Aromatic Hydrocarbons:

Aromatic hydrocarbons contain a higher percentage of carbon than the corresponding aliphatic compounds and contain at least six carbon atoms.  They are closed chain or cyclic compounds.

Benzene C₆H₆ is the simplest aromatic compound. It is regarded as the parent substance from which practically all the aromatic compounds are derived.  Hence, aromatic chemistry is the chemistry of benzene and all the compounds structurally related to it.

Aromatic hydrocarbons are defined as cyclic carbon compounds which are related to benzene and at least contain one aromatic (benzene-like) nucleus. e.g. benzene, naphthalene etc.

The general formula of arenes is c_nH_{2n – 6y}, where y is the number of the benzene ring and n is not less than 6.

The benzene ring found in aromatic compounds have following structure. It is called the benzene nucleus.

Characteristics of Aromatic Compounds:

Main characteristic features in which the aromatic compounds differ from aliphatic compounds are given below

• Aromatic compounds contain a higher percentage of carbon than the corresponding aliphatic compounds. Benzene C₆H₆ contains 2.3% of carbon while corresponding aliphatic compound hexane C₆H₁₄ contains 83.7% of carbon.
• Due to the higher percentage of carbon, aromatic compounds burn with sooty or smoky flame while aliphatic compounds burn with a luminous flame.
• Aromatic compounds are closed chain compounds whereas aliphatic compounds have open chain structures.
• The nucleus (ring of atoms) in aromatic compounds is highly stable.
• The aromatic compounds, though they contain double bonds, differ from alkenes, in fact, that they are quite stable, and do not undergo addition reactions easily. Aromatic compounds show substitution reactions like halogenation, nitration, and sulphonation.
• They give nitro derivatives when heated with concentrated nitric acid ‘(nitration). Aliphatic compounds do not form nitro derivatives easily.
• When treated with concentrated sulphuric acid, aromatic compounds undergo sulphonation which is not very common with aliphatic compounds.
• The hydroxy derivatives of aromatic compounds (phenols) are acidic in nature, while the hydroxy derivatives of aliphatic compounds (alcohols) are neutral.
• Chlorine and bromine form addition as well as substitution product with aromatic compounds.
• Aromatic amines undergo diazotisation while aliphatic amines do not undergo diazotisation.
• Aromatic halogen compounds are less reactive than alkyl (aliphatic) halides.
• Homologues of benzene get oxidized to benzoic acid, irrespective of the length of the chain.
• Friedel-Craft’s reaction, Perkin reaction, mercuration and chloromethylation are only shown by aromatic compounds.

Examples and Names of Some Arenes:

Monocyclic Aromatic Hydrocarbons:

Polycyclic Aromatic Hydrocarbons (PAH):

• PAH’s (polycyclic aromatic hydrocarbons) are likely to be some of the most potent carcinogens (chemicals that cause cancer) known to science. They are found in coal and all fossil fuels.
• PAH containing isolated benzene rings

• PAH containing fused or condensed benzene rings

PAHs as Carcinogens:

PAH’s are likely to be some of the most potent carcinogens (chemicals that cause cancer) known to science. They are found in coal, all fossil fuels, (the main compounds in coal tar) and cigarette smoke. PAH’s are flat multi-ring structures that mimic PAH’s have the perfect structure to slip in between the steps of the DNA ladder (intercalation). DNA base pairs. PAH’s adjusts themselves in between the base pairs in DNA disrupting DNA replication and causing mutations in them. Cancer is caused by a series of DNA mutations.

PHAs have been shown to cause cancer in laboratory animals, but not in Humans. There are no “safe levels” defined, so PAH’s are largely overlooked in public health considerations. They are more dangerous than asbestos.

Sources of Aromatic Hydrocarbons:

Coal:

Carbonization or Coking of a Coal:

Carbonization of coal is also known as coking of coal. The process consists of thermal decomposition of coals either in the absence of air or in a controlled atmosphere to produce a carbonaceous residue known as coke.

Carbonization can be carried out at low temperature (450-750 °C)or high temperature (above 900 °C). Low-temperature carbonization is used to produce liquid fuels while high-temperature carbonization is used to produce gaseous products.

Coke (about 70%), Coal tar (4-5%), Ammonia liquor (8-10%) and coal gas (17%) are the main products of carbonization of coal.

Coke: Coke is a high-carbon product obtained by the destructive distillation of coal. Coke is greyish-black in colour and is a hard, porous solid. The amount of carbon content in coke is so high that it is said to be an almost-pure form of carbon. The burning of a coke produces very little smoke. The most common use of coke is as a domestic and industrial fuel. It is used in a blast furnace and to manufacture steel and many other materials.

Coal tar: It is a highly viscous black liquid with an unpleasant odour. It also has an extremely unpleasant smell. Coal tar contains benzene, toluene, xylenes, naphthalene, anthracene, water (neutral substances), phenol, cresols (acidic substances) and pyridine, quinoline (basic substances). Coal tar is widely used to manufacture paints, perfumes, synthetic dyes, photographic material, drugs, explosives, insecticides, pesticides, naphthalene balls, anti-dandruff and lice-repelling shampoos, soaps and ointments.

Coal gas: This is the gaseous product obtained by carbonization of coal. It is a highly flammable gas and its main constituent is methane. It is mainly used as a domestic and industrial fuel and also used for street lighting in old days.

Ammonical Liquor: Ammonia is used for making fertilizers such as ammonium sulphate, ammonium superphosphate etc.

Coal Tar Distillation:

Coal tar is obtained as a by-product during coking of coal. Coal tar pitch is a complex chemical mixture of phenols, cresols and xylenols (which together termed as tar acids), polycyclic aromatic hydrocarbons (PAH), and heterocyclic compounds. These constituents are separated by fractional distillation of the coal tar.

e purpose of tar distillation is to 1) dehydrate the tar in the dehydration column, 2) remove the pitch from dehydrated tar in pitch column and 3) separate tar oils in a fractionating column.

The different fractions obtained are

Each fraction is redistilled for separation of the constituent and each constituent obtained is further distilled for purification.

Petroleum:

Petroleum is a mixture of different hydrocarbon molecules of different sizes and a different number of carbon atoms. Each constituent has its own boiling point. The fractional distillation is used to separate the constituent of petroleum (crude oil). The specific fractions obtained are:

Paraffines: C1-C4 (mainly Propane), C4-C12 (gasoline, petrol ether), C12-C15 (kerosene), C15-C25 (diesel), >C25 (wax, asphalt)

Cycloparaffines: Cyclohexane, Cyclopentane

Aromates: Benzene, Toluene, Ethylbenzene, Xylene

The alkanes obtained from the distillation of crude oil are heated over platinum-rhenium-alumina catalyst at 793 K at about 20 atmospheric pressure to obtain cyclic compounds.

n – Hexane → Cyclohexane  → Benzene

n-Heptane → Methyl cyclohexane → Toluene

1,3 – Butadine + Ethene  → cyclohexene  → Benzene

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Reactivity of Alkanes (Less Reactivity):

• Alkanes are saturated hydrocarbons. They do not contain any functional group. Hence they are less reactive.
• Under normal conditions, they are not acted upon by concentrated acids, alkalies, oxidising agents etc.
• Branched-chain alkanes are more reactive then straight-chain alkanes.
• Alkanes shows substitution reactions

Reactions of Alkanes:

Halogenation of Alkanes:

Reactivity: F₂ > Cl₂ > Br₂ > I₂.

