Hydrocarbons

(c) H C . + C . → H C–C1 Though in (c), CH – Cl, the one of the products is formed but free radicals are consumed and the chain is terminated. The above mechanism helps us to understand the reason for the formation of ethane as a byproduct during chlorination of methane.

. Combustion Alkanes on heating in the presence of air or dioxygen are completely oxidized to carbon dioxide and water with the evolution of large amount of heat. è - c è - c CH (g) (g) CO (g) 2H O( ); Ä H 890kJmol ( . ) C H (g) /2O (g) 4CO (g) 5H O( ) Ä H .84kJmol ( .

) → − → = − The general combustion equation for any alkane is : n 2n+ 3n C H O nCO (n )H O   →     ( . ) Due to the evolution of large amount of heat during combustion, alkanes are used as fuels. During incomplete combustion of alkanes with insufficient amount of air or dioxygen, carbon black is formed which is used in the manufacture of ink, printer ink, black pigments and as filters. CH (g) + O (g) incomplete combustion C(s)+2H O( ) ( .

) . Controlled oxidation Alkanes on heating with a regulated supply of dioxygen or air at high pressure and in the presence of suitable catalysts give a variety of oxidation products. (i) 2CH + O Cu/523K/100atm 2CH OH Methanol ( . ) (ii) CH + O Mo O ∆ HCHO + H O Methanal ( .

) (iii) 2CH CH + 3O (CH COO)Mn ∆ 2CH COOH Ethanoic acid + 2H O ( . ) (iv) Ordinarily alkanes resist oxidation but alkanes having tertiary H atom can be oxidized to corresponding alcohols by potassium permanganate. (iCH ) CH KMnO Oxidation (CH ) COH -Methylpropane -Methylpropane- - ( . ) .

Isomerisation n -Alkanes on heating in the presence of anhydrous aluminium chloride and hydrogen chloride gas isomerise to branched chain alkanes. Major products are given below. Some minor products are also possible which you can think over. Minor products are generally not reported in organic reactions.

CH (CH) ) CH Anhy, AICI / HCI n -Hexane CH CH–(CH ) –CH +CH CH –CH–CH –CH -Methylpentane -Methylpenatone ( . ) . Aromatization n -Alkanes having six or more carbon atoms on heating to 773K at - atmospheric pressure in the presence of oxides of vanadium, molybdenum or chromium supported over alumina get dehydrogenated and cyclised to benzene and its homologues. This reaction is known as aromatization or reforming .

( . ) Toluene (C H ) is methyl derivative of benzene. Which alkane do you suggest for preparation of toluene ? .

Reaction with steam Methane reacts with steam at K in the presence of nickel catalyst to form carbon monoxide and dihydrogen. This method is used for industrial preparation of dihydrogen gas CH + H 2IO Ni ∆ CO + 3H ( . ) . Pyrolysis Higher alkanes on heating to higher temperature decompose into lower alkanes, alkenes etc.

Such a decomposition reaction into smaller fragments by the application of heat is called pyrolysis or cracking . ( . ) Pyrolysis of alkanes is believed to be a free radical reaction. Preparation of oil gas or petrol gas from kerosene oil or petrol involves the principle of pyrolysis.

For example, dodecane, a constituent of kerosene oil on heating to 973K in the presence of platinum, palladium or nickel gives a mixture of heptane and pentene. C H Pt/Pd/Ni 973K C H + C H + Other Products Dodecane Heptane Pentene ( . ) . .

Conformations Alkanes contain carbon-carbon sigma ( σ ) bonds. Electron distribution of the sigma molecular orbital is symmetrical around the internuclear axis of the C–C bond which is not disturbed due to rotation about its axis. This permits free rotation about C–C single bond. This rotation results into different spatial arrangements of atoms in space which can change into one another.

Such spatial arrangements of atoms which can be converted into one another by rotation around a C-C single bond are called conformations or conformers or rotamers . Alkanes can thus have infinite number of conformations by rotation around C-C single bonds. However, it may be remembered that rotation around a C-C single bond is not completely free. It is hindered by a small energy barrier of - kJ mol – due to weak repulsive interaction between the adjacent bonds.

Such a type of repulsive interaction is called torsional strain . Conformations of ethane : Ethane molecule (C H ) contains a carbon – carbon single bond with each carbon atom attached to three hydrogen atoms. Considering the ball and stick model of ethane, keep one carbon atom stationary and rotate the other carbon atom around the C-C axis. This rotation results into infinite number of spatial arrangements of hydrogen atoms attached to one carbon atom with respect to the hydrogen atoms attached to the other carbon atom.

These are called conformational isomers (conformers). Thus there are infinite number of conformations of ethane. However, there are two extreme cases. One such conformation in which hydrogen atoms attached to two carbons are as closed together as possible is called eclipsed conformation and the other in which hydrogens are as far apart as possible is known as the staggered conformation.

