Hydrocarbons are compounds of carbon and hydrogen only. Hydrocarbons are obtained mainly from petroleum, natural gas  and coal. Examples of hydrocarbons are methane, ethane, acetylene and benzene. The important fuels like petrol, kerosene , coal gas, oil gas , CNG (compressed natural gas) , LPG (liquefied petroleum gas) etc. are all hydrocarbons or their mixtures.
The hydrocarbons are divided into two main categories , aliphatic and aromatic. The aliphatic  hydrocarbons are further classified into saturated (alkanes) , unsaturated (alkenes and alkynes) and alicyclic (cycloalkanes) hydrocarbons.
They are also known as paraffins derived from the Latin words, ‘parum’ (little) and ‘affins’ affinity, that is with little affinity. The name is justified as under normal conditions, alkanes are inert towards reagents like acids, bases, oxidising and reducing agents. However, under drastic conditions, i.e.,  at high temperature and pressure alkanes undergo different types of reactions like halogenation, nitration, sulphonation , pyrolysis etc.
Alkanes have tetrahedral structure around carbon atom with average C- C and C- H bond lengths of 154 pm and 112 pm respectively.
Alkanes form homologous series with the general formula CnH2n+2 , where n is the number of carbon atoms in the molecule. Methane CH4 with n = 1 and ethane C2H6 with n = 2 are the first two members of the series.
Cycloalkanes are cyclic hydrocarbons which form a homologous series with the general formula CnH2n, whose first member is cyclopropane ( with n = 3).
            The nomenclature  has discussed in Unit 14.
1.      Write the IUPAC names for the following structures.
2.      Write the structures for the compounds having IUPAC names,
(i)        3-Ethyl-2-methylpentane.
(ii)       3,4,8-Trimethyldecane.
3.      Write structures of different chain isomers of alkanes corresponding to molecular formula C6H14. Also write their IUPAC names.
4.      Write structures of different isomeric alkyl groups corresponding to the molecular formula C5H11. Write IUPAC names of alcohols obtained by attachment of –OH groups at different carbons of the chain.
5.      Write the IUPAC names of the following compounds.
(i)    (CH3)3CCH2C(CH3) 3
      (ii)    (CH3)2C(C2H5) 2
      (iii)    tetra-tert-butylmethane

6.    Write the structural formula of the following compounds :
i)   3, 4, 4, 5-tetramethylheptane
ii)   2,5-dimethylhexane
7.      Write the structures for each of the following compounds. Why are the given names incorrect ? Write the correct IUPAC names.
i)   2-ethylpentane
      ii)   5-ethyl-3-methylheptane
8.    Write IUPAC names of the following compounds.
9.      For the following compounds , write structural formula and IUPAC names for all possible isomers having the number double or triple bond as indicated :
(a)   C4H8 (one double bond)   (b)  C5H8 (one triple bond)
Alkanes can be prepared on a small scale in the laboratory by the following methods.
1. Hydrogenation of alkenes and alkynes
Hydrocarbons containing double or triple bonds can be hydrogenated to the corresponding alkanes. The addition of hydrogen to unsaturated compounds is called hydrogenation.
            Alkenes and alkynes are hydrogenated to alkanes by Sabatier and Senderen’s reaction. It is the addition of hydrogen to alkenes and alkynes in the presence of nickel catalyst at 570 K.

 For example, ethylene (C2H4) when mixed with hydrogen and passed over nickel catalyst at 570 K gives ethane(C2H6) and acetylene (C2H2) on hydrogenation yields ethane.

2.       From Alkyl halides
Alkyl halides are halogen derivatives of alkanes with general formula R-X. They can be used to prepare alkanes by following methods.
(a) It involves the chemical reaction between alkyl halide(usually bromides and iodides) and metallic sodium in presence of dry ether (wurtz reaction). The product is the symmetrical alkane containing twice the number of carbon atoms present in alkyl halide.

In  case, two different alkyl halides are taken in order to prepare alkane with odd number of carbon atoms, a mixture of three alkanes will be produced as follows:

Therefore, this method is not suited to prepare alkanes containing odd number of carbon atoms. Methane, however, cannot be prepared by this method.
(b) By the reduction of alkyl halides
It involves the reduction of alkyl halides (usually bromides or iodides) by suitable reducing agents such as H2/ Pd or nascent hydrogen obtained from Zn  and HCl or Zn-Cu couple and ethanol.
Reduction of iodo-derivative can be carried out by the use of hydro-iodic acid (HI) in presence of red phosphorus.

Red phosphorus removes iodine formed and pushes the reaction in the forward direction.

If not removed, iodine will convert ethane back into ethyl iodide.
(c) By the use of Grignard reagent
 Alkyl halides especially bromides and iodides react with magnesium metal in diethyl ether to form alkyl magnesium halides (RMgX). Alkyl magnesium halides are known as Grignard reagents. In the Grignard reagent, the carbon-magnesium bond is highly  polar with carbon atom being relatively more electronegative than magnesium. It reacts with water or with other compounds having active hydrogen ( the H atom attached on N, O, F or with triple bonded carbon atoms are known as active hydrogen) to give alkane.

In these reactions the alkyl group of alkyl magnesium halide gets converted into alkane. In general the reaction can be represented  as :

3. From Fatty acids (monocarboxylic acids)
(i)    By heating with soda lime
 Sodium salts of carboxylic acids on heating with soda lime give alkanes. Soda lime is prepared by soaking quick lime (CaO) with a solution of sodium hydroxide (NaOH). For example, sodium acetate which is sodium salt of acetic acid(CH3COOH) on heating with soda lime gives methane. Sodium propanoate gives ethane.

In this reaction CaO does not participitate , but helps in the fusion of the reaction mixture.
(ii) Electrolytic method ( Kolbe’s method ) : An alkane is obtained when an aqueous solution of sodium or potassium salt of carboxylic acid is electrolysed.  The reaction is known as decarboxylation reaction (Kolbe’s reaction)
2 RCOONa + 2 H2O ® R-R + 2 CO2 + 2 NaOH + H2
For example, ethane is obtained when a solution of sodium acetate is electrolysed.
2 CH3COONa + 2 H2O ® CH3CH3 + 2 CO2 + 2 NaOH + H2
At anode the carboxylate ion (RCOO- ) gives up one electron to produce free radical RCOO· which decomposes to give the alkyl radical and carbondioxide. The two such radicals combine together to yield higher alkane.

Normally these methods are useful for preparing alkanes containing even number of carbon atoms.
10. How do you account for the formation of ethane during chlorination of methane ?
The following are the physical properties of alkanes.
State : The lower members in the alkane series are gases (methane, ethane, propane and butane). Alkanes containing 5 to 17 carbon atoms are liquids, while higher members are waxy solids. Thus alkanes change from gaseous state to the solid state with an increase in molecular mass.
Non-polar nature : Alkanes are non-polar in nature. Therefore alkanes are soluble in non-polar solvents such as benzene, ether, chloroform, carbon tetrachloride etc. Liquid alkanes themselves are good solvents for other non-polar substances.
Boiling points : Lower alkanes have lower boiling points. The boiling point gradually increases with increase in the molecular mass. The increase is by 20° to 30° for each –CH2- unit added to the chain. The reason for the increase in boiling point is that the alkane molecule is non-polar in nature. These are attracted to each other by weak van der Waal’s forces. In the case of alkanes with higher molecular masses, these forces are more because of the larger surface area. The variation in boiling points of n-alkanes with increase in number of carbon atoms per molecule is shown in Fig.

Variation in boiling points of n-alkanes with increase in number of carbon atoms per molecule.
Moreover, straight chain alkanes have higher boiling point than the corresponding branched chain hydrocarbons (isomers). Greater the branching in the chain is, lower is the boiling point. This is due to the fact that branching of the chain makes the molecule compact and brings it closer to a sphere. This decreases the surface area and hence the magnitude of inter-particle van der Waal’s forces leads to the decrease in boiling point. The boiling points of the isomeric pentanes are as follows.

Melting points : The intermolecular forces in a crystal depend not only on the size of the molecues but also on how they are packed into a crystal. The rise in melting point with increase in number of carbon atoms is, therefore , not as regular as the boiling point is in the case of liquids. On plotting the melting points of        n-alkanes against the number of carbon atoms in a chain , a sawtooth pattern shown in Fig is obtained.

Variation in melting points of n-alkanes with increase in number of carbon atoms
The increase in melting point is much more in moving from an alkane having odd number of carbon atoms to higher alkane than in moving from alkane having even number of carbon atoms to the higher alkanes as given below :
m.p (K)
This implies that the molecules with an odd number of carbons do not fit well into the crystal lattice.  The carbon chains in the alkanes are zig-zag rather than straight. Now in n-alkanes, terminal methyl groups lie on the same side when the number of carbon atoms is odd and on opposite sides when the number of is even (Fig).

Representation of alkanes with even and odd number of carbon atoms
CH3 groups of odd and even carbon compounds appreciably affect the forces of interaction and hence the melting temperature. The energy required to break the crystal structure and thus melt the alkane is lesser in case of alkanes having an odd number of carbon atoms. This is because their molecules do not fit well into the crystal lattice.
Colour : Alkanes are colourless gases, liquids or solids.
Density : The density of alkanes increases with increase in chain length.
Solubility : As ‘like dissolves like’ , therefore alkanes being      non-polar in character are more soluble in non-polar solvents such as ether, carbon tetrachloride, etc. These are insoluble in water and other polar solvents.


Alkanes are quite unreactive. They do not react with usual reagents. Alkanes are inert (less reactive). Inertness of alkanes is due to the presence of strong carbon-carbon and carbon-hydrogen bonds, which are difficult to break. However, hydrogen can be substituted by other atoms or radicals under drastic reaction conditions such as high temperature, presence of ultra violet light or catalyst etc. Alkanes undergo few reactions, the most important being halogenation, oxidation and thermal cracking.
1.      Halogenation
Alkanes react with chlorine or bromine in presence of sunlight or ultraviolet light to give halogen substituted products, i.e., alkyl halides containing one or more halogen atoms. Halogenation of alkanes is a substitution  reaction. Hydrogen atoms of alkanes are replaced by halogen atoms.
Alkanes also undergoes halogenation if a mixture of alkane and halogen is heated to 600 K. For example, when a mixture of methane and chlorine is exposed to diffused sunlight or ultra violet light or heated to 600 K, the hydrogen atoms are replaced by chlorine atoms, one after another.

