Organic chemistry

Class 11 Chemistry, unit 15, book pages 284–317

Practise 63 MCQs Take the unit quiz Review 66 cards

Organic chemistry is the chemistry of carbon compounds. This unit explains why carbon forms so many compounds (catenation), how organic molecules are represented by molecular, empirical, condensed, displayed, skeletal and 3D formulae, and how functional groups and homologous series organise the compounds into families. It then teaches IUPAC naming of aliphatic compounds, the key terms of organic reactions (free radicals, electrophiles, nucleophiles, bond fission) with the mechanisms of three or four important reaction types, and finally structural isomerism.

Learning outcomes (10)

Define catenation and its importance in organic chemistry.
Recognize that hydrocarbons are the compounds of hydrogen and carbon only.
Recognize that alkanes are the simplest hydrocarbons with no functional group.
Functional groups determine the chemical and physical properties of organic compounds.
Interpret the general, structural, displayed and skeletal formulae of organic compounds.
Describe the use of systematic nomenclature of simple aliphatic organic compounds with functional groups.
Deduce the molecular and empirical formulae of a compound, given its structural, displayed and skeletal formulae.
Describe the terms saturated hydrocarbons, unsaturated hydrocarbons, heterolytic fission, homolytic fission, free radicals, initiation, propagation, termination, nucleophile, electrophile.
Explain free radical substitution, nucleophilic addition, electrophilic addition, nucleophilic substitution, elimination and redox reactions.
Describe structural isomerism and its types (chain isomerism, positional isomerism, functional groups isomerism, metamerism and tautomerism).

What is organic chemistry?

The word organic comes from the Greek word "organikos", meaning organ, and it was generalised to the organs of living organisms. The chemistry that deals with compounds containing carbon is called organic chemistry.

Organic compounds: Compounds composed of the element carbon. Carbon atoms in them are covalently bonded with one another and form the backbone of the molecule.
Inorganic compounds: Compounds made of elements other than carbon. They come from rocks and minerals.

In the early nineteenth century chemists thought that organic compounds could be obtained only from plants and animals (living or once living), and never from inorganic compounds.

The vital force theory and Wohler's synthesis of urea

In 1810, Jons Jacob Berzelius said that organic compounds originate from living things (plants and animals) and can be synthesised only from living matter. He thought that living things have a mysterious spiritual force, called the vital force (vital force theory), which was needed to prepare organic compounds. For example, urea is made inside the body of animals and, he said, cannot be synthesised from inorganic compounds in the laboratory.

In 1828 the German chemist Friedrich Wohler wanted to prepare ammonium cyanate from silver cyanate and ammonium chloride.

AgOCN (s) + NH X_{4} Cl (aq) NH X_{4} OCN (aq) + AgCl (s)

The white powder he obtained from the reaction mixture was not ammonium cyanate but urea, which had been reported earlier in urine. The mixture actually contained ammonium cyanate, which isomerised to urea at high temperature. In other words, Wohler synthesised urea, an organic compound, in the laboratory by heating ammonium cyanate, an inorganic compound.

(inorganic) NH X_{4} OCN ⟶ (organic) NH X_{2} - CO - NH X_{2}

urea Urea,

NH X_{2} CONH X_{2}

, the organic compound Wohler made from inorganic ammonium cyanate.

After the discovery of urea, Wohler wrote a letter to his mentor Berzelius, telling him that he could make urea without the use of the kidney of man or dog. He further said that ammonium cyanate is urea (they have the same formula).

Today several organic compounds are still obtained from living things, but very many organic compounds are synthesised in the laboratory from inorganic material. The chemistry of carbon is so vast that the number of compounds formed by carbon alone is greater than the number of compounds formed by all other elements in nature. Roughly speaking, there are almost 20 million organic compounds known so far.

15.1Catenation

Catenation: The ability of carbon atoms to form long chains and rings by bonding with other carbon atoms. The word comes from the Greek "catena", which means chain.

Some other elements show a limited degree of catenation: silicon forms small chains and sulphur forms small rings.

Importance of catenation in organic chemistry

Carbon is the essential part of all organic compounds, and catenation of carbon plays a significant role in organic chemistry. The existence of the large number of organic compounds in nature is due to the following reasons.

Catenation of carbon, making single, double and triple bonds.
Kinetic stability of organic compounds.
Attachment of carbon with other elements.

Carbon shows catenation because of its tendency to make strong carbon-carbon bonds and stable structures of molecules due to its small size. The presence of multiple bonds, the attachment of atoms other than carbon and hydrogen, and isomerism are the reasons behind the existence of such a large number of organic compounds in nature.

Worked example: bond energies and catenation

The bond energies of some single bonds between atoms of the same element are:

C - C

346

Si - Si

222

S - S

226

N - N

167

O - O

142

(

kJ mo l^{- 1}

). (a) By how much is the

C - C

bond stronger than the

Si - Si

bond? (b) How much energy breaks

0.25 mol

C - C

bonds? (c) Explain how the data support catenation of carbon.

Subtract the bond energies.
$346 - 222 = 124 kJ mo l^{- 1}$
Energy = moles of bonds $\times$ bond energy.
$0.25 \times 346 = 86.5 kJ$
Rank the bonds: $C - C$ (346) $>$ $S - S$ (226) $>$ $Si - Si$ (222) $>$ $N - N$ (167) $>$ $O - O$ (142). The $C - C$ bond is the strongest by a wide margin.
Explain. Catenation is the self-linking of carbon atoms into chains and rings. It needs strong bonds between atoms of the same element, so that long chains are stable and not easily broken. Carbon's small size and strong $C - C$ bonds give this, while the weaker bonds of silicon and sulphur allow only small chains and rings, and $N - N$ and $O - O$ bonds are weaker still.

124 kJ mo l^{- 1}, 86.5 kJ

15.2Representation of organic moleculesOften examined

Organic compounds have more complex structures than inorganic compounds, so they need more than one type of formula. The following types of formulae are used for organic compounds as the situation demands.

The formulae used for organic molecules
Type of formula	What it shows	Section
Molecular formula	Kinds of atoms and the actual whole-number ratio between them in a molecule	15.2.1
Empirical formula	The simplest whole-number ratio between the elements	15.2.2
Condensed structural formula	Relative positions of all atoms, without showing single bonds	15.2.3
Full structural (2D displayed) formula	All atoms and all bonds, in a plane	15.2.3
Skeletal formula	Only the carbon skeleton as lines, with functional groups	15.2.3
Stereochemical (3D) formula	Arrangement of the atoms in space	15.2.3

15.2.1Molecular formula

For convenience, scientists use specific symbols to represent elements and compounds. Such symbolic representations are used to make chemical formulae. A chemical reaction taking place in the laboratory can be shown with the help of chemical formulae on a board or notebook, so that students understand what happens in the reaction vessel. Such a representation of a chemical reaction using the formulae of the reactants and products is called a chemical equation.

Molecular formula: A formula that shows the type of atoms and the actual whole-number ratio between the atoms of different elements in a molecule of an organic compound.

Some molecular formulae of organic compounds
Compound	Molecular formula	Compound	Molecular formula
ethanol	$C X_{2} H X_{6} O$	ethene	$C X_{2} H X_{4}$
ethanoic acid	$C X_{2} H X_{4} O X_{2}$	benzene	$C X_{6} H X_{6}$
glucose	$C X_{6} H X_{12} O X_{6}$	ethyne	$C X_{2} H X_{2}$

15.2.2Empirical formulaOften examined

Organic molecules may have the same or a different ratio between the atoms in their molecular formulae. Some molecules, such as ethyne ( $C X_{2} H X_{2}$ ), have the same number of atoms of each element, whereas others, such as ethane ( $C X_{2} H X_{6}$ ), have different numbers.

Empirical formula: The symbolic representation of an organic compound which shows the simplest whole-number ratio between the elements in it.

Empirical formulae with the corresponding molecular formulae
Compound	Molecular formula	Empirical formula
ethanol	$C X_{2} H X_{6} O$	$C X_{2} H X_{6} O$
ethanoic acid	$C X_{2} H X_{4} O X_{2}$	$CH X_{2} O$
glucose	$C X_{6} H X_{12} O X_{6}$	$CH X_{2} O$
benzene	$C X_{6} H X_{6}$	$CH$
ethyne	$C X_{2} H X_{2}$	$CH$

Empirical formulae may mislead us in identifying organic compounds. For example, both glucose and ethanoic acid have the same empirical formula $CH X_{2} O$ , so it cannot tell them apart. Similarly, some compounds have the same empirical and molecular formulae, like sucrose ( $C X_{12} H X_{22} O X_{11}$ ).

Determination of empirical and molecular formulae from combustion analysis

Example 1: empirical formula from percentage composition

An organic compound has percentage composition by mass C =

38.4%

, H =

4.8%

and Cl =

56.8%

. Calculate the empirical formula of this compound.

Given

$C = 38.4 %$
$H = 4.8 %$
$Cl = 56.8 %$

To find

empirical formula

Take $100 g$ of the compound, so the masses are $38.4 g$ C, $4.8 g$ H and $56.8 g$ Cl. Calculate the moles of each element (mass divided by atomic mass).
$n_{C} = \frac{38.4}{12} = 3.2, n_{H} = \frac{4.8}{1.0} = 4.8, n_{Cl} = \frac{56.8}{35.5} = 1.6$
Divide all the moles by the smallest value (1.6) to get the ratio of atoms.
$C = \frac{3.2}{1.6} = 2, H = \frac{4.8}{1.6} = 3, Cl = \frac{1.6}{1.6} = 1$
The ratio C : H : Cl is 2 : 3 : 1, so the empirical formula has 2 carbon atoms, 3 hydrogen atoms and 1 chlorine atom.
$C X_{2} H X_{3} Cl$

C X_{2} H X_{3} Cl

Example 2: molecular formula from the empirical formula

An organic compound has empirical formula

CH

and molecular mass

78 amu

. Calculate its molecular formula.

Given

$empirical formula = CH$
$M_{r} = 78$

To find

molecular formula

The molecular formula is a whole-number multiple $n$ of the empirical formula, so first find $n$ . The empirical formula mass of $CH$ is $12 + 1 = 13$ .
$n = \frac{molecular mass}{empirical formula mass} = \frac{78}{13} = 6$
Multiply every subscript of the empirical formula by $n$ .
$molecular formula = n (empirical formula) = 6 (CH) = C X_{6} H X_{6}$

C X_{6} H X_{6} (benzene)

Worked example: empirical and molecular formula when the ratio is not whole

A compound has percentage composition C

48.65%

, H

8.11%

, O

43.24%

and relative molecular mass

74

. Find its empirical and molecular formula.

Given

$C = 48.65 %$
$H = 8.11 %$
$O = 43.24 %$
$M_{r} = 74$

To find

empirical formula
molecular formula

Take $100 g$ and divide each mass by the atomic mass.
$n_{C} = \frac{48.65}{12} = 4.054, n_{H} = \frac{8.11}{1.0} = 8.11, n_{O} = \frac{43.24}{16} = 2.702$
Divide by the smallest value ( $2.702$ ).
$C = 1.50, H = 3.00, O = 1.00$
$1.50$ is not a whole number, so do not round it. Multiply all the ratios by $2$ .
$C : H : O = 3 : 6 : 2$
The empirical formula is $C X_{3} H X_{6} O X_{2}$ , with mass $3 (12) + 6 (1) + 2 (16) = 74$ .
$C X_{3} H X_{6} O X_{2}$
Find $n$ = molecular mass divided by empirical formula mass.
$n = \frac{74}{74} = 1 \Rightarrow molecular formula C X_{3} H X_{6} O X_{2}$

The formula alone does not give the structure. Both of these compounds are $C X_{3} H X_{6} O X_{2}$ , one a carboxylic acid and one an ester.

propanoic acid

methyl ethanoate

empirical C X_{3} H X_{6} O X_{2}, molecular C X_{3} H X_{6} O X_{2}

15.2.3Structural formulaeOften examined

A structural formula tells us two things about a molecule: the way the atoms are arranged and the type of functional group in it. A molecular formula cannot do this, which is why we need structural formulae.

1. Condensed structural formula

A condensed structural formula tells us about the relative positions of all atoms in a molecule, without showing the single covalent bonds. Each carbon atom is shown individually with the hydrogen atoms attached to it. The branches, functional groups and repeating methylene groups in the middle of molecules are shown in brackets.

The condensed structural formula of acetic acid, $CH X_{3} COOH$ , shows that three hydrogen atoms are bonded to the first carbon and the remaining one is attached to one of the two oxygen atoms. It also indicates that the carboxyl group ( $COOH$ ) is the functional group of acetic acid.

Condensed structural formulae of the first ten alkanes
Alkane	Condensed structural formula	Alkane	Condensed structural formula
methane	$CH X_{4}$	hexane	$CH X_{3} (CH X_{2}) X_{4} CH X_{3}$
ethane	$CH X_{3} CH X_{3}$	heptane	$CH X_{3} (CH X_{2}) X_{5} CH X_{3}$
propane	$CH X_{3} CH X_{2} CH X_{3}$	octane	$CH X_{3} (CH X_{2}) X_{6} CH X_{3}$
butane	$CH X_{3} (CH X_{2}) X_{2} CH X_{3}$	nonane	$CH X_{3} (CH X_{2}) X_{7} CH X_{3}$
pentane	$CH X_{3} (CH X_{2}) X_{3} CH X_{3}$	decane	$CH X_{3} (CH X_{2}) X_{8} CH X_{3}$

2. Full structural formula

A full structural formula indicates all the atoms and all the bonds among them in a molecule. It is a two-dimensional formula, so it is also called the 2D displayed formula.

Figure 15.1: Full structural (2D displayed) formulae of ethane and ethanol.

3. Skeletal formula

Some organic molecules are very large and complex, so it is quite difficult to draw all their atoms. The skeletal formula is the easiest-to-draw representation for complex organic molecules. Only lines are drawn: each end and each vertex (corner) of a line is assumed to be a carbon atom, unless another atom such as oxygen or a halogen is written there. Each carbon atom is assumed to be bonded to enough hydrogen atoms to complete its valency.

In a skeletal formula, the carbon and hydrogen atoms are not shown.
Double and triple bonds between carbon atoms are shown.
The functional group is also shown.

butane Figure 15.2: Skeletal formulae. Each end and each bend of the zigzag line is a carbon atom.

butan-2-ol

4. Stereochemical formula

The 2D displayed formula gives a complete picture of all the types of bonds in a molecule but tells us nothing about the arrangement of atoms in the three dimensions of space. Chemists needed a formula that could give the structure of molecules in space, with atoms in the $x$ , $y$ and $z$ planes. This is the stereochemical (3D) formula.

Figure 15.3: Stereochemical (3D) formula of methane. A plain line is a bond in the plane of the paper, a wedge comes out towards the viewer, and a hashed line goes back behind the plane.

The following table compares the four ways of writing five compounds with three carbon atoms.

Molecular and condensed formulae of five three-carbon compounds
Name	Molecular formula	Condensed structural formula
propane	$C X_{3} H X_{8}$	$CH X_{3} CH X_{2} CH X_{3}$
propene	$C X_{3} H X_{6}$	$CH X_{3} CH = CH X_{2}$
propan-2-ol	$C X_{3} H X_{8} O$	$CH X_{3} CH (OH) CH X_{3}$
propanal	$C X_{3} H X_{6} O$	$CH X_{3} CH X_{2} CHO$
propanone	$C X_{3} H X_{6} O$	$CH X_{3} COCH X_{3}$

Figure 15.4: Displayed formulae of the compounds in the table.

propane

propene

propan-2-ol

propanal

propanone

Concept assessment exercise 15.1, question 1

Draw condensed, displayed and skeletal formulae of (i) ethene, (ii) ethanol, (iii) butanone, (iv) ethanal.

Write the condensed formula of each by writing each carbon with its hydrogens and keeping the functional group in its place.

