Aliphatic Hydrocarbons 9 Hours

I SEMESTER B. Sc. NOTES PREPARED BY DHONDIBA VISHWANATH SURYAWANSHI (DVS) GCW, KOLAR

UNIT – IV Aliphatic Hydrocarbons

9 hours: Max. Marks: 18 -20 Syllabus:

Alkanes: Sources, Nomenclature of branched chain alkanes, preparation of symmetrical andunsymmetrical alkanes- Corey- House reaction and Wurtz reaction - their merits and demerits. Conformational analysis of n-butane - Sawhorse and Newman projection formulae to be used -Energy profile diagram.Cycloalkanes: Nomenclature. Method of formation. Explanation for stability based on heat of hydrogenation data, Baeyer’s strain theory and its limitation, Sachse - Mohr theory of strain-less rings; cyclopropane ring - banana bonds.Alkenes: Preparation of alkenes by Wittig reaction-stereoselectivity. Addition of HX tounsymmetrical alkene - Markownikov’s rule and Antimarkownikov’s rule with mechanism.Reactions: Hydroboration- oxidation, reduction, oxymercuration – demercuration epoxidation. Mechanism of oxidation with KMnO4 and OsO4.Ozonolysis- mechanism and importance.Dienes: Classification- isolated, conjugated, cumulated. Structure of allene and butadiene.1,2addition and 1,4 addition reactions. Diels Alder reaction-1,3-butadiene with maleic anhydride.Alkynes: Methods of preparation - Dehydrohalogenation of vicinal and geminal dihalides; and higher alkynes from terminal alkynes. Reactions - metal ammonia reduction – significance. Oxidation with KMnO4, acidic nature of terminal alkynes.

Aliphatic hydrocarbons: A branch of organic chemistry which deals with study of open chain compounds containing carbon and hydrogen as elements called aliphatic hydrocarbons.

Classifications of aliphatic hydrocarbons: They are classified as

1) Alkanes 2) Cycloalkanes 3) Alkenes 4) Dienes 5) Alkynes.

1) ALKANES: Alkanes are the simplest organic compounds made of carbon and hydrogen only. They have the general formula CnH2n + 2, where, n= 1, 2, 3, etc. the first three members of this class can be represented as

The carbon atoms in their molecules are bonded to each other by single covalent bonds. Each carbon is again bonded to enough hydrogen atoms to give maximum covalence of 4. Since the carbon skeleton of alkanes is fully saturated with hydrogens. They are also called saturated hydrocarbons.

Alkanes contain strong C-C and C-H covalent bonds. Therefore, this class of hydrocarbons is relatively chemically inert. Hence they are sometimes referred to as paraffins.

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Sources of alkanes: Alkanes are occurred in

1) Natural gas contains about 80% methane and 10% ethane, the remaining 10% being a mixture of higher members.

2) Petroleum is the source of C5 – C20 hydrocarbons.

3) Plant waxes also contain saturated hydrocarbons having odd number of carbon atoms (C27

to C37). Bee wax contains C27H56 and C31H64.

Nomenclature of alkanes: There are two system of naming alkanes:

1) Common system: The first four members of the series are called by their common names: methane, ethane, propane and butane. The names of larger alkanes are derived from the Greek prefixes that the number of carbon atoms in the molecule. Thus, pentane has five carbons, hexane has six, and so on.

No. of carbon atoms

Name Molecular formula

Structure of the normal isomer

1 Methane CH4 CH4

2 Ethane C2H6 CH3CH3

3 Propane C3H8 CH3CH2CH3

4 Butane C4H10 CH3CH2CH2CH3

5 Pentane C5H12 CH3CH2CH2CH2CH3

6 Hexane C6H14 CH3CH2CH2CH2CH2CH3

7 Heptane C7H16 CH3CH2CH2CH2CH2CH2CH3

8 Octane C8H18 CH3CH2CH2CH2CH2CH2CH2CH3

9 Nonane C9H20 CH3CH2CH2CH2CH2CH2CH2CH2CH3

10 Decane C10H22 CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3

In the common system all isomeric alkanes have the same parent name. For examples, two isomeric C4H10 alkanes are known as butanes. The names of various isomers are distinguished by prefixes. The prefix indicates the type of branching present in the molecule.

1) Prefix n- is used for those alkanes in which all carbons are in one continuous chain. The prefix n – stands for normal.