The nature of the products formed depends upon the amount of chlorine. If excess of chlorine is used, then the product contains larger amounts of carbon tetrachloride. Smaller amounts of chlorine give more of less chlorinated products (such as methyl chloride) in this reaction.
            In case the reaction is allowed to take place for smaller duration of time, then less chlorinated products are formed in larger amounts.
            The order of reactivity of halogens with alkanes is
F2  >   Cl2  >   Br2  >  I2
Fluorine reacts with a violent explosion. The reaction with chlorine is less vigorous than fluorine and with bromine less vigourous than chlorine. The reaction with iodine is slow and reversible.

Therefore iodination is carried in presence of some oxidising agents like iodic acid (HIO3) or nitric acid which converts HI to I2 and this pushes the reaction in the forward direction.
5 HI      +  H IO3   ®   3 H2O  +  3 I2
2 HNO3 +  2 HI   ®   2 H2O  +  2 NOI2
Fluorination takes place with almost explosive violence to produce fluorinated compounds. It also involves rapture of C-C  bonds in the case of higher alkane. The reaction can be  made less violent by  dilution of fluorine with nitrogen.
The ease of substitution of a hydrogen atom by a halogen atom is :

Tertiary  >   secondary   >   primary

Mechanism of Halogenation
            Halogenation of alkanes is a free radical chain substitution. The generally accepted mechanism  for chlorination of methane is given below:
Step 1
 In the first step , Cl2 molecule breaks homolytically to give two Cl· free radicals. The step in which Cl-Cl bond homolysis occurs is called the initiation step.

Step 2
Each chlorine atom formed in the initiation step has seven valence electrons and is very reactive. Once formed, a chlorine atom abstracts a hydrogen atom from the methane as shown below :

Hydrogen chloride , one of the isolated products from the overall reacton, is formed in this step. A methyl radical is also formed.

Step 3
 Methyl radical formed in step 2, attacks a molecule of Cl2. Attack of methyl radical on Cl2 gives chloromethane, the other product of oveall reaction, along with a chlorine atom then cycles back to step 2, repeating the process.

Steps 2 and 3 are called propagation steps of the reaction and , when added together give the overall reaction. Since one initiation step can result in a large number of propagation steps , the overall process is called a free-radical chain reaction. .
Step 4 
In actual pracice, some side reactions are taking place in addition to propagation steps which reduce the efficiency of the propagation steps. The chain sequence is interrupted whenever two odd-electron species (free radicals) combine to form an even-electron species. Reactions of this type are called chain terminating steps. Some commonly observed chain terminating steps in the chlorination of methane are given below :

It may be noted that termination steps are , in general , less likely to occur than the propagation steps. Each of the termination steps requires two very reactive free radicals to encounter each other in a medium that contains far greater quantities of other materials such as methane  and chlorine molecules with with they can react. Thus some monochloromethane undoubtedly gets formed via direct combination of methyl radicals with chlorine atoms, most of it in steps 2 and 3.
 Nitration is a substitution in which a hydrogen atom of an alkane is replaced by nitro (-NO2 ) group. When alkanes having six or more than six carbon atoms boiled with fuming nitric acid, then a hydrogen atom is replaced by the nitro group. Lower alkanes cannot be nitrated by this method. However, when a mixture of such an alkane and nitric acid vapours is heated to a temperature of about 723-773 K in a sealed tube, one hydrogen atom of the alkane is substituted by nitro group. The process is called vapour phase nitration and compounds obtained are called nitroalkanes.

Since the reaction is carried out at higher temperatures, the rapture of carbon-carbon bonds occurs during the process. As a result, the reaction yields a mixture of different products. For example, nitration of propane results in the formation of mixture of four nitro compounds as shown below :

Nitration of ethane yields a mixture of nitroethane and nitromethane.

When alkanes having six or more carbon  atoms are heated with fuming sulphuric acid(H2S2O7), then hydrogen is replaced by the sulphonic acidgroup(-SO3H) group. The compound (RSO3H) thus formed is known as the alkane sulphonic acid. Lower alkanes  except those having tertiary hydrogen cannot be sulphonated.

Oxidation of alkanes gives different products under different conditions.
Alkanes readily burn with non-luminous flame in excess air or oxygen to form carbon dioxide and water with evolution of large quantity of heat. It forms the basis of the use of alkanes as fuels.
Burning in excess of oxygen :  Alkanes burn with blue flame in air or oxygen. In this process alkanes get oxidised to carbondioxide and water.
      CH4     + 2 O2    ®  CO2 + 2H2O      :  DH° = - 890.4 kJ/mol
     2C2H6  + 7 O2       ® 4 CO2 + 6 H2O  : DH° = - 1580.0 kJ/mol
The cooking gas ,which is often called L.P.G (liquefied petroleum gas) is a mixture of propane and butane.
(ii) Burning in limited supply of oxygen : Alkanes burn with a suity flame (smoky flame) . Incomplete oxidation of alkanes give carbon black (a variety of carbon used in the manufacture of printing inks and tyres).
CH4    +     O2          ®        C      +      2 H2O
                                                      lamp black
 Catalytic oxidation :
             On controlled oxidation, alkanes give alcohols which are further oxidised to aldehydes(or ketones), acids and finally carbon dioxide and water.
CH4    ®      CH3OH     ®  HCHO            ®  HCOOH   ® CO2 + H2O methane   methanol     formaldehyde    formic acid
            Alkanes undergo oxidation under special conditions to yield a variety of products. Example,

Alkanes having tertiary H can be oxidised to alcohols in presence of potassium permanganate.

When higher members of the alkane family are heated to high temperatures (700 – 800 K)  or to slightly lower temperature in presence of catalysts like alumina or silica, they break down to give alkanes and alkenes with lesser number of carbon atoms. E.g.,

The fragmentation of alkanes is also called pyrolysis or cracking . The chemical reactions taking place in cracking are mostly free radical reactions involving rupture of carbon-carbon and        carbon-hydrogen bonds.
Alkanes isomerise to branched chain alkanes  when heated with anhydrous aluminium chloride  and hydrogen chloride.
            The alkanes containing 6 or more carbon atoms when heated at about 773 K under high pressure (10-20 atm) , in presence of catalysts like oxides of chromium, vanadium or molybdenum supported over alumina  get converted into aromatic hydrocarbons. This process is called aromatisation.

Under similar conditions , n-heptane yields toluene.

            Methane reacts with steam at 1273 K in presence of nickel catalyst forming carbon monoxide and hydrogen. The method is used for industrial preparation of  hydrogen.

A  single covalent bond is formed between two atoms by axial overlap of half-filled atomic orbitals. In alkanes, the  C- C bond is formed by axial overlapping of adjacent carbon atoms. The electron distribution in the molecular orbital of sp3 - spsigma bond is cylindrical around the inter-nuclear axis.
            The single covalent bond therefore allows the freedom of rotation about it because of its axial symmetry. As a result of rotation about C- C bond, the molecule can have different spatial arrangement of atoms attached to the carbon atoms. Such different spatial arrangements of atoms which arise due to rotation around a single bond are called conformers or rotational isomers (rotomers) . The molecular geometry corresponding to a conformer is known as conformation.. The rotation around a sigma bond is not completely free ; it in fact is hindered by an energy barrier of           1 to 20 kJ mol-1. There is a possibility of weak repulsive interactions between bonds or electron pairs of the bonds on adjacent carbon atoms. Such type of repulsive interactions is referred to as   torsional strain.
In ethane (CH3CH3), the two carbon atoms are bonded by a single covalent bond and each of the carbon atom is further linked to three hydrogen atoms. If one of the carbon atom is held still and the other carbon atom is allowed to rotate around the C- C bond, a large number of different spatial arrangements of hydrogen atoms of one carbon atom with respect to the other carbon atom can be obtained.
            However, the basic structure of the molecule including   C- C bond length and H - C - H or H - C - C bond angles will not change due to rotation.
            Out of the infinite number of possible conformations of ethane, two conformations represent the extremes. These are called staggered conformation (a) and eclipsed conformation (b).
            In staggered conformation, the hydrogen atoms of the two carbon atoms are oriented in such a way so that they lie far apart from one another. In other words, they are staggered away with respect to one another.
            In eclipsed conformation, the hydrogen atoms of one carbon atom are lying directly behind the hydrogen atoms of the other. In other words, hydrogen atoms of one carbon atom are eclipsing hydrogen atoms of other.
Conformations can be represented by Sawhorse and Newmann projections.
1.  Swhorse projection
            In  this projection the molecule is viewed along the axis of the model from above and right. The central C- C is drawn as a straight line slightly tilted to the right for the sake of clarity. The line is drawn somewhat longer. The front carbon is shown  as the lower left hand carbon whereas the rear carbon is shown as the upper right hand carbon. Each carbon has three lines corresponding to three atoms/groups ( H in the case of ethane).

            The saw horse projection formulae of the two extreme conformations of ethane are shown in the Figure.
Sawhorse projections of ethane
2.  Newman projection
Newman proposed simpler formulae for representing the conformations. These are called Newman projection formulae. In Newman projection, the two carbon atoms forming the s - bond are represented by two circles, one behind the other, so that only the front carbon is seen. The hydrogen atoms attached to the front carbon atom are represented by C- H bonds from the centre of the circle. The C- H bonds of the back carbon are drawn from the circumference of the circle.
            Newman’s projection formulae for staggered and eclipsed conformations of ethane are shown in Fig.

Newman projections of ethane

It may be noted that one conformation of ethane can be converted into the other by rotation of 600 about the bond. The infinite other conformations of ethane lying between the two extremes are called skew conformations.

Relative stabilities of the conformers of ethane

            The coformers of ethane do not have the same stability. In eclipsed conformation of ethane the hydrogen atoms eclipse each other resulting in crowding, whereas in staggered conformation hydrogen atoms are as far apart as possible. The infinite number of possible intermediate conformations between eclipsed and staggered are called as skew conformations. The staggered conformation of ethane is more stable than eclipsed conformation. It is because in staggered form the H-atoms are far apart, therefore, the magnitude of repulsion is lesser than in eclipsed conformation. Hence the order of stability follows the sequence :
Staggered > gauche (skew) > eclipsed

The repulsive interaction between electron clouds , which affects stability of a conformation , is called torsional strain.  Magnitude of torsional strain depends upon the angle of rotation about C – C bond. This angle is also called dihedral angle or torsional angle. Of all the conformations of ethane, staggered form has the least torsional strain and eclipsed form, the maximum torsional strain. Thus it may be inferred that rotation around C – C bond in ethane is not completely free. The difference in energy content of the staggered and eclipsed conformation is 12.5 kJ mol-1. The variation in energy versus rotation about the bonds has been shown in Fig.

