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ALKANES

1.1 Methane

The simplest member of the alkane family is methane, $C{H_4}$ . Our approach would be to first analyse this single compound and study its reactions, then we will see how we can apply what we learn about it to any alkane.

Structure of methane: (already covered in GOC)

You have an idea about what the molecule of methane will look like, based on what you have studied in General Organic Chemistry. Each of the four hydrogen atoms is bonded to the carbon atom by a covalent bond. The hybridisation of carbon atom is sp3 in methane.

1.1.1 Source:

The principle source of alkanes is petroleum and natural gas. Methane is the main constituent of natural gas (upto 97% methane). It is also called as marsh gas. If methane is wanted in very pure from, it can be separated from the other constituents of natural gas (mostly other alkanes) by fractional distillation.

Interesting Fact

According to one theory, the origins of life go back to a primitive earth surrounded by an atmosphere of methane, water, ammonia, and hydrogen. Lightning discharges broke these simple molecules into reactive fragments (free radicals) these combined to form larger molecules which eventually yielded the complex organic compounds thus creating life. The methane generated in the final decay of once ‘living organism’– is the very substance from which the ‘living organism’ was derived.

Combustion of methane produces carbon dioxide and water. When natural gas burns, combustion of methane is the main reaction taking place. The desired result for us is the heat, but the product that is grabbing our attention these days is carbon dioxide. It is the chief of the green house gases, whose accumulation in the stratosphere is causing global warming.

The quantity of heat evolved when one mole of a hydrocarbon is burned to carbon dioxide and water is called the heat of combustion.

$6C{H_4} + {O_2}2HC \equiv CH + 2CO + 10{H_2}$

Controlled partial oxidation of methane produces acetylene, carbon monoxide and hydrogen.

(ii) Halogenation (Chlorination)

In presence of ultraviolet light or at a temperature of 250-400ºC, mixture of methane and chlorine reacts vigorously to yield hydrogen chloride and methyl chloride $(CH_{3}Cl)$ . The methyl chloride formed can itself undergo further substitution to form hydrogen chloride and $CH_{2}Cl_{2}$ , methylene chloride $\left( {{\rm{C}}{{\rm{H}}_2}\,{\rm{ = methylene}}} \right)$ . The chlorination of methane is an example of a chain reaction. A reaction that involves a series of step, each of which generates a reactive substance that brings about the next step is called chain reaction.

Mechanism:

then steps 2 and 3 repeat.

The first step is the homolytic cleavage of the chlorine - chlorine bond in presence of sunlight. The chlorine atom (free radical) formed is extremely reactive because of its tendency to gain an additional electron and thus complete its octet. Chlorine atom collides with methane molecule producing methyl free radical and hydrogen chloride. You might wonder why not chlorine atom formed after step 1, reacts in the following manner:

Step 2: $Cl \cdot + C{H_4}\xrightarrow{{}}C{H_3}Cl + H \cdot$

Theoretically, there is nothing wrong in considering the above step but in reality it doesn’t take place. Let us explore the reason by comparing the two reaction steps.

(i) $Cl \cdot + C{H_4}{{}}CH_3^{} \cdot + HCl & \Delta H = + 1kcal$

(ii) $Cl \cdot + C{H_4}{{}}C{H_3}Cl + H \cdot & \Delta H = + 20kcal$

In both the above reactions chlorine atom attacks a methane molecule. It can either attach itself to carbon and expel a hydrogen atom or vice-versa. There is thus a competition between the two reactions, the faster reaction wins. Simply looking at the $\Delta$ H values we can see that for reaction (i), $\Delta$ H is much lower than that of reaction (ii). Therefore, we can safely assume that the energy required to bring about reaction (i) is much less than that required for reaction (ii). So we see why reaction (i) proceeds faster than reaction (ii).

Note:-

The point here is not that 20 kcal is in itself too high a barrier for reaction (ii) to occur (reactions do take place which have $\Delta$ H value around 20 kcal). The point here is that a reaction with $\Delta$ H value 20 kcal cannot compete successfully with a reaction with much less $\Delta$ H value.

