ALKANESAlkanes or paraffins are saturated aliphatic hydrocarbons, that is, they contain only carbon-hydrogen (C-H) and carbon-carbon (C-C) single bonds. The general formula of alkanes is CnH2n+2, where n is the number of carbon atoms in one molecule or RH (where R represents the alkyl group CnH2n+1)

Number of carbon atoms

Prefix

1 Meth-2 Eth- 3 Prop-4 But-5 Pent-6 Hex-7 Hept-8 Oct-9 Non-10 Dec-11 Undec-12 Dodec-

Name Molecular formula

Condensed Structural formula

Methane CH4 CH4

Ethane C2H6 CH3 CH3

Propane C3H8 CH3 CH2 CH3

Butane C4H10 CH3 CH2 CH2 CH3

Pentane C5H12 CH3 CH2 CH2 CH2 CH3

Hexane C6H14 CH3 CH2 CH2 CH2 CH2 CH3

Heptane C7H16 CH3 CH2 CH2 CH2 CH2 CH2 CH3 Octane C8H18 CH3 CH2 CH2 CH2CH2 CH2 CH2 CH3 Nonane C9H20 CH3 CH2CH2 CH2 CH2 CH2 CH22 CH2 CH3

Decane C10H22 CH3 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH3

ALKYL GROUPSAn alkyl group is derived from an unbranched alkane molecule by removing one hydrogen atom from the end carbon atom of the alkane. The symbol R is used to represent an alkyl group, CnH2n+1Alkyl groups are named by dropping the suffix -ane from the parent alkane and adding the suffix -yl.

Alkane Name of alkyl group Condensed formulaMethane Methyl

Ethane Ethy

Propane Propyl

Isopropane Isopropyl

Butane Butyl

Isobutane Isobutyl

Pentane Pentyl

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QUESTIONS ANSWERS1. Write the IUPAC names for the following alkanes.

(a)

(b)

(c)

(d)

2. Write the IUP AC names for the following alkanes.

3-Methylpentane

2,4.6- Trimethyloctane(start from right to left)

3-ethyl-2,6-Dimethylheptane(alphabetically)

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NAMING CYCLOALKANESAlkanes having one or more rings of carbon atoms are called cycloalkanes.

1. Give the names for the following cydoalkanes(a) (b)

2. Name the following cydoalkanes(a)

Cyclopentane

(b)

1-Ethyl-2-metylcyclohexane

(c)

1,3 - Dimethylcyclobutane

(d)

1-ethyl-3,4-dimetylcyclohexane

(3)Name the following compounds(a) 2,3-dimethylpentane (b) 3 -ethyl- 3,5 -dimethylheptane

(c) 2,4-dimethylhexane (d) 4-ethyl-2-methylhexane

(4)Draw the structural formula of the following compounds. a) 2-methylhexane (b) 2,4-dimethylpentane

(c) 2,2,3-trimethylpentane (d) 2,2-dimethyl-S-ethyloctane

PHYSICAL PROPERTIES OF ALKANES

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1.Molecular

formula

Physical

states

Solubility Melting

point

Boliling

point

Density Viscosity Combustibility

Colourless

gases

water ethanol/ether

increases

(attractive

van der

Waals

forces

increases)

Increases

(attractive

van der

Waals

forces

increases

Increases

(molecules

are

closer to

each

other)

increases

(more

difficult

for

molecules

to glide

or flow)

decreases (as larger molecules burn with a sootier flame due to the increase percentage of carbon).

CH4 Almost

insoluble readily soluble

but solubility

decreases as

molecular

weight

increases.

C2H6

C3H8

C4H10

C5H12

Colourless

liquids

C6H14

C7H16

C8H18

C9H20

C10H22

C18 onwardsColourless

solids

2. The normal alkanes from C1 to C4 are colourless gases; C5 to C17 are colourless liquids; and from C18 onwards are colourless solids at room temperature and pressure.

3. The alkanes are almost insoluble in water, but readily soluble in ethanol and ether, the solubility diminishing with increase in molecular weight.

4. Generally, when the number of carbon atoms increases, the attractive van der Waals forces between the molecules increases.