            Changes in energy during rotation about C- C bond in ethane.
The difference in the energy of various conformations constitutes an energy barrier to rotation. But this energy barrier is not large enough to prevent rotation. Even at ordinary temperature, the molecule possesses sufficient thermal and kinetic energy to overcome the energy barrier through molecular collisions.          Thus conformations keep on changing from one form to another very rapidly and cannot be isolated as separate conformers.
The Newman conformations of propane and butane are shown below.

Newman projections of propane

Newman projections of butane (changes with dihedral angle)
11. How many eclipsed conformations are possible in butane ?
          The unsaturated hydrocarbons containg C=C bond and having general formula CnH2n are called alkenes . The simplest member of the alkene family is ethene, C2H4 , which contains 5 s and 1 p bond.  The carbon-carbon double bond is made up of a s bond and a p bond. The bond enthalpy of a C=C is 681 kJ mol-1 while carbon-carbon bond enthalpy of ethane is (348 kJ mol-1 ). As a result of this C- C bond length in ethene (134 pm) is shorter than the C- C bond length in ethane(154 pm).
            The presence of the pi(p) bond makes alkenes behave sources of loosely held mobile electrons. Therefore alkenes are attacked by reagents or compounds which are in search of electrons. Such reagents are called electrophilic reagents. The presence of weaker weaker p -bond makes alkenes unstable molecules in comparison to alkenes and thus alkenes can be changed into single bond compounds by combining with the electrophilic reagents . Strength of the double bond(681 kJ mol-1) is greater than that of a carbon-carbon single bond in ethane     (348 kJ mol-1). Orbital diagrams of ethane molecule are shown the following Figures.

Orbital picture of ethane depicting s-bonds only

Orbital picture of ethane showing formation of (a) p-bond (b) p-cloud and (c) bond angles and bond lengths
            Alkenes  generally show the types of isomerism given below:
·         Position isomerism
·         Chain isomerism
·         Geometrical isomerism
1. Position isomerism
            First two members of alkenes (ethene and propene) do not show this type of isomerism. However, butene exhibits position isomerism as but-1-ene and but-2-ene differ in the position of the double bond.
CH3CH2CH=CH2                       CH3CH=CHCH3
      But-1-ene                             But-2-ene
2.  Chain isomerism
            Butene also shows chain isomerism as isobutene has branched chain.

3.   Geometrical isomerism
Alkenes exhibit  geometrical isomerism due to restricted rotation around the C=C bond. This results in two possible arrangements  of groups attached to the two doubly bonded carbon atoms, as shown below for but-2-ene.

            This isomerism is called cis-trans isomerism. When the groups of similar nature are on the same side of the double  bond , the isomer is designated as cis and when they are on opposite sides , as trans. The necessary and sufficient condition  for geometrical isomerism is that the two groups attached to the same carbon must be different , i.e., olefins of the type abC=Cab or abC=Cax or abC=Cbx show geometical isomerism.  When two groups of of highest priority are on the same side of the double bond, the isomer is designated as Z (Zusammen in German means together) and when these groups are on opposite sides, the isomer is designated as E (Entegegen in German meaning opposite).
The priority of groups decided by sequence rules given by Cahn-Ingold and Prelog. According to these rules the atom having higher atomic number gets higher priority. Thus, between carbon(atomic number 6 ) and oxygen (atomic number 8 ), oxygen gets priority over carbon. If the relative priority of the two groups cannot be decided i.e., their bonded atoms are same , then the next atoms in the groups (and so on) are compared. Thus between the groups methyl (-CH3) and ethyl (-CH2CH5), the latter will get the priority since the next atoms in methyl is H while in ethyl it is C(carbon  has higher atomic number than hydrogen) , as shown below.
There are two  systems for naming alkenes.
1. The common system
            The common names of first few alkenes are derived from corresponding alkanes by replacement of ‘ane’ by ‘ylene’, e.g,. H2C=CH2 is ethylene.
2.  IUPAC system
            In this most commonly used system the ending ‘ane’ of corresponding alkanes is replaced by ‘ene’. The parent chain , is considered to be one corresponding to the longest chain of carbon atoms containing the double bond.  The next longest side chain may be selected provided it contains the maximum number of double bonds present. It is then numbered, starting from the end nearest to the double bond. The position of the double bond is indicated by the number of carbon atom preceding the double bond. If there are two or more double bonds the ending ‘ane’ is replaced by ‘adiene’ or ‘atriene’ etc. The remaining rules are the same as applicable to alkanes. Some examples are :

The locant for double bond may be written preceding or following the name of parent alkane or following the root name followed by the suffix , ene , diene or  triene. All the the three names, i.e.,  1,3-butadiene ,        butadiene-1,3 or buta-1,3-diene are correct for butadiene.
12.  Which of the following compounds will show cis-trans     isomerism ?
(i)     (CH3)2C=CHCH3        (ii)    CH2=CCl2
         (iii)    C6H5CH=CHCH3       (iv)    CH3CH=CBr(CH3)
13.    Classify the following as Z or E isomers.
14.    Write IUPAC names of following compounds.
         i)    (CH3)2CHCH=CHCH2CH=CHCH(C2H5)CH3
       ii)   CH2=C(CH2CH2CH3)2
15.  Calculate the number of  s and p  bonds in the above
        structures ( i – iv) .
16.  Draw cis and trans isomers of the following compounds. Also write their IUPAC names.
i)    CHCl=CHCl        ii)    C2H5C(CH3)=C(CH3)C2H5
17. Which of the following compounds will show cis-trans    isomerism ?
i)     (CH3)2C=CHC2H5
ii)     CH2=CBr2
iii)   C6H5CH=CHCH3
iv)   CH3CH=CClCH3
Some of the general methods of preparation of alkenes are :
1.      From alkyl halides
 Alkenes can be prepared from alkyl halides (preferably bromides or iodides) by treatment with alcoholic solution of caustic potash (KOH) at about 353-363 K. The reaction is known as dehydrohalogenation of alkyl halides.
          CH3CH2Br  +  KOH(alco) ® H2C=CH2+ KBr + H2O
Similarly , treatment of n-propyl bromide with ethanolic solution of potassium hydroxide produce propene.

This is an elimination reaction. The ease of dehydrohalogenation for different halides is:
  iodide > bromide > Chloride
 while for carbons, it is ,
   tertiary > secondary > primary.
i.e., a tertiary alkyl iodide is most reactive.
2.  From alcohols
 Alkenes can be prepared by dehydration (removal of water molecule) of alcohols. The two common methods for carrying out dehydration are to heat the alcohol with either alumina or a mineral acid such as phosphoric acid or Con. Sulphuric acid. In dehydration reaction the OH group is lost from a-carbon while H atom is lost by b-carbon atom creating a double bond between a and b-carbons. Some examples are :

            This is an elimination reaction , i.e., a molecule of water is eliminated.
3        From dihalogen derivatives
 Dihalogen derivatives are the derivatives of alkanes containing two halogen atoms. Alkenes can be prepared from vicinal dihalogen derivatives (having halogen atoms on adjacent carbon atoms) by the action of zinc. The process is called dehalogenation of vicinal dihalides.

4. From alkynes
            Alkynes on partial reduction with calculated amount of dihydrogen in presence of palladised charcoal partially deactivated with poisons like sulphur compounds or quinoline give alkenes. Partially deactivated charcoal is known as Lindlar’s catalyst. Alkenes thus obtained are having cis geometry. However, alkynes on reduction with sodium in liquid ammonia form trans alkenes.

Physical properties of alkenes
Alkenes as a class have physical properties similar to those of alkanes.
(i)      First three members are gases, next fourteen members are liquids and higher ones are solids.
(ii)     They are lighter than water and insoluble in water but soluble in organic solvents like benzene, ether etc.
(iii)    Their boiling points (b.p) increase with increase in the number of carbon atoms, for each added –CH2  group b.p. rises by 20° to 30°. Their b.ps. are comparable to alkanes with corresponding carbon skeleton.
(iv)    Alkenes are weakly polar. The p electrons of the double bond can be easily be polarised. Therefore , their dipole moments are higher than those of alkanes.
(v)     The dipole moments, melting points and boiling points of alkenes are dependent on the position of groups bonded to the two doubly bonded carbons. Thus,

cis-but-2-ene  with two methyl groups on the same side has a small resultant dipole while trans–but-2-ene , the bond moments cancel out.  Due to relatively high polarity of the cis-isomer compared to its trans-isomer it fits well into crystalline lattice  and therefore generally has higher m.p. than the cis-isomer.  However, this is not a general rule and therefore may be exceptions. The measurement of dipole moment is a method to assign configuration to geometrical isomers.

Chemical properties of alkenes

Alkenes contain two types of bonds viz., sigma and pi-bonds. The pi-electrons are loosely bound between carbon atoms and are quite mobile. Hence electron deficient reagents are attracted by pi-electrons. Therefore such reagents readily react with alkenes to give addition products. The presence of mobile (or loosely held pi-electrons) are responsible for the reactive nature of alkenes.
            The most important reactions of alkenes are electrophilic additions and free radical additions. Other important common types of reactions of alkenes are oxidation and polymerisation.

A. Electrophilic addition
1.  Addition of Hydrogen Halides
            Hydrogen halides (HCl, HBr, HI) readily add to alkenes forming alkyl halides. The order of reactivity of hydrogen halides  in this reaction is :
            HI > HBr > HCl
          CH2=CH+  HX ®   CH3CH2X      ( X = I, Br, Cl )
It is an electrophilic addition reaction and proceeds through the formation of carbocations. Addition of HX takes place in two steps.

In step (i) , addition of proton (electropile) to the double bond is slow, while step (ii) i.e., addition of nucleophile (X- ) to carbocation is fast. Any addition in which electrophile adds first is an electrophilic addition.
            In the case of unsymmetrical alkenes the addition takes place according to Markownikov rule, i.e., the more electronegative of the addentum adds to that carbon which contains lesser number of hydrogen atoms. Thus a molecule of HBr adds to propene in such a way that bromine adds to to the central carbon atom which has lesser number of hydrogen atom.
CH3CH=CH2 +   HBr  ®  CH3CHBrCH3
            If we look at the structure of propene it has one methyl group attached to one of the doubly bonded carbons. In this case there are two possibilities  of formation of carbocation in step (i)  i.e., the proton either adds to terminal carbon (path 1) or to the central carbon (path 2).