2) Prefix iso – is used for those alkanes which have a methyl groups attached to the second last carbon atom of the continuous chain.

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3) Prefix neo- is used for those alkanes which have two methyl groups attached to the second last carbon atom of the continuous chain.

The structural formulas of alkanes contain four types of carbon atoms:

1) A carbon atom attached to one other (or no other) carbon is called primary carbon (10 carbon).

2) A carbon atom attached to two other carbon atoms is called secondary carbon (20 carbon).

3) A carbon atom attached to three other carbon atoms is called tertiary carbon (30 carbon).

4) A carbon atom attached to four other carbon atoms is called quaternary carbon (40 carbon).

Hydrogen atoms attached to 10 , 20, 30 carbon atoms are often referred to as primary, secondary and tertiary hydrogen atoms.

Alkyl groups: An alkyl group is formed by removing one hydrogen from an alkane. They are named simply by dropping –ane from the name of the corresponding alkane and replacing it by –yl (alkane –ane + yl = alkyl).

Parent alkane Structure Alkyl group Name of alkyl groupMethane CH4 CH3- methylEthane CH3CH3 CH3CH2 - ethylPropane CH3CH2CH3 CH3CH2CH2 - propyln -Butane CH3CH2CH2CH3 CH3CH2CH2CH2- n-butylIsobutane (CH3)2CHCH3 (CH3)2CHCH2 - isobutyl

II) IUPAC system: The IUPAC system is much the same for all families of organic compounds. The steps for alkanes are as follows.

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1) Name the longest chain: The longest continuous carbon chain is chosen as the basic for the name and it written in horizontal fashion.

2) Number the longest chain: The carbon atoms in the longest chain are numbered. The numbering is started from that end which will give numbers having the lowest value to carbons carrying substituents.

3) Locate and name the constituents: Each substituent is named, and the position of each substituent is indicated by the number of the carbon atom to which it is attached.

4) Combine the longest chain and substituents into the name: The position and the name of the substituent are added to the name of the longest chain and written as one word.

Additional steps are needed when more than one substituent is attached to the longest chain.

5) Indicate the number and position of substituents: If the same substituent is present two or more times in the molecule, the number of this substituent is indicated by a prefix di-, tri-, tetra-, penta-, etc., and the location of each is indicated by a separate number. These position numbers, separated by commas, are put just before the name of the substituent, with hyphen before and after the numbers when necessary.

Preparation of symmetrical alkanes (Wurtz reaction): Higher alkanes are produced by heating alkyl halides with sodium metal in dry ether solution. Two molecules of the alkyl

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halide loss their halogen atoms as NaX. The net result is the joining of two alkyl groups to a symmetrical alkanes (R-R type) having an even number of carbon atoms.

Merits of Wurtz reaction:

1) This method is satisfactorily explained to prepare symmetric alkanes in good yield.

2) This method is satisfactorily explained to prepare higher alkanes from the lower alkyl halides.

Demerits of Wurtz reaction:

1) This method fails to prepare unsymmetrical alkanes because it very difficult to separate.

2) This method fails to prepare alkanes from alcohol or from fatty acids.

3) This reaction is fails to prepare methane.

Preparation of unsymmetrical alkanes by Corey – House method: An alkyl halide is first converted to lithium dialkyl copper, LiR2Cu. This is then treated with an alkyl halide to give an alkane.

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Merit of Corey House reaction: This method is satisfactorily explained the preparation of unsymmetrical alkanes.

Conformations of alkanes: The different spatial arrangements of a molecule that can be obtained by free rotation around carbon – carbon single bonds are called conformations.

Sawhorse representation formulae of alkanes: In sawhorse projection, the carbon atom at the left of the structure is considered to be close to the observer than in carbon at the right.

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Newman projection formulae of alkanes: In Newman projection formulae, one is viewing the C – C bond end on along the axis of connection. The front carbon atom is represented by the intersection of bonds from it, while the rear carbon appears as a circle.

Conformations analysis of n –butane: n – Butane is a somewhat complex molecule from the standpoint of conformation because there are three carbon – carbon single bonds (one central and two terminal) around which rotation can take place. If we consider rotation about the central carbon – carbon bond C2 – C3, the situation is quite similar to that in the case of ethane, except that n –butane molecule has more than one staggered and eclipsed conformations. Staggering or eclipsing in these conformations may be complete or partial as shown below.