Fluorination of alkanes:

 Reaction of fluorine with alkanes highly explosive and highly exothermic. The reaction takes place even in the dark. The reaction is so spontaneous that even the C-C bond is also broken. The reaction of fluorine with alkanes is carried out by diluting fluorine with nitrogen.

Chlorination of alkanes: 

Chlorine reacts with alkanes in presence of diffused sunlight in steps by successive replacement of hydrogen by chlorine. A mixture of mono, di, tri and tetrachlorides is obtained.

Bromination of alkanes:

A reaction of alkanes with bromine is similar to that of chlorine but the reaction is very slow. Hence it is carried out in presence of AlBr₃ as a catalyst.

CH₄ +  Br₂    →       CH₃Br     +          HBr

Methane       Methyl bromide

Iodination of alkanes:

The reaction of alkanes with iodine is reversible hence direct iodination of alkanes is not possible.

CH₄ +  I₂  ⇌      CH₃I     +     HI

Methane             Methyl iodide

If the reaction is carried out in the presence of HIO₃, HNO₃  or HgO, alkyl halides are obtained.

5CH₄ +  2 I₂  + HIO₃ →   5CH₃I  +  3H₂O

Methane                          Methyl Iodide

Mechanism of Chlorination of Alkanes:

Chlorination of alkanes is a free radical reaction. The reaction takes in the following steps.

Chain initiation: In presence of sunlight the chlorine molecule is decomposed into chlorine atoms. These atoms are highly reactive as they possess unpaired electron.

In presence of sunlight

Cl–Cl    → Cl*   +   Cl*

Chain propagation: The active chlorine atoms attack methane molecule and remove one hydrogen atom. During this process hydrogen chloride and a free methyl radical is formed.

CH₄ +  Cl *    →   CH₃*  +  HCl

Methyl radical formed reacts with another chlorine atom to form methyl chloride and an active chlorine atom. This active chlorine atom again brings about the whole sequence of similar reactions.

CH₃*   +  Cl₂   →  CH₃Cl +  Cl *

In this reaction, we can assume one active chlorine atom can bring about an unlimited number of conversions. Hence the reaction is called chain reaction or chain propagation.

Chain termination: Reaction of two species possessing unpaired electrons form neutral molecules. Thus free radicals are consumed and hence these reactions are called chain termination reactions.

Cl *   +   Cl *  → Cl₂

CH₃*   + CH₃*  → CH₃–CH₃

CH₃*   + Cl*  → CH₃Cl

Nitration of Alkanes:

There is no effect of nitric acid at normal temperature on alkanes. When nitric acid is heated with alkanes to about 423 K to 698 K nitroalkanes are obtained.

General Reaction and Examples:

Pyrolysis of Alkanes:

The decomposition of the compound using heat alone (in absence of air) is called pyrolysis In alkanes two types of pyrolysis reaction can take place. Dehydrogenation and cracking.

Cracking of Alkanes:

The decomposition of alkanes by heating to a very high temperature in the absence of air is called the cracking of alkanes. In this process, the carbon-carbon bonds are broken and a mixture of alkane and alkene is obtained.

When propane is heated to about 873 K in the absence of air, ethene and methane is obtained.

CH₃–CH2–CH₃   →   CH₂=CH₂   +   CH₄

Propane                        ethene       methane

Dehydrogenation of Alkanes:

The decomposition of alkanes by heating to a very high temperature in the absence of air is called the dehydrogenation of alkanes. In this process, the carbon-hydrogen bonds are broken and a mixture of alkanes and alkenes is obtained.

CH₃– CH₃      →    CH₂=CH₂ +  H₂

Ethane                    Ethene

CH₃–CH₂–CH₃  →  CH₃ –CH=CH₂  + CH₂=CH₂ + CH4 +   H₂

Propane     Propene         Ethene       Methane

Aromatization or Reforming Reaction of Alkanes:

Alkanes having more than five carbon atoms get cyclized to benzene and its homologues, on heating under pressure of 10 to 20 atm at about 773 K in presence of oxides of chromium, vanadium or molybdenum supported on alumina.

Combustion of Alkanes:

On burning in air alkanes combine with oxygen to form carbon dioxide and water. This reaction is highly exothermic and large amount of heat is liberated during combustion.

CH₄ +    2O₂       →   CO₂ +  2 H₂O   + Heat

C₂H₆ +    7O₂       →    4CO₂ +  6 H₂O   + Heat

Uses of Alkanes:

• Methane is the main constituent of natural gas. Natural gas is used as industrial fuel and for illumination.
• Iso-butane is liquefied petroleum gas (LPG) and is filled in steel cylinders and is used as fuel for cooking and industrial heating.
• Petrol contains a mixture of alkanes containing 6 to 12 carbons, which is used as motor fuel.
• Higher liquid alkanes like kerosene and furnace oil are used as industrial and domestic fuels.
• Diesel is used in a diesel engine.
• Propane is used as refrigerant in petroleum industry.
• A mixture of alkanes containing 20 to 30 carbons is called petroleum jelly or Vaseline and are used in ointments and cosmetics.
• Alkanes containing 17 to 20 carbons are thick liquids which are used as lubricating oils for machines.
• Alkanes are used to manufacture carbon black which is used in paints, inks, boot polish and rubber.
• Methane is used for the synthesis of ammonia
• Methane is used as starting reactant for the manufacture of methanol, methylene chloride, chloroform, carbon tetrachloride, formaldehyde etc.

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From Unsaturated Hydrocarbons:

Preparations of Alkanes From Alkenes:

Alkanes are obtained by passing the mixture of alkenes and hydrogen over Raney nickel (Ni), or platinum(Pt) or palladium (Pd) at about 473 K to 573 K, corresponding alkanes are obtained. This is hydrogenation process and is also known as Sabatier and Sanderson reaction.

General Reaction and Examples:

Preparations of Alkanes From Alkynes:

Alkanes are obtained by passing the mixture of alkynes and hydrogen over Raney nickel (Ni), or platinum(Pt) or palladium (Pd) at about 473 K to 573 K, corresponding alkanes are obtained. This is hydrogenation process and is also known as Sabatier and Sanderson reaction.

General Reaction and Examples:

Preparations of Alkanes From Alkyl Halides:

Method – I:

When alkyl halides are treated with a suitable reducing agent like a zinc-copper couple with alcohol, or aluminium – amalgam in water-alcohol or zinc in acetic acid corresponding alkanes are obtained.

General Reaction and Examples:

Method – II (By Wurtz Reaction):

When alkyl halides are treated with active metal like sodium or zinc in presence of dry ether.  corresponding alkanes are obtained. In this method, a higher alkane containing a double number of carbon atoms as compared to that in alkyl halide are obtained.

General Reaction and Examples:

From Sodium Salt of Carboxylic Acids (Fatty Acids):

When anhydrous sodium salt of fatty acid is fused with soda lime it gives alkanes containing one carbon less than the carboxylic acid. The process is called decarboxylation.

General Reaction and Examples:

Physical Properties of Alkanes:

• They are colourless, odourless, tasteless compounds.
• The first four alkanes namely, methane, ethane, propane and butane are gaseous at room temperature. Those containing 5 carbons to 17 carbons are liquids with increasing boiling points. Those containing more than 17 carbons are waxy solids.
• They are insoluble in water but soluble in organic solvents like benzene, acetone, ether etc.
• They are lighter than in air and hence float on water.
• They are bad conductors of heat and electricity.
• Boiling points of branched alkanes are lower than that of straight chain alkanes.