            The two carbocations are (CH3)2C+H, a secondary (2°) carbocation and H2C+CH2CH3   a primary (1°) carbocation.
            The secondary carbocation (CH3)2C+H is more stable than primary carbocation H2C+CH2CH3 , therefore the cation formed in step  (1) i.e., CH3C+HCH3   will be more stable and thus nucleophile (negatively charged X:- ) will add to the central carbon.
            Tertiary carbonium ion  (3°) is is even more stable than secondary(2°) and primary (1°) carbocations.  Markowinkov‘s rule can also be interpreted in terms of electronic interaction  that is ‘’electrophilic addition to a carbon-carbon double bond involves the intermediate formation of the more stable carbocation’’.
            Decreasing order of reactivity of alkenes towards electrophilic addition is :
R2C=CR’2  > R2C=CHR’ > R2C=CH2 ³ RCH=CHR > RCH=CH2 > CH2=CH2    > CH2=CHX
R and R’ are alkyl groups and X is a halogen.

Free radical addition  -  Peroxide Effect

Markowinkoff rule is general, but not universal. In 1933  M.S. Kharash discovered that, the addition of HBr to unsymmetrical alkenes in the presence of organic peroxide (R-O-O-R) takes a course opposite to that suggested by Markowinkoff . This phenomenon of anti Markonikov addition of HBr in the presence of peroxide is known as peroxide effect. For example, when propylene reacts with HBr in the presence of a peroxide, the major product is n-propyl bromide, whereas in absence of a peroxide, the major product is isopropyl bromide.

Remember that HCl and  HI  do not give anti-Markonikov’s products in the presence of peroxides.
2. Addition of sulphuric acid
Alkenes  react with sulphuric acid to produce alkyl hydrogen sulphates. Markowinkoff’s rule is followed in the case of unsymmetrical alkenes. Alkyl hydrogen sulphates on hydrolysis yield alcohols.

The overall result of the above reaction appears to be Markowinkoff addition of H2O (hydration) to double bond.

This method is used for the industrial preparation of alcohols.
3. Addition of halogens
            Halogens (chrorine and bromine only) add, at ordinary temperature and without exposure to UV light to alkene to give vicinal dihalides. The order of reactivity is :
Fluorine  >   Chlorine  >  Bromine  >   Iodine

                  CH2=CH2   +  Br2   ®    CH2BrCH2Br
                  Ethene           red        1,2-Dibromoethane (colourless)
This is a test of unsaturation , since the colour of bromine solution is decolourised.

4. Hydoboration oxidation
            Diboranes adds to alkenes to give trialkyl borane, which on oxidation gives alcohol.

The net  addition is that of a water molecule, but follows anti-Markowinkov’s addition. 
Alkenes can be oxidised by different reagents to get different products.
(a)     Burning in excess of air or oxygen gives carbon dioxide and water. The reaction is exothermic.
                    CH2=CH2  + 3 O2 ®  2 CO2 + 2 H2O
(b)     Bayer’s reagent : Bayer’s reagent is a cold dilute and weakly alkaline solution of KMnO4.  Bayer’s reagent oxidises alkenes to diols. Two hydroxyl groups are introduced, where there is a double bond in the alkene. This reaction is called hydroxylation of alkene.

(c) Hot alkaline KMnO4 : Hot alkaline KMnO4 oxidises alkene to carbonyl compounds.

In case a hydrogen atom is attached to the carbon atoms containing a double bond, the hydrogen atom is replaced by a hydroxyl group to form a carboxylic acid on oxidation with hot alkaline KMnO4.

Thus by identifying the products formed on oxidation, it can be possible to fix the location of the double bond in alkene molecule.
(d)     Ozonolysis
 Ozonolysis is a method of locating unsaturation in a hydrocarbon. Ozone reacts with alkenes (also with alkynes) to form ozonides. On reduction with   zinc / water , the products formed are aldehydes or ketones or both.
            By identifying the products, it is possible to fix the position of double bond in an alkene.

If a hydrogen is attached to the carbon atom forming the double bond, an aldehyde results ; otherwise ketones are formed.
18.  What are the products obtained when the following molecules are subjected to ozonolysis ?
(i)        1-Pentene                (iii) 2-Methyl-2-butene
      (ii)    2-Pentene              (iv) ethane
19.    Write IUPAC names of the products obtained by addition of HBr to hex-1-ene .
i)  in the absence of peroxide and
ii) in the presence of peroxide
20.    Write IUPAC names of the products obtained by the ozonolysis of the following compounds :
i)   Pent-2-ene                 ii)    3,4-Dimethylhept-3-ene
iii)  2-Ethylbut-1-ene           iv)   1-phenylbut-1-ene
21.    An alkene ‘A’ on ozonolysis gives a mixture of ethanal and pentan-3-one. Write structure and IUPAC name of ‘A’.
22.    An alkene ‘A’ contains three C – C , eight C – H  s bonds and one C – C p bond. ‘A’ on ozonolysis gives two moles of an aldehyde of molecular mass 44 u. Write IUPAC name of ‘A’.
23.    Propanal and pentan-3-one are the ozonolysis products of an alkene ? What is the structural formula of the alkene ?
The addition of hydrogen is called hydrogenation. Alkenes add hydrogen to give alkanes when heated under pressure in presence of suitable catalyst such as finely divided nickel, platinum or palladium.

In the absence of a suitable catalyst, the hydrogenation reaction is extremely slow.
7. Polymerisation
 Polymerisation is the process in which a large number of simple molecules combine under suitable conditions to form large molecule, known as macromolecule or a polymer. The simple molecules are known as monomers.
            Alkenes undergo polymerisation in the presence of catalysts. In this process, one alkene molecule links to another molecule as represented by the following reaction of ethene.
       2 CH2=CH®  -CH2-CH2-CH2-CH2 or (-CH2-CH2-)2
When ethene is heated in the presence of traces of oxygen at about 500 - 675 K under pressure, ‘n’ molecules of ethene participate in the reaction as shown below :
             n CH2=CH2    ®    (-CH2-CH2-)n
                 ethene         polyethylene or polythene
A variety of polymers are obtained by using substituted ethenes in place of ethene.
For example :

Uses of various Polymers
Polythene is used in making electrical insulators, laboratory articles like funnel, burette, beaker etc.
Polyvinyl chloride (PVC) is used in making plastic bottles, plastic syringes, rain coats , pipes etc.
Polystyrene is used in making household goods, toys and models.
Polytetrafluoroethylene is also known as teflon is inert towards the action of chemicals. It is used in making chemically resistant pipes and some surgical tubes. The high thermal stability and chemical inertness of teflon makes it advantageous in the manufacture of non-stick cooking utensils, where a thin layer of teflon is coated on the interior of the vessel.
24.   An alkene with molecular formula C7H14 gives propanone and butanal on ozonolysis.  Write down the structural formula.
            The hydrocarbons  having carbon-carbon triple bond (CºC) and general formula CnH2n-2 are called alkynes.. The simplest
member of this class is ethyne C2H2. Ethyne has a total of 3 s bonds and 2 pbonds. The triple bond is made up of one s  and      2 pbonds. C2H2 is linear molecule in which CºC has a bond strength of 823 kJ mol-1 in ethyne. It is stronger than the C=C of ethene (610 kJ mol-1) and C- C of ethane (370 kJ mol-1)
            In IUPAC system the suffix –ane of the corresponding alkane is replaced by yne , e.g.
HCºCH                            CH3CºCH
Ethyne                          propyne
CH3CºCCH3         CH3CH2CºCH
But-2-yne                          But-1-yne
The remaining rules of nomenclature are same as in the case of alkanes and alkenes.
            Ethyne and propyne have got only one structure but there are two possible structures for butyne- (i)  but-1-yne and             (ii) but-2-yne. These two compounds differ in their structures due to position of triple bond, they are known as position isomers.
25.   Write structures of different isomers corresponding to the 5th member of alkyne series. What type of isomerism is exhibited by different pairs of isomers ? 

Structure of Triple bond

            Ethyne is the simplest molecule of alkyne series. Structure of ethyne is shown in Fig.

Orbital picture of ethyne showing
(a) sigma overlaps (b) pi overlaps
                Each carbon atom of ethyne has two sp hybridized orbitals. Carbon-carbon sigma (s) bond is obtained by the head-on overlapping of two sp hybridized orbitals of two carbon atoms. The remaining sp hybridized orbital of each carbon atom undergoes overlapping along the internuclear axis with 1s orbital of each of the two hydrogen atoms forming two C – H sigma bonds. H – C – C  bond angle is 180°. Each carbon has two unhybridised p-orbitals which are perpendicular to each other as well as to the plane of the C – C  sigma bond. The 2p-orbitals of one carbon atom are parallel to 2p orbitals of other carbon atom, which undergo lateral or sideways overlapping to form two  pi (p) bonds between two carbon atoms. Thus ethyne molecule consists of one C –C  s -bond, two     C – H   s- bonds and two C – C  p- bonds. The strength of CºC bond (bond enthalpy 823 kJ mol-1) is more than those of C = C (bond enthalpy 681 kJ mol-1) and C – C bond(bond enthalpy 348kJ mol-1). The CºC (133 pm) and C – C (154 pm) . The electron cloud between two carbon atoms is cylindrically symmetrical about the internuclear axis. Thus ethyne is a linear molecule.


1.     From vicinal dihalides : Acetylene and its higher homologoues can be prepared by treatment of alcoholic alkali with vicinal dihalides.

2. By dehalogenation of tetrahalides
          From tetrahalides, the alkynes can be prepared by the action of zinc.

3. Synthesis from carbon and hydrogen
Acetylene can be prepared by passing a stream of hydrogen through electric arc struck between carbon electrodes.

