Unit 11-1

Unit 11 Fundamentals of Organic Chemistry

Introduction

（ 1 ） What is Organic Chemistry ? Organic Chemistry is the chemistry of compounds which contain the element carbon. Most of these compounds contain hydrogen and many also contain oxygen, nitrogen or other

elements. There are a few compounds containing carbon, however, which are not normally classified as organic compounds. Carbon monoxide, carbon dioxide and the metal carbonates are examples.

Although organic substances such as sugars, alcohol and vinegar had been known for thousands of years, it was not until the eighteen century that organic compounds were first isolated. One of the first scientists to study organic compounds was the self-taught chemist, Carl Wilhelm Scheele (1742-1786). He obtained and purified a number of organic acids and other organic compounds from plant and animal sources. He isolated 2-hydroxypropanoic acid (lactic acid) from milk and showed that this acid was the cause of turning milk sour.

During the eighteen century chemists believed that organic compounds could only be synthesized by means of a ‘life force’ in living cells. This was called the vitalistic theory of organic chemistry. However, in 1828, the German chemist Friedrich Wohler (1800-1882) prepared urea (carbamide) by heating an aqueous solution of ammonium cyanate :

NH4CNO CO(NH2)2

This was the first synthesis of an organic compound. It heralded the decline of the vitalistic theory. The term ‘biochemistry’ is now used for the chemistry of living things and life processes.

All living things contain organic compounds. Furthermore, many of the modern products and materials upon which we depend are organic.

Some organic compounds :Naturally occurring Synthetic carbohydrates, proteins, fats and oils, vitamins plastics, many medicines and drugs,

insecticides, many dyes

A measure of the importance of organic chemistry nowadays can be gauged from the almost exponential growth in the number of known organic compounds over the last century :Year 1880 1910 1940 1960 1970 1980 1990Number 12000 150000 500000 1000000 2000000 5500000 7000000

A knowledge of organic chemistry enables chemist to develop and manufacture drugs, agricultural chemicals, anesthetics and other chemicals whose effects on life processes are important to humans.

Unit 11-2（ 2 ） The unique nature of carbon Why is that carbon can form such a vast number of naturally occurring and synthetic compounds ? The answer lies in its unique ability to catenate : to bond with itself and form stable long-chain and ring structures. Carbon also has the ability to form single, double and triple bonds not only with itself but also with other elements such as oxygen and nitrogen.

There are three important properties of carbon that enable it to form so many stable carbon compounds :1. Catenation The ability of carbon to form strong bonds to itself means that it can form chains and rings of varying size. This property is called catenation. The stability of the single, double and triple carbon-carbon bonds can be seen by comparing the bond enthalpies in table :

Bond C-C C=C CC Si-Si Si=Si S-S C-HBond enthalpy kJmol-1

346 610 835 226 318(estimated)

272 413

Note the strength of the C-C bond compared with that of the Si-Si and S-S bonds. Note too the high

strength of the C-H bond : all but a handful of the vast number of carbon compounds also contain hydrogen. In the presence of air, carbon compounds are not stable relative to their oxidation products, carbon dioxide and water. For example, methane :

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H = -890 kJmol-1

Although methane is energetically unstable relative to its combustion products, it does not react with

air until heated to quite high temperature. This is because the reaction between methane and oxygen has a high activation energy which must be supplied before the reaction will proceed. Thus, most carbon compounds are energetically unstable in the presence of air, but kinetically stable.

2. Carbon can form four covalent bonds Carbon is also unique in its ability to hybridize. Its four bonding electrons can undergo hybridization to form :

four single bonds --

one double bond and two single bonds --

one triple bond and one single bond --

3. Carbon has a fully shared octet of electrons in its compounds Carbon atoms have no lone pairs or empty orbitals in their outer shells, so that they are unable to form dative bonds. This properties is responsible for the kinetic stability of its compounds.

Unit 11-3

Section 11 .1 Bonding and Structure

（ 1 ） Orbital hybridisation of carbon The concept of hybridisation theory involves the following points :1. Hybridisation is the mixing of pure orbitals to form new hybrid orbitals which are equivalent and have

definite orientations in space.2. Only orbitals which lie close together in energy can be used in the construction of hybrid orbitals.3. The number of hybrid orbitals always equals the number of component atomic orbitals.

1. sp3 hybridisation It is formed by one 2s orbital and three 2p orbitals of a carbon atom to give four equivalent hybrid sp3 orbitals pointing towards the vertices of a regular tetrahedron :

2. sp2 hybridisation It is formed by one 2s orbital and two 2p orbitals of a carbon atom to give three equivalent sp2 hybrid orbitals which are pointing towards the vertices of an equilateral triangle :

3. sp hybridisation It is formed by one 2s orbital and one 2p orbital of a carbon atom to give two equivalent sp hybrid orbitals of linear structure :

Unit 11-4（ 2 ） Structures and shapes of hydrocarbons 1. Saturated hydrocarbons Hydrocarbons whose molecules contain only single bonds are known as saturated hydrocarbons. In saturated hydrocarbons the outermost shell electrons of each carbon atoms have hybridised to form four equivalent sp3 orbitals which are arranged tetrahedrally about the nucleus. This geometry gives idealized bond angles of 109.5o.

Energy ___ ___ ___ ___ sp3 sp3 sp3 sp3 hybridized state

___ ___ ___ ___ 2s 2p 2p 2p excited state

___ ___ ___ ___ 2s 2p 2p 2p ground state

When an sp3 carbon atom forms bonds, it does so by overlapping each of its four sp3 orbitals (each with one electron) with orbitals from other four atoms (each orbital in turn containing one electron). Therefore it is capable of forming four single bonds. The molecular orbital of a single bond is symmetrically around the axis passing through the nuclei of two bonded atoms. It is known as a bond.

Structures and shapes of saturated hydrocarbons :

Methane CH4

In methane, each sp3 orbital of carbon overlaps with a 1s orbital of hydrogen. Each of the resultant sp3-1s molecular orbital is symmetrically around the axis passing through the nuclei of the carbon and hydrogen atoms. The C-H bonds in methane are bonds.

Structural formula condensed structural formula 3-dimensional structure (shape)

CH4

Ethane C2H6

Unit 11-5 Ethane contains two sp3 carbon atoms. These two carbon atoms form a C-C bond by the overlap of one sp3 orbital from each carbon. Each carbon atom has three remaining sp3 orbitals, and each of these overlap with a 1s orbital of a hydrogen atom to form a C-H bond.

Structural formula condensed structural formula 3-dimensional structure (shape)

C

H

C

H

H

H

H

H

CH3CH3

Propane C3H8 and Butane C4H10 Propane and butane are examples of alkanes that are sometimes called unbranched chain alkanes. Because of the tetrahedral arrangement of the bond pairs around carbon atoms, their chains are zigzagged in shape.