Any other intermediate conformation is called a skew conformation.It may be remembered that in all the conformations, the bond angles and the bond lengths remain the same. Eclipsed and the staggered conformations can be represented by Sawhorse and Newman projections . . Sawhorse projections In this projection, the molecule is viewed along the molecular axis.

It is then projected on paper by drawing the central C–C bond as a somewhat longer straight line. Upper end of the line is slightly tilted towards right or left hand side. The front carbon is shown at the lower end of the line, whereas the rear carbon is shown at the upper end. Each carbon has three lines attached to it corresponding to three hydrogen atoms.

The lines are inclined at an angle of ° to each other. Sawhorse projections of eclipsed and staggered conformations of ethane are depicted in Fig. . .

. Newman projections In this projection, the molecule is viewed at the C–C bond head on. The carbon atom nearer to the eye is represented by a point. Three hydrogen atoms attached to the front carbon atom are shown by three lines drawn at an angle of ° to each other.

The rear carbon atom (the carbon atom away from the eye) is represented by a circle and the three hydrogen atoms are shown attached to it by the shorter lines drawn at an angle of ° to each other. The Newman’s projections are depicted in Fig. . .

Fig. . Sawhorse projections of ethane Fig. .

Newman’s projections of ethane Fig. . Orbital picture of ethene depicting σ bonds only Relative stability of conformations: As mentioned earlier, in staggered form of ethane, the electron clouds of carbon-hydrogen bonds are as far apart as possible. Thus, there are minimum repulsive forces, minimum energy and maximum stability of the molecule.

On the other hand, when the staggered form changes into the eclipsed form, the electron clouds of the carbon – hydrogen bonds come closer to each other resulting in increase in electron cloud repulsions. To check the increased repulsive forces, molecule will have to possess more energy and thus has lesser stability. As already mentioned, the repulsive interaction between the 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, the staggered form has the least torsional strain and the eclipsed form, the maximum torsional strain . Therefore, staggered conformation is more stable than the eclipsed conformation. Hence, molecule largely remains in staggered conformation or we can say that it is preferred conformation.

Thus it may be inferred that rotation around C–C bond in ethane is not completely free. The energy difference between the two extreme forms is of the order of . kJ mol – , which is very small. Even at ordinary temperatures, the ethane molecule gains thermal or kinetic energy sufficient enough to overcome this energy barrier of .

kJ mol – through intermolecular collisions. Thus, it can be said that rotation about carbon-carbon single bond in ethane is almost free for all practical purposes. It has not been possible to separate and isolate different conformational isomers of ethane. .

Alkenes Alkenes are unsaturated hydrocarbons containing at least one double bond. What should be the general formula of alkenes? If there is one double bond between two carbon atoms in alkenes, they must possess two hydrogen atoms less than alkanes. Hence, general formula for alkenes is C n H 2n .

Alkenes are also known as olefins (oil forming) since the first member, ethylene or ethene (C H ) was found to form an oily liquid on reaction with chlorine. . . Structure of Double Bond Carbon-carbon double bond in alkenes consists of one strong sigma ( σ ) bond (bond enthalpy about kJ mol – ) due to head-on overlapping of sp hybridised orbitals and one weak pi ( π ) bond (bond enthalpy about kJ mol – ) obtained by lateral or sideways overlapping of the two p orbitals of the two carbon atoms.

The double bond is shorter in bond length ( pm) than the C–C single bond ( pm). You have already read that the pi ( π ) bond is a weaker bond due to poor sideways overlapping between the two p orbitals. Thus, the presence of the pi ( π ) bond makes alkenes behave as sources of loosely held mobile electrons. Therefore, alkenes are easily attacked by reagents or compounds which are in search of electrons.

Such reagents are called electrophilic reagents . The presence of weaker π -bond makes alkenes unstable molecules in comparison to alkanes and thus, alkenes can be changed into single bond compounds by combining with the electrophilic reagents. Strength of the double bond (bond enthalpy, kJ mol – ) is greater than that of a carbon-carbon single bond in ethane (bond enthalpy, kJ mol – ). Orbital diagrams of ethene molecule are shown in Figs.

. Nomenclature For nomenclature of alkenes in IUPAC system, the longest chain of carbon atoms containing the double bond is selected. Numbering of the chain is done from the end which is nearer to the double bond. The suffix ‘ ene ’ replaces ‘ ane ’ of alkanes.