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In this article, we shall study the concept of empirical formula and a molecular formula of a compound.

Empirical Formula:

The empirical formula of a substance represents the simplest relative whole number ratio of the atoms of each element contained in the molecule of the substance.

Example: the empirical formula for glucose is CH₂O. Thus the molecule of glucose contains atoms of carbon, hydrogen and oxygen in the ratio 1:2:1.

Molecular Formula:

The formula which gives an actual number of atoms of different elements present in the molecule of the compound is called molecular formula.

Example: A molecular formula for glucose is C₆H₁₂O₆. Thus the molecule of glucose contains 6 atoms of carbon, 12 atoms of hydrogen and 6 atoms oxygen.

The molecular formula is a simple multiple of the empirical formula. The relation between Empirical formula and the molecular formula is given by

Molecular formula = n x Empirical formula

Where n is a whole number and is a ratio of molecular mass to empirical formula mass.

e.g. The empirical formula for glucose is CH₂O.

Thus the empirical formula mass of glucose is (12 x 1 + 1 x 2 + 16 x 1  = 30 )

The molecular formula for glucose is C₆H₁₂O₆

Thus the molecular mass of glucose is (12 x 6 + 1 x 12 + 16 x 6  = 180 )

Now, Molecular formula mass = n x Empirical formula mass

∴ 180 = n x 30

∴ n = 6

Determination of Empirical Formulae/ Molecular Formulae:

Following steps are involved in the determination of empirical formula of organic compound.

1. Determination of atomic ratio: The percentage concentration of each element in the compound is divided by its respective atomic weight to get the atomic ratio. It is to be noted that if the sum of the percentage of constituent elements is not 100%, then the rest of the percentage is assumed to be that of oxygen.
2. Determination of simplest ratio: The ratio so obtained is divided by the smallest ratio in order to obtain the least ratio or the simplest ratio.
3. Determination of whole-number ratio: If the ratios are fractional, reduce them to the smallest possible whole numbers by multiplying throughout by suitable integer. This gives the simplest whole-number ratio.
4. Writing the empirical formula: Write down the symbols present side by side with the above numbers as a subscript to the lower right corner of each. This gives the empirical or the simplest formula.
5. Find the value of n: Find the value of the ratio of molecular mass to empirical formula mass and denote it as n
6. Find molecular formula using relation

Molecular Formula = n x Empirical Formula.

Numerical Problems:

Example – 01:

An organic substance contains 65% carbon, 3.5 % hydrogen and 9.59%  nitrogen. Find its empirical formula.

Solution:

The sum of the percentage of carbon, hydrogen and nitrogen is not 100%.

Hence the rest of the part of the compound is oxygen.

Percentage of oxygen = 100 – (65 + 3.5 + 9.59) = 21.91%

The simplest ratio of C to H to N to O is 8:5:1:2

Hence empirical formula = Ans. C₈H₅NO₂.

Example – 02:

An organic compound contains C, H, and O and is found to have 32% carbon, 4% of hydrogen, the remaining being oxygen. What is the molecular formula? If it contains six atoms of oxygen per molecule?

Solution:

The sum of the percentage of carbon and hydrogen s not 100%.

Hence the rest of the part of the compound is oxygen.

Percentage of oxygen = 100 – (32+4) = 64%

The simplest ratio is 1 :1.5:1.5 i.e. 2:3:3

Hence emperical formula of compound is C₂H₃O₃.

The compound contains six atoms of oxygen

n = Number of oxygen in molecular formula/Number of oxygen in molecular formula = 6/3 =2

Therefore, molecular formula = 2 x (Empirical formula)

Molecular formula = 2 x (C₂H₃O₃) = C₄H₆O₆

The molecular formula for the compound is C₄H₆O₆

Example – 03:

An organic compound contains 40% carbon, 6.66% of hydrogen.  What is the molecular formula of the compound? If its molecular weight is 180.

Solution:

The sum of the percentage of carbon and hydrogen is not 100%.

Hence the rest of the part of the compound is oxygen.

Percentage of oxygen = 100 – (40 + 6.66) = 53.34%

Simplest ratio = 1:2:1

Hence empirical formula = CH₂O.

Empirical formula weight = 12 x 1 + 1 x 2 + 16 x 1 = 30

n = Molecular mass / empirical formula mass = 180/30 = 6

Therefore, molecular formula = n x (Empirical formula)

Molecular formula= 6 x (CH₂O)

Molecular formula = C₆H₁₂O₆.

Therefore molecular formula of the organic compound is C₆H₁₂O₆.

Example – 04:

An organic monobasic acid contains 18.6% carbon, 1.55% of hydrogen, 55.04% chlorine, and 24.81% oxygen.  What is the molecular formula of the acid, If its molecular weight is 129?

Solution:

The sum of the percentage of carbon, hydrogen, and chlorine is not 100%.

Hence the rest of the part of the compound is oxygen.

Percentage of oxygen = 100 – (18.6 + 1.55 + 55.04) =  24.81%

Simplest ratio = 1:1:1:1

Hence empirical formula = CHClO.

Empirical formula weight = 12 x 1 + 1 x 1 + 35.5 x 1 + 16 x 1 = 64.5

n = Molecular mass / empirical formula mass = 129/64.5 = 2

Therefore, molecular formula = n x (Empirical formula)

Molecular formula = 2 x (CHClO)

Molecular formula = C₂H₂Cl₂O₂.

Therefore molecular formula of the organic acid is C₂H₂Cl₂O₂.

But the acid is monobasic acid, hence it should contain one – COOH group

i.e. the molecular formula of organic monobasic acid is CHCl₂COOH.

Example – 05:

Analysis of Vitamin C shows that it contains 40.92% carbon by mass, 4.58% hydrogen, and 54.50% oxygen. Determine the simplest formula of vitamin C.

Solution:

The simplest ratio is 1 :1.3:1 i.e. 3:4:3

Hence the simplest formula of vitamin C is C₃H₄O₃.

Science > Chemistry > Introduction to Organic Chemistry > Empirical and Molecular Formulae of Organic Compounds

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In the last artivle we have studied what is organic chemistry? Why it is termed as chemistry of carbon compounds nowaday? In this article we shall study classification of organic compounds. Depending upon the structure organic compounds are classified into two types. viz. open-chain organic compounds or aliphatic organic compounds or Acyclic organic compounds and closed chain organic compounds or cyclic organic compounds

Open Chain Organic Compounds:

Open chain organic compounds are organic compounds that contain an open chain of carbon atoms which may be straight-chain or branched-chain.

Closed Chain Organic Compounds:

Closed chain organic compounds are organic compounds that contain one or more closed chains or rings of carbon atoms. Closed chain organic compounds are further classified into two types. viz. homocyclic organic compounds or carbocyclic organic compounds. and heterocyclic organic compounds

Homocyclic Carbon Compounds: 

Homocyclic organic compounds are organic compounds that contain one or more closed rings of carbon atoms only. Homocyclic organic compounds are further classified into two types. viz. alicyclic organic compounds and aromatic organic compounds

Alicyclic Organic Compounds: 

Alicyclic organic compounds are homocyclic organic compounds that have properties similar to that of aliphatic compounds.

Aromatic Organic Compounds:

Aromatic organic compounds are homocyclic organic compounds that contain at least one benzene ring.