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The completely staggered conformation (I) called the anti form shows the methyl groups as far apart as possible. Assuming that the angle of rotation about the central C2 –C3 bond is zero for conformation I, if we rotate one of the C2 or C3 carbon atoms through an angle of 600, the eclipsed form (II) appears. In conformation (II) the methyl group attached to one carbon is at the back of hydrogen, rather than the methyl group attached to the other carbon. Rotation by another 600 leads to a staggered conformation (III) also called gauche form, in which the two methyl groups are only 600 apart. Further rotation by 600 gives rise to the fully eclipsed conformation (IV). On further rotation by 600 the gauche form (V) appears in which the two methyl groups are again 600 apart. Still further rotation by 600 leads to the eclipsed form (VI). If we rotate (VI) by another 600 thereby completing a rotation of 3600, we return to the anti form (I).

Energy profile diagram: The relative stabilities of conformations of n – butane could be seen in the following figure.

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The above figure depicts the energy changes during rotation about C2 –C3 bond in n –butane. It will be observed that the anti conformation (I) would be the most, and the fully eclipsed conformation (IV) the least stable conformation of n – butane. The gauche conformation (III) and (V) would be slightly less stable than the anti conformation (I) and the eclipsed conformation (II) and (VI) would be slightly more stable than the fully eclipsed conformation (IV).

2) Cycloalkanes: Cycloalkanes or cycloparaffins are saturated hydrocarbons in which the carbon atoms are joined by single covalent bonds to form a ring. They are also called alicyclic compounds. The prefix ali – is added because of their similarity to aliphatic compounds. The unsubstituted cycloalkanes from a homologous series with the general formula CnH2n. Where n = 3, 4, 5, 6, etc.

Examples: cyclopropane, cyclobutane, cyclohexane, etc.

Nomenclature: The IUPAC rules for naming cycloalkanes are follows:

Rule 1) Name of an unsubstituted cycloalkane is obtained by attaching the prefix cyclo – to the name of the corresponding normal alkane having the same number of carbon atoms as in the ring.

Rule 2) Substituents on the ring are named and their positions are indicated by numbers. The ring is numbered so that carbons bearing the substituents will have the lowest numbers. If possible indicate stereochemistry.

Rule 3) Naming alkyl substituted cycloalkanes. Count the number of carbon atoms in the ring and also in the largest alkyl substituent.

a) If the number of carbon atoms in the ring is equal to or greater than the number in the largest substituent, the compound is named as alkyl substituted cycloalkanes (Example -1). If

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the number of carbon atoms in the ring is less than the number of carbon atoms in the substituent. It is named as a cycloalkyl substituted alkane (Example -2).

b) If there is a tie between the number of carbons on the cycloalkane part of the molecule and the acyclic alkane part, choose the cycloalkane as the parent (main) chain.

Methods of preparation of cycloalkanes:

1) From dihalides (Freund’s method): Terminal dihalides when treated with sodium or zinc to from cycloalkanes. This reaction is an extension of Wurtz reaction and is useful for the preparation of 3- to 6 – membered rings.

2) From calcium salts of dicarboxylic acids: When the calcium or barium salts of dicarboxylic acids are heated, cyclic ketones are formed. The cyclic ketones can be readily converted into the corresponding cycloalkanes by Clemmensen reduction.

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3) From esters of dicarboxylic acids (Dieckmann reduction):

4) From aromatic hydrocarbon:

Stability of cycloalkanes:

1) Based on the heat of hydrogenation data:

Cycloalkane up to cyclopentane adds hydrogen in presence of nickel at the specific temperature.

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Higher cycloalkanes are not affected by hydrogenation reaction in presence of nickel.

In the above hydrogenation reactions breaking of one of the C – C bond in cyclopropane requires small temperature, in cyclobutane requires more temperature , in cycopentane requires still more temperature and in higher cycloalkanes i.e. in cyclohexane, cycloheptane, etc. no bond is break. This indicates that cyclopropane is least stable than the cyclobutane, which less stable than cyclopentane. Therefore stability of cycloalkanes in the following order:

Baeyer’s strain theory of stability of cycloalkanes: Adolf Baeyer proposed a theory to explain the relative stability of the first few cycloalkanes. He based his theory on the fact that the normal angle between any pair of bonds of a carbon atom is 109028’. Baeyer postulated that any deviation of bond angles from the normal tetrahedral value would impose a condition of internal strain on the ring. He also assumed that all cycloalkanes were planar and thus calculated the angle through which each of the valence bonds was deflected from the normal direction in the formation of the various rings. This he called angle strain. This determined the stability of the ring.