This causes (a) the melting point and boiling point to increase (b) the density to increase because the molecules are closer to each other (c) the viscosity to increase because it is more difficult for molecules to glide or flow (d) the combustibility to decrease as larger molecules burn with a sootier flame

due to the increase percentage of carbon.

Normal (unbranched) compound Branched compoundThe normal (unbranched) compound always has the highest boiling point.

Branched isomers have lower boiling points and lower density. Generally, the greater the branching, the lower is the boiling point.As branching increases, the alkane molecule becomesmore spherical in shape and thus, has a smaller surfacearea available for intermolecular attractions.

More spread out and more easily polarised Hence, a larger molecular surface area allows greater contact between molecules and gives rise to higher dispersion forces.

Small, compact and symmetrical molecules As the surface area decreases, contact between adjacent molecules decreases as well.As a result, the strength of the instantaneous dipole induced dipole forces decreases and hence, the boiling point decreasesBut branched isomers have higher melting points.

The boiling point of n-pentane is the highest because a straight chain isomer has the greatest surface area where two molecules can touch each other along the length of the chain. Hence, the van der Waals forces are greater

The boiling point of 2,2-dimethylpropane is the lowest since it has the most branching and the least surface area

The melting point of a substance depends on the arrangement and packing of the molecules in the crystal lattice. The melting point of 2,2-dimethylpropane is the highest among the three isomers because it is more spherical in shape and hence, can be packed as close as possible in the lattice.

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PHYSICAL PROPERTIES

BOILING POINTSThe boiling point of alkanes is lower than of other organic compounds.

Alkane molecules have low polarity, i.e. only weak instantaneous dipole-induced dipole (van der Waals) forces are present between molecules.

The boiling point rises with increasing number of atoms in the molecule.

As the number of electrons in the molecule increases (with increasing number of carbon atoms), the strength of these attractive forces also increases. More energy is needed to break these forces the molecules when boiling.

The boiling points of unbranched alkanes show a regular increase with increasing number of carbon atoms in the molecules.

MELTING POINTSThe melting point of alkanes also increases with an increase in the number of carbon atoms.

However, the increase is not as regular as that observed for the boiling points because the ability of molecules to pack into ordered patterns of solid changes as molecule size and shape changes.

There is an alternation as one progresses from an unbranched alkane with an even number of carbon atoms to the successive one with an odd number of carbon atoms .

MELTING POINTSHowever, if only the melting points of alkanes having odd number of carbon

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atoms are plotted, a smooth increase is shown. This trend is also true for alkanes having even number of carbon atomAlkanes with even number of carbon atoms pack more closely in the solid state. Hence, the attractive forces between the molecules are greater resulting in higher melting point.

DENSITYAlkanes are the least dense of all groups of organic compounds and have densities less than water.

SOLUBILITYAlkanes are almost insoluble in water because of their low polarity and their inability to form hydrogen . However, liquid alkanes are miscible with each other.

PREPARATION OF ALKANES

1. Alkanes are usually prepared through the catalytic reduction of alkenes. An alkene is converted to an alkane when passed with hydrogen over nickel heated at 200 - 300°C .

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Ethene Ethane

2. Other methods of preparation are: (a) decarboxylation of a carboxylic acid or its salt(not a useful general method for the preparation of alkanes)

(b) Wurtz reaction (The Wurtz reaction gives good yields only for even numbered carbon alkanes of high molecular weights).

When a mixture of a carboxylic acid (or its sodium salt) is heated with a soda lime (a mixture of calcium oxide and sodium hydroxide), carbon dioxide is eliminated. The process is known as decarboxylation. ( heating the anhydrous sodium carboxylate salt with soda lime (a mixture of sodium hydroxide and calcium hydroxide)) .

RCOOH + 2NaOH → RH + Na2CO3 + H2O Carboxylic acid Alkane

RCOONa + CaO → RH + CaCO3 Sodium carboxylate Alkane

For example, methane can be prepared from the decarboxylation of sodium ethanoate. CH3COONa + 2NaOH → CH4 + Na2CO3 + H2ONote:

[The alkane produced has less carbon atoms compared to the original carboxylic acid or carboxylate salt used. • Benzene can also be prepared from the decarboxylation of sodium benzoate. C6H5COONa + 2NaOH→ C6H6 + Na2CO3 + H2O

However, other sodium salts decompose to form various products. Hence, this is not ,a useful general method for the preparation of alkanes sinse the separation of the products is usually difficult. ]

CH3CH2COONa + NaOH→ C2H6 + CH4 + H2 (44%) (20%) (33%)

+ Unsaturated compounds

In a Wurtz reaction, an ethereal solution of an alkyl halide is treated with sodium to form an alkane.