4. By electrolysis of potassium salt of fumaric acid

5.       Industrial preparation : Acetylene is obtained by the action of water on calcium carbide(CaC2) . Calcium carbide is prepared by heating quick lime(CaO) with carbon at high temperature.
                    CaO  + 3 C  ®   CaC2  + CO
CaC2 + 2 H2O    ®    H-CºC-H  + Ca(OH) 2
6.       Formation of higher alkynes
  Higher alkynes can be prepared by the action of alkyl halides on sodium acetylide. Sodium acetylide can be obtained from acetylene by the action of sodamide.
          H-CºC-H   +   NaNH2  ® H-CºC-Na  + NH3
                 ethyne                     sodium acetylide
     H-CºC-Na  +  BrCH2CH3 ® H- CºC- CH2-CH3  + NaBr
Physical properties of alkynes
Alkynes have the following general properties.
1.        State : Lower members in alkynes series are gases, while higher members are liquids and solids.
2.        Colour : Alkynes have no colour.
3.        Non-polar nature :  Alkynes are non-polar in nature. Therefore, alkynes are soluble in non-polar(organic) solvents such as benzene.

Chemical properties

Alkynes are unsaturated compounds. Therefore, like alkenes, they are quite reactive. The most common type of reactions of alkyne are addition reactions. On addition, alkynes ultimately give saturated compound. Alkynes also undergo oxidation and polymerisation.
1. Hydrogenation : In the presence of a catalyst, hydrogen adds to alkynes ultimately give alkane.

2. Addition of Halogen acids : This adds on to alkynes to give a dihalide. The addition follows Markowinkoff’s rule.

3. Addition of Halogen : Halogen adds to alkynes to give halogen substituted alkanes.

4. Hydration : Alkynes react with water in the presence of mercuric sulphate and sulphuric acid to form aldehyde or ketone.
            When acetylene is bubbled through 40% sulphuric acid in the presence of mercuric sulphate(HgSO4) acetaldehyde is obtained. The reaction can be considered as the addition of water to acetylene.

5.       Reaction with alcohols, hydrogen cyanide and carboxylic acids (vinylation)
 Ethyne adds a molecule of alcohol in the presence of alkali to give vinyl ether.

With HCN , ethyne gives vinyl cyanide.

Similarly,  alkynes add acids in the presence of a Lewis acid  catalyst or Hg2+ ions to give vinyl esters. 

6. Oxidation : Alkynes can be oxidised to different products using different reagents and conditions of oxidation.
(i) Burning in air : Acetylene burns in air with suity flame, emitting a yellow light. For this reason it is used for illumination.
(ii) Burning in excess of air : Acetylene burns with a blue flame when burnt in excess of air or oxygen. A very high temperature (3000 K) is obtained by this method. Therefore, in the form of oxy-acetylene flame, it is used for welding and cutting metals.
2 H-C º C-H   +  5 O2 ®  4 CO2+ 2 H2O + heat
(iii) Degradation with KMnO4 : The oxidation of alkynes with strong alkaline potassium permanganate give carboxylic acids generally containing lesser number of carbon atoms.

(iv) Ozonolysis : Alkynes react with ozone to give ozonides which are decomposed by water to form diketones and hydrogen peroxide. Diketones are oxidised by hydrogen peroxide  to carboxylic acids by the cleavage of carbon-carbon bonds.
6.       Self addition or Polymerisation
Alkynes polymerise to give linear or cyclic compounds depending upon their temperature and catalyst used. However, these polymers are different from the polymers of alkenes as they are usually low molecular weight polymers.
When acetylene is passed through red hot tube of iron or quartz, it trimerises to benzene.

Similarly, propyne polymerises to form mesitylene.

            Polymerisation of acetylene produces linear polymer polyacetylene. It is high molecular weight conjgated polymer containing repeating units (-CH=CH- CH=CH -)n. Under proper conditions this material conducts electricity. The films of polyacetylene can be used as electrodes in batteries. Having much higher conductance than metal conductors, lighter and cheaper, batteries can be made from it.
26. How will you convert ethanoic acid into benzene ?
Acetylene forms salt like compounds because hydrogen atoms are slightly acidic. Therefore hydrogen atoms directly attached to carbon atoms linked by triple bond can be replaced by highly electropositive metals such as sodium, silver, copper etc. Salts of acetylenes are known as acetylides.
(i) Sodium acetylide is formed when acetylene is passed over molten sodium.
H-CºC-H  +  Na  ® H-CºC-Na  + ½ H2
ethyne             sodium acetylide
Sodium acetylide is also formed when acetylene is treated with sodamide (NaNH2). Sodamide is obtained by dissolving sodium metal in liquid ammonia.
Na   +   NH3    ®   NaN H2   + ½ H2
                                  (liquid)     sodamide
NaNH2  + H-CºC-Na ® Na-CºC-Na  + NH3
Other alkynes can also react in a similar manner.
NaNH2  + CH3 -CºC-H ® CH3 -CºC-Na  + NH3
(ii) Silver acetilide is obtained as a white precipitate, when acetylene is passed through ammoniacal silver nitrate solution(Tollen’s reagent).
H-CºC-H  + 2 AgNO3  +2 NH4OH
®     Ag-CºC-Ag+ NH4NO3 + 2 H2O
     silver acetylide (white ppt)
(iii)   Copper acetilide is obtained as a red precipitate when acetylene is passed through ammoniacal solution of cuprous chloride.
    H-CºC-H  + Cu2Cl2 +2 NH4OH
                                 ®Cu-CºC-Cu+ NH4Cl + 2 H2O
                          copper acetilide (red ppt)
Alkynes form insoluble silver acetylide when (alkynes having acidic hydrogen) passed through ammoniacal silver nitrate (Tollen’s reagent).
        R-Cº C-H  + Ag+®  R-Cº C- Ag  + H+
                                        white ppt
An alkyne having no hydrogen atom attached to carbon linked by a triple bond does not form acetylide e.g., CH3-CºC-CH3 (2-Butyne) as it lacks an acidic hydrogen(acetylinic hydrogen). Thus, salt formation by alkyne is shown by those alkynes which contain a triple bond at the end of the molecule (i.e., when triple bond is terminal group). Alkynes having a non-terminal triple bond do not yield acetylides. Therefore, this reaction is used for distinguishing terminal alkynes from non-terminal alkynes.

Cause of acidic character

Hydrogen attached to carbon atoms by sp-hybrid orbital is slighly acidic, the reason for acidic character is that sp-hybrid orbital has greater s-character than sp2 or sp3 hybrid orbitals. The s-character in sp3, sp2 and sp hybrid orbitals are 25, 33.3 and 50%respectively. An s-orbital tends to keep electrons closer to the nucleus than a p-orbital. It means that greater the share of s-orbital in the hybrid orbital, the nearer will be the shared pair of electrons to the nucleus of carbon atom. This makes the sp-hybrid carbon more electronegative than sp2 or sp3 hybridised carbon atom. Thus, the hydrogen attached through sp-hybid orbital acquires a slight positive charge. Therefore, it can be replaced by highly electropositive metals (such as sodium, copper, silver etc). The acidic character of hydrocarbons varies as :
Alkane   <   Alkene   <   Alkyne
However , alkynes are extremely weak acids. Compared to carboxylic acids like acetic acid, ethyne is 1020 times less acidic. Ethene is 1040 times less acidic than acetic acid.

Test for alkane, alkene and alkynes

Alkanes, alkenes and akynes can be distinguished from one another by the following tests.
            Alkanes do not decolourise bromine water. Baeyer’s reagent (alkaline KMnO4) remains unchanged on treating with alkanes.
            Alkenes and alkynes decolourise bromine water. It is used to distinguish alkanes from unsaturated compounds such as alkenes and alkynes.

Alkenes and alkynes decolourise Bayer’s reagent. Therefore, this test is used for distinguishing alkenes from alkanes.

27. How will you separate propene from propyne ?  
            The  name  alkadiene is often shortened as diene. These are unsaturated hydrocarbons having two carbon-carbon double bonds per molecule. The general formula of alkadiene is similar to alkynes CnH2n-2, hence these are isomeric with alkynes.
Classification of Dienes
            The  dienes are classified into three types depending upon the relative positions of the two double bonds.
1.    Isolated dienes
The double bonds are separated by more than one single
          bond. Examples,
          CH2=CH-CH2-CH=CH2         CH2=CH-CH2-CH2-CH=CH2
           Penta-1,4-diene                Hexa - 1,5-diene
2.       Cumulated dienes
The  double bonds between successive carbon atoms are called cumulated double bonds. E.g.
CH2=C=CH2                              CH3-CH=C=CH2
Propa-1,2-diene (allene)          Buta-1,2-diene (methylallene)
3.       Cojugated dienes
 The dienes in which double bond is alternate with single bond   are called conjugated dienes, e.g
        CH2=CH- CH = CH2       CH3- CH=CH-CH=CH-CH3
Buta-1,3-diene             Hexa-2,4-diene
Relative stabilities of Dienes
            A conjugated diene is more stable as compared with non-conjugated dienes.  The relative order of stability of diene is
      Cojugated diene > Isolated diene > Cumulated diene
Exceptional stability of conjugated dienes can be explained in terms of orbital structure. Consider  Buta-1,3-diene as an example of conjugated diene. All the four carbon atoms in Buta-1,3-diene are sp2 hybridised. Each carbon atom also has an unhybridised p-orbital which is used for p bonding. The hybrid orbitals of each carbon atom are used for the formation of s bonds. The p-orbital of C-2 can overlap equally with the p-orbitals of C-1 and C-3 . Similarly, the p-orbital of C-3 can overlap equally with the p-orbitals of C-2 and C-4 as shown in Fig.  As a result the p-electrons in conjugated dienes are delocalised.

Delocalisation of p - electrons in conjugated diens
The delocalisation of p-electrons in conjugated dienes makes them more stable because now p-electrons feel simultaneous attraction of all the four nuclei of four carbon atoms.
            This type of delocalisation of p-electrons  is not possible in isolated or cumulated dienes.
            The non-conjugated dienes behave exactly in the same way as simple alkenes except that the attacking reagent is consumed in twice the amount required for one double bond. But due to the mutual interactions of the double bonds, i.e., delocalisation of the electrons in conjugated dienes, their properties are entirely different.
Electrophilic addition
            1,3-Butadiene and HBr taken in equimolar  amounts yield two products 3-Bromo-but-1-ene and 1-Bromo-but-2-ene through 1,2 and 1,4-addition respectively.

The resonating structures of  the intermediate carbocation , formed after addition of the electrophile, i.e.,  H+ , can add to the anion in two alternative ways (path a or path b ) to yield 1,2 and 1,4 addition products.