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In cyclopropane, the three carbon atoms occupy the corners of an equilateral triangle, thus cyclopropane has C –C-C bond angle of 600. This implies that the normal tetrahedral angle of 1090 28’ between any two bonds is compressed to 600, and that each of the two bonds involved is pulled in by ½(1090 28’ - 600) = 240 44’. The value 240 44’ then represents the angle strain or the deviation through which each bond bends from the normal tetrahedral direction.

The angle strain for other cycloalkanes can be calculated in the same way. The values are given in table below. Whether the angle of strain is positive or negative, its magnitude determines the extent of strain in the ring.

No. of carbon atoms in the ring

Structure C – C- C bond angle Angle of strain

3600

½(109028’ - 600) = 240 44’

4900

½(109028’ - 900) = 90 44’

51080

½(109028’ - 1080) = 00 44’

61200

½(109028’ - 1200) = -50 16’

71280 6’

½(109028’ - 128035’) = -90 46’

81350

½(109028’ - 1350) = -120 47’

Merits (supports) of Baeyer’s strain theory:

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1) This theory satisfactorily explains the stability of small ring cycloalkanes like cyclopropane, cyclobutane and cyclopentane.

2) This theory satisfactorily calculates the enthalpy of combustion of cycloalkanes per – CH2 per group.

Demerits of Baeyer’s strain theory:

1) This theory fails to explain the stability of cyclohexane and other higher membered of cycloalkane.

2) This theory fails to explain why the ethylenic bond formed readily instead of formation of cycloalkane.

Sache – Mohr theory of strainless cycloalkanes: In order to account for the stability of cyclohexane and higher members, Sache – Mohr proposed that such rings can become free from strain if the all ring carbons are not forced into one plane, as was assumed by Baeyer. If the ring assumed a folded or puckered condition, the normal tetrahedral angles of 109028’ are retained and as a result, the strain within the ring is relieved. As shown in figure.

For example, cyclohexane can exist in two non –planar puckered conformations both of which are completely free from strain. These are called the chair form and the Boat form because of their shape. Such non – planar strain – free rings in which the ring carbons can have normal tetrahedral angles are also possible for higher cycloalkanes.

Examination of the chair form of cyclohexane reveals that the hydrogen atoms can be divided into two categories. Six of the bonds to hydrogen atoms point straight up or down almost perpendicular to the plane of the molecule. These are called axial hydrogens. The other six hydrogens lie slightly above or slightly below the plane of the cyclohexane ring, and are called equatorial hydrogens.

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Relative stability of cyclohexane: Chair – form of cyclohexane is more stable than boat – form due to following reason.

1) Under ordinary condition, cyclohexane molecule will mostly exist in chair form.

2) In boat – form of cyclohexane, two axial hydrogens present at C -1 and C -2 are like flag pole and they are very close to each other and there is repulsion between them. Hence this acquires more energy and it is unstable.

Banana bonds in cyclopropane ring: In cyclopropane, the C – C bonds are neither sigma bond (maximum SP3 – SP3 orbitals overlapping) nor pi bond (minimum SP3 – SP3 orbitals overlapping) but they are intermediate of these two. This type of bond is called banana bond or bent bond. In cyclopropane, bond angle is small this is due overlap of the sp3 orbitals of the carbon is less than the overlap of the sp3 orbitals of carbons in normal propane.

3) ALKENES: Alkenes are hydrocarbons that contain a carbon – carbon double bond (C=C) in their molecules. They have the general formula CnH2n (n = 2, 3, 4, etc.). Alkenes are commonly known as olefins (Latin, Oleum = oil; fiacre = to make) because the lower members form oily products on treatment with chlorine or bromine.

Examples:

Value n in CnH2n Structure Name

2 CH2 = CH2 Ethylene (ethane)

3 CH3CH2 = CH2 Propylene (propene)

4 CH3CH2CH2 = CH2

CH3CH = CH CH3

1 – Butene2 - Butene

Preparation of alkenes by Wittig reaction: Wittig reaction involves the preparation of olefins by the interaction of aldehydes or ketones (aliphatic or aromatic) with triphenylphosphine -alkylidines.

