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Class XI HYDROCARBONS - ALKANES / ALKENE / ALKYNES/ AROMATICS

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Q. What are Saturated and unsaturated hydrocarbons?imp SATURATED HYDROCARBONS Hydrocarbons containing carbon–carbon single bond are called saturated hydrocarbons.

e.g. alkane and cycloalkanes. ALKANE : General formula is CnH2n+2 (n is a positive whole number for all H/Cs)

CYCLOALKANES: General formula is CnH2n (n is a positive whole number for all H/Cs) UNSATURATED HYDROCARBONS Hydrocarbons containing carbon–carbon double or triple bond are called unsaturated

hydrocarbons. e.g. alkene , alkyne and aromatics. ALKENES: general formula is CnH2n

ALKYNES: general formula is CnH2n–2 AROMATICS: usually contain one or more rings of six C atoms called benzene rings (n is a positive whole number for all H/Cs) C6H2n-6

NATURE OF CARBON CHAIN PRIMARY SUFFIXES GENERIC NAME

Saturated ,C-C -ane Alkane

Unsaturated ,C=C -ene Alkene

Unsaturated ,C≡C -yne Alkyne

ALKANES: General formula: CnH2n+2 simplest organic compounds.

HYBRIDISATION: The process of merging (intermixing) of the orbital’s of similar energies so as to form

new orbital’s called hybrid orbital’s. In order to explain characteristic geometrical shapes of polyatomic molecules

concept of hybridization is used. In alkane hybridisation is sp3 . All the carbon-carbon and C-H bonds in alkanes are strong sigma (σ) bonds.

Every C-H bond length = 1.112A0 and C-C bond has a and 1.54A0 . All bond angles are regular at tetrahedral position of four bonds of carbon, the C-C chain is zig zag and not linear as usually written for sake of simplicity.

Propane Propane (3d)

A stereochemical formula is the full structural formula in terms of the 3D structural or

spatial arrangement of the atoms. Composition of methane molecule: CH4.

Geometry: of methane: Tetrahedral in structure in which carbon is central atom and four H-atoms are surrounding it in three-dimensions.

Bond Angles: H-C-H bond angles are 109.5o.

Bond Length: All C-H bonds are 1.09Ao.

CnH2n+2 n =

1 2 3 4 5 6 7 8 9 10

formula of

alkane

CH4 C2H6 C3H8 C4H10 C5H12 C6H14 C7H16 C8H18 C9H20 C10H22

name of

alkane

methane

ethane propan

e butane

pentane

hexane heptane octane nonane decane


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HYBRIDISATION IN METHANE : sp3

1(2s)+ 3(2px, 2py, 2pz) orbital’s (sp3 hybridisation) 4 (sp3 hybrid orbital’s) of 109028’ alignment forming regular tetrahedron. Each orbital has ¼s (25%) character and ¾ p (75%) character.

6C12 (ground state electronic configuration) 1s22s22px

12py12pz

0

6C*12 (excited state electronic configuration) 1s22s1px12py

12pz1

One e- from s orbital is moved from 2s orbital to 2pz orbital which is in higher energy state. The input energy for exciting an electron is available from lattice energy of formation, released when bond are formed. Initially energy has to give so as to start the

reaction. In the excited state one 2s orbital of the C* and three 2p orbitals (Px, Py, Pz) of C*

intermix to form four hybrid orbitals of sp3 hybridisation. These four hybridised orbitals aligned themselves in such a way that bond angle must be 109028’ so as to have

minimum energy and maximum stability. Orbital structure is shown in diagram. These 04 sp3 orbitals overlaps with 1s1 of four H orbitals forming sigma bonding by end to end overlap.

GEOMETRY/ SHAPE OF ETHANE:

TYPES OF CARBON ATOMS : (a) PRIMARY CARBON or 10 carbon atom

Carbon atom attached to one or no carbon atoms. CH3C(CH3)2CH(CH3)CH2CH3

Sp3


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(b) SECONDARY CARBON or 20 carbon atom 10 40 3

0 20 10 Carbon atom attached to two carbon atoms. Also known as n. (c) TERTIARY CARBON or 30 carbon atom

Carbon atom attached to three other carbon atoms. Also known as iso. (d) QUARTERNARY CARBON or 40 carbon atom.

Carbon atom attached to four other carbon atoms. Also known as neo. H atoms attached to primary, secondary, tertiary or quaternary carbon atoms are correspondingly names as primary, secondary, tertiary and quarternary hydrogen atoms.

Q. Mention primary, secondary and tertiary carbons and hydrogens in the following

Compound? 10

Ans 10 20 40 10 10

H atoms attached to primary, secondary, tertiary or quaternary carbon atoms are

correspondingly names as primary, secondary, tertiary and quarternary hydrogen atoms.

Q.In the alkane H3C – CH2 – C(CH3)2 – CH2 – CH(CH3)2, identify 1°,2°,3° carbon atoms

and give the number of H atoms bonded to each one of these. STRUCTURAL ISOMERISM:

1. CHAIN ISOMERISM: (NUCLEAR OR SKELTON ISOMERISM)

Compounds having same molecular formula but different arrangement of carbon chains (skeletons), within the molecule are called chain isomers and the phenomenon is termed as chain isomerism Example: Chain isomers of : C5H12

CHAIN ISOMERS OF THE ALKANE MOLECULAR FORMULA C5H12

(1) , , pentane, volatile colourless liquid, bpt

34oC, linear.

(2) , , methylbutane (2-methylbutane, but 2- not needed), volatile colourless liquid/gas, bpt 28oC, minimum branching.

(3), , , 2,2-dimethylpropane, colourless gas, bpt 9.5oC, maximum branching.


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2. RING CHAIN ISOMERISM : Due to difference in mode of linkage in

C atoms and the isomers may have either open chain or closed structure. e.g. C3H6 CH3-CH=CH2

3. POSITIONAL ISOMERS OF MOLECULAR FORMULA C3H7Br

(1) 1-bromopropane, , , bpt 71oC, primary

halogenoalkane,

(2) 2-bromopropane, , , bpt 59oC, secondary

halogenoalkane.

4. CONFORMATIONS IN ALKANE: Conformations or conformers or rotamers or conformational isomers:

The different arrangements (spatial) of atoms of a molecule which can be obtained by rotation around a C-C single bond are called conformations or conformers or rotamers.

CONFORMATIONAL ISOMERES: The isomers which differs in the conformation are known as conformational isomers.

REPRESENTATION OF CONFORMATIONS:

(a)Sawhorse projection representation (b) Newman projection

(a)Sawhorse projection representation In this representation molecule is viewed along the axis of the axis at an angle. C-C atom is a straight line. Front Carbon is shown as the lower left hand carbon while rear

carbon is shown as upper right hand carbon. Three C-H bonds are shown by the three lines on the each lower and upper carbon.

Sawhorse projection Newman projection formulae:

Two C-C sigma bonded atoms are viewed from the front along the C-C bond axis so that

only front C is seen. front C is seen as junction of 3C-H bond while rear Carbon is shown as circle as shown in figure above.


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CONFORMATIONS OF ETHANE: CH3- CH3 Imagine that one of the methyl group is rotated along the C-C axis keeping the rest molecule

undisturbed. An infinite no. of possible arrangements of the rotated methyl group w.r.t

undisturbed methyl group are possible, each of these possible arrangement represents a conformations. However for the sake of

convenience each rotation is done in the instalment of 600. Thus we will obtain six no. of different

conformations of ethane. Out of these six, extreme two are worth to study. Because they are different from each other while other four are similar to each other but of course different from the extreme two ends.

(a) One extreme conformations will be such in which the rear methyl group is completely

eclipsed by the front methyl group and thus only the front methyl group and thus only the front methyl group i.e. three H atoms of the methyl group nearer to the eye are visible. Such conformations is known as eclipsed conformations.

Conformations

Projections

Eclipsed Staggered Skew

Conformation in which hydrogen atoms attached

to two carbons are as closed together as

possible

Conformation in which hydrogen atoms are as far

apart as possible

Any other intermediate

conformation in between

eclipsed and sawhorse

Sawhorse projection

Newman projection

ECLIPSED CONFORMATIONS:

In this conformation, the H atom of one carbon are directly behind the other. consequently the repulsion in these atoms is maximum. In the Newmann projection formula of the eclipsed form, the H atoms are crowded i.e.

the two H atoms come close to each other.


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(b) Another extreme conformations will be such in which the rear methyl group has been rotated upside down and thus both the methyl group i.e. all the six H atoms are visible and are as far apart from each other as possible. such conformations is known as

staggered. STAGGERED CONFORMATIONS: In this conformation, H atoms of the two C atoms are

staggered wrt to one another. As they are at maximum distance apart and have minimum repulsion between them. Both the conformations are represented as below: SKEW CONFORMATIONS:

(c) The infinite no. of possible intermediate conformations between the two extreme conformations between two extreme conformations are referred to as skew

conformations. In these conformations, H atoms are close than in staggered but away than in eclipsed conformations. STABILITY OF CONFORMATIONS:

Staggered conformations in which H atoms are as far as apart as possible is thus most stable while the eclipsed conformations in which H atoms are

perfectly eclipsed is the least stable. Stability of skew conformations lie in between these are in the following order:

Staggered > Skew > Eclipsed

TORSIONAL STRAIN: weak repulsive interactions between the bonds or electron pair of bonds on adjacent C atoms. More is the torsional strain lesser is the stability.

TORSIONAL BARRIER: The difference in the energy between staggered and eclipsed conformations which is equal to 12.5KJ/mol. The energy difference is very small.

Therefore two conformations of ethane goes on changing from one form to another and consequently it is not possible to isolate the different conformations of ethane.

Q. Eclipsed conformation is less stable than staggered conformation of ethane. Explain? Ans. It is due to repulsive interactions (non bonded) between bonded pair of electrons,

i.e. electron pair which form six C-H bonds in ethane. In the staggered conformations, the electron clouds of six C-H bonds are as far as possible with the result there is minimum repulsive interactions between these electron clouds. Hence this conformations

is quite stable. On the other hand skew and eclipsed conformations, the electron clouds starts coming closer. This causes repulsive interactions which is maximum in the

eclipsed. Therefore potential energy are of order: Eclipsed > Skew > Staggered Stability : Staggered > Skew > Eclipsed


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TORSIONAL STRAIN: the repulsive interaction between the clouds which affects the solubility of a conformation is called torsional strain.

Staggered conformations has the lease torsional strain while eclipsed having maximum. Due to torsional strain, certain energy called torsional energy, is required to allow the

rotation around the C-C single bond. in other words ethane molecule having staggered conformation will have to come cross an energy barrier equivalent to the torsional energy for being converted into eclipsed conformations.

Energy difference in staggered and eclipsed is found to be 2.8Kcal/ mol which constitute the energy barrier to rotate about C-C bond i.e.for conversion of staggered to eclipsed.

This energy barrier is too small for either form to remain stable. i.e. two forms are interconvertible.

This implies that rotation about the C-C single bond in ethane is almost free for all practical purposes, and it is not possible to separate the different

conformations of ethane the mostly in staggered form. CONFORMATION IN CYCLO ALKANES:

CHAIR CONFROMATION IS MOST STABLE: Reason: (a) Adjacent H on Carbon is staggered and more distance between them,

resulting lesser repulsions and lesser energy thus more stability. (b) Energy of boat conformation is 44KJ/mol more than that of chair conformation. Therefore cyclohexane mainly exists as chair form.

Q. name two extreme type of conformationsof ethane?

Ans. stagerred and eclipsed

Q. Out of chair and boat conformations of cyclohexane? Which is more stable? Ans. chair form is more stable because of minimum repulsions and lesser energy.


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METHODS OF PREPARATION OF ALKANES:

FROM ALKENE AND ALKYNE ALKANE

A. SABATIER AND SANDEREN’S REACTION: HYDROGENATION REACTION It involves catalytic (Ni or platinum) hydrogenation of unsaturated hydrocarbons like

alkene and alkyne to alkane. Methane can’t be obtained by this method. It is a type of addition reaction.

RCH=CHR + H2 Ni 200-3000C RCH2CH2R Alkene Alkane CH2=CH2 + H2 Ni 200-3000C CH3-CH3

Ethene Pt Ethane

CHCH + 2H2 Ni 2500C or 523K CH3-CH3 Ethyne Pt Ethane

FROM ALKYL HALIDE ALKANE B. WURTZ REACTION. This reaction involves the treatment of an alkyl halide (RX) with

sodium metal in the presence of dry ether to form higher symmetrical alkane. This reaction can be used for ascending of series. For example,

R-X+2Na +X-R Dry ether R-R + 2NaX Alkyl halide Alkane

C2H5-Br + 2Na + Br-C2H5 dry ether C2H5-C2H5 + 2NaBr Ethyl bromide Butane

CH3-I + 2Na + I-CH3 dry ether CH3-CH3 + 2NaBr

Ethyl bromide Ethane LIMITATIONS OF THE REACTION: (i) Methane cannot be prepared by this reaction.

(ii) This reaction fails with 30 alkyl halides. (iii) This is not a suitable reaction to prepare higher unsymmetrical alkanes, as by-

products are also formed during the reaction and thus their separation becomes difficult.

Q. Name the products formed when methyl bromide and ethyl bromide is treated with sodium metal in ether? Ans. ethane+ propane +butane


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Q. Give mechanism of Wurtz reaction. Ans. Mechanism : (A) INTERMEDIATE FORMATION OF AN ORGANOMETALLIC COMPOUND:

CH3CH2Br + 2Na CH3CH2-Na+ + NaBr

CH3CH2-Na+ + CH3CH2Br CH3CH2CH2CH3 + NaBr

(B) INTERMEDIATE FORMATION AS FREE RADICALS. CH3CH2Br + 2Na CH3CH2

. + NaBr

CH3CH2. + CH3CH2

. CH3CH2CH2CH3 Butane

mechanism may be ionic or free radical mechanism but normally free radical is preferred. If Mechanism is free radical reaction it involves following steps: (a)Chain initiation (b)Chain propagation (c)Chain termination step.