Heterocyclic Organic Compounds: 

Heterocyclic organic compounds are organic compounds that contain at least one atom other than a carbon atom in the ring.

Homologous Series:

A series of organic compounds which have a common general formula and in which the two successive members of the series differ by – CH₂ – is known as homologous series. The individual member of the homologous series is called homologue.

Alkane Series:

Alcohol Series:

Characteristics of Homologous Series:

• Members of the same homologous series are represented by the same general formula. E.g. all alkanes are represented by the same general formula C_nH_{2n + 2}.
• They can be prepared by similar methods of preparation.
• They have the same functional group hence have a number of chemical properties in common which are called the general properties.
• They show a regular gradation in physical properties such as melting and boiling points.
• Each member of the homologous series is known as the homologue of the other elements and differs from its next higher or next lower homologue by a common difference –CH₂–
• Each member of the homologous series differs from its next higher or next lower homologue in molecular weight by 14.

Science > Chemistry > Introduction to Organic Chemistry > Classification of Organic Compounds

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In the last few articles, we have studied the methods of preparations of alkyl halides. In this article, we shall study the physical properties of alkyl halides. Some physical properties of alkyl halides are as follows:

State:

Lower members (methyl chloride, methyl bromide, methyl fluoride, ethyl bromide, ethyl chloride and ethyl bromide) are gases and higher members are liquids (Up to C18) and solids (Greater than C18).

Odour:

In the pure state, the haloalkanes up to C18 possess pleasant sweet odour. All higher haloalkanes are odourless.

Colour:

Pure haloalkanes are colourless. However, bromoalkanes and iodoalkanes on storing for long period, when exposed to light develop colour.

Boiling Points:

Haloalkanes have higher boiling points as compared to those compared to corresponding alkanes. This is due to their polarity and strong dipole-dipole attractive interaction between haloalkane molecules and greater magnitude of van der Wall’s forces.

• For the same alkyl group the boiling points of haloalkanes are in the order RCl < RBr < RI, because with the increase in the size of halogen atom the magnitude of van der Wall forces of attraction increases.
• Among isomeric alkyl halides, the boiling point decreases with an increase in branching in the alkyl group, because with branching the molecule attains a spherical shape with less surface area. As a result, interparticle forces become weaker. Hence the boiling point decreases. The order of boiling point is Primary  > Secondary >= iso > Tertiary
• For the same halogen, the boiling point increases with the increase in the molecular mass because with the increase in the size of the alkyl group the magnitude of van der Wall forces of attraction increases. i.e R – X < R -CH2-X  < R -CH2-CH2-X
• As the number of halogen in a molecule increases the boiling point of the compound increases because of the increase in the number of halogen atoms the magnitude of van der Wall forces of attraction increases. i.e. CH₃Cl < CH₂Cl₂ < CHCl₃ < CCl₄

Solubility:

Alkyl halides are polar in nature (dipole moment 2.05 to 2.15 D) but they are not able to form hydrogen bonds with water molecules. Hence they are sparingly soluble in water. But they are soluble in organic solvents like alcohols, ethers and benzene.

Density:

Alkyl chlorides are generally lighter than water, while alkyl bromides and alkyl iodides are heavier than water. The order of density is RI > RBr > RCl. Poly chlorides are heavier than water. Thus the density of alkyl halides increases with the increase in the number and atomic mass of the halogen atoms. Methyl iodide is the heaviest of all the haloalkanes.

Scientific Reasons:

Density:

Arrange the following in the order of decreasing density. 1-Chloropropane, 1-Iodopropane, 1-Bromopropane

• For the same alkyl group, the density of alkyl halides increases with the increase in the number and atomic mass of the halogen atoms. The atomic mass of I > Atomic mass of Br >Atomic mass of Cl.
• Hence boiling point of 1-Iodopropane > 1-Bromopropane > 1-Chloropropane.

Arrange each of the following set of compounds in the order of increasing densities:

CHCl₃, CH₂Cl₂, CCl₄, CH₃Cl:

• The order of density is RI > RBr > RCl. Poly chlorides are heavier than water. Thus the density of alkyl halides increases with the increase in the number and atomic mass of the halogen atoms.
• Hence, the order of densities is CH₃Cl. < CH₂Cl₂ < CHCl₃ < CCl₄.

C₂H₅Cl, C₂H₅I, C₂H₅Br:

• The order of density is RI > RBr > RCl. Thus the density of alkyl halides increases with the increase in the number and atomic mass of the halogen atoms.
• Hence, the order of densities is C₂H₅Cl < C₂H₅Br < C₂H₅I.

Which alkyl halide has the highest density and why?

• For the same alkyl group, the order of density is R-I > R-Br > R-Cl. Thus R-I will have the highest density.
• For the same halogen group, with the increase in the branching, the molecule acquires the spherical shape with less surface area. Thus the tertiary butyl group will have the smallest size.
• From the above two points, we can say that tertiary butyl iodide should have the highest density.

Boiling Points:

Which isomer of C₅H₁₁Cl has the highest boiling point? Why?

• Consider the following two isomers of C₅H₁₁Cl 

• Among isomeric alkyl halides, the boiling point decreases with an increase in branching in the alkyl group.
• 1-Chloropentane is a straight-chain isomer. It has the strongest interparticle forces. Hence it has the highest boiling point among all the isomers. While 1-Chloro-2,2-dimethylpropane has the highest number of branches in all possible isomers, hence it has the weakest interparticle forces. Hence it has the lowest boiling point among all the isomers.

Which isomer of C₄H₉Cl has the highest boiling point? Why?

• Among isomeric alkyl halides, the boiling point decreases with an increase in branching in the alkyl group.
• 1-Chlorobutane (n-Butyl chloride) is a straight-chain isomer. It has the strongest interparticle forces. Hence it has the highest boiling point among all the isomers. While 2-Chloro-2-methylpropane (tert-Butyl chloride) has the highest number of branches in all possible isomers, hence it has the weakest interparticle forces. Hence it has the lowest boiling point among all the isomers.

Arrange in the order of increasing boiling points. 

Bromobenzene. chlorobenzene, iodobenzene:

• The boiling points of mono halogen derivatives of benzene follow the order Iodo > Bromo > Chloro.
• Hence boiling point of Chlorobenzene <  Bromobenzene < Iodobenzene .

n-pentyl chloride, iso-pentyl chloride, neo-pentyl chloride:

• Among isomeric alkyl halides, the boiling point decreases with the increase in branching in the alkyl group, because with branching the molecule attains spherical shape with less surface area. As a result, interparticle forces become weaker. Hence the boiling point decreases. The order of boiling point is Primary  > Secondary >= iso > Tertiary
• Hence boiling point of neo-pentyl chloride < iso-pentyl chloride < n-pentyl chloride.

Bromomethane, Bromoform, Chloromethane, Dibromomethane:

• For the same alkyl group the boiling points of haloalkanes are in the order RCl < RBr< RI, because with the increase in the size of halogen atom the magnitude of van der Wall forces of attraction increases. For the same halogen, the boiling point increases with the increase in the molecular mass. As the number of halogen in a molecule increases the boiling point of the compound increases.
• Hence boiling point of Chloromethane < Bromomethane <  Dibromomethane < Bromoform.

