethene
Overlap	head-on, sp^2–sp^2	parallel (sideways), p–p
Where the electrons are	on the line joining the two nuclei	above and below the plane of the molecule
Strength	strong, difficult to break	weaker, easily used by reagents
Number in \ce{C=C}	one	one

Reactivity of ethene

Ethene is more reactive than ethane because it has a \pi bond, which is much weaker and more reactive than the \sigma bonds in ethane.
Similarly, ethene is more reactive than ethyne because it has a longer carbon–carbon bond distance than ethyne. This is why the \pi bond in ethene is more exposed (diffused) than the \pi bonds in ethyne.

16.6Chemical reactions of alkenesOften examined

Alkenes show addition reactions, in which a molecule is added to the alkene at the cost of the double bond. In such a reaction the double covalent bond (one \sigma and one \pi) is converted into a single covalent bond, and two new \sigma bonds are made. Alkanes cannot undergo addition because they have no multiple bond.

\ce{CH2=CH2 + X-Y -> CH2X-CH2Y}

general addition of X–Y to ethene

The important addition reactions of alkenes met in this unit are summarized below; each is explained in the sections that follow, using ethene as the example.

Addition reactions of ethene
Reaction	Reagent and conditions	Product from ethene
Hydrogenation	\ce{H2}, finely divided Ni or Pt, heat	ethane
Hydrohalogenation	\ce{HX} (for example \ce{HBr})	halogenoalkane (bromoethane)
Hydration	conc. \ce{H2SO4} (cold), then boil with water	ethanol
Halogenation	\ce{Br2} in \ce{CCl4}, room temperature	1,2-dibromoethane
Halohydration	\ce{Br2} and \ce{H2O} (HOBr)	bromohydrin, 2-bromoethanol
Epoxidation	peroxyacetic acid	epoxyethane
Ozonolysis	\ce{O3} at -78^\circ\mathrm{C} in \ce{CH2Cl2}, then Zn	methanal
Polymerization	400^\circ\mathrm{C}, 1000 atm, traces of \ce{O2}	poly(ethene)

16.6.1HydrogenationOften examined

The addition of hydrogen gas to alkenes to form the corresponding alkanes is called hydrogenation. The reaction takes place at high temperature in the presence of finely divided nickel or platinum as catalyst.

\ce{CH2=CH2 + H2 ->[Ni / Pt][heat] CH3-CH3}

ethene to ethane

16.6.2Hydrohalogenation of alkenesOften examined

In this reaction a hydrogen halide is added to the alkene across the double bond to give the corresponding halogenoalkane. When propene (\ce{CH3-CH=CH2}) is bubbled over a concentrated solution of hydrogen bromide (\ce{HBr}), a mixture of 1-bromopropane (minor product) and 2-bromopropane (major product) is obtained.

\ce{CH3-CH=CH2 + HBr -> CH3-CHBr-CH3 + CH3-CH2-CH2Br}

2-bromopropane (major)

1-bromopropane (minor)

Mechanism (electrophilic addition)

The positive hydrogen of hydrogen bromide acts as an electrophile and attacks the double bond (\pi bond) in propene. This can give two different carbocations: a primary carbocation or a secondary carbocation. The bromide ion then attacks the carbocation.

Mechanism of the addition of HBr to propene

The H–Br bond is polar, \ce{H^{\delta+}-Br^{\delta-}}. The \pi electrons of propene attack the \delta+ hydrogen; the H–Br bond breaks heterolytically and \ce{Br-} leaves.
\ce{CH3-CH=CH2 + H^{\delta+}-Br^{\delta-}}
Route (a): H adds to the end carbon \ce{CH2}, leaving the positive charge on the middle carbon. This is a secondary (2°) carbocation.
\ce{CH3-CH=CH2 + HBr -> CH3-CH^+-CH3 + Br-}
Route (b): H adds to the middle carbon, leaving the positive charge on the end carbon. This is a primary (1°) carbocation.
\ce{CH3-CH=CH2 + HBr -> CH3-CH2-CH2^+ + Br-}
The bromide ion, a nucleophile, attacks the positive carbon. The secondary carbocation gives 2-bromopropane (major).
\ce{CH3-CH^+-CH3 + Br- -> CH3-CHBr-CH3}
The primary carbocation gives 1-bromopropane (minor).
\ce{CH3-CH2-CH2^+ + Br- -> CH3-CH2-CH2-Br}

Markovnikov's rule

Markovnikov's rule: When an unsymmetrical reagent (such as H–Br) attacks an unsymmetrical alkene (such as propene), the hydrogen atom bonds to that double bond carbon which bears more hydrogen atoms. The positive part of the reagent attaches to the carbon with more hydrogens.

The rule favours the formation of the secondary carbocation over the primary carbocation, and so more 2-bromopropane than 1-bromopropane is formed. Because the reaction involves a carbocation, the rule is understood from the stability of carbocations: the secondary carbocation is more stable than the primary one and forms more readily, so the reaction follows the route of the more stable carbocation.

Stability of carbocations

Carbocations are species that lack electrons and have a positively charged carbon atom. There are three types, according to the number of alkyl groups (R) attached to the positive carbon: primary (one alkyl group), secondary (two) and tertiary (three). Their stability order is:

Figure 16.6: Stability order of carbocations: primary < secondary < tertiary. R is an alkyl group; more alkyl groups push more electron density towards the positive carbon.

Carbon is less electronegative than most atoms, so the positive charge on carbon attracts electrons from neighbouring atoms. Alkyl groups are electron-donating groups, because carbon has a lower electronegativity than hydrogen. When a carbocation is attached to alkyl groups, each alkyl group donates electrons through \sigma bonds to the positively charged carbon. This donation of electrons through \sigma bonds is called the inductive effect (here the positive inductive effect, +I). It decreases the positive charge density, which makes the carbocation more stable. The more alkyl groups there are, the more stable the carbocation.

Types and stability of carbocations
Carbocation	Alkyl groups on \ce{C+}	Example	Stability
Primary (1°)	one	\ce{CH3-CH2-CH2+}	lowest
Secondary (2°)	two	\ce{CH3-CH+-CH3}	medium
Tertiary (3°)	three	\ce{(CH3)3C+}	highest

Concept assessment exercise 16.3

What are the products of hydrochlorination of (i) 2-methylbut-2-ene and (ii) 3,3-dimethylbut-1-ene?

(i) 2-Methylbut-2-ene

2-methylbut-2-ene

The alkene is \ce{CH3-CH=C(CH3)2}. One double bond carbon (C-3) has one hydrogen and the other (C-2) has none.

The H of HCl adds to the carbon with more hydrogen atoms (Markovnikov). That gives a tertiary carbocation on C-2, which is more stable than the secondary carbocation on C-3.
\ce{(CH3)2C=CH-CH3 + HCl -> (CH3)2C^+-CH2-CH3 + Cl-}
The chloride ion attacks the tertiary carbocation.
\ce{(CH3)2C^+-CH2-CH3 + Cl- -> (CH3)2CCl-CH2-CH3}

2-chloro-2-methylbutane (major product)

The minor product, from the less stable secondary carbocation, is 2-chloro-3-methylbutane, \ce{(CH3)2CH-CHCl-CH3}.

(ii) 3,3-Dimethylbut-1-ene

3,3-dimethylbut-1-ene

The alkene is \ce{CH2=CH-C(CH3)3}. C-1 has two hydrogens and C-2 has one, so by Markovnikov\'s rule H adds to C-1 and the positive charge appears on C-2, giving a secondary carbocation (the other route would give a less stable primary one).

Addition of \ce{H+} from HCl to C-1 gives the secondary carbocation.
\ce{CH2=CH-C(CH3)3 + HCl -> CH3-CH^+-C(CH3)3 + Cl-}
The chloride ion attacks the carbocation.
\ce{CH3-CH^+-C(CH3)3 + Cl- -> CH3-CHCl-C(CH3)3}

2-chloro-3,3-dimethylbutane

\text{(i) 2-chloro-2-methylbutane (major);}\quad \text{(ii) 2-chloro-3,3-dimethylbutane}

Worked example: Markovnikov addition to 2-methylpropene

Predict the major product when 2-methylpropene, \ce{(CH3)2C=CH2}, reacts with hydrogen bromide. Explain your choice with the stability of carbocations.

Look at the two doubly bonded carbons. \ce{=CH2} carries two hydrogens; the other carbon, \ce{(CH3)2C=}, carries none.
The positive hydrogen of \ce{HBr} is the electrophile. By Markovnikov's rule it adds to the carbon with more hydrogens, \ce{=CH2}. The other carbon becomes the carbocation.
\ce{(CH3)2C=CH2 + H+ -> (CH3)3C+}
If \ce{H+} went to the other carbon, a primary carbocation \ce{(CH3)2CH-CH2+} would form. The carbocation we get has three alkyl groups on the positive carbon (tertiary), so it is far more stable, because alkyl groups push electrons towards the positive carbon.
The bromide ion attacks the positive carbon and completes the addition.
\ce{(CH3)3C+ + Br- -> (CH3)3C-Br}

2-bromo-2-methylpropane (major)

1-bromo-2-methylpropane (minor)

\begin{aligned}&\text{major product: 2-bromo-2-methylpropane,}\\ &\ce{(CH3)3C-Br}\end{aligned}

16.6.3Hydration of alkenesOften examined

Hydration of alkenes is the addition of water across the double bond between carbon atoms. If ethene reacts with water in the presence of a mineral acid, ethanol is obtained. This is the major method of preparation of ethanol on an industrial scale. The hydration of ethene takes place in two steps.

Hydration of ethene in two steps

Ethene is treated with cold concentrated sulphuric acid. The \pi electrons attack the positive hydrogen of the acid, and ethyl hydrogen sulphate is formed.
\ce{CH2=CH2 + H-O-SO2-OH -> CH3-CH2-O-SO2-OH}
The mixture is heated (boiled) with water. Ethyl hydrogen sulphate decomposes (hydrolyses) to ethanol and sulphuric acid is regenerated.
\ce{CH3-CH2-O-SO2-OH + H2O ->[boil] CH3-CH2-OH + H2SO4}

\ce{CH2=CH2 + H2O ->[H+] CH3-CH2-OH}

overall reaction

ethyl hydrogen sulphate

ethanol

16.6.4Halogenation of alkenesOften examined

Halogenation is the addition of a halogen to an alkene. At room temperature ethene reacts with bromine to give 1,2-dibromoethane. An inert solvent such as tetrachloromethane (\ce{CCl4}) is used.

\ce{CH2=CH2 + Br2 ->[CCl4] CH2Br-CH2Br}

ethene to 1,2-dibromoethane

1,2-dibromoethane

The brown colour of bromine (or bromine water) is discharged very quickly in this reaction, so bromine water can be used as a test for unsaturation in compounds.

Bromine water test. With an alkane the shaded (brown) liquid is unchanged; with an alkene the colour is discharged at once and the liquid is colourless.

Mechanism

When non-polar bromine comes close to an ethene molecule, the repulsion between the electron cloud of bromine and the \pi electrons of ethene polarizes the halogen molecule. The positive end of the bromine molecule acts as an electrophile and attacks the \pi electron cloud, producing a cyclic bromonium ion. In the second step the bromide ion acts as a nucleophile and attacks the bromonium ion to give 1,2-dibromoethane.

Mechanism of the bromination of ethene

Bromine is polarized as the molecule approaches the \pi cloud: \ce{Br^{\delta+}-Br^{\delta-}}. The \pi electrons attack \ce{Br^{\delta+}} and the Br–Br bond breaks heterolytically.
\ce{CH2=CH2 + Br^{\delta+}-Br^{\delta-} -> [CH2-CH2]Br+ + Br-}
A cyclic bromonium ion (a three-membered ring with a positively charged Br) forms, together with a bromide ion.
\ce{CH2=CH2 + Br2 -> \text{cyclic bromonium ion} + Br-}
The bromide ion attacks one of the ring carbons from the side opposite to the bromine bridge, opening the ring.
\ce{\text{bromonium ion} + Br- -> CH2Br-CH2Br}

cyclic bromonium ion

Figure 16.7: Mechanism of the bromination of ethene. Step 1 gives the cyclic bromonium ion; in step 2 the bromide ion opens the ring to give 1,2-dibromoethane.

