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

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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

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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

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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

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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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The name protein (proteios Greek  = pre-eminent or first) was first suggested, in 1838, by a Swedish chemist Berzelius to a Dutch chemist Mulder, who referred it to the complex organic nitrogenous substances found in the cells of the living beings. They are naturally occurring nitrogenous polymers of different alpha-amino acids linked by peptide (—CONH) linkage. On hydrolysis, they give a mixture of alpha-amino acids. Thus proteins are biopolymers of alpha-amino acids. They contain carbon, hydrogen, nitrogen, sulphur, and oxygen as constituent elements. They may contain cobalt, manganese, zinc, iron, copper, etc.

They are present in animals as well as plants. In the animal kingdom, it occurs in forms such as silk, wool, hair, nail, skin, haemoglobin of blood, and blood plasma. In the plant kingdom, it occurs in high concentration in seeds. Their sources are pulses, milk, eggs, fish, meat, etc. They are important because they regulate metabolic processes. They are essential constituents of all living matter.

Classification of Proteins:

Classification on the Basis of Composition:

Simple proteins:

They on acid hydrolysis give only alpha-amino acids. e.g.

• Albumins: egg albumin, Serum albumin
• Globulins: Tissue, vegetable,  and Serum globulin
• Glutenins: Glutenin in wheat.
• Protamines: Occur in the nucleic acid.

Conjugated proteins:

They on hydrolysis (acids or alkalies or enzymes) give alpha-amino acids and non-protein group. e.g.

• Glycoproteins: Contain carbohydrate as a prosthetic group. e.g. egg white, mucin of saliva
• Nucleoproteins: Contain nucleic acid as a prosthetic group. e.g. components of viruses, chromosomes, and ribosome structures.
• Chromoproteins: Contain chlorophyll as a prosthetic group. e.g. Haemoglobin
• Phosphoproteins: Contain Phosphoric acid as a prosthetic group. e.g. Casein of milk and vielline of egg yolk.
• Lipoproteins: Contain fats as a prosthetic group. e.g. membrane structure, lipids transported in the blood.
• Flavoprotein: Contain flavine Adenine Dinucleotide (FAD) as a prosthetic group. This protein is important in the electron transport chain in respiration.
• Metal-protein: Contain metal as a prosthetic group. e.g. the protein in plants nitrate reductase, which converts nitrates into nitrites.

Derived proteins:

Natural proteins undergo structural change because of heat, chemical reagents, enzymes, acids, or alkalies and form degraded products of proteins (derived).

The flow is Proteins → Proteoses → Peptones → Polypeptides → Simple peptides → amino acids

Classification Based on Structure:

Globular proteins:

• They have a globular or elliptical shape. They are also called as spheroproteins.
• They are soluble in water, acids, and bases. They form a colloidal solution with water.
• They are made up of not only primary, secondary but also tertiary and occasionally quaternary structures.
• They form enzymes, antibodies, and some hormones (insulin), etc. They are needed for the formation of chemical messengers like hormones in the body. They are essential for the formation of transporters of other particles through the membrane.
• e.g. egg albumin (egg white), Casein in milk

Fibrous proteins:

• They have a fiber-like structure. They are also called as scleroproteins.
• They are elongated strand-like structures and are usually present in the form of rods or wires.
• They are insoluble in water, weak acids, and weak bases but soluble in strong acids and alkalis.
• They have primary and secondary structures. They are made up of a single unit or structure which is repeated multiple times.
• They perform a structural function in the cell. They are needed for the formation of tough structures like connective tissue, tendons, and fibers of the muscle.
• e.g. collagen, elastin, keratin of hair, Nails, horn, feathers. fibroin of silk.

Classification Based on Functions:

Enzymic Proteins.

• They are the most varied and most highly specialized proteins and shows catalytic activity. Almost all enzymes are globular proteins.
• Enzymes catalyze a variety of reactions.
• Examples: Urease, amylase, catalase, cytochrome C, alcohol dehydrogenase

Structural Proteins.

They are usually inert to biochemical reactions. They maintain the native form and position of the organs. The cell wall and primary fibrous constituents of the cell have structural proteins.

• Collagen: It is found in connective tissue such as tendons, cartilage, a matrix of bones, and the cornea of the eye. Leather is almost pure collagen.
• Elastin: It is found in ligaments. It is capable of stretching in two dimensions.
• Keratin: It constitutes almost the entire dry weight of hair, wool, feathers, nails, claws, quills, scales, horns, hooves, tortoiseshell, and much of the outer layer of skin.
• Fibroin: It is the major component of silk fibres and spider webs.
• Resilin: The wing hinges of some insects are made of resilin, which has nearly perfect elastic properties.