Compound	Molecular formula	Condensed formula
ethene	$C X_{2} H X_{4}$	$CH X_{2} = CH X_{2}$
ethanol	$C X_{2} H X_{6} O$	$CH X_{3} CH X_{2} OH$
butanone	$C X_{4} H X_{8} O$	$CH X_{3} COCH X_{2} CH X_{3}$
ethanal	$C X_{2} H X_{4} O$	$CH X_{3} CHO$

Displayed formulae: draw every atom and every bond (a double bond for $C = C$ and $C = O$ ).

Figure 15.5: Displayed formulae for Concept assessment exercise 15.1 (question 1).

Skeletal formulae: draw only the carbon skeleton as lines and show the double bond and the oxygen.

ethene

ethanol

butanone

ethanal

Concept assessment exercise 15.1, question 2

Draw the 2D displayed formulae and skeletal formulae of (i) 2-methylpentane, (ii) 2,4-dimethylheptane, (iii) 3,3,5-trimethyloctane, (iv) 2,2-dimethylpropane.

Find the parent chain from the name: pent- (5 carbons), hept- (7), oct- (8) and prop- (3). All four are alkanes (-ane), so all carbon-carbon bonds are single.
Number the chain and attach each methyl group (a $CH X_{3}$ branch) at the position given in the name: one at C2 for (i); at C2 and C4 for (ii); two at C3 and one at C5 for (iii); two at C2 for (iv).
Complete every carbon with hydrogen atoms so that it has four bonds, then draw the bonds.

Compound	Condensed formula
2-methylpentane	$CH X_{3} CH (CH X_{3}) CH X_{2} CH X_{2} CH X_{3}$
2,4-dimethylheptane	$CH X_{3} CH (CH X_{3}) CH X_{2} CH (CH X_{3}) CH X_{2} CH X_{2} CH X_{3}$
3,3,5-trimethyloctane	$CH X_{3} CH X_{2} C (CH X_{3}) X_{2} CH X_{2} CH (CH X_{3}) CH X_{2} CH X_{2} CH X_{3}$
2,2-dimethylpropane	$C (CH X_{3}) X_{4}$

Figure 15.6: 2D displayed formula of 2-methylpentane (i).

Figure 15.7: 2D displayed formula of 2,4-dimethylheptane (ii).

Figure 15.8: 2D displayed formula of 3,3,5-trimethyloctane (iii).

Figure 15.9: 2D displayed formula of 2,2-dimethylpropane (iv).

Skeletal formulae of the four compounds:

2-methylpentane

2,4-dimethylheptane

3,3,5-trimethyloctane

2,2-dimethylpropane

15.3Functional groupOften examined

Functional group: An atom or group of atoms which is attached to carbon chains or rings and determines the characteristic properties of organic compounds.

All organic compounds have two parts: a reactive functional group and a relatively unreactive carbon network. For example, the hydroxyl group (\ce{-OH}) and the carboxyl group (\ce{-COOH}) are the functional groups of alcohols and carboxylic acids respectively.

Since chemical reactions take place at the functional group, the chemical properties of organic compounds depend on the functional group. Moreover, double and triple bonds determine the properties of alkenes and alkynes respectively, so they are also included in functional groups.

15.3.1Homologous seriesOften examined

Since the number of organic compounds is very large, we cannot study them one by one. We need to classify them into different homologous series, which help us predict the properties of all the organic compounds individually.

Homologous series: A group of organic compounds which have the same functional group but different lengths of the carbon chain.

Some important homologous series are alkanes, alcohols and aldehydes. Homologous series have the following features.

Each homologous series has its own functional group.
Each member of a homologous series differs from the next member by one methylene group (\ce{-CH2-}).
Each homologous series has its own general formula. For example, the general formulae of alkanes, alkenes and alcohols are \mathrm{C}_n\mathrm{H}_{2n+2}, \mathrm{C}_n\mathrm{H}_{2n} and \mathrm{C}_n\mathrm{H}_{2n+1}\mathrm{OH} respectively.
All members of a homologous series have similar chemical properties.
A homologous series has a gradual change in physical properties with changing size of the carbon chain.
They have a similar method of preparation.

The first four alkanes fit \mathrm{C}_n\mathrm{H}_{2n+2} and each differs from its neighbour by one \ce{CH2}
Alkane	Formula	Difference from the one before
methane	\ce{CH4}	-
ethane	\ce{C2H6}	\ce{CH2}
propane	\ce{C3H8}	\ce{CH2}
butane	\ce{C4H10}	\ce{CH2}

The important homologous series and functional groups

Important homologous series (X stands for a halogen; R for an alkyl group)
Homologous series	General formula	Functional group	Example
Alkanes (\ce{R-H})	\mathrm{C}_n\mathrm{H}_{2n+2}	no functional group	\ce{CH3-CH3}, ethane
Alkenes	\mathrm{C}_n\mathrm{H}_{2n}	double bond \ce{C=C}	\ce{CH2=CH2}, ethene
Alkynes	\mathrm{C}_n\mathrm{H}_{2n-2}	triple bond \ce{-C#C-}	\ce{H-C#C-H}, ethyne
Halogenoalkanes (\ce{R-X})	\mathrm{C}_n\mathrm{H}_{2n+1}\mathrm{X}	halo group \ce{-X}	\ce{CH3CH2Cl}, chloroethane
Alcohols (\ce{R-OH})	\mathrm{C}_n\mathrm{H}_{2n+1}\mathrm{OH}	hydroxyl group \ce{-OH}	\ce{CH3CH2OH}, ethanol
Ethers (\ce{R-O-R})	\mathrm{C}_n\mathrm{H}_{2n+1}\mathrm{O}\mathrm{C}_m\mathrm{H}_{2m+1}	ether group \ce{-O-} joining two carbons	\ce{CH3-O-CH3}, methoxymethane
Amines (\ce{R-NH2})	\mathrm{C}_n\mathrm{H}_{2n+1}\mathrm{NH}_2	amine group \ce{-NH2}	\ce{CH3NH2}, methylamine
Nitriles (\ce{R-C#N})	\mathrm{C}_n\mathrm{H}_{2n+1}\mathrm{CN}	nitrile group \ce{-C#N}	\ce{CH3CN}, ethanenitrile
Aldehydes (\ce{R-CHO})	\mathrm{C}_n\mathrm{H}_{2n+1}\mathrm{CHO}	aldehyde group \ce{-CHO}	\ce{CH3CHO}, ethanal
Ketones (\ce{R-CO-R})	\mathrm{C}_n\mathrm{H}_{2n+1}\mathrm{CO}\mathrm{C}_m\mathrm{H}_{2m+1}	ketone group \ce{-CO-} between two carbons	\ce{CH3COCH3}, propanone
Carboxylic acids (\ce{R-COOH})	\mathrm{C}_n\mathrm{H}_{2n+1}\mathrm{COOH}	carboxyl group \ce{-COOH}	\ce{CH3CH2COOH}, propanoic acid
Esters (\ce{R-COOR})	\mathrm{C}_n\mathrm{H}_{2n+1}\mathrm{COO}\mathrm{C}_m\mathrm{H}_{2m+1}	ester group \ce{-COO-}	\ce{CH3COOCH2CH3}, ethyl ethanoate
Amides (\ce{R-CONH2})	\mathrm{C}_n\mathrm{H}_{2n+1}\mathrm{CONH}_2	amide group \ce{-CONH2}	\ce{CH3CONH2}, ethanamide
Acid halides (\ce{R-COX})	\mathrm{C}_n\mathrm{H}_{2n+1}\mathrm{COX}	acid halide group \ce{-COX}	\ce{CH3COCl}, ethanoyl chloride
Arenes	\mathrm{C}_n\mathrm{H}_{2n-6m} (m = number of rings)	phenyl group \ce{C6H5-}	\ce{C6H5CH2CH3}, ethylbenzene

Concept assessment exercise 15.2

Write the general formulae and functional groups of the following homologous series: (i) alcohols, (ii) aldehydes, (iii) ketones, (iv) ethers, (v) carboxylic acids, (vi) esters, (vii) amines, (viii) nitriles.

	Homologous series	General formula	Functional group
i	alcohols	\mathrm{C}_n\mathrm{H}_{2n+1}\mathrm{OH}	hydroxyl group, \ce{-OH}
ii	aldehydes	\mathrm{C}_n\mathrm{H}_{2n+1}\mathrm{CHO}	aldehyde group, \ce{-CHO} (carbonyl carbon at the end of the chain)
iii	ketones	\mathrm{C}_n\mathrm{H}_{2n+1}\mathrm{CO}\mathrm{C}_m\mathrm{H}_{2m+1}	ketone group, \ce{-CO-} (carbonyl carbon inside the chain)
iv	ethers	\mathrm{C}_n\mathrm{H}_{2n+1}\mathrm{O}\mathrm{C}_m\mathrm{H}_{2m+1}	ether group, \ce{-O-} between two carbons
v	carboxylic acids	\mathrm{C}_n\mathrm{H}_{2n+1}\mathrm{COOH}	carboxyl group, \ce{-COOH}
vi	esters	\mathrm{C}_n\mathrm{H}_{2n+1}\mathrm{COO}\mathrm{C}_m\mathrm{H}_{2m+1}	ester group, \ce{-COO-}
vii	amines	\mathrm{C}_n\mathrm{H}_{2n+1}\mathrm{NH}_2	amine group, \ce{-NH2}
viii	nitriles	\mathrm{C}_n\mathrm{H}_{2n+1}\mathrm{CN}	nitrile group, \ce{-C#N}

15.3.2Hydrocarbons

Hydrocarbons: Compounds which have only the elements carbon and hydrogen in their molecules.

In hydrocarbons the carbon atoms act like the backbone, whereas the hydrogen atoms act like the skin of the molecule.

Alkanes are the simplest of all organic compounds because they have no functional group, just a carbon skeleton. They are the least reactive organic compounds, for two reasons.

The \ce{C-C} and \ce{C-H} bonds are very strong and difficult to break.
The \ce{C-C} and \ce{C-H} bonds are non-polar, so they cannot be attacked by polar reagents such as acids, bases, oxidizing agents and reducing agents.

Bond energies of some single bonds between atoms of the same element
Bond	Bond energy (\mathrm{kJ\,mol^{-1}})	Bond	Bond energy (\mathrm{kJ\,mol^{-1}})
\ce{C-C}	346	\ce{Si-Si}	222
\ce{N-N}	167	\ce{P-P}	201
\ce{O-O}	142	\ce{S-S}	226

Broadly, hydrocarbons are classified into two categories: saturated and unsaturated hydrocarbons.

15.3.3Saturated hydrocarbons

Saturated hydrocarbons: Hydrocarbons which have only single covalent bonds between the carbon atoms in their molecules. All alkanes and cycloalkanes are saturated hydrocarbons.

The word "saturated" is used because each carbon atom is bonded to four atoms, so the carbon atoms have reached their maximum combining capacity. The following are some examples of saturated hydrocarbons.

Figure 15.10: Saturated hydrocarbons: ethane and butane (open chain), cyclobutane and cyclopentane (rings).

Worked example: saturated or unsaturated from the molecular formula

Assuming an open chain, decide which of \ce{C7H16}, \ce{C7H14} and \ce{C7H12} are saturated hydrocarbons. Why is \ce{C5H10} not enough to decide when rings are allowed?

Compare with the open-chain alkane formula \mathrm{C}_n\mathrm{H}_{2n+2}. For n=7 it gives \ce{C7H16}, so \ce{C7H16} is a saturated alkane.
\ce{C7H14} fits \mathrm{C}_n\mathrm{H}_{2n}, the general formula of alkenes, so as an open chain it has a double bond and is unsaturated.
\ce{C7H12} fits \mathrm{C}_n\mathrm{H}_{2n-2}, the formula of alkynes, so it has a triple bond and is unsaturated.
For \ce{C5H10} the formula \mathrm{C}_n\mathrm{H}_{2n} fits both pent-1-ene (open chain, unsaturated) and cyclopentane (a ring with only single bonds, saturated). So the formula alone is not enough: the structure must be known.

pent-1-ene (unsaturated)

cyclopentane (saturated)

\text{open chain: } \ce{C7H16}\ \text{is saturated; } \ce{C7H14} \text{ and } \ce{C7H12} \text{ are unsaturated}

15.3.4Alkyl group (R-)Often examined

Alkyl group: A group obtained by removing any one hydrogen atom from an alkane. It is written \ce{R-}.

Alkyl groups are named by replacing the suffix "ane" of the alkane with "yl", which means that the root name of the corresponding alkane is followed by the suffix "yl". In the book, primary hydrogens are shown in blue, secondary in red and tertiary in green.

Alkyl groups obtained by removing one hydrogen atom (the book writes n-butane and butane for the same compound)
Parent alkane	Hydrogen removed from	Alkyl group	Name
methane \ce{CH4}	the only carbon	\ce{CH3\bond{-}}	methyl
ethane \ce{CH3CH3}	either carbon	\ce{CH3CH2\bond{-}}	ethyl
propane \ce{CH3CH2CH3}	an end (primary) carbon	\ce{CH3CH2CH2\bond{-}}	n-propyl (common name)
propane \ce{CH3CH2CH3}	the middle (secondary) carbon	\ce{CH3CH(\bond{-})CH3} or \ce{(CH3)2CH\bond{-}}	isopropyl (common name)
n-butane \ce{CH3CH2CH2CH3}	an end (primary) carbon	\ce{CH3CH2CH2CH2\bond{-}}	n-butyl (common name)
n-butane \ce{CH3CH2CH2CH3}	a second (secondary) carbon	\ce{CH3CH2CH(\bond{-})CH3}	secondary butyl (common name)
isobutane \ce{(CH3)3CH}	the tertiary carbon	\ce{(CH3)3C\bond{-}}	tertiary butyl (common name)

Primary, secondary and tertiary carbon and hydrogen

A primary carbon is further attached to one carbon atom or none; a secondary carbon is bonded further to two carbon atoms; a tertiary carbon is attached further to three carbon atoms.
Primary hydrogen atoms are attached to a primary carbon, secondary hydrogens to a secondary carbon and tertiary hydrogens to a tertiary carbon.
In a secondary alkyl group the secondary hydrogen is removed, while in a tertiary alkyl group the tertiary hydrogen is removed.
The prefix "iso" refers to any structure which has a substituent at the second-last carbon of the parent chain.

propane (parent of n-propyl and isopropyl)

isobutane (parent of tertiary butyl)

15.3.5Unsaturated hydrocarbons

Unsaturated hydrocarbons: Hydrocarbons which have one or more carbon-carbon double or triple bonds. The word "unsaturated" is used because more atoms can be added to their carbon atoms. Alkenes and alkynes are examples.

Each doubly bonded carbon is attached to three atoms, so it has the ability to bond one other atom. Similarly, a triply bonded carbon is bonded to two atoms and can bond to two more atoms.

Figure 15.11: Unsaturated hydrocarbons: propene, propyne and cyclobutene.

Unsaturated hydrocarbons show addition reactions, because more molecules can be added across the carbon atoms holding the multiple bond (double or triple bond).

15.3.6Terminology associated with naming organic compounds

For naming organic compounds we must be familiar with some important terms.

Prefix. It may be a branch or substituent attached to the longest carbon chain, for example alkyl groups and halo groups.
Stem or root name. The stem name of an organic molecule indicates the number of carbon atoms in the longest carbon chain, for example meth, eth, prop.
Suffix of the root name. An ending (suffix) to the root name, like an, en or yn, which shows whether there are only single bonds, a double bond or a triple bond in the parent chain of the molecule.
Suffix. The last part of an organic name, which indicates the family name.

Figure 15.12: The four parts of an IUPAC name and what each shows.