16.6.5Halohydration of alkenes

Halohydration is the addition of hypohalous acid (HOX) to an alkene across the double bond. The hypohalous acid is prepared when a halogen reacts with water. In this reaction the solvent (water) acts as a reactant.

\ce{Br2 + H2O <=> HOBr + HBr}

When ethene reacts with hypobromous acid (\ce{HOBr}), a halohydrin called a bromohydrin is produced. It has a Br atom on one carbon and an OH group on the next carbon.

\ce{CH2=CH2 + Br2 + H2O -> CH2Br-CH2OH + HBr}

ethene to bromohydrin (2-bromoethanol)

bromohydrin (2-bromoethanol)

16.6.6Epoxidation of alkenes

In epoxidation an alkene reacts with a peracid, such as peroxyacetic acid, to form the corresponding epoxide. Ethene reacts with peroxyacetic acid to form epoxyethane (an epoxide, a three-membered ring with one oxygen) and acetic acid.

\ce{CH2=CH2 + CH3-CO-O-OH -> C2H4O + CH3COOH}

peroxyacetic acid

epoxyethane

Equation in words: ethene + peroxyacetic acid \longrightarrow epoxyethane + acetic acid. The peracid transfers one oxygen atom to the double bond, and the peracid becomes the carboxylic acid.

16.6.7Ozonolysis of alkenesOften examined

In ozonolysis, ozone is bubbled into a very cold (-78^\circ\mathrm{C}) solution of the alkene in dichloromethane (\ce{CH2Cl2}). It cleaves the double bond and forms aldehydes and ketones. Oxygen attacks the double bond to form a molozonide (an unstable intermediate), which rearranges rapidly to the ozonide (relatively stable).

Ozonolysis of ethene

Ozone adds to the double bond and forms the unstable five-membered molozonide.
\ce{CH2=CH2 + O3 -> \text{molozonide (unstable)}}
The molozonide rearranges rapidly into the ozonide.
\ce{\text{molozonide} -> \text{ozonide}}
Ozonides cannot easily be isolated (they are not stable compounds). The ozonide is treated with a mild reducing agent such as zinc and water, which breaks it into carbonyl compounds. From ethene it gives methanal and hydrogen peroxide.
\ce{C2H4O3 + H2O ->[Zn][heat] 2HCHO + H2O2}
The zinc also reduces the hydrogen peroxide so that it cannot oxidize the aldehyde.
\ce{H2O2 + Zn -> ZnO + H2O}

molozonide (unstable)

ozonide

methanal

Finding the position of the double bond

Ozonolysis is used to determine the position of the double bond in alkenes by reasoning back from the structures of the products. Each carbon of the double bond becomes a \ce{C=O} group: a carbon with one or two hydrogens gives an aldehyde, and a carbon with no hydrogen gives a ketone.

\ce{R^1R^2C=CHR^3 ->[1.\ O3][2.\ Zn / H2O] R^1R^2C=O + O=CHR^3}

Ozonolysis products (the last two rows are worked in the exercise below)
Alkene	Products of ozonolysis
ethene \ce{CH2=CH2}	2 methanal
propene \ce{CH3-CH=CH2}	ethanal + methanal
but-2-ene \ce{CH3-CH=CH-CH3}	2 ethanal
2-methylpropene \ce{(CH3)2C=CH2}	propanone + methanal

Worked example: finding the alkene from its ozonolysis products

Ozonolysis of an alkene, followed by treatment with \ce{Zn/H2O}, gave only propanone and ethanal. Find the structure of the alkene.

Working backwards: take away the oxygen atoms of the two carbonyl compounds and join their carbons by a double bond (accent).

Each carbonyl carbon (\ce{C=O}) in a product was one of the doubly bonded carbons of the alkene. The oxygen replaced the other half of the double bond.
Write the carbonyl compounds side by side.
\ce{(CH3)2C=O}\quad\text{and}\quad\ce{CH3CH=O}
Remove the two oxygen atoms and join the two carbonyl carbons by a double bond.
\ce{(CH3)2C=CH-CH3}
Check by cutting the double bond again. \ce{(CH3)2C=} becomes propanone and \ce{=CH-CH3} becomes ethanal, so the answer is consistent with the data.

propanone

ethanal

2-methylbut-2-ene

\text{2-methylbut-2-ene, }\ce{(CH3)2C=CH-CH3}

Concept assessment exercise 16.4

(a) What are the products of ozonolysis of (i) propene and (ii) but-2-ene? (b) Name the alkenes which give the following compounds on their ozonolysis: (i) ethanal only; (ii) a mixture of methanal and propanone.

(a) Products of ozonolysis

Break the \ce{C=C} bond and put an oxygen atom (=\ce{O}) on each carbon that was in the double bond.

(i) Propene, \ce{CH3-CH=CH2}: the \ce{CH3-CH} end becomes ethanal and the \ce{CH2} end becomes methanal.
\ce{CH3-CH=CH2 ->[1.\ O3][2.\ Zn / H2O] CH3-CHO + HCHO}
(ii) But-2-ene, \ce{CH3-CH=CH-CH3}: each half is \ce{CH3-CH}, so two molecules of ethanal are formed.
\ce{CH3-CH=CH-CH3 ->[1.\ O3][2.\ Zn / H2O] 2CH3-CHO}

(b) Alkenes from the products

Join the carbons of the carbonyl groups with a double bond, removing the oxygen atoms.

(i) Ethanal only: both halves are \ce{CH3-CH=O}, so the alkene is symmetrical. Join the two carbonyl carbons.
\ce{CH3-CH=O + O=CH-CH3 -> CH3-CH=CH-CH3}\quad(\text{but-2-ene})
(ii) Methanal and propanone: join \ce{H2C=O} and \ce{O=C(CH3)2}.
\ce{H2C=O + O=C(CH3)2 -> H2C=C(CH3)2}\quad(\text{2-methylpropene})

but-2-ene

2-methylpropene

\text{(a) (i) ethanal + methanal; (ii) 2 ethanal.}\quad \text{(b) (i) but-2-ene; (ii) 2-methylpropene}

16.6.8Polymerization

In polymerization, small molecules called monomers react together to form large molecules called polymers. Ethene molecules can polymerize at 400^\circ\mathrm{C} and 1000 atm pressure in the presence of traces of oxygen (about 0.1\%) to form the polymer poly(ethene) (polyethylene).

\ce{n CH2=CH2 ->[400^\circ C][1000 atm, O2 (0.1\%)] [-CH2-CH2-]_n}

n is about 100 000

Monomer: A small molecule (here ethene) that joins with many others to form a polymer.
Polymer: A large molecule made by joining many monomer molecules; here poly(ethene), with n \approx 100\,000.

Ziegler–Natta catalyst

Ethene gives a harder, higher-melting poly(ethene) when a mixture of triethylaluminium, \ce{Al(C2H5)3}, and titanium tetrachloride, \ce{TiCl4}, is used as catalyst. This is the Ziegler–Natta catalyst.

16.7Conjugation in alkenes

The word conjugation comes from Latin and means "join together". In a conjugated molecule the unhybridized p orbitals on adjacent carbon atoms of the parent chain combine together, making a bridge along which the \pi electrons can move.

Conjugation is found in alkenes that have alternate double and single bonds. In such molecules the carbons are sp^2 hybridized and each has one unhybridized p orbital on which the \pi electrons move. The bridge of unhybridized p orbitals extends to the last double bond carbon in the molecule. The double bonds in conjugated molecules are called delocalized, because their \pi electrons move along the whole bridge of p orbitals.

hexa-1,3,5-triene a conjugated triene

Figure 16.8: Conjugation in hexa-1,3,5-triene. Every carbon is sp² hybridized and has an unhybridized p orbital; these overlap in a continuous bridge so the π electrons are delocalized over all six carbons.

Recognizing conjugation
Type of diene	Position of the double bonds	Conjugated?
hexa-1,3,5-triene, penta-1,3-diene	alternate double and single bonds	yes: p orbitals overlap along the chain
penta-1,2-diene, buta-1,2-diene	two double bonds share one carbon (cumulated)	no: there is no alternate single bond between them
but-1-ene, ethene	only one double bond	no: there is nothing to conjugate with

Concept assessment exercise 16.5

Explain which one of the following compounds shows conjugation: (i) ethene, (ii) but-1-ene, (iii) penta-1,2-diene, (iv) buta-1,2-diene, (v) penta-1,3-diene.

A compound shows conjugation only if it has alternate double and single bonds: double bond, single bond, double bond. Write the structure of each one.

Compound	Structure	Double bonds	Conjugated?
(i) ethene	\ce{CH2=CH2}	one only	no
(ii) but-1-ene	\ce{CH2=CH-CH2-CH3}	one only	no
(iii) penta-1,2-diene	\ce{CH2=C=CH-CH2-CH3}	two double bonds on the same carbon (C-2), no single bond between them	no
(iv) buta-1,2-diene	\ce{CH2=C=CH-CH3}	two double bonds on the same carbon (C-2)	no
(v) penta-1,3-diene	\ce{CH2=CH-CH=CH-CH3}	C-1=C-2, a single bond C-2–C-3, then C-3=C-4	yes

penta-1,3-diene

In penta-1,3-diene all four double bond carbons are sp^2 hybridized and their p orbitals form one continuous bridge over C-1 to C-4, so the \pi electrons are delocalized.

\text{Only (v) penta-1,3-diene shows conjugation}

16.8Redox reactions of organic compounds

The word "redox" refers to reduction and oxidation reactions. In organic chemistry redox reactions are very common. Oxidation and reduction take place simultaneously, which is why the single word "redox" is used for them.

1. Oxidation reaction

Oxidation of an organic compound is defined in four ways:

addition of oxygen;
loss of hydrogen;
increase in the number of bonds of carbon with oxygen;
bond making of carbon with an element of higher electronegativity.

Examples:

\ce{CH4 + 1/2 O2 -> CH3-OH}

methane to methanol: addition of oxygen

\ce{CH3-OH + [O] ->[K2Cr2O7][H+] H-CHO + H2O}

methanol to methanal: loss of hydrogen, more C–O bonds (the C=O bond is added)

\ce{CH4 + Cl2 ->[h\nu] CH3Cl + HCl}

methane to chloromethane: C–Cl bond with an element of higher electronegativity

2. Reduction reaction

Reduction of an organic compound is defined in four ways:

addition of hydrogen;
loss of oxygen;
decrease in the number of carbon bonds with oxygen;
bond making of carbon with an element of lower electronegativity value.

Examples:

\ce{H-CHO + 2[H] ->[LiAlH4] CH3-OH}

methanal to methanol: addition of hydrogen

\ce{CH3-CO-NH2 + 4[H] ->[LiAlH4] CH3-CH2-NH2 + H2O}

ethanamide to ethanamine: the C=O bond is lost

Oxidation and reduction of organic compounds
	Oxidation	Reduction
Oxygen	addition of oxygen	loss of oxygen
Hydrogen	loss of hydrogen	addition of hydrogen
Bonds of carbon with oxygen	increase in number	decrease in number
Bond with another element	bond making with a more electronegative element	bond making with a less electronegative element
Example reagent	\ce{K2Cr2O7} / \ce{H+}, [\ce{O}]	\ce{LiAlH4}, [\ce{H}]

16.9StereoisomerismOften examined

Molecules that have the same molecular formula but different structural formulae with respect to the arrangement of atoms in space are stereoisomers, and the phenomenon is called stereoisomerism. It is also called configurational isomerism, because it deals with the configuration (arrangement of atoms in space). There are two types of stereoisomerism:

optical isomerism;
geometric (cis–trans) isomerism.