1-Chloropropane, isopropyl chloride, 1-Chlorobutane:

• As the number of halogen in a molecule increases the boiling point of the compound increases. Among isomeric alkyl halides, the boiling point decreases with the increase in branching in the alkyl group, because with branching the molecule attains a spherical shape with less surface area. As a result, interparticle forces become weaker. Hence the boiling point decreases. The order of boiling point is Primary  > Secondary >= iso > Tertiary.
• Hence boiling point of isopropyl chloride < 1-Chloropropane < 1-Chlorobutane.

Methyl chloride, methyl bromide, methyl iodide:

• For the same alkyl group the boiling points of haloalkanes are in the order RCl < RBr < RI, because with the increase in the size of halogen atom the magnitude of van der Wall forces of attraction increases.
• Hence the order of boiling points is Methyl chloride (CH₃Cl) < methyl bromide (CH₃Br) < methyl iodide (CH₃I).

Methyl bromide, methylene bromide, bromoform:

• As the number of halogen in a molecule increases the boiling point of the compound increases.
• Hence the order of boiling points is methyl bromide (CH₃Br) < methylene bromide (CH₂Br₂) < Bromoform (CHBr₃).

Propane, n-propyl bromide, isopropyl bromide:

• Haloalkanes have higher boiling points as compared to those compared to corresponding alkanes. This is due to their polarity and strong dipole-dipole attractive interaction between haloalkane molecules.
• Among isomeric alkyl halides, the boiling point decreases with the increase in branching in the alkyl group, because with branching the molecule attains a spherical shape with less surface area. As a result, interparticle forces become weaker. Hence the boiling point decreases. The order of boiling point is Primary  > Secondary >= iso > Tertiary.
• Hence the order of boiling points is propane (alkane) < isopropyl bromide (isoalkyl halide) < n-propyl bromide (primary alkyl halide).

n-butyl chloride, iso-butyl chloride, tert-butyl chloride:

• Among isomeric alkyl halides, the boiling point decreases with the increase in branching in the alkyl group, because with branching the molecule attains a spherical shape with less surface area. As a result, interparticle forces become weaker. Hence the boiling point decreases. The order of boiling point is Primary  > Secondary >= iso > Tertiary.
• Hence the order of boiling points is tert-Butyl chloride < iso-butyl chloride < n-butyl chloride.

1-Bromopropane, isopropyl bromide, 1- Bromobutane:

• Among isomeric alkyl halides, the boiling point decreases with the increase in branching in the alkyl group, because with branching the molecule attains spherical shape with less surface area. As a result, interparticle forces become weaker. Hence the boiling point decreases. The order of boiling point is Primary  > Secondary >= iso > Tertiary. or the same halogen, the boiling point increases with the increase in the molecular mass.
• Hence the order of boiling points is isopropyl bromide  <  1-Bromopropane <  1- Bromobutane

Explain why 1-Chlorobutane has higher B.P. than 2-Chlorobutane?

Among isomeric alkyl halides, the boiling point decreases with the increase in branching in the alkyl group, because with branching the molecule attains spherical shape with less surface area. As a result, interparticle forces become weaker. Hence the boiling point decreases.

The order of boiling point is Primary  > Secondary >= iso > Tertiary. Hence 1-Chlorobutane (primary alkyl halide) has higher B.P. than 2-Chlorobutane (secondary alkyl halide).

Explain why Bromoethane has a higher boiling point than Chloroethane. OR Out of ethyl bromide and ethyl chloride which has a higher boiling point and why?

For the same alkyl group the boiling points of haloalkanes are in the order RCl < RBr < RI, because with the increase in the size of halogen atom the magnitude of van der Wall forces of attraction increases. Hence Bromoethane has a higher boiling point than Chloroethane.

Solubility:

Explain why choroform is not soluble in water although it is polar. OR alkyl halides though polar, are immiscible with water.

• A substance is soluble in water if its molecules are capable of forming hydrogen bonds with water. Chloroform molecules do not form the hydrogen bond with water.
• The energy required to break the bonds between haloalkane molecules is much larger than the energy released during the formation of the bond between haloalkane molecules and water molecules. Hence chloroform is not soluble in water although it is polar.

Alkyl halides are insoluble in water though they contain polar C-X bond. Explain.

• Alkyl halides are polar in nature but they are not able to form hydrogen bonds with water molecules. Hence they are sparingly soluble in water. But they are soluble in organic solvents like alcohols, ethers and benzene.

Dipole Moment:

Which one of the following has the highest dipole moment?

CH₂Cl₂, CHCl₃, CCl₄.

• The dipole moment of CH₂Cl₂ is the highest while that of CCl₄ is zero. Dipole moment of CH₂Cl₂ is greater than  CHCl₂ because the dipole moment of the third C-Cl bond of CHCl₃ opposes the dipole moment of the remaining two C-Cl bonds.

Science > Chemistry > Organic Chemistry > Halogen Derivatives of Alkanes > Physical Properties of Alkyl Halides

The post Physical Properties of Alkyl Halides appeared first on The Fact Factor.

In the last three articles, we have studied the methods of preparations of alkyl halides from alkanes, alkenes, alcohols. In this article, we shall study the method of the preparation of alkyl halids by halide exchange method. This method is typical for preparations of alkyl iodides/fluorides/bromides.

Preparation of Alkyl iodides by Halide Exchange Method(Finkelstein Reaction)

General Reaction:

When alkyl chlorides or bromides when treated with NaI in presence of dry acetone give alkyl iodides.

R–Cl or R–Br  +  NaI → RI + NaCl  or NaBr

Alkyl chloride / bromide   sodium iodide   →    Alkyl iodide sodium chloride / bromide

Example – 1: Preparation of Ethyl iodide (Iodoethane) from ethyl chloride (Chloroethane):

C₂H₅-Cl    +      NaI   C₂H₅I      +   NaCl

Ethyl chloride    sodium iodide    Ethyl iodide  sodium chloride

Example – 2: Preparation of Ethyl iodide (Iodoethane) from ethyl bromide (Bromoethane):

C₂H₅-Br    +      NaI   C₂H₅I      +   NaBr

Ethyl bromide    sodium iodide   Ethyl iodide       sodium bromide

Preparation of Alkyl fluorides by Halide Exchange Method(Swart’s Reaction):

The direct reaction of alkanes with fluorine is highly explosive in nature, hence it can’t be produced by direct fluorination of alkanes.

General Reaction:

When alkyl halides are treated with salts like AgF, Hg₂F₂, CoF₃, SbF₃ fluoroalkanes can be obtained.

Example – 1: Preparation of Ethyl fluoride (Fluoroethane) from ethyl chloride (Chloroethane):

C₂H₅Cl          +     AgF    →  C₂H₅F           +   AgCl

Ethyl chloride   silver fluoride   Ethyl fluoride       silver chloride

Example – 2: Preparation of Methyl fluoride (Fluoromethane) from methyl bromide (Bromoethane):

2  CH₃Br          +         Hg₂F₂     →       2 CH₃F           +   Hg₂Br₂

Methyl bromide Mercurous fluoride  Methyl fluoride  Mercurous bromide

For replacement of two or three halogens CoF₃, SbF₃ are used

Example – 3

Preparation of Alkyl bromides from silver salt of fatty acid (Borodine Hunsdiecker Reaction)

General Reaction:

When silver salt of fatty acid is refluxed with bromine in CCl4, alkyl bromide is obtained.