Structural formula condensed structural formula 3-dimensional structure (shape)

CH3CH2CH3

CH3CH2CH2CH3

Cyclohexane C6H12

Cyclohexane is an example of cyclic alkane. If the cyclohexane ring were flat, a serious bond strain would be resulted and all the hydrogen atoms on the ring would be eclipsed. There are many shapes that a cyclohexane ring can assume, the most stable one is the ‘chair’ form with minimum bond strain and all hydrogen atoms staggered.

Draw the 3-dimensional structure of 2-methylpropane which is a branched chain alkane.

2. Unsaturated hydrocarbons

Unit 11-6 Hydrocarbons whose molecules contain double bonds or triple bonds are known as unsaturated hydrocarbons.

Carbon-carbon double bond : the sp 2 hybridization

Formation of the C=C bonds The atomic orbitals of carbon can also hybridise to give three identical sp2 orbitals arranged symmetrically in a plane at an angle of 120o to each other, with the remaining electron in a p orbital at right angles above and below this plane.

Energy ___ ___ ___ ___ sp2 sp2 sp2 2p hybridized state

___ ___ ___ ___ 2s 2p 2p 2p excited state

___ ___ ___ ___ 2s 2p 2p 2p ground state

Structure and shape of the ethene C2H4 molecule : In ethene, one of the three sp2 orbitals overlaps with an sp2 orbital of a second carbon atom to form a bond. The other sp2 orbitals of the two carbon atoms each overlaps with the 1s orbital of a hydrogen atom to form four bonds. The 2p orbitals of the carbon atoms overlap above and below the plane to form a bond. A bond is formed by lateral overlap of the p orbitals and has a plane of symmetry.

A C=C double bond consists of a bond and a bond. A bond is stronger than a bond as it is formed between the lines of centres of the bonding atoms and greater orbital overlap is possible.

Shape of the ethene molecule :

Draw the 3-dimensional structure of propene and indicate the and bonds in the molecule.

Carbon-carbon triple bond : the sp hybridization

Unit 11-7

Formation of the CC bonds The hybrid orbitals formed from the 2s and one of the 2p orbitals of carbon atom consists of two identical collinear sp orbitals with lobes directed in opposite directions on either side of the nucleus. The two remaining unpaired electrons are in 2p orbitals at right angles to each other.

Energy ___ ___ ___ ___ sp sp 2p 2p hybridized state

___ ___ ___ ___ sp sp 2s 2p 2p 2p excited state

___ ___ ___ ___ 2s 2p 2p 2p ground state

Structure and shape of the ethyne C2H2 molecule : In the ethyne C2H2 molecule, one of the two sp orbitals overlap with an sp orbital of a second carbon atom to form a bond, while the two remaining sp orbitals(one on each carbon atom) form bonds with hydrogen atoms. The unhybridised 2p orbitals overlap laterally and two bonds are formed in planes at right angles above, below and on either side of the linear molecule.

A CC triple bond consists of a bond and two bonds. These bonds draw the carbon atoms closer together, making the CC bond both shorter and stronger than the carbon-carbon double and single bonds.

The following two compounds have the same molecular formula C5H8 but different structures :

Give the hybridisation states of all carbon atoms and draw the 3-dimensional structures for both compounds.

3. Aromatic hydrocarbons Organic compounds are divided into two main classes : the aliphatic and the aromatic.

Unit 11-8The molecules of both saturated and unsaturated aliphatic compounds are open-chain (acyclic) or

simple rings (cyclic) structures. The term aromatic refers to a major class of unsaturated cyclic organic compounds which includes

benzene and its derivatives. Benzene is an unsaturated, but highly stable, cyclic molecule containing six carbon atoms. Aromatic hydrocarbons are hydrocarbons which have benzene ring in their structures. They are sometimes called arenes. Structure and shape of the benzene C6H6 molecule : Benzene C6H6 is a cyclic compound with six carbon atoms joined in a ring. Each carbon atom is sp2 hybridised, and the ring is planar. Each carbon atom has one hydrogen atom bonded to it and an unhybridised 2p orbital perpendicular to the plane of the bonds of the ring. Each of these six 2p orbitals can contribute one electron for bonding. This type of p-orbital system leads to complete delocalisation of all six electrons giving rise to a highly stable structure of the benzene molecule.

It is known that all carbon-carbon bond length in benzene is the same, 0.140 nm. All six bonds are longer than C=C double bonds, but shorter than C-C single bonds. From the bond lengths plus a body of other evidence, chemists have concluded that benzene is a symmetrically molecule with six ring bonds (C-C bonds) and six bonds of C-H bonds. Instead of alternating double and single bonds, the six electrons are completely delocalised in a cloud of electronic charge above and below the ring.

bonds skeleton (C-C : sp2-sp2 ; C-H : sp2-1s) bonds (6 electrons delocalised)

Table : Correlation of bond length and bond order for benzene

Benzene is just one example of aromatic compounds that contain aromatic clouds. If an

unsaturated ring system contains 4n + 2 electrons, complete delocalisation of electrons and subsequently stabilised structure is obtained.

Example : Pyridine (C5H5N) and naphthalene (C10H8) are also classified as aromatic compounds :

Delocalisation energy of benzene An indication of the enhanced stabilization conferred by the delocalisation of the electrons is given by the enthalpy change of hydrogenation of benzene to form cyclohexane.

Unit 11-9

+ 3 H2 H = -208 kJ mol-1

If there were no delocalisation of the electrons and there were three alternative bonds in the ring,

the enthalpy change of hydrogenation would be expected to be three times that of cyclohexane :

H = -120 kJ mol-1

The difference between the theoretical, 3 x (-120) = -360 kJ mol-1, and the experimental enthalpy change of hydrogenation is the delocalisation energy and is a measure of this increased stability.

Enthalpy kJ mol-1

The enthalpy changes of combustion of benzene and cyclohexene determined experimentally are - 3267.4 kJ mol-1 and - 4128.1 kJ mol-1. Together with the data of combustion for carbon and hydrogen,

C(s) + O2(g) → CO2(g) ΔHOc = - 393.5 kJ mol-1

H2(g) + O2(g) → H2O(l) ΔHOc = - 285.8 kJ mol-1

calculate the enthalpy changes of formation of benzene and cyclohexene.

From the above calculations, the enthalpy change of formation of benzene is found to be less positive than that of cyclohexene. This shows that benzene is energetically more stable than cyclohexene despite that it has 'three double bonds' in the molecule.Section 1 1 . 2 Functional groups and homologous series

Unit 11-10

（ 1 ） Functional groups A functional group is an atom, a group of atoms, or a bonding arrangement, which is responsible for specific properties of an organic compound or class of compounds.