Frequently used root names in IUPAC nomenclature
Number of carbons	Root name	Number of carbons	Root name
one	meth-	six	hex-
two	eth-	seven	hept-
three	prop-	eight	oct-
four	but-	nine	non-
five	pent-	ten	dec-

Prefixes and suffixes of the main families
Class or family	Prefix or suffix	Example	Name
Alkanes	suffix, -ane	\ce{CH3CH2CH2CH3}	butane
Alkenes	suffix, -ene	\ce{CH3CH=CHCH3}	but-2-ene
Alkynes	suffix, -yne	\ce{CH3CH2C#CH}	but-1-yne
Halogenoalkanes	prefix, halo (fluoro-, chloro-, bromo-, iodo-)	\ce{CH3CH(Cl)CH3}	2-chloropropane
Alcohols	suffix, -ol	\ce{CH3CH(OH)CH2CH2CH3}	pentan-2-ol
Ethers	prefix, alkoxy	\ce{CH3OCH2CH3}	methoxyethane
Aldehydes	suffix, -al	\ce{CH3CH2CHO}	propanal
Ketones	suffix, -one	\ce{CH3COCH2CH2CH3}	pentan-2-one
Amines	suffix, -amine	\ce{CH3CH2NH2}	ethylamine (common name; systematic name ethanamine)
Nitriles	suffix, -nitrile	\ce{CH3CH2CN}	propanenitrile
Carboxylic acids	suffix, -oic acid	\ce{CH3CH2COOH}	propanoic acid
Acid halides	suffix, -oyl halide	\ce{CH3COCl}	ethanoyl chloride
Amides	suffix, -amide	\ce{CH3CONH2}	ethanamide
Arenes (complicated class)	suffix, -benzene; prefix, phenyl (varies with compounds)	\ce{C6H5CH3}	methylbenzene

15.4Nomenclature of aliphatic organic compoundsOften examined

Aliphatic compounds: Organic compounds which have straight-chain or branched-chain molecules. Cyclic molecules, except benzene, are also included in aliphatic compounds. They may be both saturated and unsaturated.

There are millions of organic compounds, many with quite complex structures, which are difficult to name. The International Union of Pure and Applied Chemistry (IUPAC) is the universally recognised authority on naming organic and inorganic compounds. In this topic we bring only organic compounds under consideration. IUPAC provides internationally accepted rules for giving systematic names to organic compounds.

The systematic names of all homologous series, except carboxylic acids, are written as one word (like "Salman" is one word). Carboxylic acids are written as two words, for example propanoic acid.
The non-systematic (trivial or common) names are still used for some compounds in specific situations.

15.4.1Nomenclature of alkanesOften examined

Alkanes provide the basis for the nomenclature of organic compounds, so we apply the IUPAC rules to alkanes first and then extend them to the other families of organic compounds.

Common (trivial) names of alkanes

n-pentane

isopentane

neopentane

Systematic names of alkanes

The following rules are applied while giving systematic (IUPAC) names to alkanes.

Rule 1: find the longest continuous carbon chain

Identify the longest continuous carbon chain (the parent chain) in the given compound. The number of carbon atoms in the longest chain gives the root name of the compound.

The book shows three drawings (a straight chain and two folded chains) with seven carbon atoms in the parent chain, so the root name is "hept". All the carbon-carbon bonds are single, so the compounds belong to the alkane family with the suffix "ane". The systematic name of all three is heptane. They are straight-chain molecules of heptane, identical in all respects.

heptane (all three drawings in the book)

Rule 2: identify the substituents and their positions

Identify the substituents (branches or side chains) and their position (location) and use them as the prefixes of the systematic name. In alkanes the alkyl groups are the substituents attached to the principal chain. The position of a substituent is the number of the carbon atom (the locant number) in the parent chain to which it is attached. Number the parent chain from the end where the substituents come closer and get the smallest locant.

4-methyloctane

How 4-methyloctane is named

The longest chain has 8 carbons (octane) and a methyl group is attached to it.
Numbering from the right gives the methyl locant 4, and numbering from the left gives 5. The smaller number is chosen.
Name: locant, hyphen, substituent, root name and suffix.
\text{4-methyloctane}

Rule 3: identical substituents use di, tri, tetra

If two or more identical substituents are attached to the parent chain, use the prefixes di, tri, tetra etc. with their names. If two identical substituents are bonded to the same carbon atom of the principal chain, repeat that locant. If identical substituents are bonded to different carbon atoms, write their locants separately before the prefixes di, tri, tetra etc. and before their names.

3,3-dimethylheptane

3,4-dimethylheptane

3,3,5-trimethyloctane

3,3-dimethylheptane: two methyl groups on the same carbon (C3), so the locant 3 is repeated.
3,4-dimethylheptane: two methyl groups on different carbons (C3 and C4), so the locants are written once each, separated by a comma.
3,3,5-trimethyloctane: three methyl groups (tri), two on C3 and one on C5, so the locants are 3,3,5.

Rule 4: different substituents go in alphabetical order

If there are two or more different substituents in a molecule, whether on the same or on different carbon atoms of the parent chain, write their corresponding locants separately before them and put their names in alphabetical order before the root name of the molecule.

5-ethyl-3-methyloctane

The parent chain has 8 carbons (octane). Substituents: an ethyl group on C5 and a methyl group on C3.
Alphabetical order puts ethyl (e) before methyl (m), so ethyl is named first.
\text{5-ethyl-3-methyloctane}

Worked example: naming an alkane: choose the longest chain

Name the alkane \ce{CH3CH(CH3)CH2CH(CH2CH3)CH2CH2CH3}.

the compound to be named

Rule 1: find the longest continuous chain. Starting at one methyl of the isopropyl end and going through the carbon that carries the ethyl group to the propyl end gives seven carbons. The path that goes through the ethyl group has only six. The parent is therefore heptane, and the ethyl group is a branch.
Rule 2: identify the substituents: a methyl group and an ethyl group.
Number the chain from both ends and compare the locants. From the isopropyl end the substituents are on C2 (methyl) and C4 (ethyl): \{2,4\}. From the other end they are on C4 (ethyl) and C6 (methyl): \{4,6\}. The numbering must start from the end where the substituents get the smallest locant, so number from the isopropyl end.
\{2,4\}<\{4,6\}
Rule 4: write the substituents in alphabetical order, ethyl before methyl, and join the parts with hyphens.
\text{4-ethyl-2-methylheptane}

\text{4-ethyl-2-methylheptane}

15.4.2Nomenclature of cycloalkanesOften examined

The following rules are used for naming cycloalkanes according to the IUPAC system.

Rule 1: prefix cyclo

While naming cycloalkanes, their stem name is prefixed by "cyclo".

cyclopropane

cyclopentane

cyclohexane

Rule 2: substituted cycloalkanes

In substituted cycloalkanes the names of the substituents are put before the name of the cycloalkane along with their position on the carbon ring. If there is more than one different substituent, put them in alphabetical order. One substituent should be given position 1, and the numbering should continue either clockwise or anti-clockwise so that the remaining substituents get the lowest numbers at the first point of difference.

1-ethyl-2-methylcyclopropane

1,1,3-trimethylcyclohexane

2-ethyl-1,1-dimethylcyclopentane

1-ethyl-2-methylcyclopropane: the ring has an ethyl and a methyl group on neighbouring carbons. Ethyl comes first alphabetically, so it gets position 1 and the methyl gets 2.
1,1,3-trimethylcyclohexane: two methyl groups are on the same ring carbon (position 1, locant repeated) and a third is on the carbon two steps away (position 3).
2-ethyl-1,1-dimethylcyclopentane: the carbon carrying two methyl groups is numbered 1, and the ethyl group on the next carbon gets 2, although ethyl is alphabetically first.

Concept assessment exercise 15.3

Give systematic names to the following compounds: (i) \ce{C(CH3)4}; (ii) \ce{CH3CH2CH(C2H5)CH3}; (iii) \ce{(CH3)3CH}; (iv) \ce{CH3(CH2)4C(CH3)3}; (v) \ce{(CH3)3C(CH2)3C(C2H5)3}; (vi) \ce{C(C2H5)4}; (vii) a cyclopentane ring carrying an ethyl group on one carbon and a methyl and an ethyl group together on a carbon two places away.

	Compound	Working	Name
i	\ce{C(CH3)4}	Central carbon with four methyl groups. Longest chain is 3 carbons (propane); two methyl groups on C2.	2,2-dimethylpropane
ii	\ce{CH3CH2CH(C2H5)CH3}	The carbon with the ethyl group has \ce{CH2CH3}, \ce{C2H5} and \ce{CH3} attached. Longest chain is 5 carbons (\ce{CH3CH2-CH-CH2CH3}) with a methyl group on C3.	3-methylpentane
iii	\ce{(CH3)3CH}	Longest chain is 3 carbons; one methyl group on C2.	2-methylpropane
iv	\ce{CH3(CH2)4C(CH3)3}	Longest chain: the \ce{CH3(CH2)4C} part plus one methyl of the \ce{C(CH3)3} group gives 7 carbons; two methyl groups on the carbon next to the end (C2).	2,2-dimethylheptane
v	\ce{(CH3)3C(CH2)3C(C2H5)3}	Longest chain: \ce{CH3-C-(CH2)3-C-CH2-CH3} is 8 carbons. Two methyl groups on C2 and two ethyl groups on C6 (numbering from the other end would give 3,3,7,7, which is higher). Ethyl comes before methyl alphabetically.	6,6-diethyl-2,2-dimethyloctane
vi	\ce{C(C2H5)4}	Central carbon with four ethyl groups. Longest chain is 5 carbons (pentane); a second ethyl on C3.	3,3-diethylpentane
vii	ring with \ce{C2H5} on one carbon and \ce{CH3} + \ce{C2H5} on another	Ring of 5 carbons. Number from the carbon with two substituents as 1, so the locants are 1,1,3 (the other direction gives 1,3,3). Alphabetical order: ethyl before methyl, "di" for the two ethyl groups.	1,3-diethyl-1-methylcyclopentane

(i) 2,2-dimethylpropane

(ii) 3-methylpentane

(iii) 2-methylpropane

(iv) 2,2-dimethylheptane

(v) 6,6-diethyl-2,2-dimethyloctane

(vi) 3,3-diethylpentane

(vii) 1,3-diethyl-1-methylcyclopentane

15.4.3Nomenclature of alkenes and alkynesOften examined

Alkenes and alkynes follow the same rules of IUPAC nomenclature as alkanes, but they have a functional group (the carbon-carbon double or triple bond) which must be located and put in the systematic name.

Rule 1: the parent chain must contain the multiple bond

The parent chain of alkenes and alkynes must have the double and triple bond respectively. The position of the double or triple bond must follow the root name immediately, which is followed by the suffixes "ene" for alkenes and "yne" for alkynes.

ethene

propene

Rule 2: number from the end nearer the multiple bond

Numbering must start from that side of the parent chain which is closer to the multiple bond.

4-methylpent-1-ene

5-methylhex-1-yne

4-methylpent-1-ene: the longest chain containing the double bond has 5 carbons. Numbering from the double-bond end puts the double bond at C1 (the other direction would put it at C4) and the methyl group at C4.
5-methylhex-1-yne: 6 carbons, the triple bond at C1 when numbered from the \ce{C#CH} end, and the methyl group at C5.

Rule 3: more than one multiple bond

If there is more than one multiple bond in a molecule, add "a" to the stem name and put di, tri etc. before "ene" in alkenes and "yne" in alkynes.

2-methylbuta-1,3-diene

3-methylhexa-1,5-diyne

Rules 4 to 6: substituents, mixed bonds and numbering ties

For naming and putting substituents in order, the same rules are applied as for alkanes.
If the molecule has both a double and a triple bond, use the ending "en" for the root name and the suffix "yne" for the whole molecule.
Numbering must be started from the end where the multiple bonds come closer. If the double and triple bonds are at the same distance from either end, prefer the side for numbering where the double bond comes closer (the first letter "e" of "ene" comes before the letter "y" of "yne" alphabetically).

4-methyloct-1-en-7-yne

4-ethylhex-1-en-5-yne

4-methyloct-1-en-7-yne: numbering from the double-bond end gives the multiple bonds the locants 1 and 7. Numbering from the other end would also give 1 (triple bond) and 7 (double bond), a tie, so the double bond is given the lower number.
4-ethylhex-1-en-5-yne: the same tie (1 and 5 either way) is settled in favour of the double bond, so "1-en-5-yne".

Concept assessment exercise 15.4, question 1

Draw the skeletal formulae of (i) but-1-ene, (ii) penta-1,2-diene, (iii) 5-methylhepta-1,3-diyne, (iv) hex-1-en-4-yne.

Write the parent chain from the root name and put the multiple bonds at the locants given: but = 4 carbons with a double bond starting at C1; penta = 5 carbons with double bonds at C1 and C2; hepta = 7 carbons with triple bonds starting at C1 and C3 and a methyl group on C5; hex = 6 carbons with a double bond at C1 and a triple bond at C4.

(i) but-1-ene

(ii) penta-1,2-diene

(iii) 5-methylhepta-1,3-diyne

(iv) hex-1-en-4-yne

Concept assessment exercise 15.4, question 2

Give systematic names to the following compounds: (i) \ce{CH2=CH(CH2)3C#CCH3}; (ii) \ce{CH2=CH(CH2)4CH(CH3)C#CH}; (iii) \ce{CH3CH=C=CH2}; (iv) \ce{CH3CH=C(CH3)CH(CH3)2}.

	Compound	Working	Name
i	\ce{CH2=CH(CH2)3C#CCH3}	Chain of 8 carbons. Numbering from the double-bond end gives the double bond at C1 and the triple bond at C6 (locants 1,6). The other direction gives 2,7. The lower set is chosen.	oct-1-en-6-yne
ii	\ce{CH2=CH(CH2)4CH(CH3)C#CH}	Chain of 9 carbons. Either direction gives the multiple bonds the locants 1 and 8, a tie, so the double bond gets the lower number. The methyl group is then on C7.	7-methylnon-1-en-8-yne
iii	\ce{CH3CH=C=CH2}	Chain of 4 carbons with two double bonds. Numbering from the \ce{CH2} end puts them at C1 and C2.	buta-1,2-diene
iv	\ce{CH3CH=C(CH3)CH(CH3)2}	Longest chain containing the double bond has 5 carbons; the double bond is at C2 and there are methyl groups on C3 and C4.	3,4-dimethylpent-2-ene

(i) oct-1-en-6-yne

(ii) 7-methylnon-1-en-8-yne

(iii) buta-1,2-diene

(iv) 3,4-dimethylpent-2-ene

15.4.4Nomenclature of halogenoalkanesOften examined

Halogenoalkanes: Organic compounds in which one or more hydrogen atoms of an alkane are replaced by halogen atoms. The halogen atom acts as the functional group, called the halo group (\ce{-X}), and is treated as a substituent in IUPAC nomenclature.

The common names of halogenoalkanes are alkyl halides. Halogenoalkanes are named as halo-substituted alkanes. While locating the halo group, start numbering from that end of the parent chain where the halo group comes closer. We use the prefixes fluoro-, chloro-, bromo- and iodo- before the hydrocarbon part.

iodomethane (methyl iodide)

chloroethane (ethyl chloride)

2-chloropropane (isopropyl chloride)

1-bromo-2-methylpropane (isobutyl bromide)

2-bromo-2-methylpropane (tert-butyl bromide)

Naming with more than one kind of substituent

Two different substituents, such as a halogen and an alkyl group or two different halogens, are named in alphabetical order.
2-bromo-4-chlorohexane: bromine on C2 and chlorine on C4 of a 6-carbon chain. The set of locants is 2,4 from the left, 3,5 from the right, so the left is chosen; bromo comes before chloro.
2-bromo-4-methylpentane and 4-chloro-2-methylhexane: the lowest locants are used first, and the substituents are then listed alphabetically (bromo before methyl, chloro before methyl).

2-bromo-4-chlorohexane

2-bromo-4-methylpentane

4-chloro-2-methylhexane

If a halogenoalkane has more than one halo group, we use di, tri, tetra etc. as for alkyl groups in alkanes.

2,4-dichlorohexane

2,2-dibromopentane

15.4.5Nomenclature of alcoholsOften examined

Alcohols: An important class of organic compounds which have the hydroxyl functional group (\ce{-OH}) bonded to a hydrocarbon skeleton.