Evidence for the mechanism of chlorination of methane.

(a) Methane and chlorine do not react in the dark at room temperature. (b) Reaction takes place readily in the dark at high temperatures. (c) Reaction also takes place under the influence of ultraviolet light at room temperature (d) The wavelength of light that induces chlorination is the same that causes dissociation of chlorine molecules. (e) In the light induced reaction, many molecules of methyl chloride are obtained for each photon of light that is absorbed. (f) The presence of a small amount of oxygen slows down the reaction for a period of time, after which the reaction proceeds normally (refer next topic ‘Inhibitors’)

Inhibitors:

Addition of oxygen slows down the chlorination of methane. Oxygen reacts with methyl radical to form a new free radical.

Due to formation of less reactive $C{H_3}--O--O \cdot$ radical the overall reaction slows down. Converting one radical into radical prevents the formation of many CH3Cl molecule. After all the oxygen molecules present have combined with methyl radicals, the reaction is free to proceed at its normal rate.

Concept of Transition state and Energy of activation

When two atoms react to form a product, there is some minimum amount of energy required for the reaction to take place. To understand this concept let us look closely at the following reaction.

What must happen if the above reaction is to occur? First of all chlorine atom and methane molecule must collide. Hydrogen and chorine bond will only form when they are in close contact. Note that not every collision will lead to the formation of bond between hydrogen and chlorine.

Therefore, to be effective, collision must provide a certain minimum amount of energy. This minimum energy provided by a collision for reaction to occur is called the energy of activation, $E_{act}_$ . Its source is the kinetic energy of the moving particles.

Finally, in addition to being sufficiently energetic, the collision must occur when the particles are properly oriented. Potential energy change and the progress of reaction can be studied efficiently through the diagram given below:

Note:-

There is an energy of activation for nearly every reaction where bonds are broken.

$2Cl \cdot \to C{l_2}\,\,\,{E_{act}}$

for this reaction is zero because no bonds are broken, so this reaction is expected to take place very easily.

When a chemical reaction proceeds, reactant is not immediately converted into products. It is a kind of continuous process where reactant gradually transforms into product, going through an intermediate structure which has maximum energy. This intermediate structure is known as transition state.

In the transition state the carbon-hydrogen bond is not completely broken and the hydrogen - chlorine bond has started to from but is not complete. Dashed lines in the transition state represent partly broken or partly formed bonds.

1.2 PREPARATION OF ALKANES

1.1.1 Hydrogenation of alkenes and alkynes:

Reduction of alkenes and alkynes with $H_2$ in the presence of Ni, Pt or Pd catalyst at 200-300 ºC produces alkanes:

1.1.2 Hydrolysis of Grignard Reagent:

Metallic magnesium turnings reacts with a solution of an alkyl halide in dry ethyl ether, to produce Grignard reagent, after Victor Grignard.

The delta negative charge present on R (alkyl) group attacks hydrogen atom of water molecule. Grignard reagent forms hydrocarbon with compounds containing acidic hydrogen e.g.

Reaction with terminal alkynes leads to formation of a new grignard reagent.

1.1.3 Decarboxylation of sodium salt of fatty acids:

Distillation of sodium salt of carboxylic acid with sodalime $\left( {NaOH + CaO} \right)$ produces alkane.

This reaction can be employed for decreasing the length of carbon chain.

1.1.4 Reduction of alkyl halides:

(a) Alkyl halides react with zinc and aqueous acid to produce an alkane.

$R - X\;\; + \;\;Zn + HX\; \to R - H\;\;\; + \;\;\;Zn\;{X_2}$

(b) Primary and secondary alkyl halides are readily reduced to alkenes by lithium aluminium hydride $\left( {LiAl{H_4}} \right)$ . Tertiary halides give mainly alkenes when reacted with .

Sodium borohydrides $\left( {NaB{H_4}} \right)$ reduces only secondary and tertiary halides, but not primary.

Triphenyltin hydride $\left( {P{h_3}SnH} \right)$ can be used to reduce all three types of alkyl halides.