The structure of a typical carbon compound can be considered as two parts : a saturated carbon-hydrogen ‘skeleton’, which is comparatively unreactive, and a reactive part consisting of one or more functional groups. Two examples are shown below :

Table 1 lists three simplest functional groups, which contain only carbon-carbon bonds. In the

formulae of these groups, the open-ended bond-lines represent attachment to other carbon atoms or hydrogen atoms. The name of an organic compound usually contains clues that indicate which functional groups are present.

Functional group Class names Name-clues ExampleDouble bond Alkenes -ene ethene

Triple bond Alkynes -yne Ethyne

Benzene aromaticsarenes

-benz-

phen-

methylbenzene

phenol

C C C CH

H

H H H H

HHH

C NC C C C O HH

H

H H H H

HHH

Unit 11-11

Table 2 lists functional groups which contain bonds linking carbon atoms to halogen, oxygen and nitrogen atoms. The functional groups in this table are attached only to carbon atoms, as indicated by the open-ended bond-lines.

Functional group Class names Name-clues ExamplesCarbon-halogenbonds

halogeno -X -F -Cl -Br -I

halogeno-compounds

fluoro-chloro-bromo-iodo-

chloromethane CH3-Cl

1-iodobutane CH3CH2CH2CH2-I

Carbon-oxygenbonds

hydroxyl -OH hydroxycompounds(alcohols)

-ol

hydroxy-

ethanol CH3CH2-OHhydroxyethanoic acid

ether C-O-C ethers -oxy ethoxyethane CH3CH2-O-CH2CH3

carbonyl -CO-

-CHO

carbonylcompounds :ketones

aldehydes

-one

-al

propanone

ethanal

carboxyl -COOH carboxylic acids -oic acid propanoic acid CH3CH2COOH

ester -COOR esters alkyl -oate ethyl ethanoate CH3COOCH2CH3

acyl halide -COX acyl halides -oyl halide ethanoyl chloride

amide -CONH2 amides -amide ethanamide

acid anhydride (-CO)2O

acid anhydrides -oic anhydride ethanoic anhydride

Carbon-nitrogenbonds

amino -NH2 primary amines ethylamine

-NHR secondary amines

-amineamino-

diethylamine

-NR2 tertiary amines trimethylamine

cyano -CN nitriles -nitrile ethanenitrile

nitro -NO2 nitro compounds nitro- nitroethane

Unit 11-12

（ 2 ） Homologous series and physical properties Homologous series is a series of related organic compounds in which the formula of each member differs from the preceding member by -CH2-. Members of a homologous series have the same functional group and the same general formula. They show similarities in chemical properties and trends in physical properties. The similarity in chemical properties is explained by the same functional group in each compound. The trend in physical properties is explained by the change in intermolecular forces with the increasing length of carbon chains.

Example : A homologous series of alcohols CnH2n+1OHName Molecular formula Structural formula Boiling point ℃methanol CH3OH CH3OH 65ethanol C2H5OH CH3CH2OH 78propan-1-ol C3H7OH CH3CH2CH2OH 97butan-1-ol C4H9OH CH3CH2CH2CH2OH 118pentan-1-ol C5H11OH CH3CH2CH2CH2CH2OH 138

The physical properties of organic compounds depend on both the structure of their functional groups and the length of their carbon chains.

Melting points and Boiling points Most organic compounds are covalent molecules and exist as gases, liquids or relatively low melting solids at room temperature. This is because the organic molecules are held together by intermolecular forces only. As van der Waals’ forces increase with increasing molecular size and molecular mass, both melting point and boiling point increase with increasing length of carbon chains in a homologous series.

For instance, the boiling points of the unbranched chain alkanes are shown below :Alkane CnH2n+2 Methane CH4 Ethane C2H6 Propane C3H8 Butane C4H10 Pentane C5H12

Boiling point / ℃ -164 -89 -42 0 36

Alcohols and carboxylic acids, both of which contain the -OH group, have abnormally high boiling points. This is due to intermolecular hydrogen bonding.Example methoxymethane

CH3OCH3

ethoxyethane CH3CH2OCH2CH3

Butan-1-ol C4H9OH

butanoic acid C3H7COOH

Boiling point / ℃ - 25 35 118 164

SolubilityNon-polar organic compounds such as the hydrocarbons are generally immiscible with water.

However, they do dissolve in non-polar solvents such as trichloromethane and methylbenzene. The more polar organic compounds such as the alcohols and carboxylic acids tend to be more soluble in polar solvents such as water. This is due to the formation of hydrogen bonds between the solute and solvent molecules. However, the solubility decreases rapidly as the length of carbon chains increases.

Density As the length of carbon chains and thus the relative molecular masses of organic compounds increase along a homologous series, the densities of the compounds increase. The branched chain isomers usually have lower densities than the unbranched chain isomers. This can be explained by the decrease in contact surface area between branched chain isomers and thus weaker van der Waals forces are expected.Alcohol Butan-1-ol 2-methylpropan-1-ol 2-methylpropan-2-olDensity at 20℃ 0.810 gcm-3 0.802 gcm-3 0.789 gcm-3

Section 11. 3 Systematic Nomenclature

Unit 11-13 An organic compound may have a trivial name, e.g. wood spirit; a radicofunctional name, e.g. methyl alcohol; or a substitutive name, e.g. methanol. The systematic nomenclature described here is substitutive nomenclature, since internationally this is the most favoured system.

The systematic name of an organic compound comprises three parts :(a) the beginning - one or more prefixes indicating substituent groups;(b) the middle part - the ‘stem’ or ‘root’ derived from the parent hydrocarbon;(c) the ending - one or more suffixes indicating unsaturation and the ‘principal group’.

Organic nomenclature begins with the principal group, for if one is present it determines the identity of the hydrocarbon (the stem) and the numbering of the molecule.

Table 1 Prefixes and Suffixes for some principal groupsGroup Formula ＊ Suffix Prefix Aromatic compoundCations -ammonium ----- -----

Carboxylic acids

- COOH

- (C)OOH

-carboxylic acid

-oic acid

carboxy

benzoic acid

Sulphonic acid - SO3H -sulphonic acid Sulpho -----

Esters - COOR

- (C)OOR

alkyl -carboxylate

alkyl -oate

alkoxycarbonyl

Acyl halides - COX

- (C)OX

-carbonyl halide

-oyl halide

halocarbonyl

Amides - CONH2

- (C)ONH2

-carboxamide

-amide

aminocarbonyl

benzamideNitriles - CN

- (C) N

-carbonitrile -nitrile

cyano

benzonitrileAldehydes - CHO

- (C)HO

-carbaldehyde

-al

methanoyl

benzaldehydeKetones - COR -one

oxo -----

Alcohols - OH -ol hydroxy

phenolAmines - NH2 -amine amino

phenylamine＊Note : Carbon atoms in parentheses are numbered as part of the parent hydrocarbon.