C. FRANKLAND REACTION:

2R-X + Zn R-R + ZnX2 Alkyl halide Alkane

CH3Br+ CH3Br + Zn ZnBr2 + CH3-CH3

Bromo methane Ethane

C2H5I+ CH3I + Zn ZnI2 + CH3-CH3

Iodo ethane Ethane

D. REDUCTION REACTIONS:

Common reducing agent: LiAlH4 (anhyd ether), Na/ alcohol, NaBH4 (Protic solvent, ROH, H2O), H2 PtO2+ heat. Or Ni + H2 (g), Zn/ HCl , HI, Red Phosphorus, P+HI Other Reducing agent:

Zn+ CH3COOH , Zn+ NaOH, Zn Cu couple + alcohol, aluminium amalgam and alcohol, :LiAlH4. ,

(I)REDUCTION OF ALKYL HALIDES ALKANE RX + [H] Red P/ 423K RH + HX

Alkyl halide Pressure Alkane Halogen acid

C2H5Br + 2[H] Red P 423K C2H6 + HBr C2H5I + 2[H] Red P/HI 423K C2H6 + I2

Function of red P is to remove iodine formed in the reaction by forming PI3 which is

volatile , otherwise alkane will react back to give alkyl halide. C2H5Br + 2[H] LiAlH4 C2H6 + HBr C2H5Br + 2[H] Zn/ HCl C2H6 + HBr

(II) REDUCTION OF ALCOHOLS:

ROH + 2HI Red P RH + H2O + I2 C2H5OH + 2HI Red P C2H6 + H2O + I2

(III) REDUCTION OF ALDEHYDE, KETONE, ACIDS

RCHO + 4HI Red P RCH3 + H2O + 2I2 CH3CHO + 4HI Red P CH3CH3 + H2O + 2I2

Ethanal ethane RCOR + 4HI Red P RCH2CH3 + H2O + 2I2

Ketone CH3COCH3 + 4HI Red P CH3CH2CH3+ H2O + 2I2

Propanone Propane


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RCOOH + 6HI Red P RCH3 + 2H2O + 3I2 Carboxylic acid Alkane CH3COOH + 4HI Red P CH3CH3+ 2H2O + 2I2

Propanoic Ethane

(IV) CLEMMENSEN REDUCTION: Carbonyl group of aldehydes and ketones is reduced to CH2 group on treatment with zinc

amalgam and concentrated hydrochloric acid

RCHO + 4[H] Zn-Hg, HCl RCH3 + H2O

Aldehyde Alkane CH3CHO + 4[H] Zn-Hg, HCl CH3CH3 + H2O

Ethanal Ethane CH3COCH3 + 4[H] Zn-Hg, HCl CH3-CH2-CH3 + H2O

Propanone Propane C6H5COCH3 + 4[H] Zn-Hg, HCl C6H5-C2H5 + H2O Acetophenone Ethyl benzene

Clemmensen Reduction is widely used when carbonyl compounds are sensitive to alkalies.

(V) WOLFF-KISHNER REDUCTION:

Carbonyl group of aldehydes and ketones is reduced to CH2 group on treatment with hydrazine followed by heating with sodium or potassium hydroxide in high boiling solvent such as ethylene glycol

RCHO + H2NNH2 R-CH=N-NH2 + KOH, glycol R-CH3 + N2 - H2O Hydrazone 453-473K

CH3CHO + H2NNH2 + KOH, glycol CH3-CH3 + N2 -H2O 453-473K

Aldehyde alkane CH3COCH3 + H2NNH2 + KOH, glycol CH3-CH2-CH3 + N2 -H2O 453-473K

Ketone alkane C6H5COCH3 + H2NNH2 + KOH, glycol C6H5-CH3 + N2

Benzophenone -H2O, 453-473K toluene Wolf kishner reduction is used when carbonyl compounds are sensitive to acids.

Q. How will you convert acetaldehyde to ethane and acetone to propane? Ans Wolf kishner reduction CH3CHO + H2NNH2 + KOH, glycol CH3-CH3 + N2 -H2O 453-473K

CH3COCH3 + H2NNH2 + KOH, glycol CH3-CH2-CH3 + N2 -H2O 453-473K

Or Clemmensen Reduction CH3CHO + 4[H] Zn-Hg, HCl CH3CH3 + H2O

Ethanal Ethane CH3COCH3 + 4[H] Zn-Hg, HCl CH3-CH2-CH3 + H2O

Propanone Propane


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E. DECARBOXYLATION REACTION OR SODA LIME REACTION : RCOONa + NaOH Na2CO3 + RH

Sodium salt of carboxylic acid sodium carbonate Alkane

CH3COONa + NaOH CH4 + Na2CO3 Sod. Salt of ethanoic acid Methane

Net result of this reaction is the removal of CO2 in the form of sodium carbonate from a carboxylic acid, therefore reaction is known as de-carboxylation reaction.

This reaction is used in the descending of series.

Q. sodium salt of which acid will be needed for the preparation of propane? Write chemical equation for the reaction? Ans. Butanoic acid reaction do yourself

F. KOLBE’S ELECTROLYTIC METHOD:

Alkali metal salts of carboxylic acids undergo decarboxylation on electrolysis of their aqueous solutions and form hydrocarbons having twice the number of carbon atoms present in the alkyl group of the acid.

Electrolysis of sodium salt of carboxylic acids

2CH3COO-Na+ + 2H2O CH3-CH3 +2CO2 + 2NaOH + H2 Sod salt of ethanoic acid Ethane (Anode ) at cathode

CH3COO-Na+ CH3COO- + Na+

H2O H+ + OH-

Remember: (i) methane can’t be prepared by this method. (ii) electrolysis of HCOONa produces only H2 and CO2.

(iii) unsymeetrical alkane, i.e. alkane having odd no. of C atoms cannot be prepared by Kolbe method.

CH3-CH3 symmetrical Ethane CH3CH2CH3 unsymmetrical propane

G. FROM GRIGNARD REAGENT (RMgX) RH ALKANE. RMgX + H2O dry ether RH + MgX(OH)

Alkyl halide alkane CH3MgBr + H2O dry ether CH4 + MgBr(OH) Methyl magnesiumbromide Methane

R’OH + RMgX R’H + ROMgX (alcohol to alkane) Alcohol


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CH3OH + CH3MgBr dry ether CH4 + CH3MgBr(OH) Methanol methane

Q. How are alkanes prepared by Grignard's reagent? CH3OH + CH3MgBr dry ether CH4 + CH3MgBr(OH)

Methanol methane H SPECIAL METHOD FOR PREPARATION OF ALKANE:

Al4C3+ 12H2O 4Al(OH)3 + 3CH4 ↑

Aluminium carbide methane C+ 2H2 electric arc 1473K CH4

I. COREY HOUSE REACTION:

RBr + Li ether RLi+ LiBr 2RLi + CuI Li(R)2Cu +LiI

Li(R)2-Cu+ R’Br R-R’ + RCu+ LiBr (main reaction)

USES OF ALKANES:

(a) Lower alkanes upto C4 like LPG is used as domestic fuel. (b) Alkanes are used for the manufacturing of chemicals like CH3OH, CH3Cl, CCl4 etc. (c) In the manufacturing of fuel like gasoline which is a mixture of alkane alkene and

aromatic hydrocarbons. Higher is the quantity of branched chain hydrocarbon better is the octane No, better is the fuel having anti-knocking property of fuel.

(d) Methane is used in the manufacturing of carbon black used in black paints, printer’s ink and as a filler in the rubber industry.

PHYSICAL PROPERTIES OF ALKANES: A. PHYSICAL STATE: C1 to C4 are colourless and odourless gases,

C5 to C17 are liquids and C> C18 are colourless solids at 298 K.

Reason: Hydrocarbons are nonpolar because of the covalent nature of C-C and C-H bonds. Thus, they have weak van der Waals forces.

B. BOILING POINTS: BRANCHED CHAIN HAS LOWER BP THAN STRAIGHT CHAIN.

Alkanes have low boiling points because they have weak Vander Waal’s forces. Higher alkanes with higher molecular masses have high boiling points than lower alkanes with

low molecular masses. Vander Waal’s forces surface area This is because magnitude of Vander Waal’s forces increases with increase in molecular

mass or increase in surface area. Q. Boiling points of isomeric alkanes goes on decreasing with increased branching. Why?

Ans Boiling points of branched chain alkanes are lower than the corresponding straight chain alkanes. This is because with branching the molecules become more compact and

hence the surface area decreases. E.g. n-pentane has higher bp than iso pentane. Alkane CH3CH2CH2CH2CH3 (CH3)2CHCH2CH3 (CH3)4C n-Pentane iso- Pentane neo- Pentane

Boiling point: 360C 280C 9.50C


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Q. n-pentane has higher boiling point than neo pentane. Explain. Ans. Boiling points of branched chain alkanes are lower than the corresponding straight chain alkanes. This is because with branching the molecules become more compact and

hence the surface area decreases. Therefore n-pentane has higher bp than neo-pentane.

Q. Which of the following has higher bp? (a) 2-Methylpentane (b) 2,3-Dimethylbutane (c) 2,2-Dimethylbutane Reason: least branching

C. MELTING POINTS:

Q. Alkanes with even no. of carbon atoms have high melting point as compare to alkanes with odd no. of carbon atoms why? Ans. The alkanes with even number of carbon atoms (i.e. butane, ethane ) have higher melting points as compared to the immediately next lower alkane with odd number of carbon atoms. This is because alkanes with even number of carbon atoms have

symmetrical structure and hence it results in closer packing in the crystal structure. C5H12 MP = 143.3K C6H14 MP 178.5K

D. SOLUBILITY: Alkanes are soluble in nonpolar solvents like benzene, toluene, acetone etc and are

insoluble in polar solvents like water since alkanes are nonpolar molecules. Like dissolves like principle: Polar dissolves in polar solvents and non polar in non

polar solvents. CHEMICAL PROPERTIES OF ALKANES:

Alkanes are not very reactive molecules. Most reactions require some energy input to initiate a reaction e.g. high temperature and catalyst for cracking, uv light for

chlorination or a spark to ignite them (initiating free radical reactions). Q. Why are alkanes stable as compared to alkene and alkyne? Ans. Alkanes are extremely stable and un-reactive substances because:

(a) small difference in Electronegativity between C and H. Therefore C-H bond is almost nonpolar and C-C bond is completely non polar. Attack of

nucleohile and electrophile attack on the alkane finds no site for attack. Therefore no reactive as compared to alkene and alkyne.

(b) Bond Strength C-C and C-H bonds are strong bonds: Alkanes having these two types of bonds are unaffected by acids, alkali and oxidising agents under normal conditions.

The single covalent C-C (bond enthalpy 348 kJ mol-1) and C-H (bond enthalpy 412 kJ mol-1) bonds are very strong so bond fission does not readily happen. The carbon

atom radius is small, giving a short and strong bond with other small atoms. Therefore the reactions will tend to have high activation energies resulting in slow/no reaction.

CHEMICAL REACTIONS OF ALKANES:

1. SUBSTITUTION 2. DECOMPOSITION

3. ISOMERISATION 4. OXIDATION


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CHEMICAL REACTIONS OF ALKANE:

SUBSTITUTION REACTIONS:

1. HALOGENATION: (IMP.)


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(a) CHLORINATION (B) BROMINATION (C) IODINATION It is a type of substitution reaction where H atom of alkane is substituted with Cl

atom. This reaction takes place in the presence of sunlight (hv) at a temperature of 250-4000C (673K). CH4+ Cl2 SUNLIGHT (hν) 673K CH3Cl + HCl CH3Cl+ HCl (hν) 673K CH2Cl2

CH2Cl2+ HCl (hν) 673K CHCl3 + HCl CHCl3+ HCl (hν) 673K CCl4 + HCl

In the uv light catalysed reaction of bromine and propane gases, the free radical substitution reaction can produce two initial mono-substitution products.

CH3CH2CH3 + Br2 {CH3CH2CH2Br or CH3CHBrCH3} + HBr

Relative reactivity: F2 > Cl2 > Br2 > I2

Relative reactivity of 10 20 30 H replacement by halogen: Tertiary H (30) > secondary (20) > primary (10)>CH4

LIMITATIONS: It is difficult to stop halogenations at CH3Cl. when equimolar amount of CH4 and

Cl2 is used, it gives mixture of four products viz CH3Cl, CH2Cl2, CHCl3, CCl4. MECHANISM OF CHLORINATION REACTION (FOR INFORMATION)

(i) Free Radical Substitutions e.g. alkanes undergo free radical substitution reactions CH4+ Cl2 light , hv CCl4+ HCl radicals are chlorine free radicals (Cl.)

Methane Carbon Tetrachloride MECHANISM STEPS:

(a) Chain initiation (requires energy- endothermic reaction)

Cl-Cl hv or 2500C 2Cl. radical formation Reaction with hv= photolysis, heat (Δ) thermolysis C6H5COOOC6H5 dark 2C6H5COO.

C6H5COO. C6H5. + CO2 C6H5. + 2Cl. C6H5Cl + Cl.

(b) chain propagation steps: (release energy- exothermic reaction) Cl. + H- CH3 HCl + .CH3

.CH3 + Cl-Cl CH3-Cl + .Cl (c)Chain termination step.

.Cl + .Cl Cl-Cl

.CH3+ .Cl CH3Cl

.CH3+ .CH3 CH3-CH3

Ease of substitution at various carbon atoms is of the order:

Tertiary > secondary > primary 2. NITRATION OF ALKANE BY HNO3: It is a type of substitution reaction where H atom of alkane is substituted with

–NO2. Temperature: 400-5000C (vapour phase nitration) CH3CH2CH3 + HNO3 4500C CH3CH2CH2NO2 + H2O

Propane Nitropropane

3. SULPHONATION OF ALKANE H2SO4: It is a type of substitution reaction where H atom of alkane is substituted with –SO3H Temperature: 5000C Sulphonation is possible with C>6.