RCOOAg   +    Br₂        RBr  +  CO₂ +  AgBr

Silver salt of fatty acid   bromine    alkyl bromide   silver bromide

Example -1: Preparation of Methyl Bromide (Bromomethane) from silver acetate:

CH₃COOAg   +    Br₂        CH₃Br  +  CO₂ +  AgBr

Silver acetate bromine    methyl bromide  silver bromide

Example -2: Preparation of Ethyl Bromide (Brommethane) from silver propionate:

C₂H₅COOAg   +    Br₂        C₂H₅Br  +  CO₂ +  AgBr

Silver propionate   bromine          ethyl bromide    silver bromide

Note:

Chloroalkanes can be obtained by this method but yield is very low. Iodoalkanes can not be obtained by this method because iodine forms easter with  silver salt of fatty acid.

Science > Chemistry > Organic Chemistry > Halogen Derivatives of Alkanes > Preparation of Alkyl Halides by Halide Exchange Method

The post Preparation of Alkyl Halides by Halide Exchange Method appeared first on The Fact Factor.

In the last two articles, we have studied the methods of preparation of alkyl halides from alkanes and alkenes. In this article, we shall study the preparation of alkyl halides from alcohols.

The Action of HX on Alcohols:

• The reaction is nucleophilic substitution.
• Methyl and primary alcohol undergo SN¹ mechanism while secondary and tertiary alcohols undergo SN² mechanism.
• Among halogen halides, HCl is the least reactive in nature. Because the chloride ion is a weaker nucleophile than bromide or iodide ions. Hence hydrogen chloride is mixed with anhydrous ZnCl2.
• In these reactions, anhydrous ZnCl2 not only acts as a dehydrating agent but also helps in the cleavage of C-O bond of the alcohol. ZnCl2 is Lewis acid and coordinates with the oxygen atom and thus weakens C-O bond. This results in the formation of carbocation which combines with Cl- ion to form chloroalkane. Tertiary alcohols are highly reactive hence for tertiary alcohol ZnCl2 is not required.
• This reaction is nucleophilic substitution. The stability of carbocations is of order tertiary > secondary > primary. Hence the order of reactivity also follows the same order tertiary > secondary > primary.
• The bond dissociation energy of H-X bond is of order H-Cl > H-Br > H-I. hence the reactivity of halogen acids follows the order HI > HBr > HCl.
• Unlike alkyl chlorides, the secondary and tertiary bromides and iodides cannot be obtained from their respective alcohols. It is because the secondary and tertiary alcohols on heating with concentrated H₂SO₄ undergo dehydration to form alkenes. Hence for these preparations dilute H₂SO₄ is used.

General Reaction:

R–OH  +            HX          →     R–X       +    H₂O

Alcohol     Halogen acid       alkyl halide

Order of Reactivity:

Tertiary alcohol > Secondary alcohol > Primary alcohol

and HI > HBr > HCl.

By Action of Haloacids: 

Preparation of Alkyl Chlorides:

General Reaction:

When alcohol is treated with Lucas reagent alkyl chloride is obtained. Lucas reagent is a solution of a concentrated hydrochloric acid with zinc chloride. This reaction is known as Groove’s process.

R–OH  +            HCl               R–Cl       +    H₂O

Alcohol     Conc.Hydrochloric acid     Alkyl chloride

Example – 1: Preparation of ethyl chloride (Chloroethane) from ethyl alcohol (Ethanol):

C₂H₅OH  +            HCl                C₂H₅Cl       +    H₂O

Ethyl alcohol   Conc.Hydrochloric acid       Ethyl chloride

Example – 2: (Preparation of isopropyl chloride (2-Chloropropane) from isopropyl alcohol (Propan-2-ol):

Example – 3: (Preparation of tert- Butyl chloride (2-Chloro-2-methylpropane) from tert- Butyl alcohol (2-Methylpropan-2-ol):

Preparation of Alkyl Bromides:

General Reaction:

When alcohol is treated with concentrated hydrobromic acid (NaBr + H2SO4) alkyl bromide is obtained.

R–OH  +            HBr                R–Br       +    H₂O

Alcohol    Hydrobromic acid               Alkyl chloride

Example – 1: Preparation of ethyl bromide (Bromoethane) from ethyl alcohol (Ethanol):

C₂H₅OH  +            HBr                C₂H₅Br       +    H₂O

Ethyl alcohol    Hydrobromic acid                          Ethyl bromide

Example – 2: (Preparation of isopropyl bromide (2-Bromopropane) from isopropyl alcohol (Propan-2-ol):

Example – 3: (Preparation of tert- Butyl bromide (2-Bromo-2-methylpropane) from tert- Butyl alcohol (2-Methylpropan-2-ol):

Preparation of Alkyl Iodides:

General Reaction:

When alcohol is refluxed with Potassium or sodium iodide with 95% phosphoric acid and heated, alkyl iodide is obtained.

R–OH  +     HI     R–I       +    H₂O

Alcohol    Hydroiodic acid         Alkyl iodide

Example – 1: Preparation of ethyl iodide (Iodoethane) from ethyl alcohol (Ethanol):

C₂H₅OH   +       HI      C₂H₅I   +    H₂O

Ethyl alcohol    Hydroiodic acid           Ethyl iodide

Example – 2: (Preparation of isopropyl iodide (2-Iodopropane) from isopropyl alcohol (Propan-2-ol):

By Action of Phosphorous Halides: 

• This method is suitable for preparation of primary and secondary alkyl halides.
• A good yield of tertiary alkyl halides cannot be obtained by this method.
• The reaction of an alcohol with PX₃ does not involve the formation of carbocation and usually occurs without rearrangement of the carbon skeleton.

Preparation of Alkyl Chlorides Using PCl₃:

General Reaction:

When alcohol is treated with phosphorous trichloride, alkyl chloride and phosphorous acid are obtained.

3 R-OH  +            PCl₃            3 R-Cl       +    H₃PO₃

Alcohol       phosphorous trihalide    alkyl halide    phosphorous acid

Example – 1: Preparation of ethyl chloride (Chloroethane) from ethyl alcohol (Ethanol):

3C₂H₅OH  +            PCl₃           3C₂H₅Cl       +    H₃PO₃

Ethyl alcohol   Phosphorous trichloride    ethyl chloride    Phosphorus acid

Both the products are in the liquid state and are separated by fractional distillation.

Example – 2: (Preparation of isopropyl chloride (2-Chloropropane) from isopropyl alcohol (Propan-2-ol):

Preparation of Alkyl Chlorides Using PCl₅:

General Reaction:

When alcohol is treated with phosphorous pentachloride, alkyl chloride and phosphoryl chloride (phosphorous oxychloride) are obtained.

ROH         +     PCl₅         RCl         +     POCl₃           +            HCl

Alcohol   Phosphorous pentachloride       Alkyl chloride    Phosphoryl chloride    Hydrogen chloride

Example – 1: Preparation of ethyl chloride (Chloroethane) from ethyl alcohol (Ethanol):

C₂H₅OH    +   PCl₅         C₂H₅Cl   +     POCl₃   +  HCl

Ethyl alcohol   Phosphorous pentachloride       Ethyl chloride    Phosphoryl chloride    Hydrogen chloride

Example – 2: (Preparation of isopropyl chloride (2-Chloropropane) from isopropyl alcohol (Propan-2-ol):

Preparation of Alkyl Bromides:

Unlike PCl₃, the compounds PBr₃ and PI₃ are not stable and hence they are to be prepared when they are to be used by treating bromine and iodine with red phosphorous. Compounds PBr₅ and PI₅ do not exist.