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For examples 1) When R1 = R2 =H

2) When R1 = - CH3 R2 =H

3) When R1 = - CH3 R2 = - CH3

This reaction is completely regioselective and has the double bond only in one location. This reaction is also stereo selective because one of the stereoisomer is a predominant product.

For example:-

Properties of alkenes:

1) Addition of HX to the unsymmetrical alkene: When unsymmetrical alkene is treated with unsymmetrical reagent (HX) to form two types of alkyl halides.

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Markovnikov rule: “When unsymmetrical alkene is treated with unsymmetrical reagent then the positive part of the reagent will be attack on the doubly bonded carbon atom of alkene which bears more number of hydrogen atoms and negative part will be attack on doubly bonded carbon atom which bears less number of hydrogen atoms”. This rule is called Markovnikov rule.

Example: R= - CH3, X = -Br

Mechanism: Mechanism if addition of HX to the unsymmetrical alkenes (Markovnikov’rule) takes place in following steps

1) Formation of a carbo-cation. Notice that for alkene, two carbo-cations are possible.

2) Attack of halide ion on the more stable secondary carbocation to form 2 - haloalkane

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Where R = alkyl group and X = - Cl, -Br, - I.

The order of stability of carbocations is 30¿20¿10. Addition of a reagent to an unsymmetrical alkene proceeds by way of the more stable carbocation.

Anti - Markovnikov rule: “When unsymmetrical alkene is treated with unsymmetrical reagent in presence of peroxide, then the positive part of the reagent will be attack on the doubly bonded carbon atom of alkene which bears less number of hydrogen atoms and negative part will be attack on doubly bonded carbon atom which bears more number of hydrogen atoms”. This rule is called Anti -Markovnikov rule. This rule also called Kharash peroxide effect.

Example: R= - CH3, X = -Br

Mechanism of Anti –Markovnikov rule: Alkenes reacts with HBr in presence of a peroxide by a free radical mechanism. Following steps are involved.

1) Peroxide dissociates to give alkoxy free radicals

2) Peroxide free radical attacks HBr to form a bromine free radical.

3) Attack of bromine free radical on alkene to give a primary free radical and a secondary free radical.

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The order of stability of free radicals is 30¿20¿10. Therefore, the more stable 20 free radical is formed predominantly.

4) More stable 20 free radical attacks the H – Br molecule to form anti- Markovnikov product and a bromine free radical. The bromine free radical goes back to step 3.

Where R = alkyl group

Note:1) Markovnikov addition reaction follows ionization mechanism whereas anti-Markovnikov addition proceeds by free radical mechanism.

2) The HCl and HI do not give anti –Markovnikov products in the presence of peroxides. This is because

i) The H – Cl bond (103 kcal/mole) is stronger than H – Br bond (87 kcal/mole). It is not broken by the alkoxide free radicals obtained from peroxides.

ii) The H – I bond (71 kcal/mol) is weaker than H – Br (87 kcal/mol). It is broken by the alkoxy free radicals obtained from peroxides. But the iodine free radicals so formed readily combine with each other to yield iodine molecules rather than attack the double bond in alkenes.

Reactions of alkenes:

1) Hydroboration – oxidation reaction: Alkenes reacts with borane (BH3) or diborane (B2H6) in tetrahydrofuran to give trialkylboranes, which is on treatment with alkaline hydrogen peroxide gives primary alcohol.

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2) Reduction of alkenes: Alkenes undergo catalytic reduction reaction in presence of nickel or platinum catalyst in hot condition to from alkanes

Where, R = H or alkyl group

For example: 1) R = H

2) R = - CH3

3) Oxymercuration – Demercuration of alkenes: Mercuric acetate, Hg (O2CCH3)2, and water add to alkenes in a reaction called oxymeruration. The product of oxymercuration is usually reduced with sodium borohydride (NaBH4) in a subsequent reaction called demercuration to yield an alcohol.

Oxymurcuration:

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Demercuration:

Where, R = H or alkyl group.

For, example: R = - CH3

Oxymercuration:

Demercuration:

4) Epoxidation of alkenes: Alkenes react with oxygen gas in presence silver catalyst at 250 -4000C to from epoxide.

Where, R = H or alkyl group

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For example,

5) Mechanism of oxidation of alkene with potassium permanganate (KMnO4) and Osmium oxide (OsO4): When alkene is undergo oxidation in presence alkaline solution of KMnO4 and OsO4 in pyridine to form glycol.