1.1.5 Reduction of alcohols, carbonyl compounds, acids and acid derivatives:

Red phosphorus and HI is used to reduce alcohols, aldehydes, ketones, acids and their derivatives to alkanes.

1.1.6 Wurtz Reaction:

Alkyl halide is treated with sodium in ether solution to produce alkane with greater number of carbon atoms than the alkyl halide.

The yields are best for 1º alkyl halides and least for 3º alkyl halides. In this reaction two alkyl group is coupled to produce higher alkane. Mixture of products are obtained when we use different alkyl halides.

Therefore, best yield is obtained when the alkane contains an even number of carbon atoms and is symmetrical. The reaction is believed to involve free radicals as intermediates.

1.1.7 Corey House Synthesis:

This coupling method to produce higher alkanes can be used for obtaining alkanes containing odd number of carbon atoms. Any alkyl halide (primary, secondary or tertiary) can be used.

The reaction follows ${S_N}2$ mechanism. If $R' - X$ is secondary or tertiary, elimination products are also formed. [You will study in detail about substitution and elimination mechanisms in later chapters.]

1.1.8 Kolbe’s electrolytic method:

Sodium or potassium salts of carboxylic acid is electrolysed which produces alkane.

${C_2}{H_5}COONa \to {C_2}{H_5}CO{O^ - } + N{a^ + }$

At anode:

${C_2}{H_5}CO{O^ - } \to {C_2}{H_5}COO \cdot + {e^ - }$

${C_2}{H_5}COO\, \cdot \to {C_2}{H_5} \cdot + C{O_2} \\$

Now alkyl free radical formed can undergo following steps.

(i)

(ii)

(iii)  (disproportionation reaction)

At cathode:

${H^ + } + {e^ - } \to H \cdot$

$H \cdot + H \cdot \to {H_2}$

1.3 Physical properties of Alkanes and Cycloalkanes

(a) Boiling Points: The boiling points of the unbranched alkanes show a regular increase with increasing molecular weight. Branching of the alkane chain lowers the boiling point.

Above behaviour can be explained by the Van der Waals forces. With unbranched alkanes as the molecular weight increases, molecular surface area also increases. With increasing surface area, the van dar waals forces between molecules increase; therefore, more energy is required to separate molecules from one another and cause boiling. Chain branching, on the other hand, makes a molecule more compact, reducing its surface area and with it the strength of the van der waals forces operating between it and adjacent molecules. This lowers the energy required for boiling, thus lowering the boiling point.

(b) Melting points: Unbranched alkanes do not show smooth increase in melting point with the increase of molecular weight. The increase is some what zig-zag in manner as we increase the molecular weight. Chains with even number of carbon atom melts at a higher temperature than the ones having odd number of carbon atoms.

Alkane chains with an even number of carbon atoms pack more closely in the crystalline state. As a result, attractive forces between individual chains are greater and therefore melting points are higher.

In case of branched chain alkanes, structures having high symmetry result in abnormally high melting points.

(c) Density: Of all the groups of organic compounds, alkanes and cycloalkanes are the least dense compounds. All alkanes and cycloalkanes have densities considerably less than 1.00 $g\;m{L^{ - 1}}$ (density of water at 4 ºC). That is why petroleum floats on water.

(d) Solubility: Like dissolves like. Alkanes and cycloalkanes being non polar are almost insoluble in water. They dissolve in solvents of low polarity for eg. benzene carbon tetrachloride, and other hydrocarbons.

1.4 Reactions of Alkanes

1.4.1 Halogenation:

We studied in chlorination of methane that the reaction is initialised by sunlight. When higher alkanes are used, a mixture of all possible isomeric monochlorides is obtained. Their yield ratio depends upon the reactivity of primary, secondary and tertiary hydrogen atoms:

The order in case of substitution of hydrogen is tertiary > secondary > primary.

Fluorination of alkanes is very fast and is explosive in nature. Chlorination is not vigorous but generally substitutes more than one hydrogen atoms.

Bromination is similar to chlorination but reacts slowly and selectively. Iodination is reversible but it may be carried out in presence of an oxidising agent such as $HI{O_3},HN{O_3},HgO$ etc.