Alkyl bromides are prepared by the action of bromine, in presence of red phosphorus, on alcohols.  Phosphorus bromide PBr₃, which is unstable is formed as intermediates in the reaction.

3 R-OH  +        PBr₃            3 R-Br       +    H₃PO₃

Alcohol       phosphorous tribromide    alkyl bromide    phosphorous acid

Example – 1: Preparation of ethyl chloride (Chloroethane) from ethyl alcohol (Ethanol):

3C₂H₅OH  +        PBr₃          3C₂H₅Br   +    H₃PO₃

Ethyl alcohol   Phosphorous tribromide  Ethyl bromide  Phosphorus acid

Preparation of Alkyl Iodides:

Alkyl iodides are prepared by the action of iodine, in presence of red phosphorus, on alcohols.  Phosphorus iodide PBr₃, which is unstable is formed as intermediates in the reaction.

3 R-OH  +            PI₃            3 R-I       +    H₃PO₃

Alcohol       phosphorous triiodide    alkyl iodide    phosphorous acid

Example – 1: Preparation of ethyl iodide (Iodoethane) from ethyl alcohol (Ethanol):

3C₂H₅OH  +            PI₃            3C₂H₅I       +    H₃PO₃

Ethyl alcohol   Phosphorous triiodide          ethyl iodide    Phosphorus acid

By Action of Thionyl Chloride: 

This reaction is also known as Darzen’s Procedure.

General Reaction:

When alcohol is refluxed with thionyl chloride, alkyl chloride, sulphur dioxide and hydrogen chloride are obtained. Better yield is obtained if pyridine is added in a small amount.

ROH  +  SOCl₂        RCl  + SO₂ ↑­  + HCl  ­↑

Alcohol    Thionyl chloride              Alkyl chloride

Example – 1: Preparation of ethyl chloride (Chloroethane) from ethyl alcohol (Ethanol):

C₂H₅OH  +  SOCl₂        C₂H₅Cl  + SO₂ ↑­  + HCl  ­↑

Ethyl alcohol  Thionyl chloride                 Ethyl chloride

Example – 2: (Preparation of isopropyl chloride (2-Chloropropane) from isopropyl alcohol (Propan-2-ol):

The use of thionyl chloride for preparation of alkyl chlorides is most convenient because the other products of reaction (SO2 and HCI) being gases go off  (or can be expelled during distillation very easily). and hence the method requires no special purification/separation.

C₅H₁₁OH on bromination gives compound C₅H₁₁Br. Suggest all possible structures of alcohol and bromide.

Science > Chemistry > Organic Chemistry > Halogen Derivatives of Alkanes > Preparation of Alkyl halides From Alcohols

The post Preparation of Alkyl halides From Alcohols appeared first on The Fact Factor.

In the last article, we have studied the preparation of alkyl halides from alkanes. In this article, we shall study the preparation of alkyl halides from alkenes (olefins).

The Action of HX on Alkenes:

• It is an electrophilic addition reaction.
• Carbonium ion is formed as an intermediate.
• This method is more useful to prepare secondary and tertiary alkyl halides.
• For unsymmetric alkene, Markownikoff’s rule should be applied.
• Hydrogen chloride and hydrogen iodide add according to Markownikoff’s rule.
• The addition of hydrogen bromide takes place according to Markownikoff’s rule only when the reaction is carried out in the total absence of light or oxygen or peroxides. In the presence of any one of these agents the addition of HBr takes place in exactly the reverse way and is called Peroxide e1ffect or anti-Markownikoff addition or Kharasch effect or Kharasch-Mayo effect.

General Reaction

R –CH=CH₂ +  H– X  → R –CH₂–CH₂X       OR    R –CH₂X–CH₂

Alkene      Hydrogen halide           Alkyl halide

Conversion of -C=C- (Alkenes) into -X (Alkyl halides)

Order of Reactivity for Halogen acids:

HI   > HBr > HCI

Preparation of Alkyl Halides From Symmetrical Alkenes: 

R –CH=CH–R   +        H–X      →  R –CH₂–CHX–R

Symmetric alkene         Hydrogen halide            alkyl halide

Preparation of Alkyl Chlorides / Alkyl Bromides / Alkyl Iodides:

R –CH=CH–R   +        H–Cl      →  R –CH₂–CHCl–R

Symmetric alkene        Hydrogen chloride      alkyl chloride

Example – 1: Preparation of ethyl chloride (Chloroethane) from Ethylene (Ethene):

H₂C=CH₂  +        H–Cl      →  CH₂–CH₂Cl

Ethylene      hydrogen chloride     Ethyl chloride

Example – 2: Preparation of ethyl bromide (Bromoethane) from Ethylene (Ethane):

H₂C=CH₂  +        H–Br      →  CH₂–CH₂Br

Ethylene        Hydrogen bromide    Ethyl bromide

Example – 3: Preparation of Ethyl iodide (iodoethane) from Ethylene (Ethane):

H₂C=CH₂  +        H–I      →  CH₂–CH₂I

Ethylene      Hydrogen  iodide    Ethyl iodide

Example – 4: Preparation of sec-butyl chloride (2-Chlrobutane) from Butylene (But-2-ene):

CH₃ –CH=CH–CH₃  +        H–Cl        →  CH₃ –CH₂–CH₂Cl–CH₃

butylene                Hydrogen chloride       sec-butyl chloride

Example – 5: Preparation of sec-butyl bromide (2-Bromobutane) from β-butylene (But-2-ene):

CH₃ –CH=CH–CH₃  +        H–Br        →  CH₃ –CH₂–CH₂Br–CH₃

butylene                Hydrogen bromide       sec-butyl bromide

Preparation of Alkyl Halides From Unsymmetrical Alkenes: 

R –CH=CH–R’ +  H–X  →  R –CH₂–CHX–R’  or R –CHX–CH₂–R’

Unsymmetric alkene         Hydrogen halide            alkyl halide

Markownikoff’s Rule:

When an unsymmetrical reagent (like HBr) is added to an unsymmetrical alkene, (in the total absence of oxygen and peroxide and light) then the negative part of the reagent gets attached to that unsaturated carbon atom which carries less number of hydrogen atoms.

This behaviour is explained by 1,2-hydride shift to attain greater stability of cation.

Example – 6: Preparation of isopropyl bromide (2-Bromopropane) from propene:

Example – 7: Preparation of sec-butyl bromide (2-Bromobutane) from α-butylene (But-1-ene):

Example – 8: Preparation of tert-butyl bromide (2-Bromo-2-methylpropane) from iso-butylene (2-Methylpropene):

Example – 9: Preparation of 2-Bromo-2-methylbutane from 3-Methyl-but-ene:

The common name of 3-Methyl-but-1-ene is alpha-isoamylene. The above reaction is an example of the 1,2-hydride shift to attain greater stability of cation.

Anti-Markownikoff’s Rule OR Kharasch Effect OR Kharasch Mayo Effect OR Peroxide Effect:

When an unsymmetrical reagent (like HBr) is added to an unsymmetrical alkene in presence of oxygen or peroxide or light then the negative part of the reagent gets attached to that unsaturated carbon atom which carries more number of hydrogen atoms.

