Unit 11 Fundamentals org chem

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					                                                                                                   Unit 11-1
Unit 11              Fundamentals of Organic Chemistry


(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,           plastics, many medicines and drugs,
      vitamins                                          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        1990
Number       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          CC           Si-Si       Si=Si       S-S         C-H
Bond enthalpy 346              610          835           226         318         272         413
kJmol-1                                                               (estimated)

     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 

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
sp -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)
           H                                                               H
      H C H                          CH4                                    C H
                                                                        H     H
                                                                                                 Unit 11-5
Ethane C2H6
     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)
              H   H
      H       C   C   H

           H      H

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)
          H       H   H

  H       C       C   C   H           CH3CH2CH3

          H       H   H

          H H H H
  H C C C C H                        CH3CH2CH2CH3
          H H H H

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

Carbon-carbon double bond : the sp2 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.
                                                                                                     Unit 11-7
Carbon-carbon triple bond : the sp hybridization

Formation of the CC 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 CC triple bond consists of a  bond and two  bonds. These  bonds draw the carbon atoms
closer together, making the CC bond both shorter and stronger than the carbon-carbon double and single

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

                              CH3                     H                     CH3
                   C C C                                  C CH CH C
              H               H                       H                     H

    Give the hybridisation states of all carbon atoms and draw the 3-dimensional structures for both
                                                                                                  Unit 11-8
3.   Aromatic hydrocarbons
     Organic compounds are divided into two main classes : the aliphatic and the aromatic.
     The 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
sp 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 :
                                                                                                     Unit 11-9
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.

                          +    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.

          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
          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.
                                                                                                  Unit 11-10
Section 11.2         Functional groups and homologous series

(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 :

       H      H     H     H
                                                              H     H      H      H
H       C     C     C     C      O      H
                                                       H      C      C     C      C         C     N
        H     H     H     H
                                                               H    H      H      H

     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                 Example
Double bond               Alkenes                   -ene                       ethene
                                                                                      H           H
    C C                                                                                     C C
                                                                                        H         H

Triple bond               Alkynes                   -yne                       Ethyne
   C C                                                                                  HC C H

Benzene                   aromatics                 -benz-                     methylbenzene

                                                    phen-                      phenol
                                                                                                     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            Examples
Carbon-halogen halogeno    -X            halogeno-compo fluoro-               chloromethane CH3-Cl
bonds           -F                       unds           chloro-
                  -Cl                                   bromo-                1-iodobutane
                                                        iodo-                       CH3CH2CH2CH2-I
Carbon-oxygen hydroxyl          -OH      hydroxy           -ol                ethanol
bonds                                    compounds                                  CH3CH2-OH
                                         (alcohols)        hydroxy-           hydroxyethanoic acid

                 ether      C-O-C        ethers            -oxy               ethoxyethane
                 carbonyl      -CO-      carbonyl
                                         compounds :                          propanone
                    C O                  ketones           -one

                 -CHO                    aldehydes         -al                ethanal

                 carboxyl      -COOH     carboxylic acids -oic acid           propanoic acid

                 ester         -COOR     esters            alkyl      -oate   ethyl ethanoate

                 acyl halide    -COX     acyl halides      -oyl halide        ethanoyl chloride

                 amide         -CONH2    amides            -amide             ethanamide

                 acid anhydride       acid anhydrides -oic anhydride ethanoic anhydride

Carbon-nitrogen amino          -NH2      primary amines                       ethylamine
                               -NHR      secondary         -amine             diethylamine
                                         amines            amino-
                               -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+1OH
         Name                Molecular formula Structural formula                  Boiling point ℃
         methanol            CH3OH                CH3OH                            65
         ethanol             C2H5OH               CH3CH2OH                         78
         propan-1-ol         C3H7OH               CH3CH2CH2OH                      97
         butan-1-ol          C4H9OH               CH3CH2CH2CH2OH                   118
         pentan-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 ethoxyethane               Butan-1-ol        butanoic acid
                   CH3OCH3              CH3CH2OCH2CH3 C4H9OH                   C3H7COOH
Boiling point / ℃ - 25                  35                   118               164