The position of the functional group is shown by the number of the carbon atom in the parent chain to which it is attached. The parent chain is numbered from the end where the hydroxyl group lies nearer. While naming alcohols, the "e" of the alkane is replaced by the suffix "ol". In alcohols with more than two carbon atoms in the parent chain, the locant of the hydroxyl group is written before the suffix "ol".

ethanol (ethyl alcohol)

propan-2-ol (isopropyl alcohol)

2-methylpropan-2-ol (tert-butyl alcohol)

2-methylpropan-1-ol (isobutyl alcohol)

3,3-dimethylbutan-1-ol (neohexyl alcohol)

If there is more than one hydroxyl group on the parent chain, the prefixes di, tri and tetra are added to the suffix "ol". In such cases the last letter "e" of the alkane is retained. These are polyhydric alcohols, which have more than one hydroxyl group.

ethane-1,2-diol (ethylene glycol)

3-methylbutane-1,3-diol

propane-1,2,3-triol (glycerol)

Alkenols have a double bond in the parent chain, so the suffix "en", preceded by the position of the double bond, is added to the root name, followed by "ol". For numbering, the end closer to the hydroxyl group is preferred.

ethenol (vinyl alcohol)

prop-2-en-1-ol (allyl alcohol)

Concept assessment exercise 15.5

Give systematic names to the following compounds: (i) \ce{(CH3)2CHCH(Br)CH2CH3}; (ii) \ce{CH3CH(OH)CH2CH3}; (iii) \ce{CH3CH(OH)CH(CH3)CH2CH3}; (iv) \ce{CH3CH(CH3)CH2CH(OH)CH3}; (v) \ce{CH3CH=C(OH)CH3}; (vi) \ce{CH3CH(OH)CH(Br)CH3}.

	Compound	Working	Name
i	\ce{(CH3)2CHCH(Br)CH2CH3}	Longest chain 5 carbons (pentane). Bromo on C3 and methyl on C2 when numbered from the \ce{(CH3)2CH} end (locants 2,3, not 3,4). Alphabetical: bromo, then methyl.	3-bromo-2-methylpentane
ii	\ce{CH3CH(OH)CH2CH3}	4 carbons; the OH is on C2 when numbered from the \ce{CH3CH(OH)} end (not C3).	butan-2-ol
iii	\ce{CH3CH(OH)CH(CH3)CH2CH3}	5 carbons. Number from the end that gives the OH the lowest locant: OH on C2, methyl on C3.	3-methylpentan-2-ol
iv	\ce{CH3CH(CH3)CH2CH(OH)CH3}	5-carbon chain. OH lowest locant (C2) is from the \ce{CH(OH)CH3} end; methyl then falls on C4.	4-methylpentan-2-ol
v	\ce{CH3CH=C(OH)CH3}	4 carbons with a double bond and an OH. The end closer to the OH is preferred: OH on C2, and the double bond is then also at C2.	but-2-en-2-ol
vi	\ce{CH3CH(OH)CH(Br)CH3}	4 carbons. OH takes the lowest locant (C2), so bromo is on C3.	3-bromobutan-2-ol

(i) 3-bromo-2-methylpentane

(ii) butan-2-ol

(iii) 3-methylpentan-2-ol

(iv) 4-methylpentan-2-ol

(v) but-2-en-2-ol

(vi) 3-bromobutan-2-ol

15.4.6Nomenclature of aldehydesOften examined

The aldehyde group (\ce{-CHO}) is always terminal, so the carbon atom of this group is part of the parent chain. The numbering always starts from the carbon of the aldehyde group, and that carbon gets number 1. This is why the position of the aldehyde group is not mentioned in systematic names.

While giving systematic names to aldehydes, the "e" of the alkane is replaced by the first two letters "al" of aldehydes.

methanal (formaldehyde)

ethanal (acetaldehyde)

propanal (propionaldehyde)

butanal (butyraldehyde)

Numbering of the parent chain of aldehydes is required when substituents are attached. In common names, the carbon next to the aldehyde group is alpha (\alpha), then beta (\beta) and so on.

3-methylbutanal (beta-methylbutyraldehyde)

2-ethyl-3-methylpentanal (alpha-ethyl-beta-methylvaleraldehyde)

Reading the two substituted names

3-methylbutanal: C1 is the carbon of the CHO group, C2 is \alpha, C3 is \beta. The methyl group is on C3 (\beta in the common name).
2-ethyl-3-methylpentanal: the 5-carbon chain starts at the CHO carbon (C1). The ethyl group is on C2 (\alpha) and the methyl group is on C3 (\beta). Alphabetical order puts ethyl before methyl.

15.4.7Nomenclature of ketonesOften examined

The ketone group (\ce{-CO-}) is always inside the parent chain, so it must be part of the parent chain. While numbering the principal chain, the carbon atom of the ketone group must be given a number followed by the suffix "one". Numbering should start from the end where the ketone group comes nearer.

The parent chains of the first two members of the ketone family are not numbered. All ketones are named by replacing the "e" of the alkane by "one", preceded by the position of the ketone group in the parent chain.

propanone (acetone or dimethyl ketone)

butanone (ethyl methyl ketone)

2-methylpentan-3-one (ethyl isopropyl ketone)

Concept assessment exercise 15.6

Name the following compounds according to IUPAC nomenclature: (i) \ce{HCHO}; (ii) \ce{(CH3)2CHCHO}; (iii) \ce{C6H5COCH3}; (iv) \ce{CH3COCH2CH3}; (v) \ce{(CH3)2CHCH2COCH2CH3}.

	Compound	Working	Name
i	\ce{HCHO}	An aldehyde with one carbon: root meth + al.	methanal
ii	\ce{(CH3)2CHCHO}	An aldehyde. The CHO carbon is C1; C2 carries a methyl group; the chain has 3 carbons.	2-methylpropanal
iii	\ce{C6H5COCH3}	A ketone: a 2-carbon chain (ethanone) with a phenyl group (\ce{C6H5}, a prefix) on the carbonyl carbon. In full IUPAC form the locant is added, 1-phenylethan-1-one.	phenylethanone
iv	\ce{CH3COCH2CH3}	A ketone with 4 carbons. The first two members are not numbered.	butanone
v	\ce{(CH3)2CHCH2COCH2CH3}	6-carbon chain. Numbering from the ethyl end puts C=O on C3 (from the other end it would be C4). The methyl group is then on C5.	5-methylhexan-3-one

(i) methanal

(ii) 2-methylpropanal

(iii) phenylethanone

(iv) butanone

(v) 5-methylhexan-3-one

15.4.8Nomenclature of ethers

Ethers have two carbon chains, one on each side of an oxygen atom. While giving systematic names to them, we add the prefix alkoxy to the alkane, so they are also called alkoxyalkanes. The smaller alkyl group is made the alkoxy group, while the bigger alkyl group is treated as the alkane.

methoxymethane (dimethyl ether)

methoxyethane (ethyl methyl ether)

ethoxyethane (diethyl ether)

15.4.9Nomenclature of amines

Amines are alkyl derivatives of ammonia in which the hydrogen atoms are replaced by alkyl groups. If one alkyl group is bonded to nitrogen it is a primary amine; secondary amines have two and tertiary amines have three alkyl groups attached to the nitrogen atom.

The systematic names of primary amines are alkanamines, with the "e" of the alkane removed and the suffix "amine" added.

methanamine (methyl amine)

ethanamine (ethyl amine)

propan-1-amine (n-propyl amine)

propan-2-amine (isopropyl amine)

2-methylpropan-1-amine (isobutyl amine)

In secondary and tertiary amines, the alkyl groups attached to the nitrogen atom are treated as substituents and their locant is given just by "N".

N-ethyl-N-methylpropan-1-amine

Naming a tertiary amine

The longest alkyl group on nitrogen is propyl, so the parent amine is propan-1-amine.
The other two groups on nitrogen, ethyl and methyl, are written as substituents with the locant N (one N for each group).
Alphabetical order: ethyl before methyl.
\text{N-ethyl-N-methylpropan-1-amine}

15.4.10Nomenclature of nitriles

Nitriles have the functional group \ce{-CN}, called the nitrile group. The old name of this group was cyanide, but it is no longer used in the systematic names of IUPAC nomenclature. Nitrile compounds are named by using the suffix nitrile preceded by the alkane (the letter "e" of the alkane is not omitted). While numbering the hydrocarbon chain, the carbon of the nitrile group is counted.

ethanenitrile (acetonitrile)

propanenitrile (propionitrile)

butanenitrile

Concept assessment exercise 15.7

1. Name the following compounds according to IUPAC nomenclature: (i) \ce{CH3OCH2CH3}; (ii) (\ce{C2H5})_2\ce{CHNH2}; (iii) \ce{(CH3)3N}. 2. Draw the 2D displayed formulae of (i) 2-methylpentanenitrile and (ii) N,N-dimethylethanamine.

Question 1: names

	Compound	Working	Name
i	\ce{CH3OCH2CH3}	An ether. The shorter group (methyl) becomes methoxy and the longer group (ethyl) is the alkane.	methoxyethane
ii	(\ce{C2H5})_2\ce{CHNH2}	The carbon with the \ce{NH2} group has two ethyl groups, so it is the middle of a 5-carbon chain: \ce{CH3CH2CH(NH2)CH2CH3}. The amine is on C3.	pentan-3-amine
iii	\ce{(CH3)3N}	A tertiary amine with three methyl groups on nitrogen. The parent is methanamine; the other two methyl groups are N-substituents.	N,N-dimethylmethanamine

Question 2: displayed formulae

(i) 2-methylpentanenitrile: pent = 5 carbons including the nitrile carbon (C1), so the chain is \ce{CH3CH2CH2CH(CH3)C#N} with the methyl on C2.

Figure 15.13: 2D displayed formula of 2-methylpentanenitrile.

(ii) N,N-dimethylethanamine: ethanamine (\ce{CH3CH2NH2}) with two methyl groups on the nitrogen, \ce{CH3CH2N(CH3)2}.

Figure 15.14: 2D displayed formula of N,N-dimethylethanamine.

2-methylpentanenitrile

N,N-dimethylethanamine

15.4.11Nomenclature of carboxylic acidsOften examined

The systematic names of carboxylic acids are alkanoic acids, which means that the "e" of the alkane is replaced by "oic acid". The functional group of carboxylic acids is the carboxyl group (\ce{-COOH}), which is always terminal, and its carbonyl carbon is numbered carbon 1.

methanoic acid (formic acid)

ethanoic acid (acetic acid)

propanoic acid (propionic acid)

butanoic acid (butyric acid)

pentanoic acid (valeric acid)

hexanoic acid (caproic acid)

Dicarboxylic acids have two carboxyl groups, one at each end. They are named by adding "dioic acid" to the alkane name (the "e" is kept before the consonant "d").

ethanedioic acid (oxalic acid)

propanedioic acid (malonic acid)

butanedioic acid (succinic acid)

pentanedioic acid (glutaric acid)

hexanedioic acid (adipic acid)

15.4.12Nomenclature of estersOften examined

The systematic names of esters consist of two parts, alkyl and carboxylate, with a space between them. Esters are derivatives of carboxylic acids. While naming, the alkyl part (from the alcohol side, attached to the oxygen) is written first, followed by the name of the acid with the "e" of the alkane replaced by "oate".

methyl methanoate (methyl formate)

methyl ethanoate (methyl acetate)

ethyl 2-methylpropanoate

1-methylethyl ethanoate (isopropyl acetate)

How to read an ester name

Find the acid part: the carbon chain that includes the \ce{C=O} carbon gives the "oate" name (methanoate, ethanoate, 2-methylpropanoate).
Find the alkyl group on the single-bonded oxygen: methyl, ethyl, 1-methylethyl (isopropyl).
Write the alkyl name, a space, then the "oate" name.

Concept assessment exercise 15.8, question 1

Give systematic names to the following compounds: (i) \ce{(CH3)2CH(CH2)3COOH}; (ii) \ce{(CH2COOH)2}; (iii) \ce{(COOH)2}; (iv) \ce{CH3COOCH2CH3}; (v) \ce{HCOOCH3}; (vi) \ce{CH3CH(Cl)CH2COOH}.

	Compound	Working	Name
i	\ce{(CH3)2CH(CH2)3COOH}	Number from the COOH carbon (C1): three \ce{CH2} are C2 to C4, the \ce{CH} is C5 and carries a methyl group, and the last \ce{CH3} is C6.	5-methylhexanoic acid
ii	\ce{(CH2COOH)2}	Two \ce{CH2COOH} units joined together, \ce{HOOC-CH2-CH2-COOH}: 4 carbons with a carboxyl group at each end.	butanedioic acid
iii	\ce{(COOH)2}	Two carboxyl carbons joined: 2 carbons.	ethanedioic acid
iv	\ce{CH3COOCH2CH3}	Acid part \ce{CH3COO} is ethanoate; alkyl part \ce{CH2CH3} is ethyl.	ethyl ethanoate
v	\ce{HCOOCH3}	Acid part \ce{HCOO} is methanoate; alkyl part is methyl.	methyl methanoate
vi	\ce{CH3CH(Cl)CH2COOH}	C1 is the COOH carbon, C2 is \ce{CH2}, C3 carries the chlorine, C4 is \ce{CH3}.	3-chlorobutanoic acid

(i) 5-methylhexanoic acid

(ii) butanedioic acid

(iii) ethanedioic acid

(iv) ethyl ethanoate

(v) methyl methanoate

(vi) 3-chlorobutanoic acid

Concept assessment exercise 15.8, question 2

Draw the 2D displayed formulae of (i) oxalic acid and (ii) 2-methylpropanoic acid.

(i) Oxalic acid is ethanedioic acid, \ce{HOOC-COOH}: two carbons, each with a \ce{C=O} and an \ce{O-H}.

Figure 15.15: 2D displayed formula of oxalic acid (ethanedioic acid).

(ii) 2-methylpropanoic acid: C1 is the carboxyl carbon, C2 carries a methyl group and C3 is the other \ce{CH3}: \ce{(CH3)2CH-COOH}.

Figure 15.16: 2D displayed formula of 2-methylpropanoic acid.

15.4.13Nomenclature of acid chlorides

Acid halides are derivatives of carboxylic acids. They are named by replacing the "ic acid" of the carboxylic acid name with "yl halide" (for example ethanoic acid becomes ethanoyl chloride).

ethanoyl chloride (the book gives acyl chloride as the common name)

propanoyl chloride (propionyl chloride)

butanoyl chloride (butyryl chloride)

15.4.14Nomenclature of amides

Amides are derivatives of carboxylic acids. They are given systematic names by replacing the "oic acid" part of the carboxylic acid name with the suffix amide. In the case of secondary and tertiary amides, the alkyl groups attached to nitrogen are counted as substituents at position "N".

methanamide (formamide)

ethanamide (acetamide)

propanamide (propionamide)

N-methylethanamide

15.5Terminology of organic chemical reactions

Chemical reactions of organic compounds are complex in nature, so different terms are needed to describe them. The terms are the types of reagents (free radicals, electrophiles and nucleophiles) and the types of bond breakage (homolytic and heterolytic fission), explained in the next two sections.

Terms used for organic reactions
Term	In one line
Free radical	an atom or group of atoms with an unpaired electron
Electrophile	an electron-deficient species that accepts an electron pair
Nucleophile	an electron-rich species that donates an electron pair
Homolytic fission	equal splitting of a covalent bond, giving free radicals
Heterolytic fission	unequal splitting of a covalent bond, giving ions

15.5.1Types of reagents used in chemical reactionsOften examined

i. Free radical

Free radical: An atom or group of atoms with an unpaired electron.

A free radical is quite an unstable species, so it reacts with any substance it comes across. A free radical is formed as a result of homolytic fission of molecules in different chemical reactions. Examples: the chlorine radical (\ce{Cl.}) and the methyl radical (\ce{CH3.}).

ii. Electrophile

Electrophile: An atom or group of atoms which is electron deficient. It has either a positive charge or an empty orbital, so it can easily receive an electron pair from its reaction partner (the nucleophile).