It may be remembered that first member of alkene series is : CH (replacing n by in C n H 2n ) known as methene but has a very short life. As already mentioned, first stable member of alkene series is C H known as ethylene (common) or ethene (IUPAC). IUPAC names of a few members of alkenes are given below : Structure IUPAC name CH – CH = CH Propene CH – CH – CH = CH But – l - ene CH – CH = CH–CH But- -ene CH = CH – CH = CH Buta – , - diene CH = C – CH -Methylprop- -ene CH = CH – CH – CH -Methylbut- -ene Problem . Write IUPAC names of the following compounds: (i) (CH ) CH – CH = CH – CH – CH  CH – CH – CH C H (ii) (iii) CH = C (CH CH CH ) (iv) CH CH CH CH CH CH CH – CHCH = C – CH – CHCH (i) , -Dimethyl- , -decadiene; (ii) , , , Octatetraene; (iii) - n -Propylpent- -ene; (iv) -Ethyl- , -dimethyl-dec- -ene; Problem .

Calculate number of sigma ( σ ) and pi ( π ) bonds in the above structures (i-iv). σ bonds : , π bonds : σ bonds : , π bonds : σ bonds : , π bond : σ bonds : , π bond : . . Isomerism Alkenes show both structural isomerism and geometrical isomerism.

Structural isomerism : As in alkanes, ethene (C H ) and propene (C H ) can have only one structure but alkenes higher than propene have different structures. Alkenes possessing C H as molecular formula can be written in the following three ways: I. CH = CH – CH – CH But- -ene (C H ) II. CH – CH = CH – CH But- -ene (C H ) Fig.

. Orbital picture of ethene showing formation of (a) π -bond, (b) π -cloud and (c) bond angles and bond lengths III. CH = C – CH -Methylprop- -ene (C H ) Structures I and III, and II and III are the examples of chain isomerism whereas structures I and II are position isomers . Problem .

Write structures and IUPAC names of different structural isomers of alkenes corresponding to C H . (a) CH = CH – CH – CH – CH Pent- -ene (b) CH – CH=CH – CH – CH Pent- -ene (c) CH – C = CH – CH -Methylbut- -ene (d) CH – CH – CH = CH -Methylbut- -ene (e) CH = C – CH – CH -Methylbut- -ene Geometrical isomerism: Doubly bonded carbon atoms have to satisfy the remaining two valences by joining with two atoms or groups. If the two atoms or groups attached to each carbon atom are different, they can be represented by YX C = C XY like structure. YX C = C XY can be represented in space in the following two ways : In (a), the two identical atoms i.e., both the X or both the Y lie on the same side of the double bond but in (b) the two X or two Y lie across the double bond or on the opposite sides of the double bond.

This results in different geometry of (a) and (b) i.e. disposition of atoms or groups in space in the two arrangements is different. Therefore, they are stereoisomers . They would have the same geometry if atoms or groups around C=C bond can be rotated but rotation around C=C bond is not free.

It is restricted. For understanding this concept, take two pieces of strong cardboards and join them with the help of two nails. Hold one cardboard in your one hand and try to rotate the other. Can you really rotate the other cardboard ?

The answer is no. The rotation is restricted. This illustrates that the restricted rotation of atoms or groups around the doubly bonded carbon atoms gives rise to different geometries of such compounds. The stereoisomers of this type are called geometrical isomers .

The isomer of the type (a), in which two identical atoms or groups lie on the same side of the double bond is called cis isomer and the other isomer of the type (b), in which identical atoms or groups lie on the opposite sides of the double bond is called trans isomer . Thus cis and trans isomers have the same structure but have different configuration (arrangement of atoms or groups in space). Due to different arrangement of atoms or groups in space, these isomers differ in their properties like melting point, boiling point, dipole moment, solubility etc. Geometrical or cis-trans isomers of but- -ene are represented below : Cis form of alkene is found to be more polar than the trans form.

For example, dipole moment of cis -but- -ene is . Debye, whereas, dipole moment of the trans form is almost zero or it can be said that trans -but- -ene is non-polar. This can be understood by drawing geometries of the two forms as given below from which it is clear that in the trans -but- -ene, the two methyl groups are in opposite directions, Threfore, dipole moments of C-CH bonds cancel, thus making the trans form non-polar. (ii) CH = CBr (iii) C H CH = CH – CH (iv) CH CH = CCl CH (iii) and (iv).

In structures (i) and (ii), two identical groups are attached to one of the doubly bonded carbon atom. . . Preparation .

From alkynes: Alkynes on partial reduction with calculated amount of dihydrogen in the presence of palladised charcoal partially deactivated with poisons like sulphur compounds or quinoline give alkenes. Partially deactivated palladised 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.

( . ) ( . ) iii) CH ≡ CH+H Pd/C CH =CH ( . ) Ethyne Ethene iv) CH –C ≡ CH+H Pd/C CH –CH =CH Propyne Propene ( .

) Will propene thus obtained show geometrical isomerism? Think for the reason in support of your answer. . From alkyl halides : Alkyl halides (R-X) on heating with alcoholic potash (potassium hydroxide dissolved in alcohol, say, In the case of solids, it is observed that the trans isomer has higher melting point than the cis form.