Table 2 Groups which may be named only as prefixes

Unit 11-14

Formula Prefix -F fluoro -Cl chloro -Br bromo -I iodo

-OR alkoxy -NO2 nitro

-N=N- azo R- alkyl

C6H5- phenyl

Table 3 Hydrocarbon chains

Carbon atoms Stem name Carbon atoms Stem name1 meth- 11 undec-2 eth- 12 dodec-3 prop- 13 tridec-4 but- 14 tetradec-5 pent- 15 pentadec-6 hex- 16 hexadec-7 hept- 17 heptadec-8 oct- 18 octadec-9 non- 19 nonadec-10 dec- 20 icos-

EndingsFor substituent groups -yl ExampleFor saturated compounds -aneFor unsaturated compounds :one double bond -enetwo double bonds -adiene buta-1,3-dienethree double bonds -atriene hexa-1,3,5-trieneone triple bond -ynetwo triple bonds -adiyne penta-1,4-diyne

1. The principal group, and the final suffixSome groups can be named as suffixes or prefixes, while others can be named only as prefixes (see

Table 1 and Table 2). If a group is present which can be named as a suffix, this is the principal group and it is named as the final suffix. If more than one such group is present, the one highest in Table 1 is the principal group.

Example : HOCH2CN

If no such group is present, the final suffix becomes -ane for saturated compounds, or -ene or -yne for unsaturated compounds.

Examples : CH3CH2NO2 CH2=CHCl

2. The parent hydrocarbon and the stem

Unit 11-15The parent hydrocarbon may be a chain or a ring system and its name forms the stem. Stem names

for parent chains are given in Table 3. In order of priority the parent hydrocarbon should :(a) contain the principal group if one is present, or the most principal groups if there is more than one;(b) contain the maximum number of multiple bonds if any are present; and(c) be as long as possible.

Examples :

3. Numbering atoms in the parent hydrocarbonLowest possible numbers are allocated in the following strict order of priority to :

(a) the principal group if one is presentExamples : CH2=CH-CH2-OH

(b) double and triple bonds taken together if one or more is presentExamples :

CH3-CH=CH-CCH CH3-CC-CH=CH2

COOH

CN

CHO

Unit 11-16

(c) double bonds before triple bondsExample : CH=CH-CH2-CCH

(d) prefixes taken together

Examples :

(e) prefixes taken in alphabetical orderExamples : Br-CH2-CH2-CH2-Cl

4. Adding the suffixes(a) The name of the principal group, if one is present, will be the final suffix.Example : CH3CHO

(b) If unsaturation is present, the order is double bond(s), followed by triple bond(s), followed by principal group.

Example : CH2=CH-CC-CH2NH2

(c) Each suffix (except -ane) will normally preceded by one or more numbers indicating the position in the parent hydrocarbon.

Examples : CH2=CH-CH2-CH2-OH CH3CH(OH)CH(OH)CH2CH3

5. Adding the prefixes(a) The names of all other substituents are added as prefixes in alphabetical (not numberical) order.Example : CH3CHBrCH2Cl

(b) The prefixes ‘di’, ‘tri’, ‘tetra’ etc. are ignored when considering alphabetical order.Example : CH3CH2C(CH3)2CH2CI3

6. Use of numbers, commas and hyphens(a) A number immediately precedes the name of the group or multiple bond to which it refers, and

is separated from it by a hyphen.Example : CH3CH2CHClCH3

CH3Br

NO2

CH3CHBr

CH2 Cl

Unit 11-17

(b) When there is more than one type of substituent, numbers in the middle of the name are both preceded and followed by a hyphen.

Example : CH3CHBrCH2CHClCH2OH

(c) When there is more than one of a particular substituent, the numbers indicating the position of each are separated by commas.

Example : CH3CCl3

(d) Some or all numbers may be omitted.Examples : The principal group includes a terminal carbon atom in the parent chain.

CH3CH2CH2CH2CHO HOOCCH2CH2CH2CH2COOH

Examples : A single substituent named either as a principal group or a prefix is at position 1 on a benzene or a cyclohexane ring.

Examples : No ambiguity occurs.

CH3CH2COCH3 CH3CH2CH(CH3)CH3

7. When to omit or retain an ‘e’(a) Omit ‘e’ immediately before a suffix if the suffix begins with a vowel (this includes y).Examples : CH3CH2CH2CH2COCH3 CH3CH2CH=CH-CCH

(b) Leave ‘e’ in place if the suffix begins with a consonant.Examples : CH3CH2COCH2COCH3 CH3CH2CH2CN

8. Use of bracketsThey are generally used for clarity.

Example :

Write the systematic names of the following organic compounds :

Structural formula Name1. CH2(OH)CH2CH(OH)CHC=CH2

OH

Unit 11-18

2.

3.

4.

5.

6.

7.

8.

9. CH3CH2NH2

Section 1 1 . 4 Isomerism

（ 1 ） Structural isomerism If the compounds with the same molecular formula have their atoms attached in different orders, they have different structures and are said to be structural isomers of each other.

1. Structural isomers containing the same functional groupChain isomerism The structural isomers differ in the arrangement of the carbon atoms. In general, a branched chain isomer has a lower boiling point than a unbranched chain isomer; the more numerous the branches, the lower the boiling point. This is because with branching the shape of the molecule tends to approach that of a sphere; and as this happens the surface area decreases, with the result that the van der Waals’ forces become weaker and are overcome at a lower temperature.

Example 1 : Structural isomers for C4H10

butane b.p. 0℃ methylpropane b.p. -12℃Example 2 : Structural isomers for C5H12

Unit 11-19

pentane b.p. 36℃ methylbutane b.p. 28℃ dimethylpropane b.p. 9.5℃

Position isomerism Structural isomers have the same carbon skeleton and belong to the same homologous series, but differ in the position of the functional group.

Example 1 : Structural isomers for chloropropane, C3H7Cl

1-chloropropane 2-chloropropane

Example 2 : Structural isomers for propanol, C3H7OH

propan-1-ol propan-2-ol

Example 3 : Structural isomers for disubstituted benzene, C6H4X2

Example 4 : Structural isomers for trisubstituted benzene, e.g. C6H3Br3

2. Structural isomers containing different functional groupsFunctional group isomerism The structural isomers have the same molecular formula but belong to different homologous series, i.e. they differ in the nature of the functional group.