     Non-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

     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-ol
Density at 20℃    0.810 gcm-3           0.802 gcm-3              0.789 gcm-3
                                                                                                    Unit 11-13
Section 11.3          Systematic Nomenclature
     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 groups
    Group       Formula *              Suffix            Prefix               Aromatic compound
   Cations                       -ammonium             -----                      -----

  Carboxylic      - COOH             -carboxylic acid                       benzoic acid
    acids                                                 carboxy
                  - (C)OOH           -oic acid

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

    Esters        - COOR             alkyl -carboxylate                             COOR
                  - (C)OOR           alkyl      -oate
 Acyl halides     - COX              -carbonyl halide                                COX
                  - (C)OX            -oyl halide
    Amides        - CONH2            -carboxamide                                   CO NH2
                  - (C)ONH2          -amide
    Nitriles      - CN               -carbonitrile        cyano                     CN
                  - (C) N            -nitrile
  Aldehydes       - CHO              -carbaldehyde                                  CHO
                  - (C)HO            -al
   Ketones        - COR              -one                                           -----

   Alcohols       - OH               -ol                                            OH
    Amines        - NH2              -amine               amino                     NH2

*Note : Carbon atoms in parentheses are numbered as part of the parent hydrocarbon.
                                                                                                   Unit 11-14
Table 2       Groups which may be named only as prefixes

        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 name
          1                 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-

For substituent groups             -yl                          Example
For saturated compounds            -ane
For unsaturated compounds :
one double bond                    -ene
two double bonds                   -adiene                   buta-1,3-diene
three double bonds                 -atriene                 hexa-1,3,5-triene
one triple bond                    -yne
two triple bonds                   -adiyne                  penta-1,4-diyne

1.   The principal group, and the final suffix
     Some 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
                                                                                                 Unit 11-15
2.   The parent hydrocarbon and the stem
     The 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 :         CH2         C     OH

                               CH3                                       OH

                                                                                 CH2 CH3
                                   CH2 CH2 OH

3.  Numbering atoms in the parent hydrocarbon
    Lowest possible numbers are allocated in the following strict order of priority to :
(a) the principal group if one is present
Examples : CH2=CH-CH2-OH

                      COOH                                   CHO

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

             CH3-CH=CH-CCH                          CH3-CC-CH=CH2

(c) double bonds before triple bonds
Example : CH=CH-CH2-CCH

(d) prefixes taken together
Examples :                                                         CH3
                  CH CH CH2 Cl

(e) prefixes taken in alphabetical order
Examples : Br-CH2-CH2-CH2-Cl
                                                                                                  Unit 11-16
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-CC-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

(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
                                                                                                Unit 11-17
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-CCH

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

8.  Use of brackets
    They are generally used for clarity.
Example :
                           CH3 CH2 CH2 CH2 CH CH2 CH2 CH2 CH2 CH3
                                         HC CH3

 Write the systematic names of the following organic compounds :
      Structural formula                                 Name
2.    CH3 CH2 CH2 CH CH3
3.    CH3 C CH2 CH COOH
          O      Br
4.    Br CH2 C O CH3
5.               CH3
      HO CH2 C CH CH C                C COOH
                             CH2 CH2 CH2 CH3
6.        COOH


7.       COO CH3

8.        CH2 OH

9.    CH3CH2NH2
                                                                                                  Unit 11-18
Section 11.4              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 group
Chain 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

            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
                                                                                                 Unit 11-19
2. Structural isomers containing different functional groups
Functional 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

             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              Alcohols
C4H8O              Aldehydes
C4H8O2             Carboxylic acids
                                                                                                     Unit 11-20

     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
atoms or groups.

Example 1 : Geometrical isomers for 1,2-dichloroethene

     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 -6.3
CH3-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.
                                                                                                     Unit 11-21
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 isomers
                              cis form                          trans form
                              cis form                          trans form


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 give 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 acid
m.p.                 ℃              130                           286(dec.)
density              gcm            1.590                         1.635
solubility in water g 100cm         79                            0.7
                             -3              -2
Ka1                  mol dm         1.2 x 10                      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.
                                                                                                  Unit 11-22
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-ol
     Butan-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.