C6H14 + HO-SO3H 5000C C6H13SO3H + H2O Hexane n-Hexanesulphonic acid Q. Give mechanism of sulphonation of alkanes?

Ans. Mechanism: free radical substitution (a) Chain initiation (requires energy- endothermic reaction)


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HO-SO3H 675 HO. +.SO3H

(b) chain propagation steps: (release energy- exothermic reaction) R-H +HO. R. + H2O

(c)Chain termination step. .R +.SO3H R-SO3H alkane sulphonic acid

DECOMPOSITION REACTIONS: 4A. THERMAL DECOMPOSITION OR PYROLYSIS A decomposition reaction in which higher alkanes on heating to higher temperature decompose into lower alkanes, alkenes etc. called pyrolysis

4B. CATALYTIC DECOMPOSITION OR CRACKING:

A decomposition reaction in presence of catalyst Ni, Pt etc higher alkanes are decomposed into lower alkanes, alkenes etc at high temperature.

It is used in the preparation of petrol (gasoline), petrol gas, oil gas etc. 5. ISOMERISATION: Process of conversion of one isomer of a compound to another isomer is known as

isomerisation. Reaction takes place in the presence of lewis acid like AlCl3 and at temperature of 473K

and pressure of 35atm. CH3

CH3CH2CH2CH3 Anhd AlCl3 / HCl CH3-CH-CH3

n-Butane 2-Methyl Propane or

Isomerisation is widely used to increase the octane rating of gasoline. OCTANE RATING:

Gasolines are rated on a scale known as octane rating, which is based on the way they

burn in an engine. The higher the octane rating, the greater the percentage of complex-structured hydrocarbons that are present in the mixture, the more uniformly the gasoline burns, and the less knocking there is in the automobile

engine. Thus, a gasoline rated 92 octane will burn more smoothly than one rated 87 octane.

Knocking: objectionable metallic sound produced during the working of internal combustion engine.

Knocking order :

Straight chain paraffins > branched chain paraffins > Olefins> Aromatic hydrocarbons. Knocking may be prevented by using tetraethyl lead (TEL) (C2H5)4Pb in small proportion 0.5%. Such petrol is known as leaded petrol or ethyl petrol.


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6A. AROMATIZATION OR CATALYTIC REFORMING: The conversion of aliphatic compound into aromatic compound in the presence of catalyst is known as aromatisation. E.g.

6B. REFORMING: The conversion of aliphatic compound into aromatic compound in the presence of H as hydro forming and in the presence of platinum is known as

Platforming. Cracking, isomerisation and aromatisation are widely used to increase increase the octane rating of gasoline.

7. OXIDATION REACTIONS:

7A. COMPLETE COMBUSTION REACTIONS: (Exothermic reactions) Complete combustion: CH4+ 2O2 CO2+ H2O +heat (212KCal/mol)+light

C2H6+ 7O2 4CO2+ 6H2O +heat (745KCal/mol)+light

2C4H10+ 13O2 8CO2+ 10H2O +heat +light

General reaction: CnH2n+2 + (3n+1) O2 nCO2+ (n+1) H2O +heat + light

2 7B. INCOMPLETE COMBUSTION OR PARTIAL COMBUSTION:

2CH4+ 3O2 2CO+ 4H2O +lesser heat +light 2C4H10+ 9O2 8CO+ 10H2O +heat +light

It burns with a black soot. 7C.CONTROLLED OXIDATION:

Methane to methanol (a) CH4+ O2 Cu tube CH3OH

9 : 1 100atm, 2000C Methane to methanal

(b) CH4+ O2 Mo2O3 , 2750C HCHO + H2O METHANE METHANAL

Methane to methanoic acid (c) 2CH4+ 3O2 Ag2O HCOOH + 2H2O

Methane methanoic acid Action of steam: (d) CH4 + H2O (stm) CO + 3H2

Q. Complete the reactions: (i) CH3CH3 + HNO3 vapor phase

(ii) CH4 +O2 Cu tube (iii) C2H5COONa + NaOH CaO

Ans(1) nitroethane (2) CH3OH (3) C2H6 + Na2CO3 Q. Write structures of different chain isomers of alkanes corresponding to the molecular

formula C6H14. Also write their IUPAC names? Q. Write chemical equations for combustion reaction of the following hydrocarbons: (i) Butane (II) HEXANE?

Q. In the alkane H3C – CH2 – C(CH3)2 – CH2 – CH(CH3)2, identify 1°,2°,3° carbon atoms and give the number of H atoms bonded to each one of these?

Q. What effect does branching of an alkane chain has on its boiling point? Ans. branching decreases bp.


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Q. Why is Wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms? Illustrate your answer by taking one example? Q. what is difference between conformers and isomers?

Ans. isomers cannot be changed into one another while conformations can be changed into one another.

Q.bp of 2,3-dimethylbutane is greater than 2,2-dimethylnutane? T/F Q. staggered conformations of ethane are mopre stable than eclipsed? T/F Q. Cyclohexane mainly occurs in boat conformaitions ?T/F : chair conformations

Q. sodiumpropanate on heating with soda lime gives ethane/ propane? Ans ethane Q. how do you account for the formation of ethane during chlorination of methane?

Ans. CH4+ Cl2 SUNLIGHT (hν) 673K CH3Cl + HCl (a) Chain initiation (requires energy- endothermic reaction)

Cl-Cl hv or 2500C homolytic fission 2Cl. radical formation (b) chain propagation steps: (release energy- exothermic reaction) Cl. + H- CH3 HCl + .CH3 .CH3 + Cl-Cl CH3-Cl + .Cl

(c)Chain termination step. .Cl + .Cl Cl-Cl and .CH3+ .Cl CH3Cl and .CH3+ .CH3 CH3-CH3

Ease of substitution at various carbon atoms is of the order: Tertiary > secondary > primary.

ALKENE:

Gen formula: CnH2n Hybridisation of C is sp2 Also known as olefins i.e. oil forming. Bond angle is almost 121.70 , C=C 1sigma and 1pi bond

σ bond is strong than dissociation energy 95Kcal i.e. strong

bond is weak dissociation energy is 68Kcal i.e. weak Therefore C-C in ethene are stronger than C-C in ethane because bond length in ethene

is less (1.34A0) than the C-C distance in ethane (1.54A0) Smaller is the bond length, more is the dissociation energy. Ethylene (ethene): It is used as a general anaesthetic

and for making mustard gas. Isobutene is used for production of iso-octane , a high octane fuel used for cars and aeroplanes.

HYBRIDISATION IN ETHENE Molecular formula = C2H4

Empirical formula = CH2 Molecular mass C2H4= 28 Empirical mass CH2= 14

Homologous series = alkene CnH2n

HYBRIDISATION OF ETHENE: sp2 1(2s)+ 2(2px, 2py, ) orbital’s (sp2 hybridisation) 3 (sp2 hybrid orbital’s) of 1200

6C12 (ground state E.configuration):1s22s2px12py12pz0

6C*12 (excited state electronic configuration) 1s22s1 2px12py12pz1

alignment forming trigonal planar arrangement .

Each orbital has 1/3 (33%) s character and 2/3 (67% ) p character. On un-hybridised orbital

2pz is oriented in a plane at right angle to the Plane containing three hybridised orbital’s. one e- from s orbital is moved from 2s orbital to

2p orbital which is in higher energy state. The input energy for exciting an electron is available from

lattice energy of formation, released when bond are formed. Initially energy has to give so as to start the reaction. In the excited state one 2s orbital and two (2px, 2py) orbitals of C* intermix to form 3 hybrid orbitals of sp2 hybridisation. These three sp2 hybridised


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orbitals aligned themselves in such a way that bond angle must be 1200 so as to have minimum energy and maximum stability. Orbital structure is trigonal planar. Note: 01 orbital of 2pz remains unhybridized.

When molecule of ethene is formed, three sp2 hybridised orbitals interacts with other sp2 hybrid orbital of ethene forming 1 sigma overlapping between each C of Ethene. 2 each

sp2 hybrid orbitals forms 4 sigma bond by overlapping with 1s1 orbital of 4 H atoms in end to end manner. One 2pz orbital of one C atom overlap sidewise forming pi bond with other C atom of Ethene. Therefore Bonds between C-C atom is double in nature forming

1 sigma and 1 pi (π). Others 4 C-H sigma (σ) bonds C=C bond length = 134pm, C- H bond length = 108pm,

H-C-H bond angle 117.60, H-C-C bond angle = 1210 . above orbital structure clear all the overlapping:

PHYSICAL PROPERTIES OF ALKENES: a. Physical state: The first three members are gases, the next fourteen are liquids and

the higher ones are solids. C1 to C4 are colourless and odourless gases except ethylene which has pleasant smell. C5 to C18 are liquids and C> C18 are solids. Reason: Hydrocarbons are nonpolar because of the covalent nature of C-C and C-H

bonds. Thus, they have weak van der Waals forces of attraction. Q. Why branched chain alkene is having lower bp?

b. Boiling points: The boiling points of alkenes increase with the increase in molecular

mass or no. of Carbon atoms. Boiling points of branched chain alkenes are lower than the corresponding straight chain alkenes. This is because with branching the molecules become more compact and hence

the surface area decreases. Vanderwaal forces surface area Vander Waals forces increases with increase in molecular mass or increase in surface area.

Note: Cis alkene bp will be high due to more polarity and trans alkene is less polar due to low polarity.

Trans alkene Mp will be high due to more symmetrical structure than cis alkene which is less symmetrical structure. Cis trans isomer is discussed in detail.

c. Solubility: Alkenes are soluble in nonpolar solvents like benzene CCl4 and toluene and are insoluble in polar solvents like water.

Alkene are generally non polar but certain alkene are weakly polar due to unsymmetry geometry. e.g. propene and 1-butene have dipole moment of 0.35D.

ISOMERSISM: same molecular formula but different arrangement of atoms. A. STRUCTURAL ISOMERISM:

(I) CHAIN ISOMERISM: CH3-CH2-CH=CH2 CH3-C=CH2

But-1-ene 2-Methylpropene CH3

(2) POSITION ISOMERISM: The isomerism differs in the position of the double bonds. E.g.

Two linear positional isomers of molecular formula C5H10

(1) , pent-1-ene, bpt 30oC,

(2) , pent-2-ene, cis bpt 37oC, trans bpt 36oC,


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Q.Draw different structural isomers of alkenes corresponding to molecular formula C5H10. B. GEOMETRICAL ISOMERS OF ALKENES:

Isomerism due to the difference in spatial arrangements of groups about the doubly bonded carbon atoms is known as geometrical isomerism.

Geometrical isomerism in alkenes is due to restricted rotation around carbon-carbon double bond. Example: Geometrical isomers C4H8

Trans isomer is nonpolar and has zero dipole moment where as cis isomer is polar. cis isomer has higher boiling points than trans isomer because it is polar. In case of solids, cis isomer has a lower melting point than trans isomer. This is because

trans isomer is symmetrical and fits well into the crystal lattice. Cis isomer is having less symmetry, symmetry leads to stability and thus high

Mpt. Due to poor symmetry cis molecules do not fit into crystal lattice so well.

Therefore trans isomer has higher mp as they are more symmetrical. e.g. H3C-C-H H3C-C-H ║ ║

H3C-C-H HC-C-CH3 Cis- But-2-ene Trans - But-2-ene

Dipole moment = μ > 0 Dipole moment = μ = 0 Bpt 40C (high) Bpt 10 C (low) Mpt -1390C (low) Mpt -1060C (high)

H-C-COOH H-C-COOH H3C-C-COOH H3C-C-COOH ║ ║ ║ ║ H-C-COOH HOOC-C-H H-C-COOH HOOC-C-H Maleic acid Fumaric acid Citraconic acid Mesaconic acid

(cis- isomer) (trans- isomer) (cis-isomer) (trans- isomer) H-C-Cl H-C-Cl ║ ║ H-C-Cl lC-C-H Cis-1,2-Dichloroethylene Trans -1,2-Dichloroethylene

H3C-C-H H3C-C-H ║ ║

H3C-C-H HC-C-CH3 Cis- But-2-ene Trans - But-2-ene Q. What is geometrical isomerism and what is its cause? Q. What are the necessary conditions for the geometrical isomerisation? Q. Cis alkenes show higher boiling point as compared to trans-isomer. Why? Hint polar


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HOW TO DISTINGUISH BETWEEN CIS AND TRANS: ON THE BASIS OF PHYSICAL PROPERTY:

1. DIPOLE MOMENT: 2 MELTING POINT 3. BOILING POINT (a) Lesser dipole moment of trans isomers (symmetrical alkene) while cis has non zero

dipole moment or more than the trans isomer. or otherwise Cis isomer is having higher D moment than Trans. H-C-Cl H-C-Cl ║ ║ H-C-Cl Cl-C-H

Cis-1,2-Dichloroethylene Trans -1,2-Dichloroethylene Dipole moment = μ = 1.91D Dipole moment = μ = 0

H3C-C-H H3C-C-H ║ ║

H3C-C-H HC-C-CH3 Cis- But-2-ene Trans - But-2-ene Dipole moment = μ > 0 Dipole moment = μ = 0

Dipole moment is high means Polarity of cis isomer > trans isomer. Boiling point: cis isomer has higher due to high polarity, high density and higher

Refractive index than the corresponding trans isomer (Auwers-skita rue). H3C-C-H H3C-C-H ║ ║

H3C-C-H HC-C-CH3 Cis- But-2-ene Trans - But-2-ene

Dipole moment = μ > 0 Dipole moment = μ = 0 Bp 40C (high) Bp 10 C (low)

Mp -1390C (low) Mp -1060C (high) H-C-Cl H-C-Cl ║ ║ H-C-Cl lC-C-H

Cis-1,2-Dichloroethylene Trans -1,2-Dichloroethylene Dipole moment = μ = 1.91D Dipole moment = μ = 0

Bp 600C (high) Bp 480 C (low) MP -800C (low) MP -500C (high)

The trans isomer has higher mp than cis- isomer due to symmetrical nature and more close packing of the trans-isomer.