$C{H_4} + {I_2} \to C{H_3}I + HI$

$5HI + HI{O_3} \to 3{I_2} + 3{H_2}O$

Reactivity of ${X_2}:{F_2} > C{l_2} > B{r_2} > {I_2}$

We have already studied the mechanism of halogenation of methane.

In complex alkanes, substitution of different kind of hydrogen atom gives a different isomeric product. Their yield depends on following factors:

(1) Probability factor: More the probability of a particular type of hydrogen atom more will be yield of alkyl halide corresponding to substitution of that hydrogen atom. For eg. in $C{H_3} - C{H_2} - C{H_3}$ , there are six 1º H and two 2º H. The probability of substituting a 1º H to 2º H is 6 : 2 or 3 : 1.

(2) Reactivity of hydrogen atom: The order of reactivity of hydrogen atom is 3º>2º>1º

(3) Reactivity of halogen radical: If the halogen radical is reactive then it will be less selective and yield will be more influenced by the probability factor along with the reactivity of hydrogen atoms. Halogen radical which is less reactive and more selective will yield products governed exclusively by reactivity of hydrogen atoms.

Let us consider monochlorination of propane. It has two types of hydrogen - 1º and 2º. Thus, it will result in formation of two products, one by replacement of primary hydrogen and other by replacement of secondary hydrogen. Also, the percentage of the two products formed in latter case will be more because 2º radical, formed as an intermediate, will be more stable than 1º. In practice, the following results are obtained.

The ratio of products can’t be explained on the basis of stability of free radical alone. It depends upon two factors.

(i) Number of hydrogens which can be replaced i.e. probability. (ii) Rate at which hydrogen can be replaced i.e. reactivity. In case of chlorination, the relative rate of replacement of hydrogen by chlorine free radical is:

HYDROGEN - 1º : 2º : 3º RELATIVE RATES - 1 : 3.8 : 5

The relative amounts and percentage of compounds formed is calculated using the formula Relative amount = number of particular type of hydrogens × reactivity factor of that hydrogen.

Percentage yield =

Let us calculate the relative amounts and percentage yields of products formed in case of chlorination of propane.

Relative amount of 1-chloropropane = No. of 1º H × Relativity factors of 1º H

= 6 × 1 = 6

Percentage yield of 1-chloropropane =

Percentage yield of 2-chloropropane =

Taking another case, monochlorination of isobutane:

It will again give two products.

Relative amount of 1- Chloro isobutane = No. of 1º H × Relativity factors

= 1 × 9 = 9

Relative amount of 2-Chloro isobutane = No. of 3ºH × Relativity factor

= 1 × 5 = 5

Percentage yield of 1-chloro isobutane =

Percentage yield of 2-chloro isobutane =

However, in case of bromination, the relative rates of replacement of different hydrocarbons by bromine free radical has been found to be in the ratio:

HYDROGEN - 1º : 2º : 3º

RELATIVE RATES - 1 : 82 : 1600

This is because bromine is more selective than chlorine in replacement of hydrogen. Thus, it is possible to carry out exclusive bromination at tertiary positions.

eg.

Relative amount of 1-bromoisobutane = No. of 1º H × Relativity factor

= 9 × 1 = 9

Relative amount of 2-bromoisobutane = No. of 3º H × Relativity factor

= 1 × 1600 = 1600

Percentage yield of 1-bromoisobutane =

Percentage yield of 2-bromo isobutane =

1.4.2 Oxidation:

All alkanes burn in excess of air or oxygen to form carbondioxide and water

Note:- Remember the following series where next member is obtained by oxidising the previous one.

[O]: represents oxidiation, [H]: represents reduction, depending on how strong oxidising or reducing agent is, you can obtain different products. Controlled oxidation of alkane leads to a mixture of ketones and alcohols.

1.4.3 Sulphonation:

Hydrogen atom of alkane is replaced by sulphonic acid group, $S{O_3}H$ . The order of ease of hydrogen atom is tertiary > secondary > primary.