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

Example 2 : Enantiomers for 2-chlorobutanal


     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.
                                                                                                    Unit 11-23
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
            wavelength 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 ?
                                       OH OH OH OH OH         O
                                       CH2CH CH CH CH C
   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 (-).
                                                                                                     Unit 11-24
Section 11.5              Preparation, separation and purification of organic compounds


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.

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 compounds
1. 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.
                                                                                                    Unit 11-25
(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.

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.
                                                                    Unit 11-26
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.

     The 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.
                                                                                                       Unit 11-27
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)

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)

 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 neck
13.                             14. screwcap adapter      15. air condenser        16.
                                                                                                   Unit 11-28
Section 11.6             Reaction 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.

   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

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

                 C C              A B                 C C
                                                      A B
   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 C                       C C              A B
                 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 :
                                                                                                     Unit 11-29
(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

   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 :


   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

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


   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.
                                                                                                    Unit 11-30
(5)Factors affecting reactivity of organic compounds
   A number of factors influence the breaking and formation of bonds and thus the reactivity of organic

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 :

    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 :



    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.
                                                                                                       Unit 11-31
Section 11.7                Organic acids and organic bases

(1)Relation of structural characteristics of organic compounds and their acid
1.  Acidic properties of alcohols
    The 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

Name                water          methanol       ethanol     2-methylpropan-2-ol            phenol
PKa                  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).
                          + H2 O

      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-methylphenol
pKa                  10.0            8.5            7.2              0.42             10.3
                                                                                                              Unit 11-32
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 alcohols
Acid              ethanoic acid        benzoic acid        phenol                              ethanol
              -3           -5                   -5
Ka    mol dm      1.7 x 10             6.3 x 10            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 :
                                  + NaOH(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 :
                                   + NaHCO3(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 +              H 2O             H3O+       +    R-O-
             O H                                                          O

                        +       H2 O          H3O+       +
            O                                                         O
     R C                                                      R C
           O H      +         H2 O            H3O+       +           O
                                                                       increasing stability of the oxoanion
                                                                                                       Unit 11-33
Influence of substituents, viz. alkyl and chloro groups, on acidity

Acid                           Formula                  Ka at 25℃      mol dm-3     pKa
methanoic acid                 HCOOH                    1.6 x 10-4                  3.8
ethanoic acid                  CH3COOH                  1.7 x 10-5                  4.8
propanoic acid                 CH3CH2COOH               1.3 x 10-5                  4.9
2,2-dimethylpropanoic acid     (CH3)3CCOOH              1.0 x 10-5                  5.0
chloroethanoic acid            ClCH2COOH                1.4 x 10-3                  2.85
dichloroethanoic acid          Cl2CHCOOH                5.0 x 10-2                  1.3
trichloroethanoic acid         Cl3CCOOH                 2.3 x 10-1                  0.7
benzoic acid                   C6H5COOH                 6.3 x 10-5                  4.2
4-nitrobenzoic acid            NO2 C6H4COOH             4.0 x 10-4                  3.4
4-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 :
                         H      O                CH3 O               CH3 O
                     H C C                 CH3 C C              CH3 C C             decreasing
                                                                                    acid strength
                         H      OH              H       OH           CH3 OH

          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
 Arrange the following acids in order of increasing acid strength :

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                             RCOOH
Add NaOH(aq) No reaction             Soluble, forming phenoxide ion.    Soluble, forming carboxylate ion.
Add NaHCO3(aq) No reaction           No reaction                        CO2 gas released.
                                                                                                  Unit 11-34
(2)Relation of structural characteristics of organic compounds and their base
1.   Base properties of amines
     Like ammonia, amines are weak bases. They react with strong acids to form salts.
                     dil. H2SO4(aq)
          R-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 range
ammonia              NH3                   1.8 x 10-5                 4.8         10 - 12
methylamine          CH3NH2                4.4 x 10-4                 3.4         10 - 12
ethylamine           CH3CH2NH2             5.4 x 10-4                 3.3         10 - 12
phenylmethylamine    C6H5CH2NH2            2.2 x 10-5                 4.7         10 - 12
phenylamine          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.

                        R     N H                        R      N H
                              H                                 H

     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.

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