Example 1 : Alcohols and ethers Two structural formulae can be written for the molecular formula C2H6O . Methoxymethane is an ether and exists as a gas that has been used as an aerosol propellant and a refrigerant. Ethanol is an alcohol and exist as a liquid that is used as a solvent and in alcoholic beverages.

methoxymethane b.p. - 25℃ ethanol b.p. 78℃

Example 2 : Aldehydes and ketones, e.g. C3H6O

propanal (an aldehyde) propanone (a ketone)

Example 3 : Carboxylic acids and esters, e.g. C3H6O2

Unit 11-20 propanoic acid methyl ethanoate (an ester) ethyl methanoate (an ester)

Example 4 : Acyclic and cyclic hydrocarbons All acyclic alkanes have the general formula CnH2n+2 , where n = the number of carbon atoms in the

molecule. The presence of a ring or a double bond reduces the number of hydrogen in the formula by two for each double bond or ring; that is, a compound with the general formula CnH2n contains either one double bond or one ring. For example, structural isomers for C4H8 are :

A compound with the general formula CnH2n-2 might have one triple bond, two rings, two double bonds, or one ring plus one double bond. For example, three of many possible structural isomers for C8H14 are :

Write the structural formulae for the following organic compounds :Molecular formula Structural formula

C3H8O AlcoholsEthers

C4H8O AldehydesKetones

C4H8O2 Carboxylic acidsEsters

（ 2 ） Stereo isomerism In stereoisomerism the isomers have the same molecular formula and the same structural formula, but differ in the spatial arrangement of their groups. There are two kinds of stereoisomerism : geometrical isomerism and enantiomerism.

1. Geometrical isomerism Geometrical isomers are compounds with the same molecular and structural formula, but differ in the spatial arrangement of their groups. They are not mirror images and usually arise from the rigidity of C=C bond in organic compounds.

Rigidity of C=C bond leading to cis/trans isomers In a molecule of ethene, the two carbon atoms are linked by a bond and a bond.

The shape of the orbital ensures that : (i)all six atoms in the molecule lie in one plane, (ii) the carbon atoms cannot be rotated relative to one another about the bond axis without breaking

the bond. This would require energy (about 250 kJmol-1) in the form of heat or radiation.

Because of this restriction, an alkene substituted on different sides of the double bond do not interconvert, but may have two possible structures differing in the spatial arrangement of the substituted

Unit 11-21atoms or groups.

Example 1 : Geometrical isomers for 1,2-dichloroethene

C2H2Cl2

molecular formula structural formula cis-1,2-dichloroethene trans-1,2-dichloroethene(geometrical isomers)

In the cis form, the two chlorine atoms lie on the same side of the C=C axis; in the trans form they lie on opposite sides. They differ only in the spatial arrangement of the atoms around C=C bond.

Example 2 : Structural and geometrical isomers for butene, C4H8

Structural isomers Geometrical isomers Name m.p. ℃ b.p. ℃CH2=CH-CH2-CH3 None but-1-ene -185.4 -6.3CH3-CH=CH-CH3 cis-but-2-ene -138.9 3.7

trans-but-2-ene -105.6 0.9

Trans-alkenes are usually higher melting than their cis isomers because of higher symmetry, but their boiling points differ very little.

Condition for geometrical isomerism Geometrical isomerism is possible in any molecule containing a C=C double bond with two different substituents on each carbon.Structural formula Geometrical isomersCab=Cab

cis form trans formCab=Cac

cis form trans formCab=Ccd

Caa=Cbd

Write all the structural formulae of hexene, C6H12. For each structural formula, indicate whether there are cis/trans isomers.

A typical example of geometrical isomerism : cis- and trans-butenedioic acids The two isomers contain the same functional groups, but the different spatial arrangement of the groups

Unit 11-22give rise to difference in chemical properties.

The most important chemical difference is the relative ease of anhydride formation. The cis-isomer readily forms a cyclic anhydride on heating to 170℃. The reaction is reversed on adding water.

The principal differences in the physical properties are summarized in table :Property cis-butenedioic acid trans-butenedioic acidm.p. ℃ 130 286(dec.)density gcm-3 1.590 1.635solubility in water g 100cm-3 79 0.7Ka1 mol dm-3 1.2 x 10-2 9.5 x 10-4

Ka2 mol dm-3 5.9 x 10-7 4.2 x 10-5

The cis-isomer has lower m.p., lower density and higher solubility than that of the trans-isomer.

These properties can be explained by the existence of intramolecular hydrogen bonding in the cis-isomer, which reduces the chance of intermolecular hydrogen bonding, and thus weaken interactions between molecules of the cis-isomer.

The cis-isomer is a stronger acid than the trans-isomer for the first dissociation owing to stabilization of the resulting anion by intramolecular hydrogen bonding. This order is reversed for the second dissociation. The lower acid strength of the cis-isomer for the second ionization is partly a result of the stability of the monoanion, but it also reflects the difficulty of removing the second proton from a carboxyl group which is close to the site of the anionic charge.2. Enantiomerism Enantiomers are compounds with the same molecular and structural formula, but differ in the spatial arrangement of their groups. They are non-superimposable mirror images and usually arise from the presence of chiral carbon atom in organic compounds.

Chiral carbon atom is a saturated carbon atom with four different attached atoms or groups.

Example 1 : Enantiomers for butan-2-olButan-2-ol has two enantiomers, whose molecules are non-superimposable mirror images. This is

because the four groups attached to C-2 (H, OH, CH3, CH3CH2) are all different, making the molecule asymmetric (without plane and axis of symmetry). C-2 is called an asymmetric carbon atom or a chiral carbon atom.

mirror

Structural formula of butan-2-ol A pair of enantiomers

Example 2 : Enantiomers for 2-chlorobutanal

Unit 11-23

mirror

Structural formula of 2-chlorobutanal A pair of enantiomers Two enantiomers are identical in all their normal physical or chemical properties, and differ only in their behaviour toward agencies which are themselves chiral. Historically, the existence of enantiomers was first discovered through their differing behaviour towards plane-polarized light, which has chirality. For this reason, enantiomerism is usually called optical isomerism.

Optical activity Plane-polarized light is produced by passing a beam of ordinary light through certain crystals, e.g. calcite or ‘Polaroid’. As light advances through space, they create an electromagnetic field which may be resolved into a series of sine-waves oscillating in all planes perpendicular to the direction of motion. Polarizing crystals, when correctly oriented, transmit only the components of these waves lying in a particular plane, and the emergent beam is the plane-polarized light. When a plane-polarized beam pass through a chiral medium, its plane of polarization becomes rotated about the axis of the beam. Any material which produces this rotation is said to be optically active. Optically active compounds may rotate plane-polarized light clockwise (+) or anticlockwise (-) as seen by an observer looking towards the source. Enantiomers are optically active and rotate plane-polarized light in opposite direction.

Polarimeter and measurement of optical rotation Optical rotation is measured by a polarimeter. Light from a monochromatic source is polarized by a polarizing lens or prism, then passes through the sample cell to a rotatable analyzing lens or prism coupled to a 360o scale. The analyzer and polarizer are first set at right angles, so that no light is transmitted. The sample is then introduced, in liquid form or in solution, and the analyzer is rotated until the light is once again extinguished. The angle through which it has to be turned corresponds to the angle of rotation, , produced by the sample.