2RX + 2Na →R - R + 2NaX For example, 2CH3CH2Br + 2Na → CH3CH2CH2CH3 + 2NaBr

‘Wurtz reaction: The action of sodium on alkyl halide in ether. 2R-X + 2Na ~ R-R + 2NaX For example, 2CH3-1 + 2Na ~ C2H6 + 2NaI

c. Kolbe's method: Electrolysis of a concentrated aqueous solution of the sodium salt of a carboxylic add, for example, sodium ethanoate

Cathode: 2H+(aq) + 2e- → H2(g)

Anode : 2CH3COO-(aq) → C2H6(g) + 2CO2(g) + 2e-

CHEMICAL PROPERTIES OF ALKANESAlkanes have relatively low chemical reactivity and are inert to acids, bases and most other common laboratory reagents.

This is because alkanes do not have any unshared pair of electrons. polar bond, electron-deficient atom or an atom with an expandable octet

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.Hence, alkanes are unreactive towards polar reagents such as water, H2Oand ammonia, NH3 ,

(not readily attacked by common chemical reagents which are usually highly polar compounds)

As such, alkanes react with nonpolar reagents such as bromine and oxygen under certain conditions.

This is because the carbon - hydrogen bond is only slightly polarised.Molecules of alkanes also have no unshared electrons to offer sites for attack by acids.

Why are alkanes inert ? A reactive site in a molecule usually has one or more unshared pairs of electrons, a polar bond, an electron-deficient atom or an atom with an expandable octet. Alkanes have none of these .

The general chemical reactions of alkanes are: (a) Halogenation (b) Oxidation (c) Dehydrogenation

HALOGENATION OF ALKANES 1. At room conditions, alkanes do not react with halogens as the activation energy is too high. Due to chemical inertness, alkanes do not readily react with halogens in the dark at room temperature.

2. In the presence of ultraviolet light, heat (around 4000C) or catalysts, alkanes react with halogens such as chlorine and bromine to form haloalkanes or alkyl halides(halogen atoms will substitute hydrogen atoms in the alkanes to form haloalkanes.)

RH + C12 → RCI + HCI For example, CH3CH3 + C12 → CH3CH2Cl + HCl (a) The order of ease of substitution is : tertiary hydrogen> secondary> primary. (b) The mechanism is found to be a free radical substitution.

3. Substitution is a reaction where an atom or a group of atoms in a molecule is substituted different atom or group of atoms. Since the reaction takes place in light, it is called a photochemical reaction. 3. When chlorine (a halogen) is mixed with methane and exposed to sunlight, yellowish-green colour of chlorine fades and a steamy acidic fumes of hydrogen chloride is evolved. CH4(g) + CI2(g) → CH3CI(g) + HCl(g)

Further substitution is possible, depending on the amount of chlorine and methane present.The formation of CH3Cl may be followed by the formation of dichloromethane,CH2Cl2 ,trichloromethane, CHCl3 and tetrachloromethane, CCI4. A mixture of products is obtained.

CH3CI + Cl2→ CH2Cl2 + HCl CH2Cl2 + Cl2 → CHCl3 + HCl CHCl3 + Cl2 → CCl4 + HCI

FREE RADICAL SUBSTITUTION MECHANISM OF CHLORINATION OF METHANE:FREE-RADICAL SUBSTITUTION

(a) A free radical is a chemical species which possesses an unpaired electron.

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(b) Free radicals are formed in two general ways: (i) through homolytic cleavage of bonds: A - B →A· + B· (ii) by reaction of molecules with other free radicals: A - B + C·→ A· + B - C (c) Homolytic cleavage of sigma bonds can be made to occur using heat or light (photolysis) (d) The mechanism consists of three steps; initiation, propagation and termination. Initiation (where radicals are formed)

Chain Propagation(where products are formed and the radicals are regenerated)

Termination(where radicals are removed)

(i) When a mixture of chlorine and alkane such as methane, CH4 is exposed to ultraviolet light, chlorine molecules undergo homolytic cleavage. (ii) The energy required for homolytic fission of the CI-Cl comes from the light absorbed (photochemical homolytic fission). (iii) As the two chlorine atoms in the bond are identical (the electron pair forming the bond are equally shared by the two atoms), the bond will break homolytically, i.e. one electron goes to each chlorine producing two chlorine atoms, each with one unpaired electron. Thus, each has an unpaired electron and is known as a free radical.