Note: Mpt relates to symmetry which is more in case of trans While Bpt relates to polarity which relates to dipole moment.

Q. Draw cis and trans isomer and write IUPAC naming of :

(a) CHCl=CHCl (b) C2H5C(CH3)=C(CH3)C2H5 Naming Hint: 1,2-Dichloroethene 3,4-Dimethylhex-3ene

Q. Which of the following compounds will show cis-trans isomerism? (i) (CH3)2C=CH–C2H5 (ii) CH2=CBr2 (iii)C6H5CH=CH–CH3 (iv)CH3CH=CClCH3

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

Q.Draw the geometrical isomer of hex-2-ene? Which isomer has higher bp and why? Hint: Cis -Hex-2-ene because of more polar.


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CH3

Q. which of the following will show geometrical isomerism? (a) H2C=CBr2 (b) (CH3)2C=CHCH3 (c) C6H5CH=CHCH3 (d) CH3CH-C=CHCH2CH3

(e) H3C-CH=CCl(CH3)

Q. Which will exhibit geometrical isomerism? (a) 2-Methyl but-2-en-1-ol (b) 3-Methyl but-2-en-1ol

Q. Cis-1,2-dichloroethene has less/more dipole moment than trans-1,2-dichloro ethne

Q. For the following compounds, write structural formulas and IUPAC names for all possible isomers having the number of double or triple bond as indicated: C4H8 (one double bond).

Q. Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p.

and why? Cis -Hex-2-ene because of more polar METHODS OF PREPARATION OF ALKENES:

1. DEHYDRATION OF ALCOHOL (Removal of water by H2SO4, dehydrating agent)

Alcohol + conc H2SO4 alkene + H2O

RCH2CH2OH conc H2SO4, 443K RCH=CH2 +H2O CH3CH2OH conc H2SO4, 443K CH2=CH2 + H2O Dehydration order for alcohols 3

0 > 2

0 >1

0

2. DEHYDROHALOGENATION : (1,2 ELIMINATION OR β ELIMINATION) It involves the treatment of an alkyl halide, with concentrated alcoholic solution of a

strong base like NaOH/ KOH to form alkene.


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RCH2-CH2-X + KOH (alc) Δ RCH=CH2 +KX + H2O

β CH3-CH2-CH2Br + KOH (alc) Δ CH3-CH=CH2 +KBr + H2O

(1-Bromopropane)

Ease of dehydrohalogenation in terms of alkyl halides : 30 > 20> 10 Ease of dehydrohalogenation in terms of halides: RI>RBr>RCl

alkyl iodide > alkyl bromide > alkyl chloride V Imp. Saytzeff rule: more substituted is the preferred product.

It states that “In dehydrohalogenation reactions, the preferred product is that alkene which has lesser no. of H atom on doubly bonded Carbon atoms or contains greater

number of alkyl groups attached to the doubly bonded carbon atoms”. CH3-CH2-CH2-CH2Br + KOH (alc) Δ CH3-CH=CH-CH3 + CH3-CH2-CH=CH2 + KBr + H2O 1-bromo Butane 2-Butene 1-Butene

3. DEHALOGENATION OF VICINAL DIHALIDES WITH ZN DUST IN ALCOHOL. CH2Br

+ Zn CH2=CH2 + ZnBr2 CH2Br 1,2-Dibromoethane Ethene

4. KOLBE’S ELECTROLYTIC METHOD:

Electrolysis of aqueous solution of Na or K salt of saturated carboxylic acids. CH2COONa + 2H2O electrolysis CH2=CH2 + 2CO2 + 2NaOH + H2

At anode At cathode CH2COONa (Sodium succinate) Ethene

Q. How are alkenes prepared by Kolbe's Electrolytic process? 5. PARTIAL HYDROGENATION OF ALKYNE:

RCCH + H2 Ni RCH=CH2

CHCH + H2 Ni 523K CH2=CH2 ethene (hydrogenation reactions) Using Ni catalyst or Lindlar catalyst (palladised charcoal deactivated with poison like S compounds or quinoline). This method gives cis-alkene.

Lindlar catalyst

RCH=CH2 Pd/BaSO4/C RCCH + H2 Ni RCH=CH2 S

However if Na/ liq NH3 is used as reducing agent , trans alkene is formed

known as birch reduction.

Q. What is Lindlar's catalyst? What is its use?

Ans. Lindlar catalyst (Pd palladised charcoal (C) deactivated with poison like S

compounds or quinoline). It is used for hydrogenation of alkyne to alkene (cis-alkene).

LINDLAR CATALYST: CH3 CH3

CH3-CC-CH3 + H2 Pd/BaSO4/C S C = C cis- But-2-ene

H H

Q. What is Birch reduction ? What is its use?


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BIRCH REDUCTION CH3 H

CH3-CC-CH3 + H2 Na/NH3 C =C trans- But-2-ene

H CH3

Q. What is the order of stability of alkene towards hydrogenation reactions? Ans. Relatives stability of alkene can be compared by their enthalpy of hydrogenation

when alkene reacts with H2 in presence of Ni catalyst? Ans. greater is the number of alkyl groups attached to the double bonded carbon atoms, the more stable is the alkene.

Q. Alkyne on reduction with sodium in liquid ammonia form trans alkene? Write reaction

will it shows geometrical isomerism? Hint Birch reduction, yes.

CHEMICAL PROPERTIES OF ALKENE:

C=C bond in alkene is having 1 σ sigma and 1 bond. Due to presence of bond ,

alkene are very much reactive. Since bond is a weak bond and held less firmly with the double bonded C as compared to σ bond. therefore in a reaction when electron is needed

electrons are easily available for a reaction. C=C bond act as source of electrons and thus electron loving reagent which are

deficient in electrons are most likely to attack at the site of electrons of the double

bond which are electron in nature. Further addition of elctrophilic reagents relieves the strain of the C=C double bond(1200)

by converting it into C-C (109.50). Electrophilic addition reactions are most common are the most common reactions of C=C bond. Example: Unsymmetrical alkene addition

(1-Bromopropane)

CH3-CH=CH2 + H-Br (presence of peroxides) CH3-CH2-CH2Br n- CH2=CH2 Δ, 1000atm ,organic peroxides -(CH2-CH2-)n

Ethene Polyethene Q. Why alkenes undergo electrophilic addition and not electrophilic substitution reaction? SUBSTITUTION REACTIONS:

Ethylene does not undergoes substitution reactions. However, alkene having alkyl group undergoes halogenations at higher temperature to form substituted products. Mechanism :free radical substitution reactions:

CH3CH=CH2 + Cl2 500-6000C gas phase CH2=CH-CH2Cl + HCl Propene allyl chloride

CHEMICAL PROPERTIES OF ALKENES:


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(A) ADDITION REACTIONS:

The reactions which involve the combination of two or more molecule (same or different substances) to form an addition product (adduct) without any loss of any species from

the reactants are known as addition reactions. e.g. CH3-CH=CH2 + H-Br (presence of peroxides) CH3-CH2-CH2Br

n- CH2=CH2 Δ, 1000atm ,organic peroxides -(CH2-CH2-)n Ethene Polyethene

Reaction mechanism: Free Radical addition reactions:

OTHER EXAMPLES OF ADDITION REACTIONS: I. ADDITION OF HYDROGEN (H2) (REDUCTION REACTIONS)

SABATIER AND SENDREN’S REACTION: (ALREADY DISCUSSED IN ALKANE) Catalytic addition reaction: catalytic hydrogenation reaction.

It involves catalytic (Ni or platinum) hydrogenation of unsaturated hydrocarbons like alkene and alkyne to alkane. Methane can’t be obtained by this method. It is a type of addition reaction.

CH2=CH2 + H2 Ni 2000C-300C or 473K CH3-CH3 +heat

Ethene Pt Ethane

CHCH + 2H2 Ni 2000C-3000C or 473K CH3-CH3 + heat Ethyne Pt Ethane

2. ADDITION OF HALOGEN (Br2, I2) in presence if inert solvent CCl4


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Reddish brown colour of Br2 is discharged due to formation of colourless dibromide.

NOTE: This reaction is used as a quick diagnostic test for the presence of unsatutration i.e. >C=C<. Reddish brown colour of Br2 is discharged due to formation of colourless

dibromide. Test for Unsaturation: There are two tests to know whether a compound is

unsaturated or not: a. Bromine water test: When an unsaturated compound is reacted with bromine solution in carbon tetrachloride, the reddish orange colour of bromine solution in carbon

tetrachloride is discharged.

b. KMnO4 TEST (oxidation reactions). 3. ADDITION OF HALOGEN ACID HX (HCl>HBr>HI)

CH2=CH2 + HX CH3-CH2-X

Order of reactivity of halogen acid: HI > HBr> HCl > HF

V. IMP. PEROXIDE EFFECT OR KHARASCH EFFECT. The addition of HBr (but no any other halogen acid) to unsymmetrical alkenes in the

presence of organic peroxide such as benzoyl peroxicle, (C6H5CO)2O2 or (B2H6) takes place contrary to the Markovnikov's rule, -ve part of addendum goes to that C atom of

the double bond which is having larger no. of H atoms. This is known as Peroxide effect or Kharasch effect.

CH3-CH=CH2 + H-Br (presence of peroxides) CH3-CH2-CH2Br (1-Bromopropane) Unsymmetrical alkene

Type of reaction: Free radical mechanism:

Stability order: 30>20>10 (same as free radical). Note: HF, HCl, HI do not exhibit peroxide effect and addition takes place according to Markovnikov’s rule even in the absence of peroxides. Q. Give mechanism of Kharash effect. Ans. Mechanism of kharash effect : free radical addition reaction

(a) FORMATION OF FREE RADICALS: (C6H5CO)2O2 HOMOLYSIS 2C6H5

. +2CO2

Benzoyl peroxide free radical (b) C6H5

. + H-Br homolysis C6H6 + Br.


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(c) Br. +CH3-CH=CH2 CH3.CH-CH2Br 20 free radical

CH3CH-CH2. 10 free radical

Br Stability order of free radical: 30>20>10

Therefore CH3.CH-CH2Br 20 free radical is the more stable product.

(d) CH3.CH-CH2Br + HBr CH3CH2CH2Br

20 free radical 1-Bromopropane V. IMP.

MARKOVNIKOV’S RULE: It states that when addition across an unsymmetrical alkene takes place, +ve part of addendum (molecule to be added goes to that carbon atom of

the double bond which holds the greater no. of H atoms). It takes place in the presence of mercuric acetate Hg(CH3COO)2 or B2H6 Br

CH3-CH=CH2 + HBr MARKONIKON RULE CH3-CH-CH3

Propene 2 Bromo propane Type of reaction: Electrophyllic addition reactions: Stability order: 30>20>10

This is due to formation of carbocation, carbocation stability order is 30>20>10 Order of reactivity of halogen acid: HI > HBr> HCl > HF

Q. Explain and Justify Markownikoff's rule. Q. Write IUPAC names of the products obtained by addition reactions of HBr to hex-1-

ene (i) in the absence of peroxide and (ii) in the presence of peroxide. 4. POLYMERISATION REACTIONS: Large no. of simple molecules combine to form a bigger molecules of higher molecular

mass called polymer. Small molecule is called monomer. e.g. Polymerisation of alkene in presence of organic peroxides like benzoyl peroxides,

Zieglar Natta catalyst. monomer is smallest repeating unit. (a) High density polythene (HDPE) or polyethylene

Monomer: Ethene (Ethylene). Catalyst: Zeigler Natta catalyst : (C2H5)3Al+ TiCl4

n- CH2=CH2 Δ, 1000atm , -(CH2-CH2-)n Polyethene

Ethene organic peroxides

Uses: It is used for manufacturing buckets, dustbins, bottles, toys, pipes, etc CH3

(b) Polypropylene

n CH2=CH-CH3 polymerisation -(CH-CH2-)n Propylene polypropylene (PP)

USES: milk crates, plastic buckets, seat covers, carpet fibres, ropes, heat shrinkable

wraps. Cl

n CH2=CH-Cl Polymerisation -(CH2-CH-)n

Chloroethene (vinyl chloride) Polyvinylchloride (PVC) PVC pipes, electric fitting, rain coats etc.


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(c) POLY TETRA FLUOROETHENE (TEFLON): monomer: CF2=CF2 1,2 difluoroethylene

Uses: It is used in making oil seals and gaskets and also used for non – stick surface coated utensils.

5. ADDITION OF HYPOHALOUS ACID (HOCl, HOBr, HOI): CH2=CH2 + HOCl H2O/ Cl2 CH2-CH2

Ethylene hypochlorous OH Cl

2-Chloro Ethanol (Ethylene chlorohydrin)

6. ADDITION OF COLD CONC. SULPHURIC ACID:

CH2=CH2 + HSO3H CH2-CH2

Ethylene alkyl hydrogen sulhphate H SO3H

Ethylene hydrogensulphate

7. ADDITION OF HYDRACID HCN (Nucleophilic addition reactions) CH2=CH2 + HCN CH3-CH2-CN

Propanenitrile 8. ADDITION OF WATER (H2O): CH2=CH2 + HOH CH2-CH

Ethylene water OH H Ethanol (Ethyl alcohol)

9. CATALYTIC ADDITION OF OXYGEN: CH2=CH2 + ½ O2 Ag CH2-CH2

Ethylene 525-675k O Ethylene oxide

10. ADDITION OF OZONE: V.IMP

OZONOLYSIS: O3 is added to an alkene molecule to form ozonide which in turn can hydrolysed to yield smaller carbonyl (ALDEHYDE / KETONES) fragments. The process of addition of ozone to form an ozonide and then to hydrolyse the product is known as

Ozonolyis. Ozonolysis involves the replacement of olefinic bond, >C=C<, by two carbonyl groups >C=O.