Stereoisomerism: Existence of molecules with the same molecular formula and the same order of attachment of atoms, but a different arrangement of the atoms in space.

The two types of stereoisomerism (explained in the next two sections)
	Optical isomerism	Geometric isomerism
Cause	a chiral (asymmetric) carbon bonded to four different groups	restricted rotation about a C=C bond or a ring, with different groups on each carbon
Isomers	enantiomers and diastereomers	cis and trans isomers
How they differ	rotate plane-polarized light in opposite directions	same groups on the same side (cis) or opposite sides (trans)

16.9.1Optical isomerismOften examined

Optical isomerism is shown by molecules having a carbon atom bonded to four different atoms or groups of atoms. Such a carbon is called a chiral carbon (chiral centre). A molecule having a chiral carbon is said to have chirality. A chiral carbon is also called an asymmetric carbon, because it has no symmetry.

Optical isomers have the same molecular formula but different structural formulae with respect to the arrangement of atoms or groups in space. They can be distinguished by their different effect on the plane of polarized light passed through them.

Plane-polarized light

Ordinary light has electric and magnetic fields perpendicular to each other, vibrating in all directions, and is unpolarized. When unpolarized light is passed through a polariser it vibrates in only one plane and is called polarized light.

Figure 16.9: A polariser lets through only the light that vibrates in one plane; the transmitted beam is plane-polarized light.

Enantiomers

A pair of optical isomers rotates the plane of polarized light, one clockwise and the other anticlockwise, by equal amounts. Such isomers are optically active and this isomerism is called optical isomerism. Optical isomers are also called enantiomers; they cannot be superimposed, like the left and right hands. Enantiomers are mirror images of each other, just as the right hand is the mirror image of the left hand. Lactic acid shows optical isomerism because its molecule has one chiral carbon.

Figure 16.10: The two enantiomers of lactic acid. The central carbon carries four different groups (CH₃, OH, H and CO₂H), so the molecule and its mirror image cannot be superimposed.

Worked example: spotting the chiral carbon: butan-1-ol and butan-2-ol

Which of butan-1-ol and butan-2-ol shows optical isomerism? Draw the optical isomers of the one that does.

butan-1-ol (no chiral carbon)

butan-2-ol (C-2 is chiral)

The two enantiomers of butan-2-ol. The chiral carbon (accent) carries OH, C₂H₅, CH₃ and H; the second drawing is the mirror image of the first and cannot be superimposed on it.

A molecule is optically active when a carbon is bonded to four different atoms or groups (a chiral carbon). Check each carbon in turn.
Butan-1-ol, \ce{CH3CH2CH2CH2OH}: the carbon with \ce{OH} carries two hydrogens, and each of the other carbons carries two hydrogens or three hydrogens. No carbon has four different groups, so there is no chiral carbon.
Butan-2-ol, \ce{CH3CH(OH)CH2CH3}: carbon-2 is bonded to \ce{H}, \ce{OH}, \ce{CH3} and \ce{CH2CH3}. These are four different groups, so carbon-2 is chiral.
A molecule with one chiral carbon exists as two mirror images that cannot be superimposed. These enantiomers rotate the plane of polarized light by equal amounts in opposite directions.
Draw one tetrahedral carbon with two bonds in the plane of the paper, one wedge (towards you) and one hashed bond (away from you). Swapping the wedge and hash positions gives the mirror image.

\text{butan-2-ol is optically active (chiral carbon C-2); butan-1-ol is not}

Two chiral carbons

If a molecule has two chiral carbon atoms, it shows four optical isomers, because each chiral centre can rotate the plane of polarized light. The book illustrates this with tartaric acid (\ce{HOOC-CH(OH)-CH(OH)-COOH}):

Figure 16.11: Stereoisomers of tartaric acid (2,3-dihydroxybutanedioic acid) in Fischer projection. I and II are mirror images (enantiomers); III has an internal plane of symmetry (dashed line) and is the meso form.

Isomers I and II are optically active because they are mirror images (I is dextrorotatory, (+), and II is levorotatory, (−)).
Isomer III has a plane of symmetry. It is internally compensated (the rotation of one half cancels the other half), so it is the meso form and is optically inactive.
IV is a racemic mixture, written (±): an equimolar mixture of the (+) and (−) forms (50% dextro and 50% levo). The rotations cancel, so a racemic mixture is overall optically inactive.

Diastereomers

Diastereomers are stereoisomers which are not mirror images of one another. Consider 2,3-dichloropentane, \ce{CH3-CHCl-CHCl-CH2-CH3}, which has two chiral carbons (C-2 and C-3).

Figure 16.12: The four stereoisomers of 2,3-dichloropentane in Fischer projection. I and II are enantiomers and III and IV are enantiomers; any structure of the first pair is a diastereomer of any structure of the second pair.

Structures I and II are mirror images (enantiomers); structures III and IV are mirror images (enantiomers). Structures III and IV are not mirror images of I or II, so III and IV are diastereomers of I and II.

Enantiomers and diastereomers
	Enantiomers	Diastereomers
Mirror images?	yes, non-superimposable	no
Example in 2,3-dichloropentane	I and II; III and IV	I and III; I and IV; II and III; II and IV
Effect on polarized light	equal rotation in opposite directions	not related by equal and opposite rotation

Optical activity in cyclic compounds

Substituted cyclic compounds can also be optically active, because they can have asymmetric carbon atoms. The book takes the example of 2-aminocyclohexanol. The four different groups attached to carbon 1 (the carbon bearing OH) are \ce{H}, \ce{OH}, and the two different ring paths (\ce{-(CH2)4-CH-NH2} and \ce{-CH(NH2)-(CH2)4-} seen from carbon 1).

2-aminocyclohexanol

Measuring optical activity

The optical activity of a substance is measured by an instrument called a polarimeter. A solution of known concentration of the substance is placed in the polarimeter and the degree of rotation of the polarized light (clockwise or anticlockwise) is determined. In this way the optical activity of an optically active compound is measured.

16.9.2Geometric isomerismOften examined

Geometric isomerism, also known as cis–trans isomerism, is shown by unsaturated compounds or by substituted cyclic compounds having the same molecular formula but a different arrangement of atoms in space. It arises because of restricted rotation of atoms or groups around the double bond of the chain or around the ring.

A molecule can show geometric isomerism if it has different atoms or groups bonded to each carbon of the C=C double bond (in unsaturated compounds) or to the ring carbons, with the C–C single bonds of the ring (in substituted cyclic compounds).

cis isomer: The isomer in which the same groups are on the same side of the double bond or ring (cis is Latin for "on the same side").
trans isomer: The isomer in which the same groups are on opposite sides of the double bond or ring carbons (trans is Latin for "opposite side").

Figure 16.13: Geometric isomers of but-2-ene. In the cis isomer the two CH₃ groups are on the same side of the double bond; in the trans isomer they are on opposite sides.

Figure 16.14: Geometric isomers of 1,3-dimethylcyclopentane. The CH₃ groups are on the same face of the ring in the cis isomer and on opposite faces in the trans isomer.

Figure 16.15: Geometric isomers of 1,2-dimethylcyclopropane. In the cis isomer both CH₃ groups point the same way (both wedges); in the trans isomer one points up (wedge) and the other down (hashed).

When is there no geometric isomerism?

Propene, but-1-ene and pent-1-ene do not show it, because one double bond carbon (\ce{=CH2}) has two identical atoms (two hydrogen atoms) bonded to it.
1,1-Dimethylcyclopropane does not show it, because both identical groups (the two methyl groups) are on the same carbon atom.
In cyclic compounds the groups considered must be on different ring carbons, for example 1 and 2, or 1 and 3.

Which compounds show cis–trans isomerism
Compound	Groups on the carbons	Geometric isomers?
but-2-ene	C-2: \ce{H}, \ce{CH3}; C-3: \ce{H}, \ce{CH3}	yes (cis and trans)
propene	\ce{=CH2} carbon: \ce{H}, \ce{H}	no
1,2-dimethylcyclopropane	different carbons each carry \ce{H} and \ce{CH3}	yes
1,3-dimethylcyclopentane	different carbons each carry \ce{H} and \ce{CH3}	yes
1,1-dimethylcyclopropane	both \ce{CH3} on the same carbon	no

Worked example: which alkenes show cis-trans isomerism

Which of pent-1-ene, pent-2-ene and 2-methylbut-2-ene show geometric (cis-trans) isomerism? Draw the isomers of the one that does.

pent-1-ene (no)

pent-2-ene (yes)

2-methylbut-2-ene (no)

Geometric isomers of pent-2-ene. The groups that decide the name (CH₃ and C₂H₅, accent) are on the same side in the cis isomer and on opposite sides in the trans isomer.

Geometric isomerism needs restricted rotation (the \ce{C=C} bond) and two different groups on each doubly bonded carbon.
Pent-1-ene, \ce{CH2=CH-CH2CH2CH3}: the carbon \ce{=CH2} carries two identical hydrogens, so there is no geometric isomerism.
2-Methylbut-2-ene, \ce{(CH3)2C=CH-CH3}: carbon-2 carries two identical \ce{CH3} groups, so there is no geometric isomerism.
Pent-2-ene, \ce{CH3-CH=CH-CH2CH3}: carbon-2 carries \ce{H} and \ce{CH3}; carbon-3 carries \ce{H} and \ce{C2H5}. Both carbons have two different groups, so cis and trans isomers exist.
Draw the double bond with the two carbons side by side. In the cis isomer the \ce{CH3} and \ce{C2H5} groups are on the same side; in the trans isomer they are on opposite sides.

\text{only pent-2-ene shows cis-trans isomerism}

Geometric isomers can be interconverted by heating, which breaks the \pi bond so that rotation about the bond becomes possible; the isomer is then changed to the other one (the desired one) followed by cooling.

Concept assessment exercise 16.6

Which one of the following shows geometric isomerism and why? (i) but-1-ene, (ii) pent-1-ene, (iii) hex-3-ene, (iv) methylcyclobutane, (v) 1,3-dimethylcyclohexane.

Test each compound: both carbons of the \ce{C=C} (or two different ring carbons) must each carry two different groups.