Where R = H or alkyl group

For examples 1) When R = H

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2) When R = - CH3

6) Ozonolysis of alkenes: When ozone is passed through an alkenes in an inert solvent. It adds across the double bond to form an ozonide. Ozonides are explosive compounds. They are not isolated. On warming with zinc and water (or dimethyl sulphide), ozonides cleave at the seat of the double bond. The products are aldehydes, ketones, or an aldehyde and ketones, depending on the structure of the alkene.

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Where, R1 =R2 = R3 = R4 = H or alkyl groups

For examples 1) When R1 =R2 = R3 = R4 = H

2) When R1 =R2 = H, R3 = R4 = - CH3

Mechanism of ozonolysis of alkenes: Its mechanism takes following steps

1) Oxygen gas is exposed to high voltage to form ozone (O3):

2) Ozone reacts with the alkenes to form a molozonide intermediate.

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3) The molozonide intermediate then breaks down into two fragments.

4) These two fragments then rearrange to form an ozonide intermediate

5) Reduction of the ozonide with reducing agent (zinc in dimethylsulphide) then results in cleavage of the ozonide into two carbonyl compounds.

Importance of ozonolysis of alkenes: Ozonolysis is probably best method for locating the position of the double bonds in unknown alkenes. The oxygenated carbons in carbonyl compounds obtained by ozonolysis are the ones that were joined by a double bond in the original alkenes. Suppose alkenes on ozonolysis gives the carbonyl compounds.

Joining the oxygenated carbons (marked by asterisk) by a double bond, we get the following structure of the unknown alkenes.

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4) DIENES: Alkenes containing two carbon – carbon double bonds are called dienes or alkadienes.

Examples:

Classification of dienes: Based on the position of two carbon – carbon double bonds, dienes are classified into following types –

1) Isolated diene: If the double bonds are separated by more than one single bond, the diene is called isolated diene.

For example –

2) Conjugated diene: If the double bonds are separated by one single bond, the diene is called conjugated diene.

For example –

3) Cumulated diene: If the double bonds are adjacent to each other, the diene is called cumulated diene.

For example –

Structure of allene (Propadiene): In the structure of allene, the central carbon atom is in sp hybridization. It is bonded to two other carbon atoms. The terminal carbons are sp2 hybridized.

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The formation of σ and π bonds of allene is shown by orbital structures as below.

Formation of σ (sigma) bonds

Formation of π (pi) bonds

The carbon atoms in allene are linear and the π bonds formed are perpendicular to each other. There is no delocalization of π electrons. Cumulated dienes are, therefore, less stable than the conjugated dienes.

Structure of allene

Structure of 1, 3 – Butadiene:

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All four carbon atoms in 1, 3 – butadiene are SP2 hybridized. The sp2 hybrid orbitals overlap with each other and with S orbitals of the hydrogen atoms to form C – C and C – H σ bonds. Since the bonds result from the overlap of trigonal SP2 orbitals, all carbon and hydrogen atoms lie in one plane. All bond angles are 1200.

Also, each carbon atom in 1, 3 – butadiene possesses an unhybridized P orbital. The p orbitals are perpendicular to the plane of σ bonds. The p orbital on C - 2 can overlap with the P orbitals on C – 1 and C - 3. The p orbital can C – 3 can overlap with the P orbitals on C - 2 and C - 4. In other words all four P orbitals overlaps with each other to form a large pi molecular orbital as shown in fig. each pair of pi electrons is thus attracted, not by two, but all four carbons.

The overlap of P orbitals of C – 2 and C – 3 in both directions, which allows the π electron to be spread over a large area, is referred to as delocalization. This delocalization of π electrons is responsible for greater stability of 1, 3 – butadiene.

1, 2 and 1, 4 - addition reactions of 1, 3 - butadiene:

1) Addition of halogen acids: 1, 3 – Butadiene reacts with halogen acids (HBr or HCl) to yield a mixture of two compounds. They are 3 – bromo 1 – butane and 1- bromo 2 – butane. The first product results from 1, 2 – addition to one of the double bonds. The second product results from addition to terminal (1, 4) positions with the formation of new double bonds between C -2 and C -3. This latter process is known as 1, 4 – addition. At low temperatures the 1, 2 – addition is preferred, whereas at high temperatures 1, 4 – addition predominates.

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2) Addition of halogens: 1, 3 – Butadiene reacts with halogens (Cl2 or Br2) in the presence of an inert solvent (CCl4) to give a mixture of two dibromo compounds.