O CH2=CH2 + O3 CH2Cl2 CH2 CH2 + H2O/Zn ZnO + HCHO+ HCHO

195-200K O O

The addition of Zn, checks the formation of H2O2 which can otherwise oxidise the products (aldehyde and ketone) to acids.


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Q. Give ozonolysis reaction of ethene.

Q. How is structure of alkene elucidated by ozonolysis ?

ANS. Identification of the aldehyde and / or ketone formed during the ozonolysis will give useful information about the structure of the original olefin or the position of double bond.

DETECTION OF POSITION OF DOUBLE BOND: H H

CH3-C=O O=C-CH3 join the C atom of carbonyl group CH3-CH=CH-CH3 Ethanal Ethanal But-2-ene CH3 CH3 CH3 CH3

CH3-C=O + O=C-CH3 join the C atom of carbonyl group CH3-CH=CH-CH3 Propanone (acetone) 2,3-Dimethyl but-2-ene Q. An olefinic was treated with ozone and the resulting product on hydrolysis gave 2-

pentanone and acetaldehyde. What is the structure of the original olefin? Write equations for the reactions involved?

Ans. by knowing the structure of 2-pentanone (CH3CH2CH2COCH3 ) and acetaldehyde the structure of olefin may be written as below. CH3

O CH3CH2CH2C=CHCH3 +O3 CH3CH2CH2-C CHCH3 + H2O/Zn A + B

- ZnO CH3 O O 3-Methylhex-2-ene A: CH3CH2CH2COCH3 B: CH3CHO

2-Pentanone Acetaldehyde

Q. Write IUPAC names of the products obtained by the ozonolysis of the following compounds :


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(i) Pent-2-ene (ii) 3,4-Dimethylhept-3-ene (iii) 2-Ethylbut-1-ene (iv) 1-Phenylbut-1-ene Ans(i) propanal, ethanal (ii) penan-2-one and butan-2-one (iii) pentan-3-one and methanal (iv) propanal and benzaldehyde.

Q. An alkene ‘A’ on ozonolysis gives a mixture of ethanal and pentan-3- one. Write structure and IUPAC name of ‘A’. A= 3-Ethylpentan-3-one.

Q. An alkene ‘A’ contains three C – C, eight C – H σ bonds and one C – C bond. ‘A’ on ozonolysis gives two moles of an aldehyde of molar mass 44 u. Write IUPAC name of ‘A’. Q. Propanal and pentan-3-one are the ozonolysis products of an alkene?

What is the structural formula of the alkene? Hint 3-Ethylhex-3-ene. Q. Propanal and Butan-2-one are the ozonolysis products of an alkene?

What is the structural formula of the alkene? Hint 3-Methylhex-3-ene Q. Ozonolyis of an alkene ‘X’ followed by decomposition with water and reducing agent gives mixture of two products which are isomers of formula C3H6O? What is the

structural formula and IUPAC name of the alkene? Ans. propanal CH3CHO propanone CH3COCH3

Alkene 2-Methylpent-2-ene B. OXIDATION REACTIONS:

(i) OXIDATION BY ALKALINE KMnO4 SOLUTION (BAEYER’S REAGENT) TO FORM

ethylene glycol CH2=CH2 + H2O + O CH2-CH2

Cold KMnO4 V Important: OH OH Ethylene glycol

BAEYER’S REAGENT TEST: When an unsaturated compound is reacted with cold, dilute aqueous

solution of potassium permanganate (Baeyer’s reagent, the pink colour

of KMnO4 solution is discharged. CH2=CH2 + H2O + [O] 2HCOOH

Hot KMnO4 formic acid 2HCOOH + [O] 2CO2 + 2H2O Formic acid

NOTE: Br2 water test and Baeyer’s test used for detecting the presence of double bond while ozonolysis helps in locating the position of double bond.

(ii) HYDROBORATION OXIDATION: CH3-CH2=CH2 +H2O DIBRORANE (B2H6) CH3-CH2-OH

Propene Propanol H2O is added opposite to Markonikov’s addition.

C. COMBUSTION REACTIONS:

(i) COMPLETE COMBUSTION REACTIONS: (Exothermic reactions) Complete combustion: C2H4+ 3O2 2CO2+ 2H2O +heat +light ΔH=-ve C4H8+ 6O2 4CO2+ 4H2O +heat +light ΔH=-ve

General reaction: CnH2n+ 3n O2 nCO2+ nH2O +heat +light ΔH=-ve

2 (ii) INCOMPLETE COMBUSTION OR PARTIAL COMBUSTION:

Incomplete combustion: C2H4+ 2O2 2CO+ 2H2O +Less heat +light ΔH=-ve C4H8+ 4O2 4CO+ 4H2O +Less heat +light ΔH=-ve

It burns with black soot.


31

Q. Write chemical equations for combustion reaction of the following hydrocarbons: (i) Pentene?

SUMMARY OF REACTIONS OF ETHENE ARE AS BELOW:

TYPES OF DIENES: COMPUNDS CONTAINING TWO DOUBLE BONDS ARE KNOWN AS DIENES

COMULATIVE DIENES:

CH2=C=CH2 CH3-CH=C=CH-CH3 1,2-Propadiene 2,3-Pentadiene ALLENE SYM-DIMETHYL ALLENE

CH3

CONJUGATED DIENES:

CH2=CH-CH=CH2 CH2=C-CH=CH2 1, 3- BUTADIENE 2-METHYL-1,3-BUTADIENE(ISOPRENE) ISOLATED DIENES:

CH2=CH-CH2-CH=CH2 CH2=CH-CH2-CH2-CH=CH2

1,4-PENTADIENE 1,5-HEXADIENE

Among the three classes of dienes, conjugated dienes are the most important since the chemical properties of conjugated dienes are very important from those of ordinary alkene.

Q. Which is more stable 1,4 or 1,3 Butadiene? Ans 1,3-Butadiene is more stable as: highly symmetrical and highly substituted.

Q. give example of conjugated diene? 1, 3- BUTADIENE Practice questions:

Q. Formula of Teflon? ans.–(CF2-CF2-)n Q. C7H14 on ozonolysis gives ethanal and pentan-3-one. What is the structure and IUPAC

name of the alkene?hint: CH3-CH=C(C2H5)CH2CH3 3-Ethyl pent-2-ene Q. Name two test to check the presence of double bond or triple bond in a molecule? Ans. Bayers test (decolour pink color) , Br2 water test (decolourise brown color)

Q. CH2=CH2 + Alk KMnO4 ? ans ethylene glycol. CH2(OH)-CH2(OH) Q. What is the cause of geometrical isomers of alkene?


32

Ans.Alkene containes 1 sigma bond and 1 pi bond, therefore rotation about C=C bond is restricted which rise to geometrical isomers. Q. Draw the geometrical isomers of but-2-ene and which of the two is nonpolar? ans.

trans is non polar. Q. Write the structural formulae and possible isomers of C2H2Cl2? Cis and trans1,2-

Dichloroethylene. Q. What is the order of stability of alkene? 30>20>10 stability. Q. Bromine water is used to distinguish ethene and Ethyne T/F? false as both give

reactions. Q. 1-butanol reacts with conc H2SO4 to give But-2-ene?

Q. Reaction of ethene with bromine electrophilic addition reaction. Q. Addition of HBr to propene in presence of peroxides follows ___ rule. Anti mk rule.

Q. The alkaline potassium permanganate solution is ___ reagent. Baeyers reagent. Q. Name the metal present in Grignard reagent? (Mg) RMgX Q. Why are alkene called olefins?

Q. What happen (a) when ethyl alcohol is treated with conc H2SO4 443K? (b) Ethyl bromide is treated with alcoholic KOH?

Q. How does ethylene reacts with (1) br2 (2) alkaline potassium permanganate (3) ozone (4) water Q. How will you distinguish between ethylene and ethane?

Q. Complete the reaction:

(i) CH3-CCH +H2 Pd, BaSO4, S?

(ii) CH3CH=CH-CH3 + O3 Q. Convert ethylene to acetylene? Q. Convert acetic acid to methane?

Q.

Q.Write structures of different isomers corresponding to the 5th member of alkyne series. Also write IUPAC names of all the isomers. What type of isomerism is exhibited by different pairs of isomers?

Solution 5th member of alkyne has the molecular formula C6H10. The possible isomers are:

(a) HC C– CH2 – CH2 – CH2 – CH3 Hex-1-yne

(b) CH3 –CC–CH2–CH2–CH3 Hex-2-yne

(c) CH3–CH2 -CC–CH2–CH3 Hex-3-yne

(d) 3-Methylpent-1-yne (e) 4-Methylpent-1-yne (f) 4-Methylpent-2-yne

(g) 3,3-Dimethylbut-1-yne Position and chain isomerism shown by different pairs


33

ALKYNES:

Triple bond between two C C atoms.

CC consists of 1 sigma and 2 bonds. General Formula: CnH2n-2 Hybridisation: sp

Strength: Stronger than C=C bond.

HYBRIDISATION IN AKYENE: Molecular formula = C2H2 Empirical formula = CH Molecular mass = 26 Empirical mass = 13

Common name = Acetylene Homologous series = Alkynes STRUCTURE OF ETHYNE: linear

BOND LENGTH: The C-H bond is 1.2Ao.

BOND ANGLE: HCC bond angle is 180o.

HYBRIDISATION OF ETHYNE: sp (linear)

1(2s) + 1(2px) orbitals (sp hybridisation) 2 (sp hybrid orbitals) of 1800 alignment

forming linear arrangement. Each orbital has 1/2 s (50%) character and ½ p(50%) character. Two unhybridised orbital 2py, 2pz is oriented in a plane at right angle to the

plane containing three hybridised orbitals.

6C12 (ground state electronic configuration) 1s2 2s2px12py12pz0

6C*12 (excited state electronic configuration) 1s22s1 2px12py12pz1 one e- from s

orbital is moved from 2s orbital to 2p orbital which is in higher energy state. The input energy for exciting an electron is available from lattice energy of

formation, released when bond are formed. Initially energy has to give so as to start the reaction. In the excited state one 2s orbital and one (2px ) orbitals of C* intermix to form 2 hybrid

orbitals of sp hybridisation. These two sp orbitals (sp) aligned themselves in such a way that bond angle must be 1800 so as to have minimum energy and maximum stability.

Orbital structure is linear. Note: 02 orbital of 2py and 2pz remains un-hybridized state. When molecule of Ethyne is formed, sp hybridised orbitals interacts with other sp hybrid

orbital of ethyne forming 1sigma which overlap between C-C bond of ethyne. 1 sp2 hybrid orbital of each C of Ethyne forms sigma bond by overlap with 1s1 orbital of 2 H

atoms in end to end manner (σ). Two 2py and 2pz orbital of C atom of ethyne overlap sidewise forming 2pi (π)bond with

other C atom of ethyne. Therefore bond between C-C is triple in nature 1 sigma (σ) and 2 pi (π). 2 C-H bonds forming sigma bond (σ).

CC bond length = 120pm, C- H bond length = 106pm, H-C-C-H bond angle 1800

Structures of Ethyne:

PHYSICAL PROPERTIES OF ALKYNES: C1-C4 gases C5-C12 liquids C>C13 solids


34

Colourless, odourless gases, however common acetylene has garlic odour due to presence of impurities like H2S, PH3 Phosphine gas etc. Liquid acetylene is dangerously explosive, use of storage acetylene is prohibited by law.

Acetone is therefore stored and transported in acetone solution.

a. Solubility: Alkynes are weakly polar in nature. They are lighter than water and immiscible with water but soluble in organic solvents like ethers R-O-R, carbon tetrachloride CCl4 and benzene C6H6. Hence alkynes are sparingly soluble in water

but dissolves in organic solvents like acetone, alcohols etc.

b. Melting and boiling point: The melting and boiling points of the members of the alkyne family are slightly higher as compared to those of corresponding alkanes and alkenes. This is because alkynes have a linear structure and therefore their

molecules are more closely packed as compared to alkanes and alkenes. The magnitude of attractive forces among them is higher and therefore, the melting and

boiling points are also higher. The melting point, boiling point and density of alkynes increase with increase in molar mass. This is because magnitude of van der Waals forces increases with increase in

molecular mass or increase in surface area. ETHANE ETHENE ETHYNE

Mpt (K) 101 104 191 Bpt (K) 184.5 171 198

ISOMERISM IN ALKYNE:

A. STRUCTURAL ISOMERISM: (I) CHAIN ISOMERISM:

Same molecular formula but different arrangement of carbon chains.

CH3-CH2-CHCH CH3-CCH

But-1-yne 2-Methylprop-1-yne

CH3

(2) POSITION ISOMERISM: Same molecular formula but differ in the position of the functional group, carbon-

carbon multiple bond. Two linear positional isomers of molecular formula C5H8

(1) CH3-CH2-CH2-CCH Pent-1-yne

(2) CH3-CH2-CC-CH3 Pent-2-yne

(3) FUNCTIONAL ISOMERISM: Same molecular formula but different functional groups in the molecule

diene and alkyne CH3-CC-CH3 CH3=CH-CH=CH2 Buta-1,3-diene Butyne-2

(4) RING CHAIN ISOMERISM: Due to difference in mode of linkage in C atoms and the isomers may have either open chain or closed structure. e.g. C3H4 and cyclopropene.