Compound	Structure	Reason	Geometric isomerism?
(i) but-1-ene	\ce{CH2=CH-CH2-CH3}	C-1 has two hydrogens (identical)	no
(ii) pent-1-ene	\ce{CH2=CH-CH2-CH2-CH3}	C-1 has two hydrogens (identical)	no
(iii) hex-3-ene	\ce{CH3CH2-CH=CH-CH2CH3}	each double bond carbon has \ce{H} and \ce{C2H5}	yes
(iv) methylcyclobutane	one methyl group on a ring	only one substituent, so no second ring carbon to compare	no
(v) 1,3-dimethylcyclohexane	methyl groups on C-1 and C-3	two different ring carbons, each with \ce{H} and \ce{CH3}	yes

hex-3-ene

\text{(iii) hex-3-ene and (v) 1,3-dimethylcyclohexane}

Key terms

Hydrocarbon: A compound containing hydrogen and carbon only.
Aliphatic hydrocarbon: An open chain hydrocarbon, or a closed chain one whose properties resemble the open chain ones (Greek: fatty).
Aromatic hydrocarbon: Benzene and compounds containing a benzene ring (Greek: fragrant).
Saturated hydrocarbon: A hydrocarbon with only single bonds between carbons; every carbon is bonded to four atoms (alkanes).
Unsaturated hydrocarbon: A hydrocarbon with at least one double or triple bond, to which more atoms can be added (alkenes, alkynes).
Alicyclic hydrocarbon: An aliphatic hydrocarbon with a ring of carbon atoms, with two hydrogens fewer than the matching open chain hydrocarbon.
Paraffins: Another name for alkanes: Latin parum (little) and affinis (reactivity), because they are unreactive.
Sigma (\sigma) bond: A covalent bond formed by head-on (end-on) overlap of orbitals; strong and difficult to break.
Pi (\pi) bond: A covalent bond formed by sideways (parallel) overlap of unhybridized p orbitals; weaker than a \sigma bond and more reactive.
Homolytic fission: Breaking of a covalent bond so that each atom keeps one electron, giving free radicals.
Heterolytic fission: Breaking of a covalent bond so that both bonding electrons go to one atom, giving ions.
Free radical: An atom or group with an unpaired electron, very reactive, for example \ce{Cl.}.
Free radical substitution: Reaction of an alkane with a halogen in ultraviolet light by a chain mechanism of initiation, propagation and termination.
Propagation: The chain step in which each radical used makes a new radical, so the reaction goes on.
Termination: The end of the chain, when two free radicals combine.
Elimination reaction: A reaction in which a small molecule such as \ce{H2O} or \ce{HX} is removed from a molecule, forming a double bond.
Dehydration: Removal of water; ethanol vapour over hot \ce{Al2O3} (or conc. \ce{H2SO4}) gives ethene.
Dehydrohalogenation: Removal of hydrogen and halogen as a hydrogen halide; bromoethane with hot ethanolic NaOH gives ethene.
Addition reaction: A reaction in which a molecule is added across a double bond, which becomes a single bond.
Hydrogenation: Addition of hydrogen to an alkene with a Ni or Pt catalyst to give an alkane.
Hydrohalogenation: Addition of a hydrogen halide to an alkene to give a halogenoalkane.
Electrophile: A species that is attracted to electron-rich regions and attacks the \pi bond, for example the positive hydrogen of H–Br.
Carbocation: A species with a positively charged carbon atom; primary, secondary or tertiary depending on the number of alkyl groups on it.
Inductive effect: Donation of electrons through \sigma bonds; alkyl groups donate electrons and stabilize a carbocation.
Markovnikov's rule: In the addition of an unsymmetrical reagent to an unsymmetrical alkene, the hydrogen goes to the double bond carbon that carries more hydrogen atoms.
Hydration: Addition of water to an alkene in the presence of acid (acid-catalyzed) to form an alcohol.
Halogenation: Addition of a halogen to an alkene, for example \ce{Br2} to ethene giving 1,2-dibromoethane; the brown colour is discharged.
Bromonium ion: The cyclic three-membered ion with a positively charged bromine formed in the first step of bromination of an alkene.
Halohydration: Addition of hypohalous acid (HOX) to an alkene giving a halohydrin.
Epoxidation: Reaction of an alkene with a peracid (peroxyacetic acid) to give an epoxide.
Ozonolysis: Cleavage of the double bond of an alkene by ozone to give aldehydes and ketones; used to locate the double bond.
Polymerization: Joining of many monomer molecules to form a polymer, for example ethene to poly(ethene).
Ziegler–Natta catalyst: \ce{Al(C2H5)3} with \ce{TiCl4}; gives harder, high-melting, unbranched poly(ethene).
Conjugation: The bridge of overlapping p orbitals in molecules with alternate double and single bonds, along which \pi electrons are delocalized.
Oxidation (organic): Addition of oxygen, loss of hydrogen, more bonds of carbon with oxygen, or a bond to a more electronegative element.
Reduction (organic): Addition of hydrogen, loss of oxygen, fewer bonds of carbon with oxygen, or a bond to a less electronegative element.
Stereoisomerism: Isomerism in which molecules have the same molecular formula but a different arrangement of atoms in space.
Chiral (asymmetric) carbon: A carbon bonded to four different atoms or groups.
Optical isomerism: Stereoisomerism of molecules with chiral carbon atoms; the isomers rotate plane-polarized light.
Enantiomers: Optical isomers that are non-superimposable mirror images of each other.
Diastereomers: Stereoisomers that are not mirror images of one another.
Racemic mixture: An equimolar mixture of the (+) and (−) forms; optically inactive.
Meso compound: A compound with chiral carbons but an internal plane of symmetry; optically inactive.
Polarimeter: Instrument that measures the rotation of plane-polarized light by an optically active substance.
Geometric (cis–trans) isomerism: Stereoisomerism due to restricted rotation about a C=C bond or ring, with different groups on each carbon involved.
cis / trans isomer: cis: the same groups on the same side; trans: the same groups on opposite sides.

Multiple choice questionsp. 336–337

The most likely reaction alkanes undergo is

Aelectrophilic addition
Bfree radical substitutionAnswer
Cnucleophilic addition
Delectrophilic substitution

Why

Alkanes have only strong sigma bonds and non-polar C-H bonds, and they have no pi bond, so addition reactions (A, C) are impossible. Under ultraviolet light they undergo free radical substitution, in which a hydrogen atom is replaced by a halogen, as in the chlorination of ethane. Electrophilic substitution (D) is not a reaction that alkanes undergo under normal conditions.

Note. The book prints no key for this question; the answer was worked out from the book's text on alkanes.

The condensed structural formula of the major product obtained by hydrobromination of 2-methylpent-2-ene is

A\ce{CH3CH(Br)CH(CH3)CH2CH3}
B\ce{(CH3)2C(Br)(CH2)2CH3}Answer
C\ce{(CH3)2CHCH(Br)CH2CH3}
D\ce{CH2(Br)CH(CH3)(CH2)2CH3}

Why

2-methylpent-2-ene is \ce{(CH3)2C=CH-CH2CH3}. In hydrobromination, the hydrogen of HBr adds to the carbon of the double bond that already carries more hydrogen atoms (Markovnikov's rule), which is C-3. The bromine therefore goes to C-2, and the carbocation formed at C-2 is tertiary, the most stable one available.

The product is \ce{(CH3)2C(Br)CH2CH2CH3}, 2-bromo-2-methylpentane, which is option B. Option A and option D put the bromine on a secondary carbon, and option C puts it on C-3, a secondary carbon that would need a less stable carbocation.

Note. The book prints no key for this question; the answer was worked out with Markovnikov's rule.

iii

Which one of the following is a possible propagation step in chlorination of ethane?

A\ce{C2H6 + Cl. -> C2H5Cl + H.}
B\ce{C2H5Cl + Cl. -> C2H4Cl. + HCl}Answer
C\ce{C2H5. + Cl. -> C2H5Cl}
D\ce{C2H4Cl2 + Cl. -> C2H4Cl. + Cl2}

Why

A propagation step is one in which a free radical attacks a molecule, takes a hydrogen atom from it to form HCl, and leaves a new free radical, so that the chain continues. Option B does exactly this: a chlorine radical removes a hydrogen atom from chloroethane to give HCl and the chloroethyl radical \ce{C2H4Cl.}.

Option C is termination, because two radicals combine to form a stable molecule and the chain stops. Option A produces a free hydrogen atom instead of HCl, which is not a radical that continues the chain. Option D removes a chlorine atom from the molecule rather than a hydrogen atom, which is not the propagation pattern of the book.

Note. The book prints no key for this question; the answer was worked out from the propagation steps of the book's ethane mechanism.

Which one of the following is the most stable (least reactive) hydrocarbon?

Ahex-1-ene
Bpropene
Cpenta-1,2-diene
Dhepta-1,3,5-trieneAnswer

Why

Hepta-1,3,5-triene has three double bonds in conjugation (alternate double and single bonds). In a conjugated system the pi electrons are delocalized along the whole chain, which lowers the energy of the molecule and gives it extra stability, as the book states for conjugated alkenes.

Hex-1-ene and propene each have one isolated double bond, which is more reactive. Penta-1,2-diene has two adjacent double bonds sharing one carbon, which is not a conjugated system and is less stable than a conjugated one.

Note. The book prints no key for this question; the answer rests on the book's statement that conjugation gives stability.

The most striking factor held responsible for the less reactivity of alkanes is

AI & II only
BII & III only
CI, II & III onlyAnswer
DIII & IV only

Why

The book attributes the low reactivity of alkanes to two factors: I, their non-polar bonds, and III, their strong sigma bonds. Option C is the only choice that contains both of these factors.

Option C also contains II, the absence of a functional group, which the book does not give as a reason, so C is the best available answer rather than an exact match. Factor IV is wrong, because a long C-C bond would be weaker, not stronger, and it is not a reason for low reactivity.

Note. Ambiguous as printed: the book names only two factors (non-polar bonds and strong sigma bonds, I and III), and no option lists exactly those two. Option C was chosen because it is the only one containing both. The book prints no key.

The systematic name of the following compound is

The compound in question

A2-ethylmethylcyclohexane
B2-methyethylcyclohexane
C1-ethylmethylcyclohexane
D1-ethyl-2-methylcyclohexaneAnswer

Why

The drawn compound has an ethyl group (two carbons) on one ring carbon and a methyl group on the adjacent ring carbon. The ring carbon with the ethyl group is numbered 1, the substituents are named in alphabetical order (ethyl before methyl) and the locants are as low as possible: 1-ethyl-2-methylcyclohexane, option D.

Options A and C name the two groups as one substituent, which is not a valid name. Option B gives the locant 2 for a substituent that is not on the lower-numbered carbon and uses the invalid name methyethyl.

Note. The book prints no key; the answer follows the IUPAC rules that the book's nomenclature section uses.

vii

Dehydration of propan-1-ol gives

ApropeneAnswer
Bpropane
Cpropanone
Dpropanal

Why

Dehydration is an elimination reaction in which water is removed from an alcohol. The OH group on C-1 and an H atom on C-2 are lost, and a double bond forms between C-1 and C-2: \ce{CH3CH2CH2OH ->[Al2O3][heat] CH3CH=CH2 + H2O}. The product is propene, option A.

Propane (B) would need reduction, and propanone and propanal (C, D) are formed by oxidation of propan-2-ol and propan-1-ol respectively, not by dehydration.

Note. The book prints no key for this question; the answer follows the book's dehydration of alcohols method.

viii

Which one of the following compounds can show geometric isomerism?

A2,2-dimethylbut-2-ene
B2-methylhex-1-ene
Chex-3-eneAnswer
D2-methylpropene

Why

Geometric (cis-trans) isomerism needs a C=C double bond in which each carbon carries two different groups. Hex-3-ene, \ce{CH3CH2CH=CHCH2CH3}, has an ethyl group and a hydrogen on each double-bonded carbon, so cis and trans forms exist (option C).

In options A and D, C-2 carries two identical methyl groups. In option B, C-1 is a CH2 group with two identical hydrogen atoms. Each of these has two identical groups on one double-bonded carbon, so no geometric isomers are possible.

Note. The book prints no key; the answer was worked out with the book's conditions for geometric isomerism.

A racemic mixture must contain

A90% levorotatory isomers
Bequimolar quantities of levorotatory and dextrorotatory isomersAnswer
Cmeso isomers and levorotatory isomers
D80% dextrorotatory and 20% levorotatory isomers

Why

A racemic mixture is an equimolar (50:50) mixture of the dextrorotatory (+) and levorotatory (-) enantiomers. Their rotations of polarized light cancel each other, so the mixture is optically inactive, as the book states.

Options A and D are unequal mixtures, which would still show a net rotation. Option C contains a meso compound, which is a single internally compensated substance and not a racemic mixture.

Note. The book prints no key; the answer follows the book's definition of a racemic mixture.

Which set of compounds are consecutive members of same homologous series?

A\ce{CH3OH, CH3CHO, HCHO}
B\ce{CH3COOH, C2H5COOH, C3H7COOH}Answer
C\ce{CH4, CH2Cl2, CH2Cl2}
D\ce{C2H2, C2H4, C2H6}

Why

Members of a homologous series have the same functional group and differ from the next member by one \ce{CH2} unit. \ce{CH3COOH}, \ce{C2H5COOH} and \ce{C3H7COOH} are all carboxylic acids, and each differs from the next by \ce{CH2}, so they are consecutive members (option B).