For a given substance, the value of varies directly with the number of molecules through which the light passes, i.e. with the path length of the cell and the concentration of the solution. Optical rotation is usually expressed as specific rotation, [] :

[] t =

Example : The specific rotations of the enantiomers of 1-phenylethanol measured at 27℃ with

Unit 11-24wavelength 589.3 nm are :

Racemic mixture A 1 : 1 mixture of two enantiomers is called a racemic mixture. It is optically inactive, because the opposite optical rotations of the two chiral forms cancel each other out. Racemic mixture differ in crystal structure from either pure enantiomer.

Example : A 1 : 1 mixture of (+)-alanine and (-)-alanine is optically inactive.

The structural formula of glucose (C6H12O6) is given below. How many chiral carbon atoms are there in a molecule of glucose ? Is glucose optically active ?

There are four chiral carbon atoms in a glucose molecule. Glucose is optically active and rotates plane-polarized light clockwise. The trade name of glucose is ‘dextrose’ where ‘dextro’ means clockwise (+) and ‘ose’ means sugar.

Note : The term dextro or d- was formerly used for (+), and laevo or l- for (-).Section 1 1 . 5 Preparation, separation and purification of organic compounds

（ 1 ） Preparation Heating under reflux Since many organic reactions occur slowly at room temperature, it is quite common to carry out the reaction by heating under reflux. The higher temperature speeds up the reaction.

A water condenser is attached vertically to the reaction flask. The flask contains the reaction mixture. This may be dissolved in an organic solvent. The mixture is heated to its boiling point and maintained at this temperature for the duration of the reaction. The vapour from the mixture condenses in the condenser and continuously drips back into the flask.

Anti-bumping granules A few small pieces of porcelain or pumice stones can be added to the reaction mixture to make it boil more evenly. These anti-bumping granules provide surface area for the vapour bubbles to form and thus avoid overheating and bumping of reaction mixture.

Unit 11-25

Water condenser Entry of water at the lower end of the condenser is essential so that the jacket is filled progressively up to the top. If water enters at the upper end of the jacket may drain faster than it is filled, thus becoming unevenly cooled.

Safety precautions when preparing organic compounds1. Reactions should be performed in a fume cupboard.2. Use electric heater instead of naked flames.3. Safety clothing and goggles should be worn.4. It is also important to know the location of fire-fighting equipment and how to use it.

（ 2 ） Separation and Purification On completion of a reaction the reaction flask will often contain not only the desired product but also other products, by-products of side reactions, unreacted reactants and the solvent. Various techniques are used for the separation and purification of the required product.

Simple distillation If the required product is the only volatile substance in the reaction flask, it can be separated from the other substances by distillation. The vapour is heated to its boiling point. Its vapour distils over, condenses and is collected. For liquids which boils at temperature lower than about 140℃ at atmosphere pressure, a water condenser is used. If the boiling point of the liquid is higher than 140℃ an air condenser is used. An air condenser is a straight glass tube without a jacket. Since many organic liquids are flammable, an electric heating mantle is commonly used.

Unit 11-26

Vacuum distillation Some organic liquids thermally decompose at temperatures below their boiling points at atmospheric pressure. In such case it is necessary to distil the liquid at reduced pressure in order to lower the boiling point. To do this a vacuum pump is connected to a side arm in the collection flask. This method is called vacuum distillation.

Fractional distillation If the required product is one of component of a mixture of two or more volatile liquids then fractional distillation can be used to separate the components.

Solvent extraction If the desired product is the only component in a reaction mixture which is soluble in a particular solvent then it can be isolated by solvent extraction. The technique is particularly useful for preparing an organic product from an aqueous solution containing inorganic impurities. The solution is shaken with an organic solvent which is immiscible with water but in which the product is soluble. Ethoxyethane(ether) is often used for this purpose. After extraction, ether can be distilled off in a fume cupboard by using an electric heater and water bath. The ether vapour should be led to the sink.

Drying agent When an organic compound is extracted from an aqueous solution using an organic solvent the solvent often becomes wet. Before it can be distilled off to leave the pure product the solvent must be dried. Anhydrous magnesium sulphate, anhydrous calcium chloride and sodium metal are the common drying agents. Anhydrous CaCl2 can not be used with alcohols, phenols and amines because it reacts with them.

Unit 11-27

CrystallisationThe crude solid is dissolved in the minimum amount of

hot solvent in order to give an almost saturated solution. The solution is then filtered to remove insoluble impurities and then allowed to cool. The solid crystallises out.

The crystals are filtered under reduced pressure using a Buchner funnel and flask. They are either dried or recrystallised to obtained a more pure product.

Mixed melting point determination A pure organic solid melts over a degree or two whereas an impure solid melt over a range of five or more degree Celsius. Most pure solids have characteristic sharp melting points. The presence of even a small amount of a second component can lower the melting point. This fact can be used to determine the identity of unknown organic compounds by the method of mixed melting points. The unknown compound is matched with a known compound with the same melting point. Approximately equal amounts of each are powdered and mixed together thoroughly. The melting point is determined. If it is not sharp and is lower than that of the two separate samples, then the samples are not identical. If the melting point is sharp and is not lowered then the two samples are identical.

Calculating the percentage yield of product In organic chemistry, the actual yield of purified product is often well below the theoretical yield.

% yield =

There are four main reasons for a lowering of the percentage yield :1. the reactants may not be completely pure, e.g. some moisture may be present;2. incomplete reaction : the reactants may not react completely;3. side reactions which give rise to by-products;4. loss of product during purification, e.g. small quantities of solid left on the filter paper or funnel

during recrystallization.

Often, product yields can be improved by carrying out the synthesis using an excess of one(or more) reactant.

Example : When 15 g of butan-1-ol and 10 g of ethanoic acid were refluxed together in the presence of concentrated sulphuric acid, 17.8 g of butyl ethanoate were formed. Calculate the percentage yield. (Mr of butan-1-ol =74, ethanoic acid = 60, butyl ethanoate = 116)

Unit 11-28

A set of ‘quick fit apparatus’ is shown in figure. Give the names for its components.

1. air /steam inlet tube 2. thermometer 3. 4.5. adapter with T connection 6. 7. sintered glass funnel 8. 9. stillhead 10. drying tube 11. 12. pear shaped flask

with angled neck13. 14. screwcap adapter 15. air condenser 16.

Section 1 1 . 6 R eaction mechansim

（ 1 ） Reaction types Part of the attraction of organic chemistry lies in the variety of chemical changes that occur among organic compounds. Equally fascinating is the fact that most of these reactions can be rationalized and unified around a few basic principles. An organic reaction could be generalized in the equation :

substrate + reactive species products (a functional group) conditions (a new functional group)

Most of the organic reactions can be classified as substitution, addition or elimination.