Example: the carbocation \ce{CH3+}.

iii. Nucleophile

Nucleophile: An atom or group of atoms which is rich in electrons. Since it carries either a lone pair of electrons or a negative charge, it can easily donate electrons to a molecule (the electrophile).

Example: the hydroxide ion \ce{OH-}.

Electrophile and nucleophile compared
	Electrophile	Nucleophile
Electron condition	electron deficient	electron rich
Carries	a positive charge or an empty orbital	a negative charge or a lone pair
Action	accepts electrons from the partner	donates electrons to the partner
Example	\ce{CH3+}	\ce{OH-}

15.5.2Types of bond breakage in chemical reactionsOften examined

There are different ways in which covalent bonds are broken during chemical reactions.

i. Homolytic fission

Homolytic fission: The equal splitting of the covalent bond between two atoms in a molecule. Each atom takes its own electron, as an unpaired electron.

This cleavage takes place because the electronegativity difference between the atoms is zero or near zero. Usually such fission takes place between the same atoms. It forms free radicals.

\ce{Cl-Cl -> 2Cl.}

ii. Heterolytic fission

Heterolytic fission: The unequal breakage of a covalent bond between atoms in organic molecules. As a result of this breakage we get positive and negative ions.

Such splitting of a molecule takes place between different atoms having enough electronegativity difference.

\ce{CH3-Cl -> CH3+ + Cl-}

Figure 15.17: Homolytic fission (each atom takes one electron of the bond pair) and heterolytic fission (the atom B takes both electrons).

Homolytic and heterolytic fission compared
	Homolytic fission	Heterolytic fission
Bond breakage	equal	unequal
Electrons of the bond pair	one goes to each atom	both go to one atom
Products	free radicals	a positive ion and a negative ion
Electronegativity difference	zero or near zero (usually same atoms)	enough difference (different atoms)
Example	\ce{Cl2 -> 2Cl.}	\ce{CH3Cl -> CH3+ + Cl-}

15.6Types of organic chemical reactionsOften examined

Organic compounds undergo various chemical reactions, depending on the reaction conditions and the type of reagent used. We study some important types of reactions according to the prescribed syllabus.

i. Free radical substitution reactions of alkanes

In this type of reaction a halogen free radical replaces the hydrogen atoms of alkanes successively and forms a mixture of halogenoalkanes. The reaction takes place in three steps: initiation, propagation and termination.

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

ii. Electrophilic addition reactions

This type of reaction involves the attack of an electrophile (an electron-deficient species) on a multiple bond (double or triple bond). In this way the molecule or electrophile is added to the unsaturated compound. Alkenes show this type of reaction. The multiple bond is converted to a single covalent bond.

\ce{CH3CH=CH2 + HBr -> CH3CH(Br)CH3}

iii. Elimination reactions

Elimination reactions involve the removal of atoms from adjacent carbon atoms of an organic molecule in the form of a small molecule such as \ce{H2O} or \ce{HCl}. Elimination of water from alcohols, and of hydrogen chloride from haloalkanes, gives the corresponding alkenes.

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

iv. Nucleophilic substitution reactions

In these reactions a nucleophile replaces an atom or group of atoms in a molecule.

\ce{CH3CH2Br + NaOH -> CH3CH2OH + NaBr}

v. Nucleophilic addition reactions

In this case a nucleophile is added to molecules, usually aldehydes and ketones, and the double bond between carbon and oxygen of the carbonyl group is converted to a single covalent bond.

\ce{CH3CHO + HCN ->[NaCN/HCl] CH3CH(CN)OH}

vi. Hydrolysis

A chemical reaction in which a water molecule attacks the organic compound and breaks it into different substances. Usually acids or alkalis act as catalysts in this type of reaction.

\ce{CH3COOCH2CH3 + H2O ->[H2SO4] CH3COOH + CH3CH2OH}

vii. Condensation

Condensation reactions involve the combination of two organic molecules to form a larger molecule, with or without the elimination of a small molecule like \ce{H2O} or \ce{HCl}. Since water is removed in this reaction, it is the opposite of a hydrolysis reaction.

\ce{CH3COOH + CH3CH2OH ->[H2SO4] CH3COOCH2CH3 + H2O}

viii. Oxidation reaction

In oxidation reactions organic molecules may either gain oxygen or lose hydrogen. Such reactions may involve an increase in the number of bonds of carbon with oxygen. For example, alcohols may be oxidised to the corresponding aldehydes, ketones or carboxylic acids.

\ce{CH3OH + 3[O] ->[K2Cr2O7/H2SO4] 2HCOOH}

as printed in the book

ix. Reduction reaction

A reduction reaction of organic molecules may involve loss of oxygen or gain of hydrogen. This type of reaction shows a decrease in the number of bonds of carbon with oxygen in the molecule. An example is the reduction of aldehydes or ketones to the corresponding alcohols.

\ce{CH3CHO + 2[H] ->[LiAlH4] CH3CH2OH}

The nine reaction types at a glance
Type	What happens	Reagent type
Free radical substitution	a halogen radical replaces H of an alkane	free radical
Electrophilic addition	an electrophile adds across a \ce{C=C} or \ce{C#C} bond	electrophile
Elimination	a small molecule (\ce{H2O}, \ce{HCl}) is removed from adjacent carbons, giving a \ce{C=C}	base
Nucleophilic substitution	a nucleophile replaces an atom or group	nucleophile
Nucleophilic addition	a nucleophile adds to the \ce{C=O} of an aldehyde or ketone	nucleophile
Hydrolysis	water breaks a compound into two	water (acid or alkali catalyst)
Condensation	two molecules combine, often losing water	-
Oxidation	gain of oxygen or loss of hydrogen	oxidising agent
Reduction	loss of oxygen or gain of hydrogen	reducing agent

15.6.1Free radical substitution reaction of alkanesOften examined

In this type of reaction a free radical is formed first, which then replaces the hydrogen atoms of an alkane one after the other. It takes place with halogens in the presence of sunlight. A general substitution is shown below.

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

The mechanism of this reaction has three stages: initiation, propagation and termination.

i. Initiation

In this step a halogen molecule splits homolytically in the presence of sunlight to give free radicals. The energy for this fission comes from the ultraviolet region of sunlight.

Initiation

The \ce{Cl-Cl} bond breaks homolytically; each chlorine atom takes one electron of the bond pair and becomes a chlorine free radical.
\ce{Cl-Cl ->[h\nu] Cl. + Cl.}

ii. Propagation

This step involves the attack of a halogen free radical on a methane molecule to generate a methyl free radical. The methyl free radical may then attack a chlorine molecule to generate chloromethane and a new chlorine free radical, which keeps the reaction going.

Propagation

A chlorine radical removes a hydrogen atom from methane, forming hydrogen chloride and a methyl radical.
\ce{CH4 + Cl. -> .CH3 + HCl}
The methyl radical attacks a chlorine molecule, forming chloromethane and a new chlorine radical.
\ce{.CH3 + Cl2 -> CH3Cl + Cl.}

chloromethane

If excess chlorine is used in this reaction, all the hydrogen atoms of methane are successively replaced by chlorine atoms, finally producing tetrachloromethane.

Further substitution with excess chlorine

Second hydrogen replaced.
\ce{CH3Cl + Cl. -> .CH2Cl + HCl}\qquad \ce{.CH2Cl + Cl2 -> CH2Cl2 + Cl.}
dichloromethane
Third hydrogen replaced.
\ce{CH2Cl2 + Cl. -> .CHCl2 + HCl}\qquad \ce{.CHCl2 + Cl2 -> CHCl3 + Cl.}
trichloromethane (chloroform)
Fourth hydrogen replaced.
\ce{CHCl3 + Cl. -> .CCl3 + HCl}\qquad \ce{.CCl3 + Cl2 -> CCl4 + Cl.}
tetrachloromethane

dichloromethane

trichloromethane (chloroform)

tetrachloromethane

iii. Termination

In this step any two free radicals react with one another, at any stage, to give a neutral molecule. Termination is the death of free radicals.

Termination

Two chlorine radicals combine.
\ce{Cl. + Cl. -> Cl2}
A methyl radical combines with a chlorine radical.
\ce{.CH3 + Cl. -> CH3Cl}
chloromethane
Two methyl radicals combine.
\ce{.CH3 + .CH3 -> CH3CH3}
ethane

The three stages
Stage	What happens	Free radicals
Initiation	\ce{Cl2} splits in sunlight into two chlorine radicals	formed
Propagation	a radical reacts with a molecule and a new radical is produced; the cycle repeats	one used, one formed
Termination	two radicals combine to give a stable molecule	removed

15.6.2Elimination reactionsOften examined

Elimination reactions involve the removal of atoms or groups of atoms, in the form of a small molecule, from two consecutive carbon atoms. Consequently we get an unsaturated molecule (an alkene) and a small molecule, mostly water or hydrogen chloride.

Consider the bromoethane molecule, which has a partially positive (acidic) hydrogen and a partially negative bromine atom. A base (\ce{OH-}) attacks the acidic hydrogen, making water. The bromine atom, because of its higher electronegativity, takes away the bonding electrons of the carbon and leaves the molecule as a bromide ion. As a result, elimination of hydrogen bromide takes place, giving the ethene molecule.

\ce{CH3CH2Br + OH- -> CH2=CH2 + H2O + Br-}

Figure 15.18: Elimination of hydrogen bromide from bromoethane by a base. Three curved arrows show the movement of electron pairs.

The three electron-pair movements

The base \ce{OH-} uses a lone pair to take the partly positive (\delta^+) hydrogen on the carbon next to the carbon holding bromine. This forms \ce{H2O}.
The electrons of the \ce{C-H} bond move to form a second bond between the two carbon atoms (\ce{C=C}).
Bromine is more electronegative than carbon, so the electrons of the \ce{C-Br} bond go to bromine and it leaves as the bromide ion \ce{Br-}.

15.6.3Nucleophilic substitution reactionsOften examined

In this type of reaction a nucleophile attacks the electrophilic carbon of a halogenoalkane, substituting for the leaving group (the halogen atom).

\ce{CH3CH2Br + OH- -> CH3CH2OH + Br-}

Figure 15.19: Nucleophilic substitution of bromoethane by hydroxide. The curved arrows show the nucleophile attacking carbon and the C-Br electrons leaving with bromine.

What happens

Bromine is more electronegative than carbon, so the carbon of the \ce{C-Br} bond is partly positive (\delta^+) and is the electrophilic centre.
The nucleophile \ce{OH-} donates a lone pair to this carbon, forming a new \ce{C-O} bond.
In the transition state the \ce{C-O} bond is partly formed and the \ce{C-Br} bond is partly broken (dotted bonds).
The \ce{C-Br} electrons leave with bromine as \ce{Br-}, the leaving group. The product is ethanol.

15.6.4Electrophilic addition reactionsOften examined

As the name shows, these reactions involve the attack of an electrophile on a multiple bond (the \pi bond) and convert it to a single covalent bond. Here we study the attack of hydrogen bromide on the ethene molecule to make bromoethane. The hydrogen ion acts as the electrophile in this reaction.

\ce{CH2=CH2 + HBr -> CH3CH2Br}

Figure 15.20: Electrophilic addition of hydrogen bromide to ethene. Curved arrows show the \pi electrons attacking the hydrogen and the H-Br electrons moving to bromine.

What happens

In \ce{HBr} the hydrogen is partly positive (\delta^+) and bromine partly negative (\delta^-), because bromine is more electronegative.
The electron-rich \pi bond of ethene attacks the electrophile H.
The \ce{H-Br} bond breaks and its electrons go to bromine. The double bond has become a single bond and \ce{H} and \ce{Br} have added across it, giving bromoethane.

15.7IsomerismOften examined

Organic molecules show isomerism because of the directional character of covalent bonds. Molecules having the same molecular formula but different structural formulae are called isomers, and the phenomenon is called isomerism.

Isomerism is one of the reasons behind the enormous number of organic compounds. The number of isomers increases with an increase in the number of carbon atoms. Butane (\ce{C4H10}) has two isomers and decane (\ce{C10H22}) has 75 isomers.

Broadly, there are two types of isomerism: structural isomerism and geometric isomerism. In this chapter we study structural isomerism only.

Structural isomers: Isomers which have the same molecular formula but a different arrangement of atoms (structure) in the molecule. There are five types: chain, positional, functional group, metamerism and tautomerism.

i. Chain isomerism

Chain isomerism arises in molecules having the same molecular formula but different structural formulae with respect to the length of the carbon chain.

butane

2-methylpropane

Both are \ce{C4H10}: one has a 4-carbon chain, the other a 3-carbon chain with a methyl branch.

ii. Positional isomerism

Positional isomers have the same molecular formula but a different position (location) of the functional group on the principal carbon chain.

propan-1-ol

propan-2-ol

Both are \ce{C3H8O} with an \ce{-OH} group: at the end carbon (C1) in one and at the middle carbon (C2) in the other.

iii. Functional group isomerism

Functional group isomerism arises in molecules having the same molecular formula but different functional groups.

propan-1-ol (alcohol)

methoxyethane (ether)

Both are \ce{C3H8O}: one has the hydroxyl group (\ce{-OH}), the other the ether group (\ce{-O-}).

iv. Metamerism

Metamers have the same molecular formula but different structural formulae with respect to the chain length on the two sides of a functional group. Metamers have the same functional group.

methoxypropane

ethoxyethane

Both are ethers, \ce{C4H10O}. The oxygen has \ce{CH3} and \ce{C3H7} on its two sides in one, and \ce{C2H5} and \ce{C2H5} in the other.

v. Tautomerism

Tautomerism is shown by isomers which differ with respect to the position of a hydrogen atom (mobile hydrogen). Tautomers exist in dynamic equilibrium with one another as a result of intramolecular migration of a proton. Ethyl acetoacetate exhibits this type of isomerism, which is also called keto-enol isomerism.

\underset{\text{keto form}}{\ce{CH3-CO-CH2-COO-C2H5}} \;\ce{<=>}\; \underset{\text{enol form}}{\ce{CH3-C(OH)=CH-COO-C2H5}}

keto form

enol form

The five types of structural isomerism
Type	Same	Different	Example pair
Chain	molecular formula	length of the carbon chain	butane and 2-methylpropane
Positional	molecular formula and functional group	position of the functional group	propan-1-ol and propan-2-ol
Functional group	molecular formula	functional group	propan-1-ol and methoxyethane
Metamerism	molecular formula and functional group (ether)	chain lengths on the two sides of the functional group	methoxypropane and ethoxyethane
Tautomerism	molecular formula	position of a mobile hydrogen (dynamic equilibrium)	keto and enol forms of ethyl acetoacetate

Concept assessment exercise 15.9, question 1

Draw 2D displayed formulae of the possible structural isomers of (i) pentane and (ii) propan-1-ol.

(i) Pentane is \ce{C5H12}. Arrange five carbon atoms in every possible way without changing the formula: one straight chain of five, one chain of four with a methyl branch on the second carbon, and one chain of three with two methyl branches on the middle carbon. A methyl branch on the end carbon of a 4-chain just makes a 5-chain again, so there are three isomers in all.

Figure 15.21: The three structural isomers of pentane, \ce{C5H12}.

pentane

2-methylbutane

2,2-dimethylpropane

(ii) Propan-1-ol is \ce{C3H8O}. Moving the \ce{-OH} to the middle carbon gives propan-2-ol (positional isomer). Putting the oxygen between the carbons instead gives methoxyethane (functional group isomer, an ether).

Figure 15.22: Propan-1-ol and its structural isomers propan-2-ol (positional) and methoxyethane (functional group).

Concept assessment exercise 15.9, question 2

Examine the concept of isomerism in organic compounds.