Option A mixes an alcohol, an aldehyde and another aldehyde, which are different functional groups. Option C repeats \ce{CH2Cl2} and includes an alkane. Option D differs by \ce{H2} at each step, not by \ce{CH2}, so it is not a homologous series.

Note. The book prints no key; the answer follows the definition of a homologous series.

Short questionsp. 336–337

i3 marks

Justify why the second step in halogenation of alkanes is named propagation.

Solution

In the second step, a chlorine free radical attacks an ethane molecule and takes a hydrogen atom from it, giving HCl and an ethyl free radical: \ce{C2H6 + Cl. -> C2H5. + HCl}. The ethyl radical then reacts with a chlorine molecule to give chloroethane and a new chlorine free radical: \ce{C2H5. + Cl2 -> C2H5Cl + Cl.}

This step is called propagation because the chlorine radical is regenerated at the end of it. Each radical formed starts the next reaction, so the substitution progresses like a chain and continues as long as there is ethane and chlorine. Only the initiation step (the splitting of Cl2 by ultraviolet light) needs light to start the chain, and the termination step ends it.

ii3 marks

Draw the structure of ethane molecule.

Solution

Ethane, \ce{C2H6}, has two carbon atoms joined to each other by a single covalent (sigma, \delta) bond, and each carbon is also bonded to three hydrogen atoms by single covalent bonds. Each carbon is sp^3 hybridized, so the four bonds around it point towards the corners of a tetrahedron, with H-C-H and H-C-C angles of 109.5^\circ. The molecule is therefore not planar.

Ethane, \ce{C2H6}: skeletal structure (each carbon carries three H atoms, not drawn)

iii3 marks

Draw stereochemical formula of propene.

Solution

Propene, \ce{CH2=CH-CH3}, has a double bond between C-1 and C-2 and a single bond between C-2 and C-3. C-1 and C-2 are sp^2 hybridized and lie in one plane with bond angles of about 120^\circ. C-3 is sp^3 hybridized, so its three hydrogen atoms are arranged tetrahedrally: one points towards the viewer (solid wedge), one points away (dashed) and one lies in the plane of the paper.

Stereochemical formula of propene. The double bond and the C-2 to C-3 bond lie in the plane; the methyl carbon has a tetrahedral arrangement of its hydrogen atoms.

Propene: skeletal structure

iv3 marks

Alkenes shows addition reactions while alkanes cannot. Why?

Solution

Addition reactions need a site where a reagent can attach. An alkene has a carbon-carbon double bond made of one strong sigma bond and one weaker pi (\pi) bond. The pi electrons are loosely held above and below the bond, so an electrophile or other reagent can attack them; the pi bond breaks and two new single bonds form, which is an addition reaction.

An alkane has only sigma bonds, which are strong and held closely by the nuclei, and no pi electrons. Its carbon atoms are already bonded to four atoms, so no atom can be added to them. Alkanes therefore cannot add other atoms and react only by substitution under special conditions (ultraviolet light or high temperature).

v3 marks

Halogen is a non-polar molecule. How can it act as electrophile in halogenation of alkenes?

Solution

Although \ce{Br2} is non-polar, when it comes close to an alkene the electron cloud of the double bond repels the electron cloud of the bromine molecule. The Br-Br bond is polarized: one bromine atom becomes \delta^+ (electron deficient) and the other \delta^-. The \delta^+ bromine is the electrophile, and it attacks the pi electron cloud of the double bond.

The pi electrons form a bond to the electrophilic bromine, which produces a cyclic bromonium (holonium) ion, \ce{H2C-CH2} bridged by \ce{Br+}. In the second step, the bromide ion \ce{Br-} attacks this ion from the opposite side and gives 1,2-dibromoethane, \ce{BrCH2CH2Br}.

vi3 marks

How can we prepare poly(ethene) by polymerization reaction?

Solution

Poly(ethene), also called polythene, is made by polymerization: many small ethene molecules (monomers) join together to form one long chain. Ethene is heated to 400^\circ\mathrm{C} at a pressure of 1000\ \mathrm{atm} in the presence of traces of oxygen (about 0.1%), which start a free radical chain reaction.

n\ce{CH2=CH2} \xrightarrow[\text{traces of } \ce{O2}]{400\,^\circ\mathrm{C},\ 1000\ \mathrm{atm}} \ce{-[CH2-CH2]-_{n}}

poly(ethene), with n about 100,000

Ziegler-Natta catalyst (triethylaluminium, \ce{Al(C2H5)3}, with titanium(IV) chloride, \ce{TiCl4}) gives a better product: the poly(ethene) is harder, has a higher melting point and is not branched, because the catalyst causes no free radical formation.

vii3 marks

Explain how conjugation bring stability to alkenes?

Solution

Conjugated alkenes have alternate double and single bonds, for example hexa-1,3,5-triene, \ce{H2C=CH-CH=CH-CH=CH2}. In each carbon of the conjugated system the sp^2 hybridized carbons form sigma bonds, and each carbon keeps one unhybridized p orbital. These p orbitals overlap sideways along the whole chain, so the pi electrons are not confined to one double bond but are delocalized over the bridge of unhybridized p orbitals.

Spreading the pi electrons over several carbons lowers the energy of the molecule and so makes it more stable than an isolated double bond system. This delocalization is the reason conjugated alkenes are less reactive than alkenes whose double bonds are separated.

viii3 marks

What is the difference between structural and stereoisomerism? Support your answer with 2D and 3D formulae.

Solution

Structural isomers have the same molecular formula but different connectivity, that is, the atoms are joined in a different order. For example, butane and 2-methylpropane are both \ce{C4H10} but differ in their carbon skeleton. Stereoisomers have the same molecular formula and the same connectivity, but their atoms are arranged differently in space, so they are also called configurational isomers.

The 2D formulae of the two stereoisomers of but-2-ene, \ce{C4H8}, are shown below (cis, with both \ce{CH3} groups on the same side of the double bond; trans, with them on opposite sides). In 3D, the cis molecule has both methyl groups on one face of the planar \ce{C=C} unit, and the trans molecule has them on opposite faces. Both have the same connectivity, so they are stereoisomers, not structural isomers.

Butane (structural isomer of 2-methylpropane)

2-methylpropane

cis-but-2-ene

trans-but-2-ene

ix3 marks

Define stereoisomerism. How can we differentiate between enantiomers and diastereomers?

Solution

Stereoisomerism is the phenomenon in which molecules have the same molecular formula and the same connectivity of atoms but a different arrangement of the atoms in space. It has two types: optical isomerism (enantiomers) and geometric isomerism (cis-trans).

Enantiomers are optical isomers that are mirror images of each other which cannot be superimposed, like a left and right hand. They rotate the plane of polarized light by equal amounts in opposite directions. In the book's example of 2,3-dichloropentane, structures I and II are enantiomers, and so are III and IV.

Diastereomers are stereoisomers that are not mirror images of each other. In the same example, structures III and IV are not mirror images of I or II, so they are diastereomers of I and II. Diastereomers have different physical properties and their optical activity is not simply opposite.

Feature	Enantiomers	Diastereomers
Relation	Mirror images that cannot be superimposed	Stereoisomers that are not mirror images
Optical activity	Equal and opposite rotation of polarized light	Not simply opposite, each has its own rotation
Example (2,3-dichloropentane)	I and II; III and IV	I (or II) with III (or IV)

x3 marks

All chiral molecules are not necessarily optically active but all asymmetric molecules are optically active. Justify this statement by an example.

Solution

As printed, the statement is reversed. A chiral molecule is one that cannot be superimposed on its mirror image, and the book defines chirality as the property that gives optical activity: every chiral molecule rotates the plane of polarized light. The sentence is therefore correct only if the two parts are swapped and the word asymmetric is used for molecules with a chiral carbon.

The corrected statement is: all chiral molecules are optically active, but not all molecules with asymmetric (chiral) carbon atoms are optically active. Example: lactic acid has one chiral carbon and is optically active. 2,3-dichloropentane has two chiral carbons, and structures I and II (enantiomers) are optically active, but structure III has a plane of symmetry, so it is achiral and internally compensated. It is optically inactive (a meso form) even though it contains two chiral carbons.

Note. Misprint in the book's question: the two halves of the statement are inverted. The answer explains the correct statement and gives the book's examples.

Long questionsp. 336–338

Generally, hydrocarbons are less reactive. Arrange alkanes, alkenes and alkynes in their reactivity order and justify.

Solution

Reactivity order

\text{alkene} > \text{alkyne} > \text{alkane}

Alkenes are the most reactive, alkynes are less reactive than alkenes but much more reactive than alkanes, and alkanes are the least reactive hydrocarbons.

Alkanes (least reactive)

Alkanes have only sigma (\delta) bonds. These are strong, because the bonding electrons lie very close to the nuclei and are held strongly.
The C-H bonds are non-polar, because the electronegativities of carbon (2.5) and hydrogen (2.1) differ too little to make them polar, so acids, bases, oxidizing and reducing agents do not react with them.
Alkanes react only by free radical substitution in ultraviolet light, or by cracking at high temperature.

Alkenes (most reactive)

An alkene has a sigma bond and a weaker pi (\pi) bond. The pi electrons are weakly held and easily available to an attacking reagent.
The double bond is therefore converted into a single bond by addition reactions, for example with \ce{Br2} or \ce{HBr}. Ethene is more reactive than ethane for this reason.

Alkynes (intermediate)

An alkyne has two pi bonds, so it is unsaturated and can add reagents. It is therefore more reactive than an alkane.
Ethyne is less reactive than ethene. The book's reason is that the carbon-carbon distance in ethene is longer than in ethyne, so the pi bond of ethene is more exposed (diffused) than the pi bond of ethyne, and it attracts reagents more easily.

Summary

Hydrocarbon	Bonds between carbon atoms	Reactivity
Alkane	One sigma bond per pair (strong, non-polar)	Least reactive; substitution only
Alkene	One sigma and one weaker pi bond	Most reactive; addition reactions
Alkyne	One sigma and two pi bonds	Intermediate; addition, less readily than alkenes

Note. The book states that ethene is more reactive than ethyne because its pi bond is more exposed; the order given follows the book.

Define primary, secondary and tertiary carbocations. Justify their order of reactivity in relation with hydrochlorination of pent-1-ene.

Solution

Definition

A carbocation is a species in which a carbon atom carries a positive charge and has only six electrons in its outer shell. It is therefore electron deficient, and the positive carbon tends to attract electrons from neighbouring atoms.

Carbocations are classified by the number of alkyl groups attached to the positive carbon:

Primary carbocation: one alkyl group, \ce{R-CH2+}.
Secondary carbocation: two alkyl groups, \ce{R2CH+}.
Tertiary carbocation: three alkyl groups, \ce{R3C+}.

Order of stability (and of ease of formation)

Alkyl groups are electron donating (positive inductive effect). They push electrons through the sigma bonds towards the positive carbon, which lowers its positive charge density. The more alkyl groups attached, the more the charge is spread out and the more stable the carbocation:

\ce{R3C+} > \ce{R2CH+} > \ce{RCH2+}\qquad(\text{tertiary} > \text{secondary} > \text{primary})

Hydrochlorination of pent-1-ene

Pent-1-ene is \ce{CH2=CH-CH2CH2CH3}. The hydrogen of \ce{HCl} (an electrophile) can add to either carbon of the double bond, so two carbocations can form.
If H adds to C-1, the carbocation forms at C-2, which is a secondary carbocation \ce{CH3CH+CH2CH2CH3}. This is more stable, so it forms more readily.
\ce{CH2=CH-R + H+ -> CH3-CH+-R}
If H adds to C-2, the carbocation forms at C-1, which is a primary carbocation \ce{CH2+CH2CH2CH2CH3}. It is less stable and forms less readily.
The chloride ion then attacks the more stable secondary carbocation, so 2-chloropentane is the major product (following Markovnikov's rule), and 1-chloropentane is only a minor product.
\ce{CH3CH(Cl)CH2CH2CH3}\ (\text{major})

Therefore the order of ease of formation of the carbocations is secondary carbocation > primary carbocation, and this order decides which chloroalkane is formed in greater amount.