Q. Why Alkyne do not shows geometrical isomerism? Ans. Alkyne involves sp hybridisation , thus angle is 1200.

Shape is linear, thus do not show geometrical isomerism


35

METHODS OF PREPARATION of ALKYNES:

1. HYDROLYSIS OF CALCIUM CARBIDE (LAB METHOD)

By the action of water on calcium carbide (CaC2) placed in a flask CaC2 + 2H2O C2H2 + Ca(OH)2

Ethyne

:CC:2- + 2HOH H-CC-H + 2OH- Base + acid Conjugated acid + Base

2. BY DEHYDROHALOGENATION OF VIC-DIHALIDES:

CH2=CH2 + Br2 CH2-CH2 + KOH(alc) CH2 +KOH (alc) CH

Br Br CHBr CH Ethene Ethylene bromide Vinyl bromide Acetylene

CH3 + KOH (alc) CH2 +KOH (alc) CH

CHBr2 CHBr CH Ethylidene bromide Vinyl bromide Acetylene

3. BY DEHYDROHALOGENATION OF VIC-tetra HALIDES:

With active metals like Zn, Mg etc and CH3OH as solvent. Br Br Br Br

H-C-C-H + Zn H-C=C-H + Zn HC=C-H

-ZnBr2 -ZnBr2 Ethyne

Br Br 1,1,2,2-Tetrabromoethane

4. BY KOLBE ELECTROLYTIC METHOD: CHCOONa + 2H2O electrolysis CH=CH + 2CO2 + 2NaOH + H2

║ At anode At cathode CHCOONa Sodium fumarate

5. BY HEATING IODOFORM OR CHLOROFORM WITH SILVER POWDER OR ZINC :

CHCl3 + 6Ag + Cl3HC Δ heat HC≡CH + 6AgCl white ppt 6. BERTHOLOT SYNTHESIS:

2C+ H2 electric arc CHCH


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V. Imp. 7. PREPARATION OF HIGHER ALKYNES FROM SODIUM ACETYLIDE:

CHCH + Na Liquid NH3 475K HCC-Na + 1/2H2 Acetylene Sodium acetylide

HCC-Na + BrCH3 H-CC-CH3 + NaBr Sodium Acetylide Propyne

HCC-H + 2Na liquid NH3 NaCC-Na + 2CH3Br CH3-CC-CH3 2-Butyne

NOTE: This reaction is used to convert lower alkyne to higher alkyne.

R-CC-H + 2Na liquid NH3 R-CC-Na + 2CH3Br R-CC-CH3 + NaBr Lower alkyne Higher alkyne

CHEMICAL PROPERTIES OF ALKYNES: Alkyne undergoes addition reaction

(a) Electrophilic addition reactions with electron deficient electrophile. E.g. addition of Br2 ,Cl2, I2 etc.

(b) Nucleophilic addition reactions with electron rich nucleophiles.e.g. hydration of alkyne, addition of HCN, NaHSO3, RMgX, Carboxylic acids and alcohols are examples of

nucleophilic additions.

Q. Why CC triple bond is less reactive than C=C toward electrophillic addition reactions?

Alkynes are more reactive than alkane and alkene due to the following reasons:

(i) Presence of electrons:

Alkynes are characterised by CC bonds (1σ, 2). Due to presence of electrons which are loosely held in alkynes, electrophile (+ve ) charge species can attack on the electron

rich electrons hence undergo electrophilic addition reactions.

(ii) Hybridisation: This is because in alkynes C atom is sp hybridised and thus has more s character of

carbon than that of sp2 hybridised carbon of alkene. Now since greater is the s character

of C atom, more strongly will be the attraction for electrons and hence lesser will be the availability for reaction.

ACIDIC NATURE OF ALKYNES: Q. Why alkynes are acidic?

Hydrogen atoms of ethyne attached to triply bonded carbon atom are acidic in

nature. Example: HC=CH + Na HC=C-Na+ + ½ H2

HC=C-Na+ + Na Na+ C-C-Na+ + ½ H2

s character in alkynes is maximum i.e.50%.

Therefore electrons constituting this bond are more strongly held by the Carbon nucleus (sp hybridised C is having more electronegative).With the result the H present on such a

carbon atom (C-H) can be easily removed as a proton. The acidic nature of differently hybridised C of C-H bonds are as follows:

RC-H > R=CH2 > R-CH3 sp sp2 sp3

Being most electronegative C in alkyne, the carbon of acetylene is best able to accommodate the electron pair in the anion left after the proton is lost.


37

In other words, anion of acetylene is most stable and hence more acidic. While anion of ethane is the least stable (sp3 carbon is the least electronegative) and hence

least acidic.

CHCH + :B- HB + HCC:- (most stable)

CH3-CH3 + :B- HB + H3C-CH2: - (least stable)

Presence of acidic H atoms: H atom or atoms attached to the triply bonded C atom can be easily removed by means

of a strong base and hence acetylene or 1-alkyne sre considered as weak acids.

R-CC-H + base R-CC- + H-Base Acidic alkyne

RELATIVE ACIDIC CHARACTER :

H-O-H R-OH > CHCH > CHCR > :NH3 > CH2=CH2 > CH3-CH3 Water alcohol alkyne alkyne ammonia alkene alkanes

RELATIVE BASIC CHARACTER (CONJUGATE BASE): -OH -OR < -CC-R < -NH2 <-CH=CH2 <-CH2-CH3

REACTION SUPPORTING ACIDIC CHARACTER:

CHCH + Na Liquid NH3 475K HCC-Na + 1/2H2 Acetylene Sodium acetylide

HCC-Na + BrCH3 H-CC-CH3 + NaBr Sodium Acetylide Propyne

CHCH +Cu2Cl2 + 2NH4OH Cu-CC-Cu + 2NH4Cl + 2H2O Ammonical cuprous chloride copper acetylide (red ppt)

CHCH + 2AgNO3 + 2NH4OH Ag-CC-Ag + 2NH4NO3 + 2H2O Ammonical silver nitrate silver acetylide (white ppt.)

The above reaction is used for detecting the presence of Acetylenic H atom. For example butyne-1 have acetylenic H while 2-butyne do not. Acidic H atoms

gives red and white ppt.

Q. Which of the following is acidic? (a) 2-Butene (b) 2-Butyne (c) 1-Butyne (d) 1-Butene Q. Propyne is less acidic than acetylene? True. Q. When ______is passed through red hot iron tube at 873K, benzene is formed.

Ethyne. Q. Alkynes are _____ acidic than alkene. (less/ more)

REACTIONS OF ALKYNE:


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ADDITION REACTIONS:

1. ADDITION OF HYDROGEN:

SABATIER & SANDEREN’S REACTION: (Already discussed in Alkane / Alkene)

It involves catalytic (Ni or platinum) hydrogenation of unsaturated hydrocarbons like alkene and alkyne to alkane. Methane can’t be obtained by this method.

CHCH + 2H2 Ni 2500C OR 5230K CH3-CH3 + heat Ethyne Ethane

COMMON CATALSYT USED ARE:

(a) Lindlar catalyst: Palladium supported over barium sulphate partially poisoned by quinoline (b) Brown catalyst: nickel boride

LINDLAR CATALYST: CH3 CH3

CH3-CC-CH3 + H2 Pd/BaSO4/C S C = C cis- But-2-ene

H H BIRCH REDUCTION:

CH3 H

CH3-CC-CH3 + H2 Na/NH3 C =C trans- But-2-ene

H CH3

Used for producing trans-But-2-ene.

2a. ELECTROPHYLIC ADDITION OF HALOGENS: Cl2, Br2, I2

CHCH + Cl2 CHCl + Cl2 CHCl2

║ CHCl CHCl2


39

Acetylene Acetylenedichloride Acetylene tetrachloride (Westron) 1,2-Dichloroethene 1,1,2,2-tetrachloroethane 2b. ADDITION OF HALOGEN ACIDS:

Order of reactivity HI > HBr > HCl

CHCH + HCl CH2 + HCl CH3

║ CHCl CHCl2

Acetylene vinyl chloride Ethylidene chloride 2c. ADDITION OF HYPOHALOUS ACIDS:

CHCH + HOCl CHOH + HOCl CH(OH)2 -H2O CHO

║ CHCl CHCl2 CHCl2

Acetylene Vinyl chloride Dichloroacetaldehyde

3. NUCLEOPHILIC ADDITION REACTIONS: 3a. ADDITION OF WATER HYDRATION:

HC≡CH (Acetylene) + H-OH HgSO4 CH2=CH-OH CH3-CHO ethanal

H2SO4 (1% HgSO4, 42% H2SO4 333K)

3b. ADDITION OF HCN TO ALKYNE

HC≡CH (Acetylene) + HCN Ba(CN)2 CH2=CH-CN (Vinyl cyanide or Acrylonitrile)

3c. ADDITION OF ACETIC ACID:

HC≡CH (Acetylene) + CH3COOH HgSO4 CH2=CH-OCOCH3 + H2O CH3-CHO ethanal H2SO4 + C2H5-OH ethanol

3d. ADDITION OF ARSENIC TRICHLORIDE:

HC≡CH (Acetylene) + ClAsCl2 CHCl=CHAsCl2 Lewsite (poisonous gas)

Antidote: BAL (British Anti Lewsite)

4. OZONOLYIS OF ALKYNE : O CH=CH + O3 CH--------CH + H2O H2O2 + CHO

O--------O CHO 5. OXIDATION REACTION: (a) Oxidation with alkaline potassium permanganate: COOH

CHCH + 4[O] Alkaline KMnO4 298K COOH (Oxalic acid)

Ethane1,2- dioic acid


40

Cold KMnO4 at 298K CH part is oxidised to –COOH

CR part is oxidised to R–CO- group

(b) Oxidation with acidic KMnO4 or K2Cr2O7 COOH

CH=CH + 4[O] K2Cr2O7+ H2SO4 2CH3COOH

COOH Ethane1,2- dioic acid

With acidic KMnO4 at high temperature:

CH part is oxidised to –COOH

CR part is oxidised to RCOOH.

6. COMBUSTION REACTIONS:

CnH2n-2 + (3n-1) O2 nCO2+ (2n-2) H2O + heat 2 2

2CHCH + 5O2 4CO2+ 2H2O Used in welding purpose in oxyacetylene flame. It has a flame temperature of 30000C

7. ISOMERISATION REACTIONS:

CH3-CH2-CCH + alc KOH , heat CH3-CH=C-CH3 1-Butyne 2-Butyne 8. POLYMERISATION REACTIONS:

(a) CYCLIC POLYMERISATION:

Alkynes when passed through a red hot iron tube, undergo cyclic polymerisation to from aromatic hydrocarbons.

3CHCH red Hot Cu tube benzene or C6H6

V. Imp 9. FORMATION OF METALLIC DERIVATIVES:

CHCH +Na liquid NH3 HCC-Na+ 1/2H2

HCC-Na + BrCH3 H-CC-CH3 + NaBr Sod. Acetylide Propyne

HCC-H + 2Na liquid NH3 NaCC-Na + 2CH3Br CH3-CC-CH3

2-Butyne

CHCH +Cu2Cl2 + 2NH4OH Cu-CC-Cu + 2NH4Cl + 2H2O Ammonical cuprous chloride copper acetylide (red ppt)

CHCH + 2AgNO3 + 2NH4OH Ag-CC-Ag + 2NH4NO3 + 2H2O Ammonical silver nitrate Silver acetylide (white ppt.)

The above reaction is used for detetcting the presence of Acetylenic H atom. for example butyne-1 have acetylenic H atom gives red and white ppt. Q. Distinguish between butyne-2 and butyne-1?

Ans. Butyne-1 have acidic H atom which can give red and white ppt. with ammonical cuprous chloride and ammonical silver nitrate silver nitrate solution, respectively while

butyne-2 does not give such ppt.

CH3-CH2-CHCH +Cu2Cl2 + 2NH4OH CH3-CH2-CH2-CCu + 2NH4Cl + 2H2O

butyne-1 Ammonical cuprous chloride copper butynlide (red ppt)

CH3-CHCH-CH3 +Cu2Cl2 + 2NH4OH no reaction


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Or

CH3-CH2-CHCH + AgNO3 + NH4OH CH3-CH2-CH2-CAg + NH4NO3+ 2H2O Butyne-1 Ammonical silver nitrate Silver butylnide (White ppt)

CH3-CHCH-CH3 +AgNO3 + 2NH4OH no reaction

10. OTHER REACTIONS OF ETHYNE:

(a) HCCH + N2 ELECTRIC ARC 2HCN

(b) 2HCCH + S CH-CH ║ ║ CH CH

S

(c) 2HCCH + NH3 CH-CH +H2 ║ ║ CH CH

N H

INDUSTRIAL APPLICATIONS OF ETHYNE (ACETYLENES)

1. It is used in form of oxyacetylene flame for welding, cutting and cleaning iron & steel. 2. Acetylene is used for illumination purposes in hawker’s lamps and light house. 3. It is used for artificial ripening of fruits.

4. It is used for the production of several chemicals like acetaldehyde (CH3CHO), acetic acid (CH3COOH), ethyl alcohol (C2H5OH), acetone (CH3COCH3), benzene etc.

5. It is used for the production of Westron (acetylene tetrachloride) and Westrosol (trichloroethylene) , industrial solvents for rubber, fats and varnishes. 6. It is used in the preparation of vinyl chloride CH2=CH-Cl, vinyl acetate CH2=CH-

COOCH3 and vinyl ethers CH2=CHOR used in the production of plastics aand artificial rubber. E.g. Polyvinylchloride(PVC).

7. It is used in the preparation of acrylonitrile (CH2=CH-CN) used for the production of synthetic rubber (Buna-N) and synthetic fibre (orlon). 8. It is used in the preparation of chloroprene and butadiene used in synthetic rubber

industry. 9. It is used for the production of Lewsite (Cl-CH=CHAsCl2) , a powerful poisonous gas.