Definition: molecules having the same molecular formula but different structural formulae are called isomers, and the phenomenon is isomerism.
Cause: the directional character of covalent bonds lets the same atoms be joined in different arrangements.
Importance: isomerism is one of the reasons for the enormous number of organic compounds. The number of isomers rises with the number of carbon atoms: butane has 2 isomers and decane has 75.
Kinds: broadly structural isomerism (different arrangement of atoms) and geometric isomerism. Structural isomerism has five types: chain, positional, functional group, metamerism and tautomerism.
Examples: butane and 2-methylpropane (chain); propan-1-ol and propan-2-ol (positional); propan-1-ol and methoxyethane (functional group); methoxypropane and ethoxyethane (metamerism); keto and enol forms of ethyl acetoacetate (tautomerism).
Discovery: Wohler found that ammonium cyanate \ce{NH4OCN} and urea \ce{NH2CONH2} have the same formula but different structures.

Worked example: drawing and naming the structural isomers of C4H10O

Draw all the structural isomers of \ce{C4H10O} that are alcohols or ethers, name them, and name the type of isomerism in four of the pairs.

Find the two families that fit \ce{C4H10O}: alcohols \mathrm{C}_n\mathrm{H}_{2n+1}\mathrm{OH} with n=4, and ethers \mathrm{C}_n\mathrm{H}_{2n+1}\mathrm{O}\mathrm{C}_m\mathrm{H}_{2m+1} with n+m=4.
Alcohols: put \ce{-OH} on a four-carbon chain at C1 or C2, then on a branched three-carbon chain. This gives four alcohols.
Ethers: split the four carbons on the two sides of oxygen as 1+3 (propyl straight or branched) or 2+2. This gives three ethers.
Check that every structure has 4 carbons and the formula \ce{C4H10O}.

butan-1-ol

butan-2-ol

2-methylpropan-1-ol

2-methylpropan-2-ol

1-methoxypropane

2-methoxypropane

ethoxyethane

Pair	Type of isomerism	Reason
butan-1-ol and butan-2-ol	positional	same \ce{-OH} group, different position on the chain
butan-1-ol and 2-methylpropan-1-ol	chain	different length of the carbon chain
butan-1-ol and ethoxyethane	functional group	alcohol against ether
1-methoxypropane and ethoxyethane	metamerism	both ethers, different chain lengths on the two sides of oxygen

7\ \text{isomers: 4 alcohols and 3 ethers}

Formula sheet

Moles of an element from percentage composition

n=\dfrac{\text{mass of the element}}{\text{atomic mass}}=\dfrac{\%\ \text{by mass}}{\text{atomic mass}}

Take 100\,\mathrm{g} of the compound, so the percentage by mass of each element equals its mass in grams. Divide each by the smallest value to get the atom ratio (the empirical formula).

Unit: \mathrm{mol}

Multiple of the empirical formula

n=\dfrac{\text{molecular mass}}{\text{empirical formula mass}}

n is the whole number by which the empirical formula must be multiplied to give the molecular formula.

no unit

Molecular formula from the empirical formula

\text{molecular formula}=n\,(\text{empirical formula})

For example 6(\ce{CH})=\ce{C6H6}.

no unit

Alkanes

\mathrm{C}_n\mathrm{H}_{2n+2}

n is the number of carbon atoms. No functional group.

no unit

Alkenes

\mathrm{C}_n\mathrm{H}_{2n}

One \ce{C=C} double bond (open chain).

no unit

Alkynes

\mathrm{C}_n\mathrm{H}_{2n-2}

One \ce{C#C} triple bond (open chain).

no unit

Alkyl groups

\mathrm{C}_n\mathrm{H}_{2n+1}

An alkane minus one hydrogen atom (R).

no unit

Halogenoalkanes

\mathrm{C}_n\mathrm{H}_{2n+1}\mathrm{X}

X is a halogen atom.

no unit

Alcohols

\mathrm{C}_n\mathrm{H}_{2n+1}\mathrm{OH}

Hydroxyl group \ce{-OH}.

no unit

Ethers

\mathrm{C}_n\mathrm{H}_{2n+1}\mathrm{O}\mathrm{C}_m\mathrm{H}_{2m+1}

n and m are the carbon numbers of the two alkyl groups.

no unit

Amines

\mathrm{C}_n\mathrm{H}_{2n+1}\mathrm{NH}_2

Primary amines.

no unit

Nitriles

\mathrm{C}_n\mathrm{H}_{2n+1}\mathrm{CN}

Nitrile group \ce{-C#N}.

no unit

Aldehydes

\mathrm{C}_n\mathrm{H}_{2n+1}\mathrm{CHO}

Aldehyde group \ce{-CHO}.

no unit

Ketones

\mathrm{C}_n\mathrm{H}_{2n+1}\mathrm{CO}\mathrm{C}_m\mathrm{H}_{2m+1}

Ketone group \ce{-CO-}.

no unit

Carboxylic acids

\mathrm{C}_n\mathrm{H}_{2n+1}\mathrm{COOH}

Carboxyl group \ce{-COOH}.

no unit

Esters

\mathrm{C}_n\mathrm{H}_{2n+1}\mathrm{COO}\mathrm{C}_m\mathrm{H}_{2m+1}

Ester group \ce{-COO-}.

no unit

Amides

\mathrm{C}_n\mathrm{H}_{2n+1}\mathrm{CONH}_2

Amide group \ce{-CONH2}.

no unit

Acid halides

\mathrm{C}_n\mathrm{H}_{2n+1}\mathrm{COX}

X is a halogen atom.

no unit

Arenes

\mathrm{C}_n\mathrm{H}_{2n-6m}

m is the number of rings (the book's form).

no unit

Key terms

Organic chemistry: The chemistry which deals with compounds having carbon.
Organic compounds: Compounds composed of the element carbon, with carbon atoms covalently bonded to one another and forming the backbone; carbon should be directly bonded to hydrogen and carbon.
Inorganic compounds: Compounds made of elements other than carbon, which come from rocks and minerals (including \ce{CO2}, \ce{CO}, carbonates, hydrogencarbonates, cyanides and cyanates).
Vital force theory: Berzelius's idea that organic compounds can only be made by living matter, which has a mysterious vital force. Wohler's synthesis of urea disproved it.
Catenation: The ability of carbon atoms to form long chains and rings by linking with other carbon atoms.
Molecular formula: Shows the type of atoms and the actual whole-number ratio between the atoms in a molecule.
Empirical formula: Shows the simplest whole-number ratio between the elements in a compound.
Condensed structural formula: Shows the relative positions of all atoms without showing single covalent bonds; branches and functional groups are put in brackets.
Full structural formula (2D displayed formula): Shows all atoms and all bonds among them in a plane.
Skeletal formula: Only lines are drawn; each end and corner is a carbon atom and hydrogens on carbon are not shown; multiple bonds and functional groups are shown.
Stereochemical formula (3D formula): Shows the arrangement of atoms in space using plain, wedge and hashed bonds.
Functional group: An atom or group of atoms attached to a carbon chain or ring which determines the characteristic properties of an organic compound.
Homologous series: A group of organic compounds with the same functional group but different lengths of carbon chain, each member differing from the next by \ce{-CH2-}.
General formula: A formula that gives the molecular formula of all members of a homologous series with n carbon atoms (for example \mathrm{C}_n\mathrm{H}_{2n+2} for alkanes).
Hydrocarbons: Compounds containing only carbon and hydrogen.
Saturated hydrocarbons: Hydrocarbons with only single covalent bonds between carbon atoms (alkanes and cycloalkanes).
Unsaturated hydrocarbons: Hydrocarbons with at least one carbon-carbon double or triple bond (alkenes and alkynes).
Alkyl group: A group obtained by removing one hydrogen atom from an alkane, \mathrm{C}_n\mathrm{H}_{2n+1}, named by replacing -ane with -yl.
Primary, secondary and tertiary carbon: A carbon attached to one carbon atom or none, to two carbon atoms, or to three carbon atoms respectively.
Aliphatic compounds: Open-chain and closed-chain organic compounds, except aromatic compounds.
IUPAC: The International Union of Pure and Applied Chemistry, the authority that gives rules for systematic names.
Prefix, root name and suffix: The three parts of a systematic name: substituents, the number of carbons in the parent chain, and the family name (with an, en or yn for the type of C-C bond).
Locant: The number of the carbon atom of the parent chain to which a substituent or functional group is attached.
Free radical: An atom or group of atoms with an unpaired electron, formed by homolytic fission (for example \ce{Cl.} and \ce{CH3.}).
Electrophile: An electron-deficient atom or group of atoms, with a positive charge or an empty orbital, which accepts electrons (for example \ce{CH3+}).
Nucleophile: An electron-rich atom or group of atoms, with a lone pair or a negative charge, which donates electrons (for example \ce{OH-}).
Homolytic fission: Equal splitting of a covalent bond so that each atom takes one electron, giving free radicals.
Heterolytic fission: Unequal breakage of a covalent bond in which one atom takes both electrons, giving positive and negative ions.
Free radical substitution: A reaction in which a halogen free radical replaces hydrogen atoms of an alkane successively; it has initiation, propagation and termination steps.
Initiation: The step in which a halogen molecule splits homolytically in sunlight to give free radicals.
Propagation: The step in which a free radical attacks a molecule to give a product and a new free radical, so that the chain continues.
Termination: The step in which two free radicals combine to give a neutral molecule, ending the chain.
Electrophilic addition: Attack of an electrophile on a multiple bond, converting it to a single bond (for example \ce{HBr} adding to ethene).
Nucleophilic addition: Addition of a nucleophile to the carbonyl group of an aldehyde or ketone, turning \ce{C=O} into a single bond.
Nucleophilic substitution: Replacement of a leaving group (such as a halogen) in a molecule by a nucleophile.
Elimination reaction: Removal of a small molecule (\ce{H2O} or \ce{HCl}) from adjacent carbon atoms, giving an alkene.
Hydrolysis: A reaction in which water attacks an organic compound and breaks it into different substances.
Condensation reaction: Combination of two organic molecules to form a larger one, with or without loss of a small molecule such as \ce{H2O}.
Oxidation (organic): Gain of oxygen or loss of hydrogen; an increase in the number of carbon-oxygen bonds.
Reduction (organic): Loss of oxygen or gain of hydrogen; a decrease in the number of carbon-oxygen bonds.
Isomerism: The existence of molecules with the same molecular formula but different structural formulae.
Structural isomerism: Isomerism due to a different arrangement of atoms in the molecule; chain, positional, functional group, metamerism and tautomerism.
Chain isomerism: Same molecular formula, different length of carbon chain (for example butane and 2-methylpropane).
Positional isomerism: Same molecular formula, different position of the functional group on the carbon chain (for example propan-1-ol and propan-2-ol).
Functional group isomerism: Same molecular formula, different functional groups (for example propan-1-ol and methoxyethane).
Metamerism: Same molecular formula and functional group, different chain lengths on the two sides of the functional group (for example methoxypropane and ethoxyethane).
Tautomerism: Isomers that differ in the position of a mobile hydrogen atom and exist in dynamic equilibrium, as in keto-enol isomerism.

Multiple choice questionsp. 314–316

The book prints no answer key for these multiple choice questions. Each key below was worked out from the unit.

Which one of the following molecules needs no number in its systematic name?

A\ce{CH3(CH2)4CHO}Answer
B\ce{CH3CH=CHCH2CH3}
C\ce{CH3CH2CH2OH}
D\ce{CH3CO(CH2)3CH3}

Why

\ce{CH3(CH2)4CHO} is hexanal. The aldehyde carbon is always carbon 1 of the parent chain, so its position is never written, and the answer is A. The other three need a locant: \ce{CH3CH=CHCH2CH3} is pent-2-ene (the double bond must be located), \ce{CH3CH2CH2OH} is propan-1-ol (the -\ce{OH} group must be located) and \ce{CH3CO(CH2)3CH3} is hexan-2-one (the ketone group must be located).

Which one of the following molecules has chiral carbon?

AStructure A
BStructure B
CStructure C
DStructure DAnswer

Why

Structure D is 3-methylcyclohexan-1-ol. Carbon 1, which carries the -\ce{OH} group, has four different groups attached: H, -\ce{OH}, and two ring arms. One arm reaches the methyl-bearing carbon in two bonds (through carbon 2), the other in four bonds (through carbons 6, 5 and 4). Because the two arms differ, carbon 1 is chiral.

The others have no chiral carbon. In A (methylcyclobutane) and C (cyclopentanol) the two ring arms from the substituted carbon are identical. In B (3-methylpentane) the branched carbon carries two identical ethyl groups.

Note. Chirality (a carbon bonded to four different groups) is not defined in this unit. The answer applies that rule directly, and the book prints no key for this question.

iii

Propene is converted to propane by heating with hydrogen gas in presence of nickel as catalyst. What type of chemical reaction does it belong to?

\ce{H3C-CH=CH2 + H2 ->[Ni] H3C-CH2-CH3}

Aelimination
BadditionAnswer
Chydrolysis
Doxidation

Why

Two hydrogen atoms add across the C=C double bond, which becomes a single bond. This is an addition reaction (hydrogenation). Elimination removes small molecules from a molecule, hydrolysis splits a bond with water, and oxidation adds oxygen or removes hydrogen from carbon, so A, C and D are wrong.

Sorbitol is natural compound with sweet taste and is used as substitute for sucrose by food industry. How many chiral carbons are present in sorbitol?

Sorbitol

A3
B4Answer
C5
D6

Why

Sorbitol is \ce{HOCH2(CHOH)4CH2OH}, a six-carbon chain with an -\ce{OH} group on each carbon. The two end carbons each carry two H atoms, so they are not chiral. Carbons 2, 3, 4 and 5 each carry H, -\ce{OH} and two different chain arms, so each is chiral. The answer is 4, option B.

Note. Chirality is not defined in this unit; the count uses the rule that a carbon bonded to four different groups is chiral. The book prints no key for this question.

Which homologous series has highest ratio of hydrogen to carbon atoms in its general formula?

Acarboxylic acid
BalcoholsAnswer
Chalogenoalkanes
Daldehydes

Why

Compare the H:C ratio of each general formula, with n carbon atoms:

\text{Carboxylic acid: } \ce{C_{n}H_{2n}O2}\;\Rightarrow\;\dfrac{H}{C}=2 \qquad \text{Aldehyde: } \ce{C_{n}H_{2n}O}\;\Rightarrow\;\dfrac{H}{C}=2\\ \text{Halogenoalkane: } \ce{C_{n}H_{2n+1}X}\;\Rightarrow\;\dfrac{H}{C}=2+\dfrac{1}{n} \qquad \text{Alcohol: } \ce{C_{n}H_{2n+1}OH}\;\Rightarrow\;\dfrac{H}{C}=2+\dfrac{2}{n}

Alcohols give the largest ratio, 2+\dfrac{2}{n}, which is above the value for halogenoalkanes (2+\dfrac{1}{n}) and above the value 2 for acids and aldehydes. The answer is B.

Which one of the following hydrocarbons has no functional group?

A\ce{C2H2}
B\ce{C6H6}
C\ce{C3H6}
D\ce{C4H10}Answer

Why

Alkanes are the only family of organic compounds with no functional group. \ce{C4H10} is butane, which has only C-C and C-H single bonds, so it has none. \ce{C2H2} (ethyne) has a triple bond and \ce{C3H6} (propene) has a double bond, and the book counts double and triple bonds as functional groups of alkenes and alkynes. \ce{C6H6} is benzene, an arene whose ring is listed as a group of its own in the book's table of functional groups.

vii

How many sigma (\sigma) bonds are present in the given molecule?

\ce{CH2=CHCH2COOCH=CH2}

A10
B13
C15Answer
D17

Why

The molecule has 16 atoms (6 carbon, 8 hydrogen and 2 oxygen) and no ring. Each pair of directly bonded atoms shares one sigma bond, and the molecule is a single chain of atoms, so the number of sigma bonds is one less than the number of atoms: 16-1=15. Counting bond by bond gives the same answer:

\text{C-H: } 8,\qquad \text{C-C: } 4,\qquad \text{C-O: } 3 \;\Rightarrow\; \text{total} = 15

Each double bond contributes one sigma bond and one extra pi bond, and the pi bonds are not counted. The answer is C.