27. A conjugated alkadiene having molecular formula C13H22 on ozonolysis yielded ethyl methyl ketone and cyclohexanal. Identify the diene, write its structural formula and give its IUPAC name.
            The term  aromatic (Greek ; aroma means fragrance) was first used in compounds having pleasant odour although structure was not known. Now the term aromatic is used for a class of compounds having  a characteristic stability despite having unsaturation. These may have one or more benzene rings (benzanoid)  or may not  have benzene ring (non-benzanoid) . Benzanoid compounds include benzene and its drivatives  having aliphatic  side chains (arenes) or polynuclear hydrocarbons, e.g. naphthalene, anthracene etc.

The nomenclature of aromatic compounds discussed in Unit 12.
Aromatic hydrocarbons or arenes are the compounds of carbon  and hydrogen with at least one benzene type ring (hexagonal ring of carbons) in their molecules. We have hydrocarbons of benzene series (containing benzene rings) and so on.
The molecular formula of benzene is C6H6 . It contains eight hydrogen atoms less than the corresponding parent hydrocarbon, i.e., hexane(C6H14). It took several years to assign a structural formula to benzene because of its peculiar properties.
Kekule Structure
In 1865 Kekule suggested the first structure of benzene. In this structure, there is a hexagonal ring of carbon atoms distributed in a symmetrical manner, with each carbon atom carrying one hydrogen atom. The fourth valence of the carbon atom is fulfied by the presence of alternate system of single and double bond as shown.

The above formula had many draw backs as described below :
(i)       The presence of three double bonds should make the molecule highly reactive towards addition reactions. But contrary to this, benzene behaves like saturated hydrocarbons.
(ii)      The carbon-carbon bond lengths in benzene should be 154 pm (C-C) and 134 pm (C=C) . It implies that the ring should not be regular hexagon but actually all carbon-carbon distances in benzene have been found to be 139 pm.
(iii)     Finally two isomers should result in a 1,2-disubstituted benzene as shown in Fig

While Kekule formula could not explain the differences in properties between benzenes and alkenes based on this structure, he explained the lack of isomers as in figure by postulating a rapid interchange in the postion of the double bond as shown below:

Orbital picture of Benzene
According to orbital structure , each carbon atom in benzene assumes sp2 hybrid orbitals lying in one plane and oriented at an angle of 120º. There is one unhybridised p-orbital having two lobes lying perpendicular to the plane of hybrid orbitals for the axial overlap with 1s-orbital of hydrogen atom to form C- H sigma bond. The other two hybrid orbitals are used for axial overlap of similar orbital of two adjacent carbon atoms on either side to form C- C sigma bonds. The axial overlapping of hybrid orbitals to form C- H and C- C bonds have been shown in Fig.

Sigma bond formation in benzene
As is evident, the frame work of carbon and hydrogen atoms is coplanar with H-C-C or C-C-C bond angle as 120º. The unhybridised p-orbital on each carbon atom can overlap to a small extent with p-orbital   of the two adjacent carbon atoms on either side to constitute pi bonds as shown in Fig.

Side-wise orverlapping of p-orbitals
The molecular orbitals containing pi-electrons spreads over the entire carbon skeleton as shown in Fig.

Orbital picture of  benzene
The spreading of pi-electrons in the form of ring of pi electrons above and below the plane of carbon atoms is called delocalisation of pi-electrons. The delocalisation of pi-electrons results in the decrease in energy and hence accounts for the stability of benzene molecule.
                The delocalised structure of benzene accounts for the X-ray data (all C- C bond lengths equal) and the absence of the type of isomerism shown in Fig ( Page 18 ). Furthermore molecular orbital theory predicts that those cyclic molecules which have alternate single and double bonds with ( 4 n + 2 ) ( where n = 0, 1, 2, ….) electrons in the delocalised p-cloud are particularly stable and have chemical properties different from other unsaturated hydrocarbons.

Resonance structure of Benzene

Structure of benzene can also be explained on the basis of resonance. Benzene may be assigned following structures A and B . Structure A and B have same arrangement of atoms and differ only in electronic arrangement. Any of these structures alone cannot explain all properties of benzene.

Resonance structure of benzene
According to these structures , there should be three single bonds (bond length 154 pm)  and three double bonds (bond length 134 pm) between carbon atoms in the benzene molecule. But actually, it has been found that all the carbon-carbon bonds in benzene are equivalent and have a bond length 139 pm. Structures A and B are known as resonating or cannonical structures of benzene. The actual structure of benzene lies somewhere between A and B may be represented as C referred to as resonance hybrid.
            To indicate two structures which are resonance forms of the same compound, a double headed arrow is used as shown in Fig (above).
            The resonance hybrid is more stable than any of the contributing (or canonical) structures. The difference between the energy of the most stable contributing structure and the energy of  the resonance hybrid is known as resonance energy.
            In the case of benzene, the resonance hybrid (actual molecule) has 147 kJ/mol less energy than either A or B. Thus, resonance energy of benzene is 147 kJ/mol. It is this stabilization due to resonance which is responsible for the aromatic character of benzene.


            Benzene forms a number of mono, di or poly-substituted derivatives by replacement one, two or more hydrogen atoms of the ring by other monovalent atoms or groups. In many cases, the derivative of benzene exist in two or more isomeric forms. The isomerism of these derivatives is discussed as follows.

Mono-substituted Derivatives

            The molecule of benzene is symmetrical and the six carbon atoms as well as the hydrogen atoms occupy similar positions in the molecule. If atom of hydrogen is substituted by a monovalent group or a radical (say methyl group), resulting mono substitution product exist in one form only. The position assigned to the substituent group does not matter because of the equivalent positions of the six hydrogen atoms. Thus, we have only one compound having the formula C6H5X where X is a monovalent group.
Disubstituted Derivatives
The various positions in the monosubstituted derivatives are not equivalent with respect to the position already occupied by the substituent. For example, taking the positions of the substituent as number 1, the other positions are as shown :

A close examination at the structures reveal that :
(i)       Positions 2 and 6 are equivalent and are called ortho(-o).
(ii)      Positions 3 and 5 are equivalent and are called meta(-m) with respect to the position 1.
(iii)     Position 4 is called para (p) with respect to the position 1.
For example in the case of dimethyl benzene (CH3)2C6H4 ,  commonly known as xylene.  There can be three  xylenes depending upon the positions of methyl groups as shown below :

Besides , the three possible dimethyl benzenes, a fourth isomer, ethyl benzene is also known.

In the case of naphthalene, even monosubstituted compounds display positional isomerism as in 1-Methyl and 2-Methyl naphthalenes.

The term aromatic was first used for a group of compounds having pleasant odour. These compounds have properties which are quite different from those of the aliphatic compounds. The set of these properties is called aromatic character or aromaticity. Some typical properties of aromatic compounds are :
·         These are highly unsaturated compounds, but do not give addition reactions easily.
·         These give electrophilic substitution reactions very easily.
·         These are cyclic compounds containing five, six or seven membered rings.
·         These  molecules are flat  (planar).
·         These are quite stable compounds.
The aromaticity in benzene is considered to be due the presence of six delocalised p-electrons.
            The modern theory of aromaticity was given by Eric Huckel in 1931. According to this theory, for a compound to exhibit aromaticity , it must have the following properties.

·  Delocalisation of p-electrons of the ring.
·  Planarity of the molecules. To permit sufficient or total delocalisation of p-electrons, the ring must be planar to allow cyclic overlap of the p-orbitals.
HUCKEL RULE or ( 4 n + 2 ) RULE
            This rule states that for a compound to exhibit aromatic character, it should have a conjugated , planar cyclic system containing 4 n + 2 ( where n = 1, 2, 3. … ) delocalised p-electrons forming a cyclic cloud of delocalised p-electrons  above and below the plane of the molecule. This is known as Huckel rule of (4n + 2 ) p-electrons.
Benzene , naphthalene, anthracene and phenanthrene are aromatic as they contain ( 4 n + 2 ) i.e., 6, 10 , 14 p-electrons in a conjugated cyclic array. The cyclopentadiene and cyclooctatetrene are non-aromatic as instead of (4n+2) p-electrons, these have 4n p-electrons. Moreover they are non-planar.

28. Predict which of the following systems would be aromatic and
       why ?

1.       From alkyne
Alkynes polymerize at high temperatures to yield arenes e.g. benzene is obtained from ethyne.

2.       Decarboxylation of aromatic acids
In laboratory benzene is prepared by heating sodium benzoate with sodalime.

3.       Reduction of Benzene diazonium salts
In presence of hypophosphorus acid, benzene diazonium chloride is convered to benzene (diazo group is replaced by H)

4.       Friedel Craft’s reaction
Benzene can yield alkyl benzene by treating with alkyl halide in presence of anhydrous aluminium chloride.

5.       Wurtz-Fittig  reaction
Arenes can be obtained by the action of sodium metal on a mixture of aryl halide and alkyl halide in ether

6.       From Grignard reagents
Arenes can be prepared by reacting Grignard reagent  with alkyl halide, e.g.

Physical properties      
Aromatic hydrocarbons are usually colouless , insoluble in water but soluble in organic solvents. They are inflammable , burn with sooty flame and have characteristic odour. They are toxic and carcinogenic in nature. Their boiling points increase with increase in molecular mass.
Chemical properties     
Aromatic hydrocarbons are unsaturated cyclic compounds. They are also known as arenes.
            Benzene and its homologoues are better solvents as they dissolve a large number of compounds. This is because the electron clouds makes it polar to some extent and therefore even polar molecules are attracted towards it. This helps in dissolving in them a large number of compounds.
            The reason for this inertness of benzene and its homologues is due to the presence of pi-electron clouds above and below the plane of the ring of carbon atoms. Therefore, nucleophilic species (electron rich species such as Cl-, OH-, CN- etc ) cannot attack the benzene ring due to repulsion between the negative charge on the nucleophile and delocalised pi-electron clouds. However, electrophiles (such as H+, Cl+, NO2+ etc) can attack the benzene ring and for this reason benzene and its homologues can be replaced by nitro(-NO2), halogen(-X), sulphonic acid group        (-SO3H) etc.
            However, arenes also undergo a few addition reactions under more drastic conditions, such as increased concentration of the reagent, high pressure, high temperature, the presence of catalyst etc.