Geometrical or cis-trans isomerism is also shown by alkenes of the types XYC = CXZ and XYC = CZW Problem . Draw cis and trans isomers of the following compounds. Also write their IUPAC names : (i) CHCl = CHCl (ii) C H CCH = CCH C H Problem . Which of the following compounds will show cis-trans isomerism?

(i) (CH ) C = CH – C H cis -But- -ene ( µ = .33D) trans -But- -ene ( µ = ) ethanol) eliminate one molecule of halogen acid to form alkenes. This reaction is known as dehydrohalogenation i.e., removal of halogen acid. This is example of β -elimination reaction , since hydrogen atom is eliminated from the β carbon atom (carbon atom next to the carbon to which halogen is attached). ( .

) Na ture of halogen atom and the alkyl group determine rate of the reaction. It is observed that for halogens, the rate is: iodine > bromine > chlorine, while for alkyl groups it is : tertiary > secondary > primary. . From vicinal dihalides : Dihalides in which two halogen atoms are attached to two adjacent carbon atoms are known as vicinal dihalides .

Vicinal dihalides on treatment with zinc metal lose a molecule of ZnX to form an alkene. This reaction is known as dehalogenation . CH Br–CH Br + Zn CH =CH + ZnBr ( . ) CH CHBr–CH Br + Zn CH CH=CH +ZnBr ( .

) . From alcohols by acidic dehydration : You have read during nomenclature of different homologous series in Unit that alcohols are the hydroxy derivatives of alkanes. They are represented by R–OH where, R is C n H 2n+ . Alcohols on heating with concentrated sulphuric acid form alkenes with the elimination of one water molecule.

Since a water molecule is eliminated from the alcohol molecule in the presence of an acid, this reaction is known as acidic dehydration of alcohols . This reaction is also the example of β -elimination reaction since –OH group takes out one hydrogen atom from the β -carbon atom. ( . ) .

. Properties Physical properties Alkenes as a class resemble alkanes in physical properties, except in types of isomerism and difference in polar nature. The first three members are gases, the next fourteen are liquids and the higher ones are solids. Ethene is a colourless gas with a faint sweet smell.

All other alkenes are colourless and odourless, insoluble in water but fairly soluble in non-polar solvents like benzene, petroleum ether. They show a regular increase in boiling point with increase in size i.e., every – CH group added increases boiling point by – K. Like alkanes, straight chain alkenes have higher boiling point than isomeric branched chain compounds. Chemical properties Alkenes are the rich source of loosely held pi ( π ) electrons, due to which they show addition reactions in which the electrophiles add on to the carbon-carbon double bond to form the addition products.

Some reagents also add by free radical mechanism. There are cases when under special conditions, alkenes also undergo free radical substitution reactions. Oxidation and ozonolysis reactions are also quite prominent in alkenes. A brief description of different reactions of alkenes is given below: .

Addition of dihydrogen: Alkenes add up one molecule of dihydrogen gas in the presence of finely divided nickel, palladium or platinum to form alkanes (Section . . ) . Addition of halogens : Halogens like bromine or chlorine add up to alkene to form vicinal dihalides.

However, iodine does not show addition reaction under normal conditions. The reddish orange colour of bromine solution in carbon tetrachloride is discharged when bromine adds up to an unsaturation site. This reaction is used as a test for unsaturation. Addition of halogens to alkenes is an example of electrophilic addition reaction involving cyclic halonium ion formation which you will study in higher classes.

( . ) ( . ) . Addition of hydrogen halides: Hydrogen halides (HCl, HBr,HI) add up to alkenes to form alkyl halides.

The order of reactivity of the hydrogen halides is HI > HBr > HCl. Like addition of halogens to alkenes, addition of hydrogen halides is also an example of electrophilic addition reaction. Let us illustrate this by taking addition of HBr to symmetrical and unsymmetrical alkenes Addition reaction of HBr to symmetrical alkenes Addition reactions of HBr to symmetrical alkenes (similar groups attached to double bond) take place by electrophilic addition mechanism. CH =CH +H–Br CH –CH –Br ( .

) CH –CH=CH–CH +HBr CH –CH–CHCH Br ( . ) Addition reaction of HBr to unsymmetrical alkenes (Markovnikov Rule) How will H – Br add to propene ? The two possible products are I and II. ( .

) Markovnikov, a Russian chemist made a generalisation in after studying such reactions in detail. These generalisations led Markovnikov to frame a rule called Markovnikov rule. The rule states that negative part of the addendum (adding molecule) gets attached to that carbon atom which possesses lesser number of hydrogen atoms. Thus according to this rule, product I i.e., -bromopropane is expected.