Note. Sigma and pi bonds are not defined in this unit. The answer uses the standard rule that each bond has one sigma component and that double and triple bonds add pi components.

viii

The systematic name of the following compound is

Compound

A4-ethenyl-3-methylhexane
B4-ethyl-3-methylhex-5-ene
C4-methyl-3-ethylhex-1-ene
D3-ethyl-4-methylhex-1-eneAnswer

Why

The longest chain that contains the double bond has six carbons: \ce{CH2=CH-CH(C2H5)-CH(CH3)-CH2-CH3}. Numbering from the end nearest the double bond puts it on carbon 1 and gives the substituents locants 3 (ethyl) and 4 (methyl). Substituents are cited in alphabetical order (ethyl before methyl), so the name is 3-ethyl-4-methylhex-1-ene, option D.

A is wrong because the parent chain must contain the double bond, so it is not a hexane with an ethenyl group. B numbers the double bond from the wrong end. C describes the same molecule as D but cites methyl before ethyl, which breaks the alphabetical rule.

Note. Options C and D describe the same compound; only D follows the alphabetical order. Option B is printed as '3-methyhex-5-ene' in the book; the spelling is corrected here.

The common name of butanedioic acid is

Aoxalic acid
Bvaleric acid
Csuccinic acidAnswer
Dmalonic acid

Why

Butanedioic acid is \ce{HOOC-CH2-CH2-COOH}: four carbons and two carboxyl groups. Its common name is succinic acid (option C), as the book's table of carboxylic acids shows. Oxalic acid is ethanedioic acid (two carbons), malonic acid is propanedioic acid (three carbons), and valeric acid is pentanoic acid (five carbons, one carboxyl group).

Which one of the following is incorrect statement about homologous series?

Aits adjacent members differ by one methylene (-\ce{CH2-}) group
Bits members share same functional group
Cits members have similar chemical properties
Dits members have same alkyl groupsAnswer

Why

Members of a homologous series share the same functional group, so they have similar chemical properties, and adjacent members differ by one -\ce{CH2-} group. The alkyl group does change along the series, becoming longer by one -\ce{CH2-} at each step (for example methanol, ethanol and propan-1-ol). So the incorrect statement is D.

Short questionsp. 316–317

i2026 Annual I, Set B, Q.2 (vi) OR2026 Annual I, Set D, Q.2 (vi) OR2025 Annual I, Set B, Q.2 (i) OR2025 Annual I, Set D, Q.2 (iv)3 marks

Define catenation and justify that this property of carbon is responsible for so large number of organic compounds.

Solution

Catenation is the ability of carbon atoms to link with one another to form long chains and rings. Silicon forms only small chains and sulphur only small rings, so the property is strongest in carbon.

Carbon-carbon bonds are strong (the C-C bond energy is 346\,\mathrm{kJ\,mol^{-1}}), and the atoms can form single, double and triple bonds, which gives many different skeletons.
Organic compounds are kinetically stable, so they do not decompose easily.
Carbon attaches to hydrogen, oxygen, nitrogen, halogens and other elements, and isomerism gives several structures for the same molecular formula.

Together these reasons explain why there are almost 20 million known organic compounds, far more than the compounds of all other elements.

ii3 marks

How was vital force theory rejected by Friedrich Wohler?

Solution

Berzelius held the vital force theory: organic compounds come only from living things, so they cannot be made from inorganic materials in the laboratory. In 1828 Wohler mixed silver cyanate with ammonium chloride to obtain ammonium cyanate:

\ce{AgOCN + NH4Cl -> NH4OCN + AgCl}

On heating, ammonium cyanate (inorganic) isomerized to urea, \ce{NH2-CO-NH2}, which is an organic compound. Urea had previously been reported in urine, but Wohler made it from an inorganic salt with no living matter involved. This showed that no vital force was needed, and Wohler rejected the theory.

\ce{NH4OCN ->[\Delta] NH2-CO-NH2}\qquad \text{(inorganic)} \to \text{(organic)}

iii3 marks

Differentiate between saturated and unsaturated hydrocarbons. Give three examples for each.

Solution

Saturated hydrocarbons have only single covalent bonds between their carbon atoms. All alkanes and cycloalkanes belong to this group. Unsaturated hydrocarbons have one or more carbon-carbon double or triple bonds (alkenes and alkynes). Because the multiple bond can open up to add other atoms, unsaturated hydrocarbons show addition reactions.

Feature	Saturated hydrocarbons	Unsaturated hydrocarbons
Bonds between carbon atoms	Single bonds only	At least one double or triple bond
Three examples	Ethane (\ce{CH3CH3}), butane (\ce{CH3CH2CH2CH3}), cyclobutane	Propene (\ce{CH3CH=CH2}), propyne (\ce{CH3C#CH}), cyclobutene
Typical reaction	Free radical substitution (alkanes)	Addition across the multiple bond

iv3 marks

Define empirical formula. What are the empirical formulae of ethanol, benzene and ethanoic acid?

Solution

The empirical formula shows the simplest whole-number ratio of the atoms of each element in a molecule. It is found by dividing the subscripts of the molecular formula by their common factor.

Compound	Molecular formula	Empirical formula
Ethanol	\ce{C2H6O}	\ce{C2H6O}
Benzene	\ce{C6H6}	\ce{CH}
Ethanoic acid	\ce{C2H4O2}	\ce{CH2O}

Ethanol is already in its simplest ratio (2:6:1). Benzene divides by 6 to give CH, and ethanoic acid divides by 2 to give \ce{CH2O}. Ethanoic acid has the same empirical formula as glucose, so an empirical formula alone cannot identify a compound.

v3 marks

Deduce the molecular and empirical formulae of the following compounds.

Solution

(i) Compound (i)

(i)

The skeleton has four carbon atoms and one C=C double bond, so it is but-1-ene, \ce{CH2=CH-CH2-CH3}. Counting the hydrogen atoms gives 8, so the molecular formula is \ce{C4H8}. Dividing by 4 gives the empirical formula \ce{CH2}.

\ce{C4H8},\quad \text{empirical formula } \ce{CH2}

(ii) Compound (ii)

(ii)

The ring has five carbon atoms, each with two hydrogen atoms, and no multiple bond. It is cyclopentane, \ce{C5H10}. Dividing by 5 gives the empirical formula \ce{CH2}.

\ce{C5H10},\quad \text{empirical formula } \ce{CH2}

(iii) Compound (iii)

(iii)

The four-membered ring has one C=C double bond. The two double-bonded carbons carry one hydrogen each and the other two carbons carry two hydrogens each, so the total is 6 hydrogen atoms. It is cyclobutene, \ce{C4H6}. Dividing by 2 gives the empirical formula \ce{C2H3}.

\ce{C4H6},\quad \text{empirical formula } \ce{C2H3}

vi3 marks

What is the importance of functional group? Discuss.

Solution

A functional group is the atom or group of atoms that takes part in chemical reactions and gives an organic compound its characteristic properties. The carbon network to which it is attached is relatively unreactive, so the functional group controls how the molecule behaves.

It decides the chemical properties. Alcohols (group -\ce{OH}) can be oxidized to aldehydes, ketones or carboxylic acids; alkenes and alkynes undergo addition across their multiple bonds; aldehydes and ketones undergo nucleophilic addition.
It classifies compounds into homologous series. Members that share a functional group have similar chemical properties, so the behaviour of a whole family can be predicted from one member.
It controls the naming and the family of a compound. Alkanes, which have no functional group, are the least reactive of all organic compounds.

vii3 marks

What is alkyl group? Name and draw 2D displayed structural formulae of all possible alkyls obtained from different isomers of propane and butane.

Solution

An alkyl group (R-) is the group left when one hydrogen atom is removed from an alkane. Its name replaces the suffix -ane with -yl. Propane has two kinds of hydrogen (on the end carbons and on the middle carbon), n-butane has two kinds, and isobutane (2-methylpropane) has two more, so six different alkyl groups can be obtained. The * in each formula marks the bond to the rest of the molecule.

n-propyl From propane (primary H)

isopropyl From propane (secondary H)

n-butyl From n-butane (primary H)

sec-butyl From n-butane (secondary H)

isobutyl From isobutane (primary H)

tert-butyl From isobutane (tertiary H)

Removing a primary hydrogen (on a primary carbon, which is bonded to one other carbon atom or none) gives n-propyl, n-butyl and isobutyl. Removing a secondary hydrogen gives isopropyl and sec-butyl, and removing the tertiary hydrogen of isobutane gives tert-butyl.

viii3 marks

Give systematic names to the following compounds.

Solution

(i) \ce{CH3CH2COCl}

The -\ce{COCl} group is an acid chloride. Its carbon is carbon 1 and the chain has three carbons, so the name is the acid name with -ic acid replaced by -yl chloride: propanoyl chloride (propionyl chloride).

propanoyl chloride

(ii) \ce{CH3CN}

The -\ce{CN} carbon is counted in the chain, so the molecule has two carbons. It is a nitrile, and the e of ethane is kept before the suffix -nitrile: ethanenitrile (acetonitrile).

ethanenitrile

(iii) \ce{CH3OCH3}

This is an ether with two methyl groups. The smaller alkyl group becomes the alkoxy prefix (methoxy) and the bigger group is treated as the alkane (methane), giving methoxymethane (dimethyl ether).

methoxymethane

ix3 marks

Each of the following names is wrong. Draw structures based on them and correct the names.

Solution

(a) 1-methylpentane

Drawn from the wrong name

A methyl group on carbon 1 of pentane is part of the longest chain, so the chain has six carbons. The structure is \ce{CH3(CH2)4CH3}, which is hexane, and this is the correct name.

hexane

(b) 2-ethylbutane

Drawn from the wrong name

An ethyl group on carbon 2 of butane makes a longer chain of five carbons (through the ethyl group). The longest chain is pentane with a methyl group on carbon 3, which is \ce{CH3CH2CH(CH3)CH2CH3}. Its name is 3-methylpentane.

Correct structure

3-methylpentane

Drawn from the wrong name

With only one substituent on the ring, that carbon is always position 1, so the locant is not written. The name is methylcyclohexane.

methylcyclohexane

(d) 3,3-methyl-4-ethylheptane

Drawn from the wrong name

Two methyl groups on carbon 3 must be written as dimethyl, and the substituents are cited alphabetically (ethyl before methyl). Numbering from the end nearest the substituents gives the locants 3, 3 and 4, so the name is 4-ethyl-3,3-dimethylheptane.

4-ethyl-3,3-dimethylheptane

x3 marks

Define electrophile and nucleophile with two examples each.

Solution

An electrophile is an electron-deficient species, with a positive charge or an empty orbital, that accepts an electron pair from a reaction partner. Two examples are the hydrogen ion \ce{H+} (the electrophile in the addition of HBr to ethene) and the carbocation \ce{CH3+}.

A nucleophile is an electron-rich species with a lone pair of electrons or a negative charge, which donates an electron pair to an electrophile. Two examples are the hydroxide ion \ce{OH-} (which replaces Br in bromoethane) and ammonia \ce{NH3}, which is neutral but has a lone pair on nitrogen.

xi3 marks

Define chain isomerism. How can you differentiate between metamerism and positional isomerism?

Solution

Chain isomerism occurs when molecules with the same molecular formula have different carbon chains, that is, a different chain length or branching. Example: butane, \ce{CH3CH2CH2CH3}, and 2-methylpropane, \ce{(CH3)2CHCH3}, are both \ce{C4H10}.

Metamerism and positional isomerism both have the same molecular formula and the same functional group. In metamers the functional group (for example, an ether -\ce{O-}) sits between two alkyl groups of different lengths: methoxypropane and ethoxyethane are both \ce{C4H10O}. In positional isomers the functional group is on a different carbon of the same chain: propan-1-ol and propan-2-ol are both \ce{C3H8O}.

To tell them apart, look at the functional group. If it sits between two alkyl groups whose lengths differ, the pair is metamers. If the same chain carries the functional group at a different carbon, the pair is positional isomers.

xii3 marks

Draw and name all the structural isomers of \ce{C3H6Br2}.

Solution

Two bromine atoms can be placed on propane in four different ways, which give four structural isomers of \ce{C3H6Br2}:

1,1-dibromopropane

1,2-dibromopropane

1,3-dibromopropane

2,2-dibromopropane

The positions of the two bromines are (1,1), (1,2), (1,3) and (2,2). Putting both bromines on the same carbon gives 1,1- and 2,2-dibromopropane. Putting them on different carbons gives 1,2- (neighbouring carbons) and 1,3- (the two ends of the chain).

xiii3 marks

What are aliphatic compounds? Give examples.

Solution

Aliphatic compounds are all organic compounds except aromatic compounds. They include open-chain compounds (straight or branched), and cyclic compounds, except those containing a benzene ring. Benzene and its derivatives are aromatic, not aliphatic.

Open-chain examples: butane (\ce{CH3CH2CH2CH3}), propene (\ce{CH3CH=CH2}) and ethyne (\ce{HC#CH}). Cyclic examples: cyclopentane, cyclohexane and cyclobutene.

Long questionsp. 317

35 marks

Define skeletal and 2D displayed structural formulae. Give three examples of each.

Solution

Definitions

A skeletal formula is a line drawing in which each end and each corner of a line is a carbon atom, unless another atom (oxygen, a halogen or nitrogen) is written. Carbon and hydrogen atoms are not shown; each carbon is assumed to carry enough hydrogen atoms to complete its valency. Double and triple bonds and functional groups are shown.

A 2D displayed formula shows every atom and every bond of the molecule in a flat (x and y) plane. It shows the arrangement of the bonds, but not the shape of the molecule in space.

Examples

Skeletal formulae of three compounds:

Butane

Butan-2-ol

Propanal

2D displayed formulae of three compounds, with every atom and bond shown:

2D displayed formulae of ethane, ethanol and ethene. Every atom and every bond is drawn.

43 marks

Derive molecular formula of a compound with molecular mass 26 amu and empirical formula CH.

Solution

Given

M_r = 26\,\mathrm{amu}
\text{empirical formula} = \ce{CH}

To find

molecular formula

Mass of one empirical unit, taking C = 12 and H = 1.
M_\text{CH}=12+1=13\,\mathrm{amu}
Number of empirical units in one molecule (the multiple n).
n=\dfrac{M_r}{M_\text{CH}}=\dfrac{26}{13}=2
Multiply the empirical formula by n.
(\ce{CH})_2=\ce{C2H2}

The compound is ethyne (\ce{HC#CH}), which the book lists with empirical formula CH.

\ce{C2H2}

Note. The book prints no answer for this question.

55 marks

What is homologous series? Write its features.

Solution

Definition

A homologous series is a group of organic compounds that have the same functional group but a different length of carbon chain. Adjacent members differ by one -\ce{CH2-} (methylene) group.

Features

Each homologous series has its own functional group.
Each member differs from the next by one -\ce{CH2-} (methylene) group.
Each series has its own general formula, from which the number of hydrogen atoms for a given number of carbon atoms can be found.
All members have similar chemical properties, because they share a functional group.
The physical properties change gradually as the carbon chain gets longer.
The members have a similar method of preparation.

Homologous series	General formula	Example
Alkanes	\ce{C_{n}H_{2n+2}}	Ethane, \ce{CH3CH3}
Alkenes	\ce{C_{n}H_{2n}}	Propene, \ce{CH3CH=CH2}
Alcohols	\ce{C_{n}H_{2n+1}OH}	Ethanol, \ce{CH3CH2OH}

68 marks

What is free radical substitution reaction of alkanes? Explain bromination of methane with mechanism.

Solution

Definition

In a free radical substitution reaction of an alkane, a halogen free radical (an atom or group with an unpaired electron) replaces the hydrogen atoms of the alkane one after another, which gives a mixture of halogenoalkanes. The reaction needs a halogen molecule and sunlight, and it takes place in three steps: initiation, propagation and termination.