The chemical reaction which involves the replacement of an atom or group of atoms from organic molecule by some other atom or group with out changing the structure of the remaining part of the molecule is called substitution reaction. The new group which finds place in the molecule is called substituent and the product formed is referred to as substitution product.
According to experimental evidences SE ( S = substitution ;             E = electrophilic) reactions are supposed to proceed via the following three steps :
·       Generation of electrophile
·       Formation of carbocation intermediate
·       Removal of proton from the carbocation intermediate
(a)   Generation of electrophile  E+
During chlorination, alkylation and acylation of benzene , anhydrous AlCl3 , being a Lewis acid helps in generation of electrophile Cl+ , R+ , RC+O (acylium ion) respectively by combing with the attacking reagent.

In case of nitration, the electrophile , nitronium ion +NO2 is produced by transfer of a proton (from sulphuric acid) to nitric acid in the following manner.

In the process of generation of nitronium ion , sulphuric acid serves as an acid and nitric acid as a base. Thus it is a simple acid-base equilibrium.
(b)   Formation of Carbocation (arenium ion )
            Attack of electrophile results in the formation of s-complex or arenium ion in which one of the carbon is sp3 hybridised.

The arenium ion gets stabilized by resonance.

Sigma complex or arenium ion loses its aromatic character because of delocalization of electrons stops at sp3 hybridised carbon.

(c) Removal of proton
            To restore the aromatic character, s-complex releases proton from sp3 hybridised carbon on attack by [AlCl4]- (in case of halogenation , alkylation and acylation ) and [HSO4]- (in case of nitration.

1.      Halogenation
 Chlorine or brominre react with benzene in the presence of Lewis acids like ferric chloride or aluminium salts of corresponding halogens, which act as catalysts to give chlorobenzne or bromobenzne.
The function of Lewis acids like AlCl3, FeCl3, FeBr3 etc is to carry halogen to aromatic hydrocarbons. Hence, they are called halogen carriers.
            Fluorine is very reactive and hence this method is not suited for the preparation of fluorobenzene.
            Iodobenzene  is obtained by heating iodine and nitric acid, mercuric oxide etc.

2.      Sulphonation
 The replacement of hydrogen atom of benzene by a sulphonic acid group(-SO3H) is known as suphonation. The reaction is carried out by treating an arene with concentrated sulphuric acid containing dissolved sulphur trioxide or with chlorosulphonic acid.

3.      Nitration
 A nitro group (-NO2) can be introduced into the benzene ring using nitrating mixture ( a mixture of concentrated sulphuric acid and concentrated nitric acid)

4. Friedel Craft’s reaction
 On treatment with an alkyl halide or acid halide (acyl halide ) in presence of anhydrous aluminium chloride as catalyst, benzene forms an alkyl or acyl benzene as described below. The reaction is known as Friedel crafts alkylation or  acylation  respectively.
5. Oxidation 
 Arenes can be oxidised to different products depending upon their structure and conditions of the reaction.
(i) Combustion : Aromatic hydrocarbons burn with luminous and sooty flame.
     2 C6H6+ 15 O2    ®  12 CO2 + 6 H2O
   2 C6 H5 CH3 + O2 ® 7  CO2 + 4 H2O
(ii) Side chain oxidation : Arenes with side chain on oxidation with strong oxidising agents such as alkaline potassium permanganate, give carboxylic acid. However, a hydrocarbon without a side chain remain unaffected.

In case , there are more carbon atoms in the side chain, on oxidation all the carbon atom gets oxidised to carbon dioxide except, the one that is directly attached to the aromatic ring.

(iii) Catalytic oxidation : Benzene can be catalytically oxidised to an aliphatic compound, maleic anhydride. Benzene when heated with excess of air at 800 K in the presence of  vanadium pentoxide(V2O5) gives maleic anhydride.

Addition Reactions of Arene

1.       Addition of Chlorine
  Benzene and its homologues undergo some addition reactions similar to alkynes. However, extremely drastic conditions are required for carrying out addition reactions in arenes. For example, benzene can be chlorinated in presence of sunlight to form benzne hexachloride, BHC.

ii.  Addition of hydrogen
Hydrogen adds on to benzene when heated (475 K) under pressure in presence of a nickel catalyst to cyclohexane (hexa hydrobenzene). Similarly, toluene gives methyl cyclohexane ( hexa hydro toluene).

iii)  Addition of ozone
Benzene slowly reacts with ozone to form triozonide. The triozonide on hydrolysis with water gives glyoxal.



            The arrangement of substituents on the benzene ring is termed as orientation. The term is often used for the process of determining the position of the substituents on the benzene ring. Whenever substitution is made in the benzene ring , the substituent can occupy any position –all position being equivalent.  The position taken by the second substituent depends upon the nature of the substituent already present in the ring. In other words, the substituent already present on the ring directs the incoming substituent. This is called directive influence of the substituents. Depending up on the directive influence of the substituent already present in the ring, the substituents are classified into following groups.

Ortho and para directing groups
The substituents that direct the incoming substituent to ortho (o) and para (p) positions relative to theirs are called ortho and para directing groups.
The ortho and para directing substituents permit the electrophilic substitution at the ortho and para positions.  If the substituent S is an ortho and para directing, then the overall reaction may be described as :

The proportion of the ortho and para disubstituted products depends upon the reaction conditions.
The ortho and para directing substituents are arranged in the decreasing directing influence as follows:

            The ortho and para directing substituents have    ‘electron-donating influence’ on the aromatic ring. Thus ortho and para directing substituents increase the electron density on the ring and therefore are called activating groups (or activators). Activating groups enhance the rate of reaction.

Ortho and para directing but deactivating groups
            The only exception to the above rule is the halogen substituents. The halogens are ortho and para directing but deactivate the benzene ring relative to benzene.
For example, if we carry out nitration of toluene, a mixture of ortho and para nitro toluene is formed.

Meta directing Groups

                The substituents that direct the incoming substituents to meta(m-) relative to theirs are called meta directing groups. The meta directing groups permit the electrophilic substitution at meta position. If S is the meta directing group, then the overall reaction may be described as :

The meta-directing substituents are arranged in the decreasing directing influences as follows :

            The meta directing substituents have ‘electron-withdrawing influence ‘ on the benzene ring. Thus meta directing substituents decrease the overall electron density on the ring and are therefore called deactivating groups(or deactivators) . The deactivating groups decrease (or lower) the rate of reaction.
For example, nitration of benzoic acid produces m-nitro benzoic acid.

Examples of Directing influence of substituents
It is seen that  :
·         -OH (phenolic) group is o- and p-directing.
·         -NO2 group is meta-directing

The o- and p-directing influence of –OH group in phenol
                Phenolic ( -OH) group has lone pairs of electrons on its oxygen  atom. The lone pairs of electrons on the O atom of the –OH group interact with p-electrons of the benzene ring and give rise to various resonance forms as shown below.

This shows that  the electron density is more concentrated  on the two ortho- and para- positions. The electrophilic substitution takes place at these positions.
The m-directing influence on –NO2 group
            The –NO2 group is a meta directing and deactivating group. The more electronegative O atoms in–NO2 group withdraw electronic density from N atom and places a slight positive charge on it. This atom pulls electrons from the benzene ring and gives rise to various resonance forms as shown below.

resonance in nitrobenzene
These resonance structures show that the electron density at o- and p-positions is generally reduced. As a result  , electrophilic attack is at meta-position.
The o- and p- directing but deactivating effect of halogen atom
            Halogen atoms are o- and p- directing but deactivating substituents. This anomalous behaviour of halogen atoms (when present in the ring) is due to two opposing effects operating at the same time.
            A halogen substituent , due to its strong electronegative character , pulls electron from the ring due to inductive effect. This decreases the electron density on the ring.
            Due to the availability of lone-pairs of electrons, the halogen atom releases electrons to the ring giving rise to various resonance structures.

Resonance structures of chlorobenzene
            Due to poor 2p (C) - 3p (Cl) ovelap , the halogen atom of the ring is not a good electron releasing substituent.  As a result , in the case of halogen substituent , the inductive effect predominates and the benzene ring gets deactivated despite o- and p- directing effect of halogen substituent.
            Polynuclear  aromatic hydrocarbons contain more than one benzene rings and have two carbon atoms shared by two or three aromatic rings. Anthracene and phenanthrene have two pairs of carbon atoms shared by two two rings, each pair is shared by a different pair of rings.

            Coal tar is the main source of naphthalene (6-10%) , anthracene (1%)  and phenanthrene.
Naphthalene is obtained from the middle oil fraction                     (b.p 443 – 503 K) of coal tar by cooling. Crude naphthalene is washed successively with dil H2SO4 , sodium hydroxide and water. The dried sample is then purified by sublimation.
            Naphthalene is used as moth-balls to protect woollenns. It is also used  for  the manufacture of phthalic anhydride, 2-naphthol, dyes etc.
Anthracene is obtained by cooling green oil fraction (b.p. 543-633K) of coal tar distillation. The crude sample of anthracene is purified by washing successively with solvent naphtha (it removes phenanthrene) and pyridine.
Phenanthrene is obtained from its solution in solvent naphtha (obtained during purification of crude anthracene crystals) by evaporation.
            All chemical substances are believed to be harmful in one way or other. On finding that prolonged exposure to coal tar could cause skin cancer, it was discovered that the high boiling fluorescent  fraction of coal tar was responsible for causing cancer. This fraction of coal tar  was found to contain 1,2-benzanthracene. Later some more compounds were also found to be carcinogenic. The names and formulae of some carcinogenic compounds are given below.
                There is no rule available so far to predict the carcinogenic activity of any hydrocarbon or its derivatives. However, the number of groups like –CH3, -OH, -CN, -OCH3 etc. has been found to influence carcinogenic activity of compounds.