3) Addition of hydrogen: 1, 3 – Butadiene reacts with hydrogen in the presence of a catalyst to give a mixture of 1- butane and 2 – butane.

4) Addition of water: 1, 3 – Butadiene reacts with water in the presence of a H2SO4 to give a mixture of 1- butan-3-ol and 2 – butan-1-ol.

5) Polymerization: 1, 3 – Butadiene polymerizes in the presence of a peroxide to give polybutadiene (Buna rubber)

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The structure of the polymer suggests that 1, 4 –addition predominates

Diels –Alder reaction: This involves the treatment of 1,3 –butadiene(conjugated diene) with an alkene or alkyne. No catalyst is required.

The alkene or alkyne used in Diels – Alder reaction is referred to as dienophile (Diene –lover). The product of Diels – Alder reaction is called the adduct. The net result is the formation of two new σ bonds and new π bond at the expense of the three original π bonds.

For example -

6) Diels –Alder reaction of 1, 3 –butadiene with maleic anhydride:

5) ALKYNES: Alkenes are hydrocarbons that contain a carbon – carbon triple bond (C≡C) in their molecules. They have the general formula CnH2n -2 (n = 2, 3, 4, etc.). These compounds are highly unsaturated even than alkenes.

Examples:

Value n in CnH2n-2 Structure Name

2 Acetylene (ethene)

3 Propyne (propyne)

4 1 – Butyne2 - Butyne

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Methods of preparation of alkynes:

1) By dehydrohalogenation of vicinal dihalides: Compounds that contain halogen atoms on adjacent carbon atoms are called vicinal dihalides or vic – dihalides. Alkynes are obtained by treatment of vicinal dihalides with alcoholic KOH followed by sodium amide.

Where R = H or alkyl group.

This method is useful method since the vicinal dihalides are readily prepared from alkenes by the addition of halogens.

For example: 1) When R = H

2) When R = - CH3

2) By dehydrohalogenation of gem dihalides: Compounds that contain two halogen atoms on same carbon atom are called gem dihalides. Alkynes are obtained by treatment of gem dihalides with alcoholic KOH solution.

Where R = H or alkyl group.

For example: 1) When R = H

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2) When R = - CH3

3) Preparation of higher alkynes from terminal alkynes: This method contains two steps –

Step 1) Preparation of sodium salt of terminal alkynes: Sodium salts of terminal alkynes are prepared by treatment of 1 – alkynes with sod amide in liquid ammonia.

Step 2) Preparation of higher alkynes: Sodium salt of alkynes is treated with primary alkyl halide to form higher alkynes.

Where R = H or alkyl group, R1 = Alkyl group and X = -Br, - Cl

For example: When R = H, R1 = - CH2CH3 and X = -Br

Step 1) Preparation of sodium acetylides:

Step 2) Preparation of higher alkynes:

Reactions of alkynes:

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1) Metal ammonia reduction of alkynes: To form a trans- alkenes, when alkynes is undergo reduction with sodium metal in liquid ammonia. During the reduction, two hydrogens must be added to the alkynes with anti – stereochemistry.

Where R = R1 = Alkyl group

For example: When R = R1 = - CH3

Significance of this reaction: This method is used to convert alkynes to trans –alkenes

2) Oxidation of alkynes with KMnO4: The oxidation of alkynes with alkaline potassium permanganate cleaves the molecule at the site of the triple bond to form carboxylic acids.

Acetylene under these condition yields oxalic acid. This degradation reaction of alkynes is useful in determining the structure of alkynes.

Where, R = R1 = H = alkyl group

For examples 1) When R = - CH3, R1 = H

2) When R = - CH3, R1 = H

3) When R = - CH3 = R1

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Acidic nature of terminal alkynes: Terminal alkynes are weak acidic in nature compare to strong base. This is due following reason.

1) Terminal hydrogen of 1 – alkyne can be readily removed by means of strong base. Hence 1- alkyne is considered as weak acids.

2) Terminal alkyne is more acidic than the alkene and alkane becausePKa value of these increases in that order.

PKa = 62 PKa=45 PKa=26

3) Greater the S – character in a hybrid orbital containing a pair of electrons, the basic is that pair of electrons and the more acidic in nature. S – Character is increases in the order of

Therefore terminal alkynes are more basic in nature the alkenes which is more than alkanes.

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Aliphatic Hydrocarbons 9 Hours