Example – 10: Preparation of n-propyl bromide (1-Bromopropane) from propene:

Example – 11: Preparation of n-butyl bromide (1-Bromobutane) from α-butylene (But-1-ene):

Notes:

• This reaction is a free radical addition and exothermic reaction.
• H-Cl has a bond energy of 103 Kcal/mol which is stronger than H-Br bond energy (87 Kcal/mol). Hence there is no breaking up of the H-Cl bond due to peroxide free radicals.
• H-I has a bond energy of 78 Kcal/mol. It forms free radicals easily but instead of attacking the double bond, the iodine radicals formed combine with each other to form an iodine molecule.

Science > Chemistry > Organic Chemistry > Halogen Derivatives of Alkanes > Preparation of Alkyl halides From Alkenes

The post Preparation of Alkyl halides From Alkenes appeared first on The Fact Factor.

In this article, we shall study methods of preparation of alkyl halides from alkanes. Alkyl halides can be prepared from alkanes by their halogenation.

General Reaction:

R – H  +  X– X  →  R – X     +     H – X

Alkane    Halogen    Alkyl halide  Halogen acid

Conversion of -H (Alkanes) into – X (Alkyl halides)

Order of Reactivity for Halogens: 

F₂ > > Cl₂ > Br₂ > I₂

Benzylic, allylic > tertiary > secondary > primary > vinylic, aryl

Preparation of Alkyl Chlorides From Alkanes:

General Reaction:

When alkane is treated with chlorine in presence of ultraviolet light or diffused sunlight alkyl chlorides or chloroalkanes are obtained. The reaction gives a mixture of all possible chloroalkanes.

R – H  +  Cl– Cl  R – Cl     +     H – Cl

Alkane         Chlorine                Alkyl chloride Hydrogen chloride

As the reaction is taking place in the presence of ultraviolet light or diffused sunlight the reaction is known as photohalogenation of alkanes.

Example –1: Preparation of methyl chloride (Chloromethane) from methane

CH₄  +   Cl₂  CH₃Cl       +      HCl

Methane      Chlorine         Methyl Chloride    Hydrogen Chloride

Example –2: Preparation of ethyl chloride (Chloroethane) from ethane

C₂H₆     +    Cl₂  C₂H₂Cl         +      HCl

Ethane       Chlorine                   Ethyl  Chloride       Hydrogen Chloride

Preparation of Alkyl Bromides From Alkanes:

General Reaction :

When alkane is heated with bromine in presence of anhydrous AlBr3 as catalyst alkyl bromide or bromoalkane is obtained.

R – H  +  Br– Br    R – Br     +     H – Br

Alkane         Bromine                Alkyl bromide    Hydrogen bromide

Example- 1: Preparation of ethyl bromide (Bromoethane) from ethane:

C₂H₆     +    Br₂    C₂H₂Br         +      HBr

Ethane     Bromine               Ethyl  bromide Hydrogen  bromide

Example- 2: Preparation of methyl bromide (Bromomethane) from methane:

CH₄  +   Br₂   CH₃Br       +      HBr

Methane      Chlorine         Methyl Chloride   Hydrogen Chloride

Preparation of Alkyl Iodides From Alkanes:

With iodine reaction is reversible. Thus we can not get a good yield of alkyl iodide. Hence direct iodination is difficult.

R – H      +  I – I     ⇌   R – I     +       H – I

Alkane         Iodine      Alkyl iodine     Hydrogen iodine

If the reaction is carried out in the presence of iodic acid HIO₃, or mercuric oxide HgO, or dilute HNO₃ which can oxidise HI formed and the lodoalkane can be obtained. The iodination reaction stops at mono-iodo state.

General Reaction Using Iodic Acid HIO₃:

5 R-H   +  2 I₂ +   HIO₃  →    5 R-I  +  3 H₂O

Alkane   Iodine  Iodic acid    Alkyl iodide

General Reaction Using Mercuric Oxide HgO:

2R-H  +  2 I₂  +    HgO      →       2 R-I  +  Hg I₂ + H₂O

Alkane   Iodine  mercuric oxide        Alkyl iodide

General Reaction Using Dilute Nitric Acid HNO₃:

8R-H  + 4 I₂    +  HNO₃  →       8R-I   + 3H₂O    +   NH₃

Alkane   Iodine    nitric acid     Alkyl iodide                ammonia

Example-1: Preparation of ethyl iodide (Iodoethane) using HIO₃:

5 C₂H₆   +  2 I₂ +   HIO₃  →    5 C₂H₅I  +  3 H₂O

Ethane   Iodine  Iodic acid         Ethyl iodide

Example-2: Preparation of ethyl iodide (Iodoethane) using HgO:

2C₂H₆ +  2 I₂  +    HgO      →       2 C₂H₅I  +  HgI₂ + H₂O

Ethane    Iodine  mercuric oxide     Ethyl iodide

Example-3: Preparation of ethyl iodide (Iodoethane) using HNO₃:

8C₂H₆ + 4 I₂    +  HNO₃  →       8C₂H5I   + 3H₂O    +   NH₃

Ethane   Iodine  Nitric acid             Ethyl iodide        Ammonia

Drawbacks/Disadvantages of direct halogenation (Photohalogenation):

After forming alkyl halide the reaction proceeds further and poly-substitution takes place and gives a mixture of di, tri, tetra etc. haloalkanes.

CH₄  +   Cl₂ CH₃Cl  (Chloromethane) +  HCl

CH₃Cl   +   Cl₂ → CH₂Cl₂ (Dichloromethane) +  HCl

CH₂Cl₂  +   Cl₂ →  CHCl₃  (Trichloromethane) +  HCl

CHCl₃  +   Cl₂  →  CCl₄  (Tetrachlromethane) +  HCl

This reaction continues till all halogens in the alkane are replaced one by one by chlorine or bromine. Thus we get mixture of di, tri, tetra etc. haloalkanes.

The mixture contains less amount of alkyl chlorides or bromides. Chlorination and bromination reactions are not selective. Hence they may give isomers of the monohalogen derivatives of alkanes. Besides the separation of constituents is difficult.

Iodination of alkanes is reversible reaction it requires additional reagents like mercuric oxide or iodic acid or nitric acid for obtaining alkyl iodes. Hence direct halogenation is not the suitable method for the preparation of alkyl halides.

Notes:

• Thermal chlorination of alkanes is called Hare and Mc Bee reaction. It is carried about at 673K.
• Diffused sunlight causes homolytic fission of halogen, hence the reaction is a free radical substitution.
• If alkanes are used in excess, the major product is monohalogen derivatives of alkanes. i.e. alkyl halides.
• The ease of substitution of different types of a hydrogen atom is Benzylic, allylic > tertiary > secondary > primary > vinylic, aryl
• Thus, an alkane containing a tertiary hydrogen atom would be more reactive towards substitution reaction than with only primary hydrogen atom or primary and secondary hydrogen atoms. Isobutane would undergo substitution more rapidly than propane or ethane.
• Alkyl fluorides are not prepared by fluorination of alkanes, because the reaction is highly explosive in nature.
• Propane on chlorination gives a mixture of 2- Chloropropane (55 %) and 1- Chloropropane (45 %)
• Butane on chlorination gives a mixture of n-butyl chloride (28 %) and sec-butyl chloride (72 %)
• iso-Butane on chlorination gives a mixture of tert-butyl chloride (64 %) and isobutyl chloride (36 %)
• Bromination of alkanes is electrophilic substitution reaction, as it forms AlBr4 ion and Br+ ion. Br+ ion takes part in substitution.

Science > Chemistry > Organic Chemistry > Halogen Derivatives of Alkanes > Preparation of Alkyl Halides From Alkanes

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