Note. The question asks about reactivity; the answer uses the book's point that the more stable carbocation is formed more readily, because the book gives the stability order.

Define Markovnikov's rule. Give mechanism of hydrobromination of 2-methylhex-2-ene.

Solution

Markovnikov's rule

When an unsymmetrical reagent such as \ce{HBr} adds to an unsymmetrical alkene, the hydrogen atom of the reagent bonds to the double-bonded carbon that already carries more hydrogen atoms. The bromine goes to the other carbon. The rule is obeyed because the more stable carbocation is formed as the intermediate.

Mechanism

Structure of the alkene. 2-methylhex-2-ene is \ce{(CH3)2C=CH-CH2CH2CH3}. C-2 carries two methyl groups and C-3 carries one hydrogen atom, so the hydrogen adds to C-3.
\ce{(CH3)2C=CH-CH2CH2CH3}
Electrophilic attack. The \delta^+ hydrogen of \ce{HBr} is attracted to the pi electrons and bonds to C-3, which produces a carbocation on C-2.
\ce{(CH3)2C=CH-R + H+ -> (CH3)2C+-CH2-R}
The carbocation on C-2 is tertiary, since it carries three alkyl groups. This is the most stable carbocation possible here, so this route is favoured.
Nucleophilic attack. The bromide ion \ce{Br-} attacks the carbocation and forms the C-Br bond.
\ce{(CH3)2C+-R + Br- -> (CH3)2CBr-R}
Product. The major product is 2-bromo-2-methylhexane.
\ce{(CH3)2C(Br)CH2CH2CH2CH3}

Result

\ce{(CH3)2C(Br)CH2CH2CH2CH3}

2-bromo-2-methylhexane (major product)

\ce{(CH3)2C(Br)CH2CH2CH2CH3}

Hydration of alkenes is acid-catalyzed reaction. Justify and exemplify this statement by propene.

Solution

Hydration of alkenes

Hydration is the addition of water across the double bond of an alkene, which gives an alcohol. It is catalyzed by a strong acid, usually sulphuric acid. The acid supplies the proton (\ce{H+}) that starts the reaction, and it is regenerated at the end, so only a small amount is needed.

Why it is acid-catalyzed

Water is too weak an electrophile to attack the double bond on its own. The \ce{H+} from the acid is the electrophile that adds to the pi bond.
The carbocation formed is attacked by water (or by the hydrogen sulphate ion) and then the proton is lost, so the acid is regenerated and can catalyze more reaction.

Example: propene

Step 1: addition of sulphuric acid. Cold concentrated \ce{H2SO4} adds across the double bond. The hydrogen goes to C-1 (more hydrogen atoms), and the hydrogen sulphate group goes to C-2, giving isopropyl hydrogen sulphate. The carbocation intermediate is secondary, which is more stable than the primary one.
\ce{CH3CH=CH2 + H2SO4 -> CH3CH(OSO3H)CH3}
Step 2: hydrolysis. On boiling with water, the hydrogen sulphate ester is broken down. The alcohol forms and the sulphuric acid is released again, so it acts as a catalyst.
\ce{CH3CH(OSO3H)CH3 + H2O ->[\text{boil}] CH3CH(OH)CH3 + H2SO4}
Overall, propene is converted to propan-2-ol, and the acid is regenerated.
\ce{CH3CH=CH2 + H2O ->[H2SO4] CH3CH(OH)CH3}

Propan-2-ol, the major product (Markovnikov's rule)

\ce{CH3CH=CH2 + H2O ->[H2SO4] CH3CH(OH)CH3}

Give complete mechanism of ozonolysis of but-1-ene.

Solution

Principle

Ozonolysis breaks the carbon-carbon double bond of an alkene. Ozone is bubbled through a very cold solution of the alkene in dichloromethane (at about -78^\circ\mathrm{C}), and the resulting ozonide is then treated with a mild reducing agent (zinc, with water and heat). The products show where the double bond was.

Mechanism

Attack of ozone. Ozone adds to the double bond of but-1-ene, \ce{CH2=CH-CH2CH3}, with the oxygen atoms attaching to both double-bonded carbons. This gives an unstable intermediate called the molozonide, a five-membered ring containing three oxygen atoms.
\ce{CH2=CH-CH2CH3 + O3 -> molozonide}
Rearrangement. The molozonide rearranges rapidly to the more stable ozonide, which is also a five-membered ring of two carbons and three oxygen atoms (-O-O-O- unit). Ozonides are not easily isolated, so the next step is carried out directly.
Reduction of the ozonide. The ozonide is treated with zinc in water with heating. The ring breaks at the carbon-carbon bond, each carbon becomes a carbonyl group, and hydrogen peroxide is also formed.
\ce{ozonide + Zn + H2O -> HCHO + CH3CH2CHO + H2O2}
Destruction of the peroxide. The zinc also reduces the hydrogen peroxide.
\ce{H2O2 + Zn -> ZnO + H2O}

Products

The CH_2 end of the double bond gives methanal, \ce{HCHO}.
The CH-CH_2CH_3 end gives propanal, \ce{CH3CH2CHO}.

Thus the double bond of but-1-ene lies between C-1 and C-2, because one product is a methanal and the other is a propanal. Ozonolysis is therefore used to find the position of a double bond from the structures of the products.

\ce{CH2=CH-CH2CH3 ->[O3][Zn/H2O] HCHO + CH3CH2CHO}

Define geometric isomerism. Apply this concept on open chain alkenes and cycloalkanes.

Solution

Definition

Geometric (cis-trans) isomerism is a form of stereoisomerism shown by compounds with the same molecular formula and connectivity but a different arrangement of groups in space. It arises because rotation about a C=C double bond, or about a C-C bond in a ring, is restricted. The groups on the same side of the double bond or ring are called the cis isomer, and those on opposite sides the trans isomer.

Conditions

The two groups on one double-bonded carbon must be different (or the two groups on one ring carbon must be different from those on the other carbon of the bond).
For cyclic compounds, the substituents must be on different ring carbons, such as 1,2 or 1,3, and not both on the same carbon (a 1,1 compound cannot show geometric isomerism).

Open chain alkenes

Alkene	Geometric isomerism	Reason
but-2-ene	Yes	Each C of C=C carries H and CH3
pent-2-ene	Yes	Each C of C=C carries H and an alkyl group
but-1-ene	No	C-1 carries two H atoms
2-methylbut-2-ene	No	C-2 carries two CH3 groups

cis-but-2-ene

trans-but-2-ene

Cycloalkanes

In a ring, the cis isomer has the two groups above the plane of the ring and the trans isomer has one group above and the other below. The book's examples are cis- and trans-1,2-dimethylcyclopropane (groups on C-1 and C-2) and cis- and trans-1,3-dimethylcyclopentane (groups on C-1 and C-3). A compound such as 1,1-dimethylcyclopropane cannot show geometric isomerism because both groups are on the same carbon.

Interconversion

Geometric isomers can be interconverted by heating, which breaks the pi bond so that rotation becomes possible; on cooling, the desired isomer is formed.

Note. The book's Concept Assessment Exercise 16.6 asks which of but-1-ene, pent-1-ene, hex-3-ene, methylcyclobutane and 1,3-dimethylcyclohexane show this isomerism; the answer is hex-3-ene and 1,3-dimethylcyclohexane.

Design a synthetic route for the preparation of the following compounds. What starting materials would you use, and what reactions mechanisms are involved? a. Bromoethane b. 1,2-dibromoethane

Solution

a. Bromoethane, \ce{CH3CH2Br}

Starting material: ethanol, which is converted to ethene, which is then converted to bromoethane by hydrobromination. Ethene itself can be used directly if it is available.

Step 1: dehydration of ethanol. Ethanol is passed over hot aluminium oxide (or heated with concentrated sulphuric acid). Water is removed and ethene is formed. This is an elimination reaction.
\ce{CH3CH2OH ->[Al2O3][heat] CH2=CH2 + H2O}
Step 2: hydrobromination of ethene. \ce{HBr} adds across the double bond. The \ce{H+} attacks the pi bond to give the carbocation \ce{CH3CH2+}, and then \ce{Br-} attacks the carbocation. Ethene is symmetrical, so only one product forms.
\ce{CH2=CH2 + HBr -> CH3CH2Br}

Free radical halogenation of ethane is not suitable, because it gives a mixture of chlorinated or brominated products, including dibromo and polybromo compounds.

b. 1,2-dibromoethane, \ce{BrCH2CH2Br}

Starting materials: ethene (from ethanol, as above) and bromine, \ce{Br2}, in an inert solvent such as tetrachloromethane.

Step 1: the bromine molecule is polarized by the electron cloud of the double bond, so the \delta^+ bromine acts as the electrophile. The bromine attacks the pi electrons to form a cyclic bromonium ion.
\ce{CH2=CH2 + Br2 -> \text{cyclic bromonium ion} + Br-}
Step 2: the bromide ion attacks the bromonium ion from the opposite side and opens the ring to give 1,2-dibromoethane.
\ce{BrCH2CH2Br}
Overall the reaction is an addition of bromine to ethene: \ce{CH2=CH2 + Br2 -> BrCH2CH2Br}, carried out in \ce{CCl4} at room temperature.

Summary

Target	Starting material	Reactions and mechanism
\ce{CH3CH2Br}	Ethanol (via ethene)	Dehydration (elimination), then hydrobromination (electrophilic addition)
\ce{BrCH2CH2Br}	Ethene and \ce{Br2}	Halogenation (electrophilic addition through a bromonium ion)

a. (i) List three characteristics of homologous series. (ii) Draw the possible structural isomers of \ce{C6H14}. b. Ethanol undergoes dehydration in the presence of sulphuric acid. (i) Identify this reaction and the product formed. (ii) Evaluate the role of sulphuric acid. (iii) Write one major commercial use of the product.

Solution

a (i) List three characteristics of homologous series.

All members have the same general formula, for example \ce{C_nH_{2n+2}} for alkanes.
Successive members differ by one \ce{CH2} unit.
Members have the same functional group and therefore similar chemical properties.
Physical properties (for example boiling point) change gradually along the series.

a (ii) Draw the possible structural isomers of \ce{C6H14}.

\ce{C6H14} has five structural (chain and branching) isomers. They differ only in how the six carbon atoms are joined:

Hexane

2-methylpentane

3-methylpentane

2,2-dimethylbutane

2,3-dimethylbutane

b (i) Identify this reaction and the product formed.

This is a dehydration reaction, which is an elimination reaction: water is removed from ethanol and a C=C double bond forms. The product is ethene.

\ce{CH3CH2OH ->[{conc.} H2SO4][heat] CH2=CH2 + H2O}

\text{Ethene, } \ce{CH2=CH2}, \text{ and water}

b (ii) Evaluate the role of sulphuric acid.

Sulphuric acid is a catalyst and a dehydrating agent. Its \ce{H+} protonates the OH group of ethanol, so the leaving group becomes \ce{H2O} rather than \ce{OH-}, which makes the elimination possible. The acid takes up the water that is removed, and it is not used up, so only a small amount of it is needed.

\ce{CH3CH2OH + H+ -> CH3CH2OH2+ ->[-H2O] CH3CH2+ ->[-H+] CH2=CH2}

b (iii) Write one major commercial use of the product.

Ethene is used to manufacture poly(ethene), commonly called polythene, by polymerization. Polythene is used for plastic bags, containers and packaging.