Substitution In a substitution reaction, an atom or group of atoms of an organic compound is replaced by another species.

C A B C B A

Addition In an addition reaction, atoms or groups add to the adjacent carbons of a multiple bond.

Unit 11-29

C C A B C CA B

Elimination An elimination reaction involves the removal of a pair of atoms or groups from adjacent carbon atoms. This necessarily results in the formation of a multiple bond.

C CA B

C C A B

（ 2 ） Breaking covalent bonds and reaction intermediates Since organic compounds are predominantly covalent compounds, their reactions inevitably involve the breaking and formation of covalent bonds.

Homolytic fission In homolytic fission the two shared electrons in the bond are split equally between the two atoms.

Examples :

The resulting species are known as free radicals. They are reaction intermediates and many only exist for a split second. A free radical is defined as an atom or group of atoms which possesses one or more unpaired electrons.

Heterolytic fission In heterolytic fission the two shared electrons in the bond are split unequally between the two atoms. One of the atoms keeps both electrons. As a result it acquires a negative charge. The other atom is deficient in one electron and thus has a positive charge. A species which contains a carbon atom with a negative charge is known as a carbanion. One with a positive charge on a carbon atom is called a carbocation (or carbonium ion).

Examples :

（ 3 ） Types of reactive species which attack organic compounds Free radicals Free radicals are formed whenever a covalent bond is broken homolytically. For example, chlorine free radicals are formed when the element chlorine is exposed to ultraviolet light or heated to temperature above 400℃ :

A free radical is a highly reactive species and will immediately attack a substrate for completing its octet.

Electrophiles An electrophile is defined as a species which is electron seeking, i.e. it seeks out an electron-rich centre in the substrate for its attack. Examples are :

Electrophile

Unit 11-30

An electrophile usually contains an atom possessing a positive charge or a partial positive charge. It must be able to form a strong bond with a carbon atom, otherwise there cannot be a stable reaction product.

Nucleophiles A nucleophile is defined as a species which seeks out a part of substrate that is positively charged for its attack. Examples are :

Nucleophile

A nucleophile usually contains an atom possessing a negative charge or a partial negative charge. It must be able to form a strong bond with a carbon atom.

（ 4 ） Reaction mechanisms A chemical equation describes what happens, whereas a reaction mechanism describes how it happens. Consider, for example, the possible ways, or reaction mechanisms, by which hydrogen bromide can add to ethene :

Do the hydrogen bromide add simultaneously ? Does the hydrogen bond first, followed by the bromine ? Or is it possible that the bromine bonds first, followed by the hydrogen ? Do the hydrogen and bromine add as neutral or charged species ? Are any short-lived intermediate species formed during the steps of the reaction ? A reaction mechanism answers these questions in describing the step-by-step process of the reaction. It is postulated and then supported by experiment.

（ 5 ） Factors affecting reactivity of organic compounds A number of factors influence the breaking and formation of bonds and thus the reactivity of organic compounds.

Inductive effect Inductive effect applies only to single covalent bonds between unlike atoms. Such bonds are polarized due to the different electronegativities of the two atoms.Example :

This shift in electron density from one atom to another and the resultant polarization of the bond is known as the inductive effect. Most atoms and groups are more electronegative than carbon and thus withdraw electrons from carbon. This withdrawal of electrons is called the negative inductive effect (-I effect) or electron-withdrawing effect.Examples :

Unit 11-31

However, some atoms and groups are less electronegative than the carbon atom and thus donate electrons to the carbon atom. This is known as the positive inductive effect (+I effect) or electron-releasing effect. Alkyl groups are known to have a positive inductive effect. The effect increases with the number of alkyl groups substituted on the carbon atom :

Mesomeric effect (Resonance effect) Mesomeric effect occurs in molecules with multiple bonds and operate through orbitals. This effect results in the redistribution of electrons and leads to the stabilization of molecules and some radicals and carbon ions.Example : Hydrolysis of 3-bromopropene using aqueous alcoholic sodium hydroxide

Mechanism :

1.

2.

The intermediate carbocation is stabilized by delocalisation of electrons and thus activation energy for the first step (the r.d.s.) is lowered.

Steric effect The most important example of steric effect is steric hindrance. Steric hindrance can occur when large groups on a molecule ‘get in the way’ and thus hinder the reaction. An example is provided by the 2,6-dimethylbenzoic acid :

Because of the presence of the adjacent methyl groups, the carboxylic group is not free to rotate about its bond between the carbon atom and the benzene ring. It is thus fixed in a position at right angles to the benzene ring. As a result of these steric factors, the 2,6-disubstituted benzoic acids are resistant to normal methods of esterification. Section 1 1 . 7 Organic acids and organic bases

（ 1 ） Re lation of structural characteristics of organic compounds and their acid properties

1. Acidic properties of alcoholsThe chemistry of the alcohols is characterized by the reactions of their functional group, i.e. the

hydroxyl group. Alkoxy-hydrogen fission releases a proton and thus alcohol functions as an acid.

R-O-H + H2O

The degree of polarity of the -OH group is dependent upon the electron-releasing or withdrawing powers of the group to which it is attached. Electron-releasing groups inhibit the withdrawal of electrons away from the hydrogen atom of the hydroxyl group and impair its facility to release a proton. Since all alkyl groups are electron-releasing, it is therefore the acidic strength of alcohols is in the order :

Acidic strength : primary alcohols secondary alcohols tertiary alcohols

Unit 11-32

Name water methanol ethanol 2-methylpropan-2-ol phenolPKa 14 15.5 16 18 10

Alcohols, being much weaker acids than water, do not react aqueous solutions of sodium hydroxide, sodium carbonate or sodium hydrogencarbonate. They are neutral to litmus paper.

Comparison of the acidic properties between alcohols and phenol The greater acidity of phenol is explained by the formation of phenoxide ion which is stabilized by delocalisation of electrons (resonance).

+ H2O

Phenol is soluble in aqueous sodium hydroxide, with which it forms a solution of sodium phenoxide.

However, its inability to liberate carbon dioxide from hydrogencarbonate enables phenol to be distinguished from carboxylic acids.

Substitution of electron-withdrawing groups in the benzene ring of phenol, especially in the 2- and 4- positions, enables the ring in turn to withdraw more electrons from the oxygen atom, thus stabilizing the phenoxide ion further and promoting the ionization process.

phenol 2-chlorophenol 2-nitrophenol 2,4,6-trinitrophenol 4-methylphenolpKa 10.0 8.5 7.2 0.42 10.3

2. Acidity of carboxylic acids Carboxylic acids are weak acids, dissociating only slightly in aqueous solution :

RCOOH(aq) + H2O(l) RCOO-(aq) + H3O+

(aq)

These equilibria lie heavily to left hand side. For example, in 0.10 M ethanoic acid, only about 1.3%

of acid molecules dissociate into ions.