SUMMARY OF REACTIONS OF ETHYNE:


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DIFFERENCE IN ALKANE ALKENE AND ALKYNE HYDROCARBONS :

Hydrocarbons Alkyne Alkene Alkane

Hybridisation of carbon to which H

atom is attached

Sp sp2 sp3

Percentage s character

of carbon

50%

(maximum)

66.6 % 25% (minimum)

Electronegativity of

carbon atom

Highest Less than alkyne and

more than alkane

Lowest

Extent of attraction of

hydrogen atoms of C-H bonds towards C

Highest Less than alkyne and

more than alkane

Lowest

Ease of liberation of H atoms as protons

Highest Less than alkyne and more than alkane

Lowest

Acidic character

Highest Less than alkyne and more than alkane

Lowest

Bond length C-C all containing 2C atoms. Bond strength KJ

154pm 148pm 138pm

348 383 433

Bond length C-H Bond strength KJ

111pm 110pm 108pm

436 443 507

Multiplicity of bonds C-C C=C CC

Q. Write structures of different isomers corresponding to the 5th member of alkyne series. Also write IUPAC names of all the isomers. What type of isomerism is exhibited by different pairs of isomers?


43

Solution 5th member of alkyne has the molecular formula C6H10. The possible isomers are: (a) HC ≡ C – CH2 – CH2 – CH2 – CH3 Hex-1-yne (b) CH3 – C ≡ C – CH2 – CH2 – CH3 Hex-2-yne

(c) CH3 – CH2 – C ≡ C – CH2– CH3 Hex-3-yne (d) 3-Methylpent-1-yne

(e) 4-Methylpent-1-yne (f) 4-Methylpent-2-yne (g) 3,3-Dimethylbut-1-yne

Q. How will you convert ethanoic acid into benzene?

Hint: ethanoic acid sod salt of ethanoic acid CH4 CH3Cl by chlorination wurtz reaction, ethane chlorination C2H5Cl + KOH elimination C2H4 Br2 CH2Br-CH2Br +

2KOH CHCH Aromatisation C6H6 Q. Following compounds, write structural formulas and IUPAC names for all possible

isomers having the number of double or triple bond as indicated:(a)C5H8(one triple bond) Ans. 1-Pentyne, Pent-2-yne, 3-Methylbut-1-yne.

Q. How would you distinguish propene from propyne? Ans. alkyne are acidic while alkene do not.

Propyne gives reaction with ammonical silver nitrate solution (tollens reagent) and give white ppt while propene do not.

CH3-CHCH + AgNO3 + NH4OH CH3-CH2-CAg + NH4NO3+ 2H2O Propyne Ammonical silver nitrate Silver Propylnide (White ppt)

CH3-CH=CH2 + AgNO3 + NH4OH no white ppt. Q. How will you convert acetaldehyde from acetylene?

Hint: hydration of water in presence of HgSO4+ H2SO4. Q. Which of the following hydrocarbon reacts with Na in liquid ammonia? (a) Propyne (b) But-1ene (c) But-1-yne(d) But-2-ene

Q. What happens when water is dropped on calcium carbide? Ethyne is produced. Q. Name the reagent : 1,2-Dibromoethane + X Acetylene ans . alc KOH.

Q. Write the chemical reaction for the combustion reaction of hexyne? 2C6H10 + 17O2 12CO2 + 10H2O

Q. 1,1,2,2-tetrachloroethane + Zn, alcohol + heat A iron tube 675K B What is A and B? Hint A= Ethyne B= benzene

Q.Propyne + H2, Pd, BaSO4 quinoline A + O3, Zn/ H2O B + C What is A , B ? ans A= CH3-CH=CH2 B= CH3CHO C= HCHO Q. convert: Ethyne to methane: hint ethyne ethanal ethanoic acid soda lime CH4

Q. convert ethene to Ethyne? Hint : + Br2 + KOH alc Q. Ethane to Ethyne: hint: ethane chloroethane KOH ethene + br2 + 2KOH alc.

Q. Ethyne to 2-butyne: + Na liq NH3 + 2CH3Br 2-butyne + 2NaBr. Q. suggest a method to separate a mixture of ethane , ethene and Ethyne?

Ans. bubble through ammonical silver nitrate solution, white ppt will separate alkyne. Treat white ppt with HCl alkyne can be separated. Balance mixture containing alkene

and alkane is passed through Br2, CCl4 then vic dihalides ppt will be formed which can be separated by treating with Zn as ZnBr2 and alkene can be separate. Balance remains only alkane.Thus alkane,alkene & alkyne are separated by this combination of methods.

Q.2- butyne on catalytic reduction with lindar catalyst gives: ---alkene ans. cis alkene

Q. electrolysis of potassium fumarate gives _________(alkane/ alkene/ alkyne) Q. hydration of acetylene gives ---(ethanol/ ethanal). Q. But-2yne cannot react with tollens reagent? T/F

Terminal alkynes are acidic in nature and undergo ionic reactions.


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Q. Chloroform on heating with silver powder gives Ethyne/ ethene? Q. Ethyne on ozonolysis gives ________ glycol/ glyoxal () Q. convert Ethyne to benzene? Hint Cyclic polymerisation reactions.

Q. complete the reactions:

(a) HC CH + alk KMnO4 hint Oxalic acid

(b) CH3C CH + hot KMnO4 hint ketone will be formed. Q. Distinguish between Buty-2-yne and But-1-yne? But-1-yne gives white ppt with TR. Q. Why cylco propane more reactive than propane? Bond angle in cyclo propane is 600

while 1090 in alkane therefore more repulsions, more strain , more energy least stable and hence reactive.

Important topics in alkane , alkene and alkyne:

conformations: butane anti>skew>eclipsed >fully eclipsed Ethane and propane: staggered are more stable than eclipsed.

Chair conformations are more stable than boat for cyclohexane. Cis more polar than trans (more symmetrical) geometrical isomer Lindlar catalyst: Pd palladised + (C) deactivated with S /quinoline) cis alkene

Birch reduction : trans alkene. Na/NH3

Bayers reagent : alkaline KMnO4

Markonikov rule, saytzeff rule, peroxide effect, Ozonolysis of alkene, testing between alkane and alkene, alkene and alkyne. Acidic character of alkyne.

Hybridisation of Ethane , Ethene, Ethyne. AROMATIC COMPOUNDS:

These molecules contain a benzene ring based a 'special' cyclic C6 system, which is an unsaturated ring (BUT not an alkene system) e.g.


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or or

Benzene C6H6, Methylbenzene(Toluene) C6H5CH3 or Phenylethene('Styrene'), 2,3-Dichloromethylbenzene,

C6H5Cl, C6H4Cl2, , ,

chlorobenzene, 1,2-dichlorobenzene, or 1,3-dichlorobenzene, or 1,4-dichlorobenzene,

or C7H6O3, Phenol. 2-Chlorophenol 3-Hydroxybenzoic acid 2-Hydroxybenzoic acid

(o-chlorophenol)

C7H8O, C6H7NO,

3-Methylphenol 2-Aminophenol, 4-Aminophenol 2-Nitrophenol (m-methylphenol, meta-cresol)

C7H6O, or , C7H6O2, , , ,

benzaldehyde, benenecarboxylic acid, benzoic acid

C6H5NO2,

Aniline benzene diazonium chloride. nitrobenzene, benzenesulphonic acid, C6H6SO3, C6H5SO3H, C6H5SO2OH, (or benzenesulphonic acid)

(1) , (2) , (3) ,


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Methyl-2-nitrobenzene 3-aminobenzoic acid 4-hydroxybenzamide

ISOMERSISM:

POSITION ISOMERISM IN ETHYL BENZENE

C8H10 (1) (2) (3) (4)

(1) Ethylbenzene, (2) 1,2-Dimethylbenzene, (3) 1,3-Dimethylbenzene, (4) 1,4-Dimethylbenzene

m pt -94oC, m pt -25oC, m pt -47 oC, m pt 14oC,

b pt 136oC, b pt 144oC , b pt 139oC , b pt 137oC, all colourless liquids.

Ortho (Substituent at 1,2) Meta(Substituent at 1,3) Para(substituent at 1,4) POSITION ISOMERISM IN METHYL BENZENE SULPHONIC ACID IN

CH3C6H4SO2OH (C7H8SO3)

1) , (2) and (3) 2-Methylbenzenesulphonic acid 3 Methylbenzenesulphonic acid 4 Methylbenzenesulphonic acid Ortho- methylbenzenesulphonic acid meta- methylbenzenesulphonic acid para methylbenzenesulphonic acid

POSITIONAL ISOMERS OF XYLENE

C8H10 or , 1,2-Dimethylbenzene, 1,3-Dimethylbenzene 1,4-Dimethylbenzene

Ortho xylene Meta xylene Para-xylene POSITIONAL ISOMERS OF C9H12

C9H12, C9H12,

1-Ethyl-2-methylbenzene, 1-Ethyl-3-methylbenzene, 1-Ethyl-4-methylbenzene, Q. Draw the structure of p-nitroaniline, o-Ethylanisole, 2,3-Dibromo-1-phenylpentane,

4-Ethyl-1-fluoro-2-nitrobenzene?Benzene, Methylbenzene (Toluene), Naphthalene, Biphenyl, 1,2-

Dimethylbenzene (o-Xylene)

BENZENE: Molecular formula = C6H6 Empirical formula = CH

Molecular mass = 78 Empirical formula mass = 13

Percentage of carbon = 93.6% Percentage of hydrogen = 6.4%

ORBITAL STRUCTURE OF BENZENE:


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Kekule Structural formula of benzene : It is an aromatic hydrocarbon. ALL bond angles 120o, symmetrical hexagonal ring,

planar molecule

HYBRIDISATION IN BENZENE: This is the most accurate concept of the structure of benzene. The structure of benzene molecule is best described in terms of molecular orbital theory.

According to this theory, all the C-atoms in benzene are sp2-hybridized. Two sp2-

hybrid orbitals of each C-atom overlap with two sp2-hybrid orbital of two other C-atoms to form sigma bonds. In this way there are six sigma bonds are formed between six C-

atoms which are 120o apart. Remaining six sp2-orbital of six C-atoms overlap with 1s orbital of six H-atoms individually to form six sigma bonds. Since sigma bond results from the overlap of above said planar orbital, all H and C atoms are in the same plane

and their generate a hexagonal ring of C-atoms.

The overlap of these 2pz-orbital result in the formation of a fully delocalized pi-bond, which extends all over the six C-atoms of benzene nucleus. The molecular orbital

approach clearly indicates that these six electrons could be found anywhere in highly

delocalized manner. As a result of delocalization, a stronger -bond and a more stable benzene molecule is obtained which undergo substitution reactions more frequently than

addition reactions.

BOND LENGTH ANALYSIS IN BENZENE C-C length in alkane = 1.54oA, C=C length in alkene = 1.34oA But in benzene, C-C length = 1.397oA.

This shows that in benzene single and double bonds have quite extraordinary character as they do not resemble to alkane and alkene in bond lengths. That is why benzene

shows a behaviour of saturated as well as an unsaturated hydrocarbon simultaneously. MODERN REPRESENTATION OF BENZENE

With the help of molecular orbital behaviour we conclude that benzene has a regular hexagonal structure with an inscribed circle. A hexagon with alternate double and single

bonds therefore benzene undergoes addition reactions but it does not discharge bayers reagent which is pink in colour which can be discharged by unsaturated hydrocarbons while not by saturated hydrocarbons.

C6H6 + H2 Ni catalyst 2000C Cyclohexane or C6H12 Benzene undergoes electrophilic substitution reactions like nitration, sulphonation,

acylation which are not shown by unsaturated hydrocarbons but shown by saturated hydrocarbons etc .


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Conclusion: benzene give reactions of both saturated hydrocarbons and unsaturated hydrocarbons because of presence of alternate single and double bond.

Q. How does the result support Kekulé structure for benzene?

RESONANCE IN BENZENE: RESONANCE:(Heisenberg) It is the phenomenon in which two or more structures can be written for the true

structure of a molecule, but none of them is said to represent it uniquely.

The true structure of the molecule is said to be a resonance hybrid of the various possible alternative structures which themselves are known as resonating structures or canonical structures. Kekule structures 1, 2 are contributing structures along with

deawar structures, structure (r) is the resonance hybrid of the structures (1) and (2).

(1) (2) (r)

E.g. C-C bond length in benzene is 1.39A0 as against the value of 1.54A0 for C-C single bond and 1.34A0 for C-C double bond. (1A0 =10-10m)


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Q. Why is benzene extra ordinarily stable though it contains three double bonds?

Ans. due to resonance . in benzene all the 6 electrons of three double bonds gets

delocalised resulting the stability of the molecule.

NECESSARY CONDITIONS FOR AROMATICITY: a. Planarity: The ring should be planar due to delocalisation of electron cloud. In simple words, all the carbons of the ring should be sp2 hybridised.

b. Complete delocalisation of the electrons in the ring: The electrons should be completely delocalized above and below the plane of the molecule. In simple words, in the ring alternate single and double bonds should exist.

c. Hückel Rule: Presence of (4n + 2) electrons in the ring where n is an integer (n = 0, 1, 2, . . .)

Benzene C6H6: n=1 Presence of electrons = (4n + 2) = 6 electrons.

Naphthalene: n=2 Presence of electrons = (4n + 2) = 10 electrons.

Anthracene: n=3 Presence of electrons = (4n + 2) = 14 electrons.


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Q. What are the necessary conditions for any system to be aromatic? POLYNUCLEAR AROMATIC RING COMPOUNDS:

Examples of fused ring of arenes or polynuclear aromatic ring compounds:

Naphthalene , C10H8 consists of two fused aromatic rings

Anthracene or C14H10 consists of three fused rings.

Naphthalene and anthracene are two of the simplest examples of what are

called polynuclear aromatic compounds or in this specific case polycyclic aromatic hydrocarbons(PAHs), also known as polycyclic arenes or polyaromatic hydrocarbons.