Initiation: Cl2 → CI· + CI·

iv) (Methane does not undergo cleavage because the C - H bond dissociation enthalpy is 412 kJ mol-1, compared to the CI - CI bond dissociation enthalpy of 242 kJ mol-1)It is easier to break the CI-Cl bond (+242 kJ mol-1]) than the C-H bond (+412 kJ mol-

1]).

Thus. in the initiation step, only chlorine radicals are produced although both the chlorine and methane molecules are exposed to sunlight.

(i) The chlorine radical (or atom) formed (possess an unpaired electron ) is very reactive. It is reactive enough to break the C-H bond when it attacks the methane molecule, producing hydrogen chloride and a methyl free radical. CH3· which then combine with CI2, molecules to form the product, chloromethane: CH3C1. CH4 + CI· → HCI + CH3·

(ii) The methyl free radical can react with a chlorine molecule to produce chloromethane CH3 C1, and a new chlorine free radical.

CH3· + Cl2 →CI· + CH3Cl

(iii) These two steps, 2 and 3, enable the reaction to continue. In step 2, a chlorine free radical is used up. Meanwhile, in step 3, a new chlorine free radical is produced which allows repetition of step 2. (iv) Steps 2 and 3 constitute a chain reaction and are called chain-propagating steps.

(i) Thousands of molecules of chloromethane are formed for every photon of light absorbed. The yield is due to the chain reaction - steps 2 and 3. The reason why the yield is not higher is that free radicals can combine with each other and bring the chain to an end -termination steps. The possible reactions that may terminate the chain are as folIows :

(ii) These, or any other termination steps will remove the free radicals and bring the chain reaction to a stop.

(iii) The reaction terminates when any two free radicals collide.

Termination: CI·. + CI· → Cl2

CI· + CH3· → CH3ClCH3·+ CH3· → CH3CH3

The overall equation for the formation of chloromethane is : CH4 + Cl2 → HCI + CH3CI

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OXIDATION OF ALKANES

(a) At room conditions, alkanes do not react with oxygen. (b) All alkanes readily burn in excess of air or oxygen to form carbon dioxide and water.

(c) Controlled oxidation in limited oxygen and under various conditions, leads to different products such as acids, mixed ketones and alcohols. (d) Oxidising agents such as potassium permanganate readily oxidise a tertiary hydrogen atom to a hydroxyl group. (CH3)3CH + [O] → ( CH3)3COH

1. Alkanes do not react with air at room temperature but if heated, they burn readily to give carbo dioxide and water in an exothermic reaction. For example, the combustion of propane:

The exothermic reaction accounts for the use of many alkanes as fuels.

1. Liquid and solid alkanes need to be vaporised first before they burn. For example, liquid is vaporised in the internal combustion engine before it burns.

~ If the supply of air is limited, the combustion of alkanes is incomplete, producing carbon monoxide and carbon (in the form of soot) along with partially oxidised hydrocarbons. For example… incomplete combustion of propane:

C3H8(g) + 7/2 O2(g) 3CO(g) + 4H2O(g)

2. Carbon monoxide is a poisonous gas which bonds to the iron of haemoglobin in the blood i:l preference to oxygen which could result in death. Hence, all gas burning appliances should be adequately ventilated

'Combustion reactions can be used to determine the molecular formulae of hydrocarbons. ·2' A measured volume of a hydrocarbon is mixed with a large excess of oxygen and the total volume

is noted. :'1 The mixture is then sparked until combustion is complete. C I After cooling to the original conditions of temperature and pressure and when all the water formed

has condensed, the volume of the remaining gas is measured. i I The volume of carbon dioxide produced is found from the decrease in the volume after shaking

the yielded mixture with concentrated potassium hydroxide. e, If the molecular formula of the hydrocarbon is CvHI" the combustion reaction is represented

by the following equation: . y y

CXHy + (x + 4)02 -7 xC02 + "TH20

~ V + (x + ~)V -7 xv. where V = volume

~ Volume of carbon dioxide produced = x(volume of hydrocarbon used) ................................ I