Comparison of the acidity of carboxylic acids with alcoholsAcid ethanoic acid benzoic acid phenol ethanolKa mol dm-3 1.7 x 10-5 6.3 x 10-5 1.0 x 10-10 1.0 x 10-16

Like phenols, carboxylic acids react with metals and alkalis, giving salts. Alcohols do not react with

alkalis. For examples :

OH

Unit 11-33a.

COOH

(s)

+ NaOH(aq)

b.

OH

(aq)

+ NaOH(aq)

However, carboxylic acids are much stronger acids than phenols, and this is shown by their ability to

liberate carbon dioxide from a hydrogencarbonate (a weak base). Indeed, this reaction can be used to distinguish carboxylic acids from phenols. For examples :a.

COOH

(s)

+ NaHCO3(aq)

b.

OH

(aq)

+ NaHCO3(aq)

Carboxylic acids, phenols and alcohols have an acidic nature. They are all oxoacids, that is, they lose a proton via the fission of an -OH bond. This is more likely to happen if the O-H bond breaks easily and the negative charge on the oxoanion is delocalised, thereby increasing the stability of the oxoanion and the acid strength. In carboxylic acids and phenols, the acyl (RCO-) and phenyl (C6H5-) groups respectively have the ability to pull electron cloud towards itself, thus weakens the O-H bond and encourages the loss of a proton. Of more importance is the stability of the oxoanions. Electron delocalisation occurs in each carboxylate and phenoxide ion, thereby spreading out its negative charge. The greater oxoanion stability is achieved in the case of carboxylate ion where there are two oxygen atoms to share the negative charge.

R-OH + H2O H3O+ + R-O-

O H

+ H2O H3O+ +

O

R C

O

O H + H2O H3O+ + R C

O

Oincreasing stability of the oxoanion

Influence of substituents, viz. alkyl and chloro groups, on acidity

Acid Formula Ka at 25℃ mol dm-3 pKamethanoic acid HCOOH 1.6 x 10-4 3.8ethanoic acid CH3COOH 1.7 x 10-5 4.8propanoic acid CH3CH2COOH 1.3 x 10-5 4.92,2-dimethylpropanoic acid (CH3)3CCOOH 1.0 x 10-5 5.0

Unit 11-34chloroethanoic acid ClCH2COOH 1.4 x 10-3 2.85dichloroethanoic acid Cl2CHCOOH 5.0 x 10-2 1.3trichloroethanoic acid Cl3CCOOH 2.3 x 10-1 0.7benzoic acid C6H5COOH 6.3 x 10-5 4.24-nitrobenzoic acid NO2 C6H4COOH 4.0 x 10-4 3.44-methylbenzoic acid CH3 C6H4COOH 4.6 x 10-5 4.3

The strength of a carboxylic acid, Y-COOH, depends on the nature of the Y group. If the Y group has an electron-releasing effect, the O-H bond will be strengthened and the carboxylate ion will be encouraged to accept a proton. Thus, the acid dissociates to a lesser extent and will be weaker. For example, the acid strength decreases as the alkyl chain is lengthened and with number of alkyl groups :

C CO

OH

HH

HC C

O

OH

CH3CH3

HC C

O

OH

CH3CH3

CH3

decreasingacid strength

pKa 4.8 4.9 5.0

Conversely, increasing the electron-withdrawing power of the Y group both assists proton loss and stabilises the resulting carboxylate ion. Hence the corresponding acid increases in acid strength. For example, the strengths of the monohalogenated ethanoic acids increases with the electronegativity of the halogen atom :

H-CH2COOH I-CH2COOH Br-CH2COOH Cl-CH2COOH increasing acid strength

pKa 4.8 3.2 2.9 2.85

Similarly, the electron-withdrawing power of the Y group increase as more chlorine atoms are attached. Arrange the following acids in order of increasing acid strength :a. 1. ClCH2CH2COOH 2. ClCH2COOH 3. ClCH2CH2CH2COOH

b. 1. 2-iodopropanoic acid 2. 2-methylpropanoic acid 3. propanoic acid 4. 2-bromopropanoic acid 5. 2-chloropropanoic acid 6. 3-iodopropanoic acid

c. 1. 2,4-dichlorobenzoic acid 2. benzoic acid 3. 4-methylbenzoic acid 4. 4-chlorobenzoic acid

Test : To distinguish alcohol, phenol and carboxylic acid

Test R-OH C6H5OH RCOOHAdd NaOH(aq) No reaction Soluble, forming phenoxide ion. Soluble, forming carboxylate ion.Add NaHCO3(aq) No reaction No reaction CO2 gas released.

（ 2 ） Re lation of structural characteristics of organic compounds and their base properties

1. Base properties of amines Like ammonia, amines are weak bases. They react with strong acids to form salts.

dil. H2SO4(aq)

Unit 11-35R-NH2 R-NH3

+(aq)

or dil. HCl(aq)

Example : CH3CH2NH2(aq) + HCl(aq) →

2. Comparison of the basic strength of ammonia, primary aliphatic amines and phenylamine When tested with universal indicator, aqueous solutions of primary aliphatic amines and ammonia are found to be weak bases. Base dissociation constants, Kb, for primary amines represent the following equilibrium :

RNH2(aq) + H2O(l) RNH3+

(aq) + OH-(aq)

Remember the higher Kb and the lower pKb, the stronger the base.

Amine Formula Kb at 25℃ mol dm-3 pKb pH rangeammonia NH3 1.8 x 10-5 4.8 10 - 12methylamine CH3NH2 4.4 x 10-4 3.4 10 - 12ethylamine CH3CH2NH2 5.4 x 10-4 3.3 10 - 12phenylmethylamine C6H5CH2NH2 2.2 x 10-5 4.7 10 - 12phenylamine C6H5NH2 4.2 x 10-10 9.4 about 7

The values in table are typical of a general pattern in basicity, namely,

increasing basic strength Primary aliphatic amines and phenylmethylamine are bases of strength stronger than ammonia.

In primary aliphatic amines, alkyl groups are able to push electrons away from themselves, and this has two effects. Firstly, an alkyl group can assist protonation by increasing electron density on nitrogen. Secondly, an alkyl group will help stabilize the resulting alkylammonium ion by dispersing its positive charge. Both effects would aid proton-acceptance, making the base stronger.

NRH

H

NRH

HH

In phenylamine, the lone pair of electrons on nitrogen is held in a p orbital. This can overlap

sideways with benzene’s electron ring. Due to this delocalisation, the lone pair is less available for dative covalent bonding with a proton. Thus, phenylamine is a poor proton-acceptor, making it a very weak base.

Note that phenylmethylamine, though aromatic, is a slightly stronger base than ammonia. This is because its -NH2 group is in an aliphatic side chain and not bonded directly to the benzene ring.