SOURCES OF AROMATIC COMPOUNDS: 1. Aromatic compounds from coal (Anthracite %age free C=90% , Bituminous C=70%,

lignite C=40%, peat C10-15%) by destructive distillation of coal. Destrucitve distillation of coal gives coal gas, coal tar. 2. Aromatic compounds from petroleum by catalytic reforming or hydro forming.

PREPARATION OF BENZENE:

1. FROM ACETYLENE:

Alkynes when passed through a red hot iron tube, undergo cyclic polymerisation to from

aromatic hydrocarbons.

3CHCH red Hot Cu tube 873K benzene or C6H6 Similarly propyne can be converted to mesitylene.

2. FROM SODIUM BENZOATE (LABORATORY METHOD) SODA LIME REACTION:

DECARBOXYLATION REACTION AS CO2 IS REMOVED: C6H5COONa + NaOH CaO C6H6 + Na2CO3 Sod. Benzoate LIME Benzene

3. FROM PHENOL:

C6H5OH Zn dust C6H6 + ZnO


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4. FROM CHLOROBENZENE: (REDUCTION) C6H5Cl + 2[H] Ni-Al alloy C6H6 + HCl

NaOH

5. FROM BENZENE SULPHONIC ACID: C6H5SO3H + H2O 150-2000C C6H6 + H2SO4

Benzenesulphonic acid Benzene 6. FROM BENZENE DIAZONIUMCHLORIDE:

(a) C6H5N2Cl + 2[H] SnCl2/ NaOH C6H6 + HCl + N2

Benzenediazonium chloride Benzene (b) C6H5N2Cl + H3PO2 + H2O Cu+1 ions

C6H6 + H3PO3 + HCl + N2

Benzenediazonium chloride Benzene 7. AROMATIZATION OR CATALYTIC REFORMING:

The conversion of aliphatic compound into aromatic compound in the presence of catalyst is known as aromatisation. E.g.

REFORMING: The conversion of aliphatic compound into aromatic compound in the presence of H as hydro forming and in the presence of platinum is known as platforming.

Cracking, isomerisation and aromatisation are widely used to increase increase the octane rating of gasoline.

8. FROM PHENYL MAGNESIUM HALIDE: RMgX + H2O RH + MgX(OH)

C6H5MgBr + H2O C6H6 + MgBr(OH)

9. WURTZ FITTIG REACTION: Formation of alkyl benzene C6H5Br + 2Na + CH3Br Wurtz fittig reaction C6H6CH3+ 2NaBr

dry ether USES OF BENZENE:

1. Benzene is used in dry cleaning and as a solvent for oils, fats, rubber etc. 2. It is use in manufacturing of nitrobenzene which is used in medicine and dyes.

3. It is also used to make gammexene (an insecticide). 4. When mixed with petrol, it may be used as motor fuel.

CHEMICAL PROPERTIES OF BENZENE:

(A) ELECTROPHILIC SUBSTITUTION REACTIONS:


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Q. Why does benzene undergo electrophilic substitution reactions easily and nucleophilic

substitutions with difficulty? ANS. Orbital structure shows benzene has delocalised pi electrons lying above and below

the benzene ring is loosely held and thus these sites are available for the attack of electrophile (e deficient species). Therefore benzene undergoes electrophilc addition reactions with ease as compared to nucleophilic addition reaction.

(B) ADDITION REACTIONS

(C) OXIDATION REACTIONS:


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CHEMICAL PROPERTIES OF AROMATIC HYDROCARBONS:

A. ELECTROPHILIC SUBSTITUTION REACTIONS: (I) HALOGNENATION (CHLORINATION, BROMINATION, IODINATION)

Chlorination: C6H6 + Cl2 An AlCl3 310-320K C6H5Cl + HCl (Electrophile: Cl+)


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(II) SULPHONATION Sulphonation: C6H6 + HSO3 (H2SO4) C6H5SO3 + H2O (Electrophile : +SO3H)

(III) NITRATION

Examples: Nitration of benzene: C6H6 + HNO3 C6H5NO2 + H2O (Electrophile: +NO2) (IV) ALKYLATION OR FRIEDEL CRAFT ALKYLATION REACTION:

Friedal craft reaction: C6H6 + CH3Cl An AlCl3 C6H5CH3 + HCl (Electrophile: CH3+)

(V) ACYLATION OR FRIEDEL CRAFT ACYLATION REACTION: Friedal craft acylation: C6H6 + CHCOCl An AlCl3 C6H5COCH3 +HCl Electrophile: (CH3CO+)

MECHANISM:(a)Formation of Electrophiles(b)formation of complex, (c) release of proton MECHANISM OF NITRATION REACTION:

Examples: Nitration of benzene: C6H6 + HNO3 C6H5NO2 + H2O (Electrophile: +NO2)

Q. Give mechanism of sulphonation of benzene? OR Q. What is the mechanism for sulphonating / sulfonating benzene? or methyl benzene? ANS. Benzene: C6H6 + SO3 ==> C6H5SO2OH

or C6H6 + H2SO4 ==> C6H5SO2OH + H2O Benzene is heated with concentrated sulphuric acid or even better, 'fuming' sulphuric

acid, which has a higher sulphur trioxide content and more efficient at introducing the sulphonic acid group into the benzene ring.


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Step (1) Sulphur trioxide is formed (or already present). It is a

powerful electrophile, i.e. electron pair acceptor because of the effect of the three very electronegative oxygen atoms bonded to the central sulphur atom. Step (2) An electron pair from the delocalised pi electrons of the benzene ring forms a

C-S bond with the electron pair accepting sulphur trioxide forming a second highly unstable carbocation. It is very unstable because the stable electron arrangement of the

benzene ring is partially broken to give a 'saturated' C (top right of ring). Step (3) A hydrogensulphate ion removes a proton and the complete benzene ring is reformed giving the anion of the aromatic sulphonic acid.

Step (4) is a proton transfer to give the sulphonic acid.

(VI) GATTERMAN KOCH REACTION: Benzene or its derivatives on treatment with carbon monoxide and HCl in presence of

anhydrous aluminium chloride or cuprous chloride (CuCl) gives benzaldehyde or substituted benzaldehydes + HCl

B. ADDITION REACTIONS:

(a) Addition of hydrogen: (catalytic reaction) C6H6 + 3H2 Ni (2000C) C6H12

Benzene Cyclohexane

(b) Addition of halogen: chlorination

C6H6 + 3Cl2 sunlight (hv) C6Cl6 + 6HCl Benzene BHC (Benzene hexachloride)

BHC is a well known insecticide under name of gammexane or lindane.

NOTE: If halogenations is done in presence of AlCl3 it hexachlorobenzene.


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(c) Addition of Ozone (O3):

C6H6 + 3O3 Zn/H2O 3 CHO

CH2Cl2

CHO Ethanedial (Glyoxal) COUPLING OF BENZENE:

2 C6H6 + RED HOT IRON TUBE C6H5-C6H5

Benzene 700-8000C Biphenyl C. CONTROLLED OXIDATION by KMnO4:

Benzene is un-reactive towards potassium permanganate (KMnO4) under normal conditions. However oxidation of benzene in presence of V2O5 as catalyst at 5000C gives

maleic anhydride. Benzene + 9/2 O2 V2O5 5000C CHCOOH

║ heat Maleic anhydride +H2O CHCOOH OXIDATION OF ALKYL BENZENE BY: From alkyl benzenes: Aromatic carboxylic acids can be prepared by vigorous oxidation

of alkyl benzenes with chromic acid or acidic or alkaline potassium permanganate

ETARD REACTION:

Chromyl chloride ( CrO2Cl2) oxidizes methyl group to a chromium complex, which on hydrolysis gives corresponding benzaldehyde

COMBUSTION REACTIONS: Arenes are inflammable liquids and burn with a sooty flame to produce CO2 and H2O. (a) 2C6H6 +15O2 12CO2+ 6H2O

Q. Write chemical equations for combustion reaction of the following hydrocarbons: (i) Toluene?


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THE DIRECTIVE INFLUENCE OF VARIOUS GROUPS:

Directive influence of a functional group in mono substituted benzene: A. ORTHO PARA DIRECTING GROUPS:

Groups which direct the incoming group to o & p positions.

Activating groups: Groups which increase the electron density on ortho and para positions of the benzene ring. E.g. : –NH2, –NHR, –NHCOCH3,-OH, –OCH3, –CH3, –C2H5,

-SH, CH2Cl, Ph-, X- F-,Cl-, Br-, I-

Deactivating groups: Groups which decrease the electron density on ortho and para positions of the benzene ring. E.g. :-X Halogens are moderately deactivating groups because they have strong – I effect (electron withdrawing effect) which overpowers +R

effect .Due to this the overall electron density on benzene ring decreases. It makes further substitution difficult. However, due to resonance the electron density on o– and

p– positions is greater than that at the m-position. Hence, they are also o– and p– directing groups.

As a general rule substituent having atleast one lone pair of e- linked to a benzene ring is o & p-director, however an alkyl group having no lone pair of electrons is also o and p-

director.


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B. META DIRECTING GROUPS The groups which direct the incoming group to meta position. Example: –NO2, –CN, -NC, -COCl, -CONH2, –CHO, –COR, –COOH, –COOR, –SO3H,

COOCOR(as a general rule, substituent containing multiple (=or≡) bond is meta director.


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CARCINOGENS: Most Poly nuclear aromatic compounds are carcinogenic in nature. E.g. working in coal

prolonged exposure suffers from skin cancer. Major component of this fraction is : 1,2-Benzanthracene.

Others are 3-methylcholanthracene, 1,2-Benzpyrene etc. Number and position of the functional group such as –CH3,-OH, -CN, -OCH3 etc. Such polynuclear hydrocarbons are formed by incomplete combustion of organic matter such

as coal, petroleum and tobacco etc. When these carcinogens enter human body, they undergoes various biochemical reactions and finally damage DNA and ultimately leads to

cancer.

Q. Convert ethanoic acid to benzene? Hint: ethanoic acid CH4 by Soda Lime CH3Cl Ethane by Wurtz, C2H5Cl, alc KOH C2H4 + Br2 C2H4Br2+ 2KOH Ethyne

polymerC6H6.

Q. How will you convert benzene into (i) p-nitrobromobenzene (ii) m- nitrochlorobenzene (iii)p-nitrotoluene

(iv) acetophenone? Hint (i) benzene bromination bromobenzene + Nitration o+ p nitro bromo benzene separated by fractional distillation

(ii) benzene + HNO3 nitro benzene + Cl2 anhd AlCl3 m nitro chlorobenzene (iii)benzene + CH3Cl Friedel Craft Reaction toluene + HNO3 o , p nitro toluene

fractional distillation p nitro toluene. (iv)FCR C6H6+. CH3COCl acetophenone.

Q. Complete the reaction:

(a) C6H6+ O3 A Zn, H2O B A= benzene triozonide B=glyoxal. (b) complete the reaction B MnO2/ H+ C6H5CH3 + KMnO4/ H

+ A hint A= benzoic

acid, B= benzaldehyde (c) difference between alkylation and acylation? Write FCR (d) what is halogen carrier? FeCl3 , AlCl3 lewis acidetc

(e) what is huckel rule? See notes (f) which is meta director?-NO2, -SO3H, -Cl, -OH, -NH2, -CHO

Q. how many isomers are possible for mono substituted and disubstituted benzene? Hin t mono -1, di- 3 (o/p/m)

Q. why benzene is extraordinarily stable although it contains 3 double bond? Ans. resonance. delocalisation of pi electrons.

Q. arrange in decreasing order of acidic acharacter (a) ethne (b) benzene (c) n hexane a>b>c ?


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Q. nitration of nitrobenzene mainly gives meta product?T/F

Q. chlorobenzene on reduction with Ni-Al alloy gives toluene? False/ benzene is formed.

Q. all the C in benzene is __ hybridised? And. sp Q. when ___________ is passed through red hot cu tube _____ is produced. Ans.

Ethyne / benzene.

Q. electrophile in sulphonation of benzene is ____ (SO3+)

Q. nitration of bromo benzene or bromination of nitrobenzene gives same/different

bromo nitrobenzene. Different.

Q. nitration of benzene is electrophilic substitution reaction. Q. IUPAC name C6H5-CH2-CH2-CH=CH2 4-Phenyl-But-1-ene

Q. o- C6H5(OH)(CH3) iupac name =2-methyl phenol

Q. How would you convert the following compounds into benzene?

(i) Ethyne (ii) Ethene (iii) Hexane Hint (i) red hot cu (ii) ethene to Ethyne red hot cu benzene (iii) aromatisatio reaction n hexane Cr2O3/ V2O5 cyclohexane aromatisation benzene.

Q. Arrange the following set of compounds in order of their decreasing

relative reactivity with an electrophile, E+ (a) (i)Chlorobenzene,(ii) 2,4-dinitrochlorobenzene, (iii)p-nitrochlorobenzene hint i>iii>iii

(b) (i)Toluene,(ii) p-H3C-C6H4–NO2, (iii)p-O2N–C6H4–NO2. ii>i>iii>iv Q. Out of benzene, m–dinitrobenzene and toluene which will undergo nitration most

easily and why? Toluene > benzene> m-dinitrobenzene (CH3 eltron releasing gp while –NO2 electron withdrawing group)

Q. Suggest the name of a Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene. Ans. anhd FeCl3, SnCl4, BF3 etc.

Q. arrange the following set of compounds in order of decreasing reactivity with an

elctrophile? C6H5NO2, C6H5-OCH3 , C6H5Cl Ans. C6H5-OCH3 > C6H5Cl > C6H5NO2,

Q. Write down the products of ozonolysis of 1,2-dimethylbenzene (o-xylene). arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also give reason

for this behaviour. ANS METHYL GLYOXAL,1,2 DIMETHYLGLYOXAL

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