1.4.4 Nitration:

Alkanes react with nitric acid to produce nitro alkanes, where hydrogen atom is replaced by nitro group $\left( {N{O_2}} \right)$ .

ALKENES

Alkenes are unsaturated hydrocarbon which contain the carbon-carbon double bond. Alkenes are also known as olefins. The simplest olefin, ethene, was called olefiant gas (Latin: oleum-oil + facers - to make) because gaseous ethene $\left( {{C_2}{H_2}} \right)$ reacts with chlorine to form ${C_2}{H_4}C{l_2}$ , a liquid (oil).

Interesting Fact

Alkenes play a very important role in our daily life. It is the alkene and its isomeric structures which are involved in the chemistry of vision. Actual analysis of how we see things is complex that involves the physics of the eye and biology of the brain. But, there is one event that is purely a chemical transformation and which initiates the whole viewing ‘process’. The event is so simple and so eligant that it has been adopted as the basis of vision in every form of animal life. When light strikes rhodopsin (a conjugated protein in the retina of mammals) it does just one thing: it transforms the cis form of retinal (part of rhodopsin) into its trans form.

Every time you see something i.e. light falls on your retina the first step is the above transformation from cis to trans that then (after a series of chemical reactions) generates the nerve impulse that let us see.

Source:

Petroleum and natural gas chiefly contain alkane, which are the primary source of organic chemicals. By cracking of alkanes we obtain many reactive substances: benzene, toluene, xylene and the smaller alkenes; ethylene, propylene and the butylenes.

Interesting Fact

Ehylene is the organic compound consumed in the largest amount by the chemical industry-and it ranks fifth among all compounds, following only sulphuric acid, lime, ammonia, and oxygen.

2.1 Preparation

2.1.1 Dehydrohalogenation of alkyl halides.

When alkyl halides is treated with a hot concentrated alcoholic solution of a strong base like potassium hydroxide, elimination of hydrogen halide takes place leading to the formation of alkene.

In this reaction, the two leaving groups are lost from the adjacent carbon atoms. The carbon atom from which halogen is lost is called as 1 or carbon atom the carbon atom from which hydrogen is lost is called as 2 or $\beta$ carbon. Therefore this type of elimination is also known as 1,2-elimination or $\beta$ -elimination. When more than one $\beta$ -carbon is present then more than one product is formed.

The major alkene is generally the stable one; (substituted alkene is more stable). The alkene having maximum number of alkyl group attached to is the major product. This is known as saytzeff rule.

If the size of the base is large, then hydrogen is lost from the less substituted b-carbon and in this case less stable alkene becomes the major product. This is known as Hoffmann’s rule.

Note:- The detailed mechanism for elimination will be discussed in the ‘Alkyl Halide’ chapter.

2.1.2 Dehydration of alcohols:

This is also an example of 1,2 elimination but in this case the elimination is catalyzed by acid. Alcohol is converted into an alkene by dehydration: elimination of a molecule of water.

Some of the reagents which can be used for bringing about dehydration of alcohol are conc. ${H_2}S{O_4},\;{H_3}P{O_4},\;{P_2}{O_5},\;A{l_2}{O_3}\;{\rm{or}}\;B{F_3}$

Ease of dehydration of alcohols: 3º>2º>1º

The above order can be explained once we look into the mechanism of the dehydration of alcohols.

(1)

(2)

(3)

The first step is the protonation of the alcoholic group, after which the water molecule is lost with the formation of a carbonium ion. $[{H_2}O$ is a better leaving group than –OH therefore the reaction is catalysed by acid]. Since the carbonium ion is formed as an intermediate therefore the ease of dehydration order 3º > 2º > 1º can be explained by the relative stabilities of the carbonium ions. More stable the carbonium ion, more easily it will be formed.

Note:- In any mechanism where carbocation (or carbonium ion) is formed as an intermediate there is always a possibility of rearrangement of carbocation where an atom or a group of atoms migrate from a carbon atom to another carbon atom so that a more stable carbocation is formed. Cyclic ring expansions can also take place in order to form more stable carbocation.