~ Volume of oxygen used = (x + ~ ) (volume of hydrocarbon used) .......................................... ' 2. Thus, the values of x and y can be obtained by solving the two equations abo\eillcipal component of petrol is an isomer of octane, 2,2,4-trimethylpentane, or more commonly h--=: as iso-octane . •• : ::}:-aw the condensed structural formula of iso-octane . • ',"":-ite a balanced equation for the complete combustion of iso-octane .

Test yourself 6 pg 55ane is used as a fuel in camping gas stoves. Write balanced equations for the combustion of ~3l1e in .i plentiful supply of air (oxygen) '·rb· a limited supply of air

Example 7 15 cm3 of a gaseous hydrocarbon was mixed with 65 cm3 of oxygen (excess) and the mixture was sparked. On cooling to the original conditions of temperature and pressure, the volume measured was 50 cm 3, of which 30 cm3 was absorbed by concentrated potassium hydroxide solution. Determine the molecular formula of the

10

hydrocarbon .

Solution:

Assume that the molecular formula of the hydrocarbon is C H . x y

y y T us, C,Hy + (x + 4)02 ~ xC02 + 2"'H20 Volume of CO2 produced = 30 cm3 Volume of CO2 produced = x(volume of hydrocarbon used) => 30 =x(15)

:. x = 2 Volume of oxygen used = 65 - (50 - 30) = 45 cm3

Volume of oxygen used = (x + ~ ) (volume of hydrocarbon used) => 45 =

(2 + ~)(15)

.. y=4

That is, the molecular formula of the hydrocarbon is C2H4•

DEHYDROGENATION(a) Alkanes can lose hydrogen when heated at a high temperature in the presence of a catalyst to form carbon black (used in the manufacture of rubber tyres, black paints and black shoe polish) and alkenes. (b) Many alkenes can be manufactured by high temperature dehydrogenation processes, using a chromium oxide-alumina catalyst.

10 cm3 of a gaseous hydrocarbon was mixed with 100 cm3 of oxygen (excess) and the mixture sparked. On cooling to the original conditions of temperature and pressure, the volume measured 85 cm3, of which 20 cm3 was absorbed by concentrated potassium hydroxide solution. Determine molecular formula of the hydrocarbon.

PG 56 TEST YOURSELF 7

Cycloalkanes, CI1H2n (a) Cycloalkanes exhibit similar chemical properties as alkanes, such as free radical substitution and oxidation. Br + HBr (b) However, cycloalkanes with smaller carbon rings such as cyclopropane may be more reactive due to the angle shain in the compound. Larger sized rings (C > 8) behave almost the same as straight-chained alkanes.

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Catalytic cracking of alkanesDraw cis and trans isomers of alkenes(propene and butane)Enantiomers(mirror image)Calculations on half life, molarity(MaVa = a MbVb b

pH =-log[H+]

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Crude Oil as a Source of Energy and Chemicals

1. Crude petroleum contains alkanes (straight and branched chains, from CI to C40), cycloalkanes and aromatic hydrocarbons. Natural gas consists chiefly of the first six alkanes.

2. Except for the low-boiling hydrocarbons, no attempt is made to separate the individual hydrocarbons. The crude oil is fractionated by continuous distillation into four main fractions, i.e. petrol. kerosene, heavy oil and lubricating oil.

Fraction Approximate Boiling Use composition point (0C)

Light petrol C5 - C7 20 - 100 Solvent Petrol C6 - CII 70 - 200 Motor fuel

Kerosene CI2 - CI6 200 - 300 Lighting Heavy oil CI3 - CIS > 300 Fuel oil

Lubricating oil CI6 - C20 > 300 Lubricants

Grease, vaseline CIS - C22 > 300 Pharmaceutical

preparations Paraffin wax C20 - C30 > 300 Candles, waxed paper

Bitumen C30 - C40 > 300 Asphalt tar

USES OF ALKANES:

(a) Most of the alkanes readily burn in excess of air or oxygen and are usually used as fuels.

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