The condensed formulae of six compounds, A to F, are given. Each compound has four carbon atoms in its molecule. A. \ce{CH3CH=CHCH3}; B. \ce{CH3CH2COCH3}; C. \ce{CH2=CHCH2CH3}; D. \ce{CH3CH2CH(OH)CH3}; E. \ce{HOCH2CH2CH2CH2OH}; F. \ce{CH3CH2OCH2CH3}. a. (i) What is the empirical formulae of all these compounds? (ii) Draw skeletal formulae of all these compounds. (iii) Which two compounds show different types of isomerism? (iv) What type of isomerism can each compound show? b. Compound D can be converted to C. (i) What type of reaction is this? (ii) What reagent is used in this reaction? (iii) What is formed when compound E undergoes the same reaction in excess of the same reagent?

Solution

a (i) What is the empirical formulae of all these compounds?

A: \ce{CH2} (from \ce{C4H8})
B: \ce{C2H4O} (from \ce{C4H8O})
C: \ce{CH2} (from \ce{C4H8})
D: \ce{C2H5O} (from \ce{C4H10O})
E: \ce{C2H5O} (from \ce{C4H10O2})
F: \ce{C2H5O} (from \ce{C4H10O})

a (ii) Draw skeletal formulae of all these compounds.

The skeletal formula shows the carbon skeleton as lines, with the functional group drawn in:

A: but-2-ene

B: butan-2-one

C: but-1-ene

D: butan-2-ol

E: butane-1,4-diol

F: diethyl ether

a (iii) Which two compounds show different types of isomerism?

Compound A, but-2-ene, shows geometric (cis-trans) isomerism because each double-bonded carbon has a hydrogen and a methyl group. Compound D, butan-2-ol, shows optical isomerism because C-2 is a chiral carbon with four different groups (H, OH, CH3 and C2H5). So A and D show different types of isomerism.

A and D

a (iv) What type of isomerism can each compound show?

A (but-2-ene): geometric (cis and trans); structural (position isomer but-1-ene, chain isomer 2-methylpropene).
B (butan-2-one): functional group isomer (butanal); no geometric or optical isomerism.
C (but-1-ene): structural (position isomer but-2-ene, chain isomer 2-methylpropene); no geometric isomerism, because C-1 carries two H atoms.
D (butan-2-ol): optical (two enantiomers); structural (functional isomer diethyl ether, position isomer butan-1-ol).
E (butane-1,4-diol): structural only; no chiral carbon and no double bond.
F (diethyl ether): functional group isomer of butan-2-ol and butan-1-ol; no optical or geometric isomerism.

b (i) What type of reaction is this?

Compound D, butan-2-ol, is converted to C, but-1-ene. This is a dehydration reaction, an elimination reaction in which water is removed and a C=C double bond forms.

Elimination (dehydration)

b (ii) What reagent is used in this reaction?

The reagent is concentrated sulphuric acid, which can be replaced by phosphoric acid or by aluminium oxide heated strongly, as in the book's methods for dehydration of alcohols. Elimination can also give but-2-ene; the product the question asks for is C, but-1-ene.

\ce{CH3CH(OH)CH2CH3 ->[{conc.} H2SO4][heat] CH2=CHCH2CH3 + H2O}

Concentrated H2SO4 (heat)

b (iii) What is formed when compound E undergoes the same reaction in excess of the same reagent?

Compound E, butane-1,4-diol, has an OH group at each end of the chain. With excess concentrated sulphuric acid and heat, both OH groups are removed as water in two successive elimination steps, giving a conjugated diene.

\ce{HOCH2CH2CH2CH2OH ->[excess {conc.} H2SO4][heat] CH2=CH-CH=CH2 + 2H2O}

\ce{CH2=CH-CH=CH2}\ \text{(buta-1,3-diene)}

Note. The book prints no key. Part a (iii): the book's answer is taken as A and D (geometric and optical). Part b (iii) assumes double dehydration to buta-1,3-diene; intramolecular ether formation (tetrahydrofuran) is also possible with acid, and the book does not discuss it.

FBISE 2025 and 2026 questions

Questions from the FBISE HSSC-I 2026 (1st Annual) and FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2026 Annual I, Set B, MCQ 71 mark

Which major product is formed when Propane reacts with concentrated HBr?

A2-bromo propane
BPropyne
C1-bromo propane
DPropaneAnswer

Why

Propane is an alkane. Its C-H and C-C bonds are non-polar and strong sigma bonds, so alkanes do not react with polar reagents such as acids, bases, oxidizing and reducing agents. Concentrated \ce{HBr} is a polar reagent, so there is no reaction and propane remains unchanged (option D).

2-bromopropane (A) is the tempting option, but it is the major product when propene, not propane, adds \ce{HBr} by Markovnikov's rule. Propyne (B) would need loss of hydrogen, and 1-bromopropane (C) is only the minor product from propene.

Note. The question is worded for propane, so the answer by the book's reasoning (alkanes are unreactive towards polar reagents) is D. If propene was meant, the answer would be A, 2-bromopropane.

2026 Annual I, Set B, MCQ 91 mark

Free radical substitution occurs in:

AAlkene
BAlcohol
CAlkaneAnswer
DCarboxylic Acid

Why

Free radical substitution is the characteristic reaction of alkanes, for example chlorination of ethane in ultraviolet light, where a halogen free radical replaces a hydrogen atom. Alkanes are non-polar and have only sigma bonds, so they do not react with ionic or polar reagents, but they do react with free radicals.

Alkenes (A) mainly undergo electrophilic addition at the \pi bond. Alcohols (B) and carboxylic acids (D) react through their polar functional groups, by substitution at the \ce{-OH} group, oxidation or esterification, not by a radical chain.

2025 Annual I, Set B, MCQ 111 mark

Identify the step in the mechanism of free radical chlorination of ethane: \ce{CH3-CH3 + Cl^{.} -> CH3-CH2^{.} + HCl}

AInitiation
BPropagationAnswer
CSubstitution
DTermination

Why

In \ce{CH3-CH3 + Cl. -> CH3-CH2. + HCl} a radical (chlorine) reacts with a molecule (ethane) and a new radical (ethyl) is formed. A step that uses one radical and produces another keeps the chain going, so this is propagation.

Initiation (A) makes radicals from a molecule, \ce{Cl2 -> 2Cl.}. Termination (D) combines two radicals into a molecule. Substitution (C) names the overall reaction, not a step of the mechanism.

2026 Annual I, Set B, Q.2 (v)3 marks

Explain the formation of but-1-ene from: (a) Butan-1-ol (b) 1-Bromo butane

Solution

(a) From butan-1-ol

Butan-1-ol, \ce{CH3CH2CH2CH2OH}, has its \ce{-OH} on C-1, so removing \ce{OH} from C-1 and \ce{H} from C-2 puts the double bond between C-1 and C-2.
This is dehydration, an elimination reaction. The alcohol vapour is passed over hot aluminium oxide (or heated with concentrated \ce{H2SO4} or \ce{H3PO4}). Water is removed.
\ce{CH3-CH2-CH2-CH2-OH ->[Al2O3][heat] CH3-CH2-CH=CH2 + H2O}

But-1-ene

\ce{CH3CH2CH2CH2OH ->[Al2O3][heat] CH3CH2CH=CH2 + H2O}

(b) From 1-bromobutane

1-Bromobutane, \ce{CH3CH2CH2CH2Br}, has \ce{Br} on C-1 and a \beta-hydrogen on C-2.
Heating with ethanolic \ce{NaOH} removes \ce{HBr} (dehydrohalogenation, a \beta-elimination). The \ce{NaOH} takes the \ce{HBr} as \ce{NaBr} and \ce{H2O}.
\ce{CH3-CH2-CH2-CH2-Br + NaOH(ethanolic) ->[heat] CH3-CH2-CH=CH2 + NaBr + H2O}
Aqueous \ce{NaOH} must not be used, because it gives the alcohol by substitution.

\ce{CH3CH2CH2CH2Br + NaOH(ethanolic) ->[heat] CH3CH2CH=CH2 + NaBr + H2O}

Note. The book's solved example prepares but-1-ene by exactly these two methods. Part (a) is also dehydration of an alcohol (unit 18) and part (b) is an elimination reaction of an alkyl halide (unit 17).

2026 Annual I, Set B, Q.2 (vii) OR3 marks

Describe the mechanism of formation of chloroethane from ethane and chlorine.

Solution

Free radical substitution of ethane

Chlorine reacts with ethane in ultraviolet light by a free radical chain mechanism, in three stages.

Initiation. Ultraviolet light is absorbed by a chlorine molecule and splits the Cl-Cl bond by homolytic fission, so each atom keeps one electron. Two chlorine free radicals are formed.
\ce{Cl-Cl ->[h\nu] 2Cl.}
Propagation, step 1. A chlorine radical attacks an ethane molecule, removes a hydrogen atom and forms hydrogen chloride and the ethyl free radical.
\ce{CH3-CH3 + Cl. -> CH3-CH2. + HCl}
Propagation, step 2. The ethyl radical removes a chlorine atom from a chlorine molecule. Chloroethane is formed and a new chlorine radical is produced, which repeats step 1.
\ce{CH3-CH2. + Cl-Cl -> CH3-CH2-Cl + Cl.}
Termination. Two free radicals combine, so the radicals are used up and the chain stops. Any of these can happen.
\begin{gathered}\ce{Cl. + Cl. -> Cl2}\\ \ce{CH3-CH2. + Cl. -> CH3-CH2-Cl}\\ \ce{CH3-CH2. + .CH2-CH3 -> CH3-CH2-CH2-CH3}\end{gathered}
Overall reaction.
\ce{C2H6 + Cl2 ->[h\nu] C2H5Cl + HCl}

\ce{C2H6 + Cl2 ->[h\nu] C2H5Cl + HCl}

2025 Annual I, Set B, Q.2 (x) OR1x3 = 3 marks

State Markovnikov's rule by giving an example and its mechanism.

Solution

Markovnikov's rule

When an unsymmetrical reagent such as \ce{HBr} adds to an unsymmetrical alkene such as propene, the hydrogen atom bonds to the doubly bonded carbon that already has more hydrogen atoms, and the positive part of the reagent goes where the hydrogens are more. The more stable carbocation is the intermediate, so the major product has the halogen on the carbon with fewer hydrogens.

Example and mechanism: propene and HBr

The \ce{H-Br} bond is polar, \ce{H^{\delta+}-Br^{\delta-}}. The \pi electrons of propene attack the \delta+ hydrogen, which is the electrophile, and \ce{Br-} leaves.
\ce{CH3-CH=CH2 + H^{\delta+}-Br^{\delta-}}
The hydrogen adds to the end carbon \ce{CH2}, which has more hydrogens. This leaves the positive charge on the middle carbon, a secondary carbocation. It is more stable than the primary carbocation that the other route would give, because two alkyl groups release electrons to it.
\ce{CH3-CH=CH2 + HBr -> CH3-CH^+-CH3 + Br-}
The bromide ion, a nucleophile, attacks the carbocation.
\ce{CH3-CH^+-CH3 + Br- -> CH3-CHBr-CH3}
The major product is 2-bromopropane; the minor product is 1-bromopropane, from the primary carbocation.

2-Bromopropane (major)

1-Bromopropane (minor)

\ce{CH3-CH=CH2 + HBr -> CH3-CHBr-CH3}

2025 Annual I, Set B, Q.5 (i) OR2 marks

Why is ethene more reactive than ethane?

Solution

Ethene has a \ce{C=C} double bond made of one strong \sigma bond and one weaker \pi bond. The \pi electrons lie above and below the plane of the molecule and are loosely held and exposed, so electrophiles can attack them easily and ethene undergoes addition reactions.

Ethane has only strong, non-polar \sigma bonds. They are held tightly between the nuclei and give no electrons for attack, so ethane reacts only by free radical substitution in ultraviolet light. Ethene is therefore more reactive than ethane.

2025 Annual I, Set B, Q.5 (ii) OR2 marks

Write down the reaction of ethene with \ce{H2SO4/H2O}

Solution

This is hydration of ethene (acid-catalysed addition of water). It takes place in two steps and gives ethanol.