Mechanism of bromination of methane

Initiation. In sunlight the bromine molecule splits homolytically, so each bromine atom takes one electron and two bromine free radicals form.
\ce{Br2 ->[h\nu] 2Br.}
Propagation (step 1). A bromine radical takes a hydrogen atom from methane, giving a methyl radical and hydrogen bromide.
\ce{CH4 + Br. -> CH3. + HBr}
Propagation (step 2). The methyl radical takes a bromine atom from a bromine molecule, giving bromomethane and a new bromine radical. This radical goes back to step 1, so the chain continues.
\ce{CH3. + Br2 -> CH3Br + Br.}
Termination. Any two free radicals meet and form a neutral molecule, so the chain stops.
\begin{gathered}\ce{Br. + Br. -> Br2}\\ \ce{CH3. + Br. -> CH3Br}\\ \ce{CH3. + CH3. -> C2H6}\end{gathered}

Overall reaction

\ce{CH4 + Br2 ->[h\nu] CH3Br + HBr}

If bromine is in excess, the remaining hydrogen atoms of bromomethane are replaced in the same way by further free radical substitution.

76 marks

Define nucleophilic substitution and electrophilic addition reactions. Give examples of each.

Solution

Nucleophilic substitution

In a nucleophilic substitution reaction, a nucleophile (an electron-rich species such as \ce{OH-}) attacks the carbon of a halogenoalkane and replaces the leaving group (the halogen atom).

\ce{CH3CH2Br + NaOH -> CH3CH2OH + NaBr}

The hydroxide ion donates a lone pair to the carbon that carries the partial positive charge in bromoethane. The C-Br bond breaks, and bromide leaves as the leaving group, so ethanol forms.

Bromoethane

Ethanol

Electrophilic addition

In an electrophilic addition reaction, an electrophile attacks a multiple bond, usually the pi bond of an alkene. The double bond becomes a single bond and the two parts of the reagent add across the two carbon atoms.

\ce{CH2=CH2 + HBr -> CH3CH2Br}

Here the electrophile is the hydrogen ion from HBr, which carries a partial positive charge. The pi electrons of ethene attack it, and the bromine then joins the other carbon to give bromoethane.

Ethene

Bromoethane

86 marks

What is structural isomerism? Explain functional group and tautomerism with examples.

Solution

Structural isomerism

Structural isomers have the same molecular formula but a different arrangement of atoms, so they have different structural formulae. The book describes five types: chain, positional, functional group, metamerism and tautomerism.

Functional group isomerism

Functional group isomers have the same molecular formula but different functional groups. Propan-1-ol (an alcohol) and methoxyethane (an ether) are both \ce{C3H8O}:

Propan-1-ol

Methoxyethane

Tautomerism

Tautomers are isomers that differ in the position of a mobile hydrogen atom. They exist in dynamic equilibrium, because a proton migrates within the molecule. Ethyl acetoacetate is the example in the book, and this type is also called keto-enol isomerism:

Keto form

Enol form

\ce{CH3COCH2COOC2H5 <=> CH3C(OH)=CHCOOC2H5}

96 marks

Explain the difference between homolytic and heterolytic fissions of covalent bonds. How can homolytic fission initiate the halogenation reaction of alkanes?

Solution

Homolytic and heterolytic fission

Feature	Homolytic fission	Heterolytic fission
Sharing of the bonding pair	Equal: each atom takes one electron	Unequal: both electrons go to one atom
Products	Free radicals (unpaired electrons)	Positive and negative ions
Usual bond	Between identical atoms, for example Cl-Cl	Between atoms of different electronegativity, for example C-Cl
Example	\ce{Cl2 -> 2Cl.}	\ce{CH3Cl -> CH3+ + Cl-}

Homolytic fission happens when the electronegativity difference is zero or nearly zero. Heterolytic fission happens when the two atoms differ enough in electronegativity to pull the bonding pair to one side.

Initiation of halogenation of alkanes

Halogenation of an alkane, for example chlorination of methane, starts with homolytic fission of the halogen molecule in sunlight. The energy comes from the ultraviolet region of sunlight.

\ce{Cl2 ->[h\nu] 2Cl.}

Each chlorine radical then takes a hydrogen atom from methane, giving the methyl radical, and the chain continues as in the free radical substitution mechanism. Without this initiating step there would be no radicals to start the reaction.

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.

2025 Annual I, Set B, Q.2 (i) OR1+2 = 3 marks

What is meant by catenation? What is its importance in organic chemistry?

Solution

Catenation is the ability of carbon atoms to link with one another to form long chains and rings. The word comes from the Greek "catena", which means chain. Silicon forms only small chains and sulphur only small rings, so the property is strongest in carbon.

Importance in organic chemistry. Catenation is the main reason for the very large number of organic compounds, for three reasons.

Carbon makes strong C-C bonds (bond energy 346\,\mathrm{kJ\,mol^{-1}}) because of its small size, and the atoms can join by single, double and triple bonds, which gives chains, branched chains and rings of many shapes and lengths.
Organic compounds are kinetically stable, so the long chains do not break down easily.
Carbon also attaches to hydrogen, oxygen, nitrogen, halogens and other elements, and the same atoms can be arranged in different ways (isomerism), so one molecular formula can stand for several compounds.

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

Which property of carbon is responsible for existence of large number of organic compounds in nature and how? Explain.

Solution

The property is catenation, the self-linking of carbon atoms to form long chains and rings. Carbon is small, so its C-C bonds are strong (346\,\mathrm{kJ\,mol^{-1}}), and this gives stable chains and rings.

Carbon makes strong C-C bonds (bond energy 346\,\mathrm{kJ\,mol^{-1}}) because of its small size, and the atoms can join by single, double and triple bonds, which gives chains, branched chains and rings of many shapes and lengths.
Organic compounds are kinetically stable, so the long chains do not break down easily.
Carbon also attaches to hydrogen, oxygen, nitrogen, halogens and other elements, and the same atoms can be arranged in different ways (isomerism), so one molecular formula can stand for several compounds.

2026 Annual I, Set B, Q.56 marks

Consider an organic compound with molecular formula \ce{C5H10O}. It exists in different isomeric forms. Draw possible isomers. Also write types of structural isomerism shown by them.

Solution

Possible isomers of \ce{C5H10O}

\ce{C5H10O} has one degree of unsaturation (one double bond or one ring), so it can be a carbonyl compound, an unsaturated alcohol or ether, or a ring compound. The main isomers are drawn below.

Aldehydes (functional group -\ce{CHO}):

Pentanal

2-Methylbutanal

3-Methylbutanal

2,2-Dimethylpropanal

Ketones (functional group >\ce{C=O}):

Pentan-2-one

Pentan-3-one

3-Methylbutan-2-one

Other functional groups (some further examples):

Pent-4-en-1-ol (unsaturated alcohol)

Cyclopentanol (cyclic alcohol)

1-Ethoxyprop-1-ene (unsaturated ether)

Oxane (cyclic ether)

Types of structural isomerism shown

Type	Example from the isomers above
Chain isomerism (different carbon chain or branching)	Pentanal, 2-methylbutanal, 3-methylbutanal and 2,2-dimethylpropanal; also pentan-2-one and 3-methylbutan-2-one
Positional isomerism (different position of the same functional group)	Pentan-2-one and pentan-3-one (the >\ce{C=O} group is on C2 or C3)
Functional group isomerism (different functional groups)	Pentanal (aldehyde) and pentan-2-one (ketone); pent-4-en-1-ol (alcohol); 1-ethoxyprop-1-ene (ether)
Metamerism (different alkyl groups on the two sides of the same group)	Pentan-2-one (\ce{CH3} and \ce{C3H7} on the carbonyl) and pentan-3-one (\ce{C2H5} on both sides)
Tautomerism (mobile hydrogen, dynamic equilibrium)	Pentan-2-one is in keto-enol equilibrium with pent-2-en-2-ol

\ce{CH3-CO-CH2CH2CH3 <=> CH3-C(OH)=CH-CH2CH3}

Note. More isomers than those drawn exist (for example other cyclic ethers and alcohols); the chain, positional, functional group and carbonyl isomers shown are enough for full marks.

2025 Annual I, Set B, Q.5 (b)2 marks

How can one identify the functional groups in the following compounds by chemical tests? Elaborate. \ce{CH3-CH2-OH}

Solution

Ethanol contains the hydroxyl group (-\ce{OH}) of an alcohol. It is identified by these tests.

Sodium test. Add a small piece of sodium: hydrogen gas bubbles off (the gas burns with a pop), which shows an -\ce{OH} group.
\ce{2CH3CH2OH + 2Na -> 2CH3CH2ONa + H2 ^}
Acidified potassium dichromate. Warm with \ce{K2Cr2O7/H+}: the orange solution turns green, because the primary alcohol is oxidised to ethanal (and then ethanoic acid).
\ce{CH3CH2OH ->[K2Cr2O7/H+] CH3CHO ->[K2Cr2O7/H+] CH3COOH}
Esterification. Warm with ethanoic acid and a few drops of concentrated \ce{H2SO4}: a sweet fruity smell of ethyl ethanoate shows an alcohol.
\ce{CH3CH2OH + CH3COOH <=>[{conc.} H2SO4] CH3COOC2H5 + H2O}

2025 Annual I, Set B, Q.5 (c)2 marks

How can one identify the functional groups in the following compounds by chemical tests? Elaborate. \ce{CH3-CH=CH2}

Solution

Propene contains a carbon-carbon double bond (\ce{C=C}), so it is unsaturated and shows addition reactions. It is identified by two tests.

Bromine water test. Shake with bromine water: the orange-brown colour disappears at once, because bromine adds across the double bond.
\ce{CH3CH=CH2 + Br2 -> CH3CHBrCH2Br}
Baeyer's test. Shake with cold dilute alkaline potassium permanganate: the purple colour fades and a brown precipitate of \ce{MnO2} forms, as the alkene is oxidised to a diol.
\ce{3CH3CH=CH2 + 2KMnO4 + 4H2O -> 3CH3CH(OH)CH2OH + 2MnO2 v + 2KOH}

A saturated hydrocarbon such as propane does not decolourise either reagent in the cold, so the tests separate alkenes from alkanes.

2026 Annual I, Set D, MCQ 71 mark

Electrophile is a specie which is:

AElectron pair donor
BProton acceptor
CProton donor
DElectron pair acceptorAnswer

Why

An electrophile is an electron-deficient species, with a positive charge or an empty orbital, so it accepts an electron pair from its reaction partner (the nucleophile). Examples are \ce{H+} and the carbocation \ce{CH3+}.

Option A describes a nucleophile, which donates an electron pair. Options B and C use the Bronsted-Lowry idea of proton acceptor and donor, which is a different definition; an electrophile is defined by electron pairs, not protons.

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

Why does carbon form a large number of organic compounds? Give three reasons.

Solution

Carbon forms a very large number of compounds for three reasons. The first is catenation; the other two are the stability of its compounds and its bonding with other elements.

Carbon makes strong C-C bonds (bond energy 346\,\mathrm{kJ\,mol^{-1}}) because of its small size, and the atoms can join by single, double and triple bonds, which gives chains, branched chains and rings of many shapes and lengths.
Organic compounds are kinetically stable, so the long chains do not break down easily.
Carbon also attaches to hydrogen, oxygen, nitrogen, halogens and other elements, and the same atoms can be arranged in different ways (isomerism), so one molecular formula can stand for several compounds.

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

What is catenation? Explain its importance in Organic chemistry.

Solution

Importance in organic chemistry. Catenation is the main reason for the very large number of organic compounds, for three reasons.

Carbon makes strong C-C bonds (bond energy 346\,\mathrm{kJ\,mol^{-1}}) because of its small size, and the atoms can join by single, double and triple bonds, which gives chains, branched chains and rings of many shapes and lengths.
Organic compounds are kinetically stable, so the long chains do not break down easily.
Carbon also attaches to hydrogen, oxygen, nitrogen, halogens and other elements, and the same atoms can be arranged in different ways (isomerism), so one molecular formula can stand for several compounds.

2025 Annual I, Set D, Q.2 (xiv)1.5+1.5 = 3 marks

Draw position isomers and functional group isomers of \ce{C3H7-OH}.

Solution

The formula is \ce{C3H8O}.

Position isomers

Same functional group (-\ce{OH}), on a different carbon of the three-carbon chain:

Propan-1-ol

Propan-2-ol

Functional group isomer

The same formula with a different functional group, the ether group -\ce{O-}:

Methoxyethane

Propan-1-ol and propan-2-ol are alcohols, so they differ only in the position of -\ce{OH}. Methoxyethane has the ether group instead of the hydroxyl group, so it is a functional group isomer of both.

2026 Annual I, Set D, Q.56 marks

Consider an organic compound with molecular formula \ce{C5H11O}. It exists in different isomeric forms. Draw possible isomers. Also write type of structural isomerism shown by them.

Solution

Possible isomers of \ce{C5H12O}

Alcohols (-\ce{OH} group), eight isomers:

Pentan-1-ol

Pentan-2-ol

Pentan-3-ol

2-Methylbutan-1-ol

3-Methylbutan-1-ol

2-Methylbutan-2-ol

3-Methylbutan-2-ol

2,2-Dimethylpropan-1-ol

Ethers (-\ce{O-} group), six isomers:

1-Methoxybutane

2-Methoxybutane

1-Methoxy-2-methylpropane

2-Methoxy-2-methylpropane

1-Ethoxypropane

2-Ethoxypropane

Types of structural isomerism shown

Type	Example from the isomers above
Chain isomerism	Pentan-1-ol, 2-methylbutan-1-ol and 2,2-dimethylpropan-1-ol (straight chain and branched chains)
Positional isomerism	Pentan-1-ol, pentan-2-ol and pentan-3-ol (the -\ce{OH} group is on C1, C2 or C3)
Functional group isomerism	Pentan-1-ol (alcohol) and 1-methoxybutane (ether)
Metamerism	1-Methoxybutane (\ce{CH3} and \ce{C4H9} around oxygen) and 1-ethoxypropane (\ce{C2H5} and \ce{C3H7} around oxygen)

Note. The formula is printed as \ce{C5H11O} in this set, which is a misprint of \ce{C5H12O}; solved for \ce{C5H12O}.

Exam focus

Define catenation and explain why carbon forms such a large number of organic compounds (three reasons). (short)
Asked in all four new-syllabus papers: 2026 Set B Q.2 (vi) OR, 2026 Set D Q.2 (vi) OR, 2025 Set B Q.2 (i) OR and 2025 Set D Q.2 (iv). It is also book short question i.
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Draw and classify the structural isomers of a given formula (\ce{C5H10O}, \ce{C5H12O}) and name the type of isomerism. (long)
A 6-mark Section C question in both 2026 sets (Set B Q.5 and Set D Q.5), so it is the most expensive unit 15 item. Practise chain, positional, functional group and metamerism.
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Draw position isomers and functional group isomers of \ce{C3H7-OH}. (short)
Asked in 2025 Set D Q.2 (xiv); the book repeats the same pair (propan-1-ol, propan-2-ol, methoxyethane) in the isomerism section and long question 8.
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Define electrophile and nucleophile; know that an electrophile accepts an electron pair. (mcq)
2026 Set D MCQ 7 asked it directly, and the book has it as short question x.
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Identify the functional group of a compound by a chemical test (alcohol, alkene). (short)
2025 Set B Q.5 asked tests for \ce{CH3CHO}, ethanol and propene (parts a to c, 2 marks each); the tests are given with the functional groups.
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Classify hydrocarbons as saturated or unsaturated and recognise them from the formula. (mcq)
2026 Set D MCQ 14 asked for the saturated compound among \ce{C6H14}, \ce{C6H12}, \ce{C6H10}, \ce{C6H6}; also book short question iii.
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Define and compare the types of structural isomerism: chain, positional, functional group, metamerism, tautomerism. (long)
Book short question xi and long question 8, and the base for the two 2026 isomer questions.
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Name organic compounds by IUPAC rules and recognise the functional group. (short)
Book exercise MCQs i, viii, ix and short questions viii and ix are on naming; naming was not asked directly in the 2025 or 2026 papers but is needed to write isomer names.
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