29.   Bring out the following conversions :
(i)       Methane to ethane           
(ii)      Ethyne to methane
(iii)     Ethane to ethene      
(iv)     Ethane to butane
30.        Suggest a method to separate a mixture of ethane, ethene and ethyne.
31.        Describe a method to distinguish between ethene and ethyne.
32.        What is Grignard reagent ? How it is prepared ?
33.        How is propane prepared from Grignard reagent ?
34.        Arrange the following in the increasing order of boiling points : Hexane, heptane, 2-Methylpentane, 2,2-Dimethyl pentane
35.        How would you obtain :
i)   ethene from ethanol
 ii)   ethyne from ethylene dibromide
36.        Write equations for the preparation of propyne from ethyne.
37.        How are the following conversions carried ?
(i)       propene to propane
(ii)      ethyne to ethane
(iii)     ethanol to ethene
(iv)     sodium acetate to methane
(v)      benzene to nitrobenzene
38.   How will you convert benzene to :
(i)       Toluene                
(ii)      Acetophenone
(iii)     Nitrobenzene       
(iv)     Ethylbenzene
(v)      Benzoic acid
39.    What happens when :
(i)       1-Bromopropane is heated with alcoholic KOH.
(ii)      2-Propanol is heated with alumina at 630 K.
(iii)     Benzene is treated with a mixture of Con. Sulphuric acid and nitric acid.
(iv)     Ethene is treated with an alkaline solution of cold KMnO4.
40.        Write chemical equations for combustion reaction of the following hydrocarbons ?
i)   butane   ii)  pentene    iii)  hexyne    iv)  Toluene
41.        Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p and why ?
42.        Why is benzene extra ordinarily stable though it contains three double bonds ?
43.        Explain why the following systems are not aromatic ?

44.        How will you convert benzene into :
i)   p-nitrobromobenzene     ii)   m-nitrochlorobenzene
iii) p-nitrotoluene                iv)  acetophenone
45.        In the alkane H3CCH2C(CH3)2CH2CH(CH3)2 ,  identify 1°, 2° , 3° carbon atoms and give the number of H atoms bonded to each one of these.
46.        Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism.
47.        Write ozonolysis products of 1,2-dimethylbenzene (o-Xylene). How does the result support Kekule structure for benzene ?
48.        Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also give reason for this behaviour.
49.        Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty ?
50.        How would you convert the following compounds into benzene ?
i)   ethyne  ii)   ethane    iii)   hexane
51.        Write the structures of all the alkenes which on hydrogenation give 2-methylbutane.
52.        Arrange the following set of compounds in order of decreasing relative reactivity with electreophile E+,
(a) chlorobenzene , 2,4-dinitrochlorobenzene ,
(b)  toluene , p-CH3C6H4-NO2 , p-O2N-C6H4-NO2
53.        Out of benzene, m-dinitrobenzene and toluene which will undergo nitration most easily and why ?
54.        Suggest the name of a Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene.
55.        Why Wurtz reaction is not preferred for the preparation of alkanes containing odd number of carbon atoms ? Illustrate your answer by taking one example.
1.         Why carbon forms a large number of compounds ? Write a note on optical isomers of tartaric acid.
2.         Bring out the following conversions :
(i)      Methane to ethane           
(ii)     Ethane to ethane
(iii)   Ethyne to methane           
(iv)    Ethane to butane
3.         Suggest a method to separate a mixture of ethane, ethene and ethyne.
4.         Describe a method to distinguish between ethene and ethyne.
5.         What is Grignard reagent ? How it is prepared ?
6.         How is propane prepared from Grignard reagent ?
7.         Arrange the following in the increasing order of boiling points : Hexane, heptane, 2-Methylpentane, 2,2-Dimethyl pentane
8.         How would you obtain :
(i)      ethene from ethanol
(ii)     ethyne from ethylene dibromide
9.         Write equations for the preparation of propyne from ethyne.
10.      How are the following conversions carried ?
(i)       propene to propane
(ii)      ethyne to ethane
(iii)     ethanol to ethene
(iv)     sodium acetate to methane
(v)      benzene to nitrobenzene
11.      How will you convert benzene to :
(i)       Toluene                
(ii)      Nitrobenzene      
(iii)     Benzoic acid
(iv)     Acetophenone
(v)      Ethylbenzene
12.      What happens when :
(i)      1-Bromopropane is heated with alcoholic KOH.
(ii)     2-Propanol is heated with alumina at 630 K.
(iii)   Benzene is treated with a mixture of Con. Sulphuric acid and nitric acid.
(iv)    Ethene is treated with an alkaline solution of cold KMnO4.
13.      What is coal ?
14.      What are the natural sources of hydrocarbons ?
15.      What is petroleum ?
16.      Compare the composition of coal and petroleum.
17.      Give the origin of coal.
18.      Give the origin of petroleum.
19.      How are aromatic hydrocarbons obtained from coal ?
20.      What do you understand by pyrolysis and discuss it with coal
21.      Give a brief account of petroleum refining. Name the various useful products obtained from it.
22.      What is straight-run gasoline ? Describe the principle of obtaining straight-run gasoline from petroleum.
23.      Explain the following processes:
(i)        Cracking                
(ii)       Reforming
24.      Explain the term ‘knocking’ What is the relationship between the structure of a hydrocarbon and knocking ?
25.      Explain the term ‘knocking’. A sample of petrol produces, the same knocking as a mixture containing 30% n-heptane and 70% iso-octane. What is the octane number of sample.
26.      Describe different methods to improve the quality of a fuel used in gasoline engine.
27.      How can you obtain aliphatic hydrocarbon from coal ?
28.      Describe two methods by which petroleum can be obtained artificially from coal.
29.      Write a note on synthetic petrol.
30.      Discuss various methods for laboratory preparation of alkanes.
31.      How can you obtain alkane from (I) Unsaturated hydrocarbons (ii) alkyl halides (iii) carboxylic acids
32.      Describe the laboratory preparation of methane.
33.      Decribe various methods for laboratory preparation of alkenes.
34.      How can alkenes be prepared from : (i) alcohols (ii) alkyl halides ?
35.      Decribe the laboratory preparation of ethene ?
36.      Give the methods of preparation of ethyne.
37.      Describe the laboratory preparation of acetylene.
38.      Give the general methods of preparation of higher alkynes.
39.      Give the important chemical reactions of alkanes.
40.      Give the important chemical properties of alkenes.
41.      Give the important chemical properties of alkynes.
42.      Give important uses of (i) ethene (ii) ethyne
43.      Write a note on halogenation of alkanes.
44.      Give the addition reactions of benzene.
45.      Discuss the halogenation of benzene.
46.      What is sulphonation ? Discuss the sulphonation of benzene.
47.      Describe a method to distinguish between ethene and ethyne.
48.      Why does ethene decolourise bromine water ; while ethane does not do so ?
49.      Give one example each of (i) an addition reaction of chlorine (ii) substitution reaction of chlorine
50.      Explain the term ‘polymerisation’ with two examples.
51.      Give one chemical equation in which chlorine adds to a hydrocarbon by substitution.
52.      What is Grignard reagent ? How is propane prepared from a Grignard reagent ?
53.      Give the reason for the following :
(i)       The boiling points of hydrocarbons decrease with increase in branching.
(ii)      Unsaturated compounds undergo addition reactions.
54.      Account for the following :
(i)       Boiling points of alkenes and alkynes are higher than the corresponding alkanes.
(ii)      Hydrocarbons with odd number of carbon atoms have low melting points than those with even number of carbon atoms.
(iii)     The melting points of cis-isomer of an alkene is lower than that of trans-isomer.
55.      Why does acetylene behave like a very weak acid ?
56.      Acetylene is acidic in character. Give reason.
57.      Write a reaction of acetylene which shows its acidic character.
58.      What is Baeyer’s test ? Give its utility.
59.      What happens when bromine water is treated with : (i) ethylene (ii) acetylene ? What is its utility ?
60.      Explain the terms (i) acetylation (ii) alkylation
61.      Write notes on (i)  Friedel-Craft’s reaction (ii) Addition reaction.  (iii) Markowinkoff’s reaction.
62.      Explain the following with examples      
(i)        Wurtz reaction          
(ii)       Kolbe’s electrolytic method             
(iii)      Jhydrogenation 
(iv)      Dehydration
63.      Write the equations for the preparation of propyne from acetylene.
64.      How are the following conversions carried out ?
(i)       Propene to propane.      
(ii)      acetylene to ethane
(iii)     Ethanol to ethene          
(iv)     Methane to ethane
(v)      Propene to 2-Bromopropane
(vi)     Methane to tetrachloromethane
65.      How will you convert :
(i)       Acetylene into acetaldehyde.    
(ii)      Methane to ethane
(iii)     acetic acid to methane.
(iv)     ethene to ethanol.
(v)      acetylene to acetic acid.
(vi)     2-Chlorobutane to 1-butene.
(vii)    Ethylene into glyoxal.
(viii)   Phenol to benzene
66.      How will you convert :
(i)       Ethane to ethene          
(ii)      Ethyl iodide to ethane
(iii)     Ethyl iodide into butane
(iv)     Ethyl alcohol into ethyne
(v)      Propyl chloride to propene
67.      What happens when :
(i)       1-Bromopropane is heated with alcoholic KOH.
(ii)      2-Propanol is heated with alumina at 630 K
(iii)     Benzene is treated with a mixture of Con. Sulphuric acid and nitric acid.
(iv)     Ethane is treated with alkaline pot. Permanganate soln (cold).
(v)      benzene is treated with bromine in presence of aluminium bromide as catalyst.
68.      What happens when :
(i)       Ethyl bromide is treated with alcoholic KOH.
(ii)      Ethylene dibromide is treated with zinc dust.
(iii)     Propene reacts with water in presence of a mineral acid.
(iv)     Ethylene is passed through alkaline KMnO4.Acetylene is hydrated in presence of mercuric sulphate and dil Sulphuric acid.
(v)      Ethanol is heated with Con. Sulphuric acid.
(vi)     Ethene is passed through Baeyer’s reagent
(vii)    Ethyne is passed through ammoniacal silver nitrate.
(viii)   Methyl bromide is treated with sodium.
(ix)     Ethanol is treated with HI acid.
(x)      Ozone is passed in ethylene in an organic solvent.
69.      Use Markowinkoff’s rule to predict the product of reaction of
(i)      HCl with CH2C(Cl)=CH2
(ii)     HCl with CH2CH=C(CH3) 2
70.      Write chemical equation describing the general mechanism for electrophilic substitution in benzene ring.
71.      Why are o- and p-directing groups called activating groups, whereas m-directing groups are deactivating groups ?
72.      Name polynuclear hydrocarbon obtained from coal tar.
73.      Name some carcino genic compounds.
74.       What effect does branching of alkane chain has its boiling  point ?
75.      What happens when :
i)  1-Bromopropane is heated with alcoholic KOH.
ii)  2-Propanol is heated with alumina at 630 K.
iii) Benzene is treated with a mixture of Con. Sulphuric acid and nitric acid.
iv)   Ethene is treated with an alkaline solution of cold KMnO4.


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