Ethene is treated with cold concentrated sulphuric acid. The \pi electrons attack the positive hydrogen of the acid and ethyl hydrogen sulphate is formed.
\ce{CH2=CH2 + H-O-SO2-OH -> CH3-CH2-O-SO2-OH}
The mixture is boiled with water. Ethyl hydrogen sulphate is hydrolysed to ethanol and sulphuric acid is given back, so it behaves as a catalyst.
\ce{CH3-CH2-O-SO2-OH + H2O ->[boil] CH3-CH2-OH + H2SO4}
Overall reaction.
\ce{CH2=CH2 + H2O ->[H+] CH3-CH2-OH}

\ce{CH2=CH2 + H2O ->[H2SO4] CH3CH2OH}

2025 Annual I, Set B, Q.5 (iii) OR2 marks

Write down the reaction of ethene with \ce{Cl2/H2O}

Solution

This is halohydration. Chlorine reacts with water to give hypochlorous acid, which adds to the double bond, with the solvent water acting as a reactant.

Chlorine and water give hypochlorous acid.
\ce{Cl2 + H2O <=> HOCl + HCl}
The \ce{OH} group adds to one carbon and \ce{Cl} to the other, so a chlorohydrin is formed.
\ce{CH2=CH2 + Cl2 + H2O -> CH2Cl-CH2OH + HCl}

2-Chloroethanol (a chlorohydrin)

\ce{CH2=CH2 + Cl2 + H2O -> CH2Cl-CH2OH + HCl}

2026 Annual I, Set D, MCQ 91 mark

Which condition is necessary to initiate a free radical substitution reaction?

ALow Temperature
BHigh Temperature
CUV radiationsAnswer
DAbsence of light

Why

The chain is started when the halogen molecule splits into two free radicals by homolytic fission, \ce{Cl2 ->[h\nu] 2Cl.}. The energy for this is supplied by ultraviolet radiation, so UV light is the condition used in the book (option C).

Absence of light (D) is the opposite condition, because the reaction does not start in the dark at room temperature. Low temperature (A) gives no energy to break the Cl-Cl bond. A very high temperature (B) would also give radicals, but the condition named for the free radical substitution of ethane in the book is ultraviolet light.

2026 Annual I, Set D, MCQ 141 mark

Which of the following is a saturated organic compound?

A\ce{C6H14}Answer
B\ce{C6H12}
C\ce{C6H10}
D\ce{C6H6}

Why

A saturated hydrocarbon has only single bonds. Alkanes have the general formula \ce{C_nH_{2n+2}}, and for n=6 this gives \ce{C6H14}, so option A is saturated.

\ce{C6H12} (B) fits \ce{C_nH_{2n}}, so it has one double bond (an alkene) or a ring; \ce{C6H10} (C) has two fewer hydrogens again (two double bonds or one triple bond); \ce{C6H6} (D) is benzene with double bonds. All of these are unsaturated.

2025 Annual I, Set D, MCQ 141 mark

What is produced when ethene is reacted with cold, dilute solution of \ce{KMnO4}?

A\ce{CH3-CH2-OH}
B\ce{CH3-CHO}
C\ce{CH3-COOH}
D\ce{HO-CH2-CH2-OH}Answer

Why

Cold, dilute, acidified (or neutral) \ce{KMnO4} adds two \ce{OH} groups, one on each carbon of the double bond, and converts ethene into the diol ethane-1,2-diol, \ce{HO-CH2-CH2-OH}. The purple \ce{MnO4-} is reduced to the nearly colourless \ce{Mn^2+}.

Ethanol (A) is made from ethene and steam, not by \ce{KMnO4}. Ethanal (B) and ethanoic acid (C) would need hot, concentrated oxidation that breaks the double bond, not the cold dilute reagent.

2026 Annual I, Set D, Q.2 (v)1.5+1.5 = 3 marks

How can ethene be prepared from: (a) Ethanol (b) Bromoethane

Solution

(a) From ethanol

Ethanol is dehydrated: the \ce{-OH} group and a hydrogen from the next carbon are removed as water and a double bond forms. Ethanol vapour is passed over hot aluminium oxide, or ethanol is heated with concentrated \ce{H2SO4} at 180\,^\circ\mathrm{C} (or with \ce{H3PO4}).
\ce{CH3-CH2-OH ->[Al2O3][heat] CH2=CH2 + H2O}

\ce{CH3CH2OH ->[Al2O3][heat] CH2=CH2 + H2O}

(b) From bromoethane

Bromoethane is heated with ethanolic \ce{NaOH} (or \ce{KOH}). This is dehydrohalogenation, a \beta-elimination, in which \ce{HBr} is removed.
\ce{CH3-CH2-Br + NaOH(ethanolic) ->[heat] CH2=CH2 + NaBr + H2O}
Aqueous alkali would give ethanol by substitution, so the solvent must be ethanol.

\ce{CH3CH2Br + NaOH(ethanolic) ->[heat] CH2=CH2 + NaBr + H2O}

Note. Part (a) is dehydration of an alcohol (unit 18) and part (b) is an elimination reaction of an alkyl halide (unit 17).

2026 Annual I, Set D, Q.2 (vii) OR1.5+1.5 = 3 marks

Give all of the possible reactions among free radicals in termination step of methane and chlorine.

Solution

In the chlorination of methane the chain stops (termination) when two free radicals meet and combine. The radicals present are \ce{Cl.} and \ce{CH3.}, so the possible combinations are:

Two chlorine radicals combine to give a chlorine molecule.
\ce{Cl. + Cl. -> Cl2}
A methyl radical combines with a chlorine radical to give chloromethane.
\ce{CH3. + Cl. -> CH3Cl}
Two methyl radicals combine to give ethane.
\ce{CH3. + .CH3 -> CH3-CH3}

2025 Annual I, Set D, Q.2 (x)3 marks

Briefly explain the factors which affect the relative stability of primary, secondary and tertiary carbocation.

Solution

A carbocation has a carbon with a positive charge and only six electrons. It is stabilised when the positive charge is spread out or reduced. The relative stability of primary, secondary and tertiary carbocations depends on the following factors.

Number of alkyl groups. Carbocations are classed by the number of alkyl groups on the positive carbon: primary (one), secondary (two) and tertiary (three). The more alkyl groups, the more stable the ion.
Inductive effect of the alkyl groups. Carbon is less electronegative than hydrogen, so an alkyl group is an electron-donating group. Each alkyl group pushes electrons through the \sigma bond towards the positive carbon (the positive inductive effect, +I). This lowers the positive charge density and makes the ion more stable.
Order of stability. Tertiary has three electron-releasing groups, secondary two and primary one, so
\text{tertiary}>\text{secondary}>\text{primary}\quad\left[\ce{(CH3)3C+}>\ce{(CH3)2CH+}>\ce{CH3CH2+}\right]

Carbocation	Alkyl groups on \ce{C+}	Example	Stability
Primary (1°)	one	\ce{CH3-CH2-CH2+}	lowest
Secondary (2°)	two	\ce{CH3-CH+-CH3}	medium
Tertiary (3°)	three	\ce{(CH3)3C+}	highest

2025 Annual I, Set D, Q.3 OR3+4 = 7 marks

Why are alkanes chemically less reactive? Describe the mechanism of free radical chlorination of ethane to form ethyl chloride.

Solution

(a) Why are alkanes chemically less reactive? (3 marks)

Alkanes are also called paraffins (parum means little, affinis means reactivity). Their low reactivity has two reasons.

Factor	What it means	Result
Non-polar bonds	The electronegativity of C is 2.5 and of H is 2.1, so the difference is only 0.4 and the C-H and C-C bonds are non-polar.	No positive or negative end for polar reagents; acids, bases, oxidizing and reducing agents do not react.
Strong \sigma bonds only	The \sigma electrons lie close to the nuclei, between them, and are held strongly. There is no \pi bond.	The electrons cannot easily be used in a reaction.

Alkanes therefore react only with free radicals, for example in ultraviolet light, or on heating (combustion and cracking).

(b) Describe the mechanism of free radical chlorination of ethane to form ethyl chloride. (4 marks)

Chlorine and ethane react in ultraviolet light by a free radical chain reaction in three stages.

Initiation. Ultraviolet light is absorbed by a chlorine molecule and splits the Cl-Cl bond by homolytic fission, so each atom keeps one electron. Two chlorine free radicals are formed.
\ce{Cl-Cl ->[h\nu] 2Cl.}
Propagation, step 1. A chlorine radical attacks an ethane molecule, removes a hydrogen atom and forms hydrogen chloride and the ethyl free radical.
\ce{CH3-CH3 + Cl. -> CH3-CH2. + HCl}
Propagation, step 2. The ethyl radical removes a chlorine atom from a chlorine molecule. Chloroethane is formed and a new chlorine radical is produced, which repeats step 1.
\ce{CH3-CH2. + Cl-Cl -> CH3-CH2-Cl + Cl.}
Termination. Two free radicals combine, so the radicals are used up and the chain stops. Any of these can happen.
\begin{gathered}\ce{Cl. + Cl. -> Cl2}\\ \ce{CH3-CH2. + Cl. -> CH3-CH2-Cl}\\ \ce{CH3-CH2. + .CH2-CH3 -> CH3-CH2-CH2-CH3}\end{gathered}
Overall reaction.
\ce{C2H6 + Cl2 ->[h\nu] C2H5Cl + HCl}

Figure: the propagation steps form a cycle. Each step uses one radical and makes a new one, so the chain continues.

\ce{C2H6 + Cl2 ->[h\nu] C2H5Cl + HCl}

Exam focus

Describe the free radical chlorination of ethane (or methane) in three steps: initiation, propagation and termination, with an equation for each. (long)
The most repeated topic. 2025 Set D Q.3 OR (7 marks), 2026 Set B Q.2 (vii) OR, 2026 Set D Q.2 (vii) OR (termination), 2025 Set B MCQ 11 and both 2026 MCQs on free radical substitution (Set B MCQ 9, Set D MCQ 9). Book short question 1 and MCQ 3.
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Why are alkanes less reactive? Explain with non-polar bonds and strong sigma bonds. (short)
Part (a) of 2025 Set D Q.3 OR, and the idea behind 2026 Set B MCQ 7 (propane with \ce{HBr}: no reaction). Book exercise MCQ 5 and long question 1.
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State Markovnikov's rule and give the mechanism of the addition of \ce{HBr} to propene, with the role of carbocation stability. (long)
2025 Set B Q.2 (x) OR, 2025 Set D Q.2 (x) (stability of carbocations) and 2026 Set B Q.2 (iii) (a), which is solved in unit 17. Book long questions 2 and 3.
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Prepare an alkene (ethene, but-1-ene) by dehydration of an alcohol and by dehydrohalogenation of a bromoalkane. (short)
Both 2026 sets: Set B Q.2 (v) (but-1-ene) and Set D Q.2 (v) (ethene). Book SLOs 12 and 13.
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Reactions of ethene: hydration with \ce{H2SO4/H2O}, halohydration with \ce{Cl2/H2O} and cold dilute \ce{KMnO4} (diol). (short)
2025 Set B Q.5 (ii) OR and (iii) OR, 2025 Set D MCQ 14, and part (i) of 2025 Set D Q.2 (iv) OR (solved in unit 18).
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Why is ethene more reactive than ethane? Explain with the \pi bond. (short)
2025 Set B Q.5 (i) OR. Book short question 4 asks why alkenes show addition while alkanes cannot.
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Identify saturated and unsaturated hydrocarbons from the general formulae \ce{C_nH_{2n+2}}, \ce{C_nH_{2n}}, \ce{C_nH_{2n-2}}. (mcq)
2026 Set D MCQ 14.
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Stereoisomerism: geometric (cis-trans) isomerism and optical isomerism, with structures. (long)
Not in the 2025 or 2026 paper, but it is an SLO and the book has short questions 8, 9 and 10 and long question 6 on it.
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