Transport or Carrier Proteins:

• Certain proteins, especially in animals, are involved in the transport of many essential biological factors to various parts of the organisms.
• Hemoglobin of erythrocytes carries oxygen to tissues. The blood plasma contains lipoproteins, which carry lipids from the liver to other organs. Ceruloplasmin transports copper in the blood.

Nutrient and Storage Proteins:

Ovalbumin is the major protein of egg white. The milk protein, casein stores amino acids. The seeds of many plants store nutrient proteins, required for the growth of the germinating seedlings. Ferritin, found in some bacteria and in plant and animal tissues, stores iron.

Contractile or Motile Proteins:

• They give an ability to contract, move about, and change shape to cells. Actin and myosin function in the contractile system of skeletal muscle and also in many nonmuscle cells.
• Microtubules are built up of Tubulin.

Defense Proteins:

• They defend organism against invasion by other species or protect them from injury.
• Immunity: The antibodies (or immunoglobulins), the specialized proteins made by the lymphocytes of vertebrates, can precipitate or neutralize invading bacteria, viruses or foreign proteins from another species.
• Blood Clotting: Fibrinogen and thrombin, although enzymic, are blood-clotting proteins that prevent loss of blood when the vascular system is injured.

Regulatory Proteins:

• They regulate the cellular or physiological activity. They are called hormones.
• For example, insulin regulates sugar metabolism and growth hormone which is required for bone growth.
• The cellular response to many hormonal signals is often mediated by a class of GTP-binding proteins called G-proteins.
• Some of them bind to DNA and regulate the biosynthesis of enzymes and RNA molecules involved in cell division.

Toxic proteins.

Snake venom, bacterial toxins, and toxic plant proteins are toxic. They have defensive functions.

Science > Chemistry > Biomolecules > Proteins and Their Classification

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Organic chemistry is a branch of chemistry, which studies carbon compounds. Organic substances are substances derived from organisms like plants and animals. In the olden days it was assumed that these substances cannot be prepared in laboratories. Berzelius assumed that some vital force was necessary for the production of these organic compounds. He assumed that living beings possess this vital force (life force). Hence it is not possible to produce organic substances in laboratories.  Now the word organic has no significance as these substances can be prepared in laboratories.

In 1828 Wohler synthesized first organic compound, urea in the laboratory by heating an inorganic compound ammonium cyanate. It is molecular rearrangement process.

In 1845 Kolbe synthesized acetic acid, from its elements. In 1856 Berthelot synthesized methane in the laboratory.

Carbon compounds are compounds in which carbon is combined with elements such as hydrogen, oxygen, nitrogen, halogens, sulphur, phosphorous and few metals. They may be derived from living organisms like plants and animals or they may be prepared in laboratories (Synthetic substances).

The simplest organic compounds are hydrocarbons. Other organic compounds can be derived from hydrocarbons. Hence in simple words, we can define organic chemistry as the study of hydrocarbons and their derivatives.

The Necessity of Separate Branch for Study of Carbon Compounds:

Organic chemistry is a branch of chemistry, which studies carbon compounds. The necessity of separate branch of carbon compounds can be explained as follows.

• Organic chemistry is the largest part of chemistry.
• The number of carbon compounds far exceeds the number of compounds of all the remaining hundred and odd elements put together. Organic compounds comprise 90% of all known compounds. It is due to the special capability of carbon called catenation and the formation of isomers of the compounds.
• Carbon compounds form the basis of modern chemical industries.
• The food materials contain nutrients like fats, carbohydrates, proteins are carbon compounds.
• The fuels like petrol, diesel, compressed natural gas (C.N.G.), liquefied petroleum gas (L.P.G.) are carbon compounds.
• Fibres like wool, silk, cotton, jute, and synthetic fibres like nylon, terylene, polyester are carbon compounds.
• Paints, varnishes, dyes, perfumes, insecticides, fertilizers, soil conditioners, plastics, detergents, drugs are carbon compounds.

Thus there is no aspect of our material life that is not touched by organic chemistry

Carbon forms a Large Number of Compounds:

The number of carbon compounds far exceeds the number of compounds of all the remaining hundred and odd elements put together. It can be explained as follows.

• The atomic number of carbon is 6. The electronic configuration of carbon is 2,4. Thus each atom of carbon contains 4 electrons in the outermost orbit.
• The carbon atom can share its electrons with the atoms of different elements like H, O, N, S, P, Cl, Br, I, etc.
• Carbon possesses a property called catenation by which it has the capacity to form a direct bond with another atom of carbon to form chains or rings of different sizes and shapes. Two carbon atoms can form a single bond, double bond, or triple bond among themselves.
• Carbon has the capacity to get attached directly to active metals like magnesium to form organic metallic compounds.
• Compounds of carbon are obtained in different isomeric forms.

Catenation:

Catenation is the capacity to form a direct bond with another atom of the same element to form chains or rings of different sizes and shapes. Example bond formation between two carbon atoms. Besides carbon following elements show the capacity of catenation.

The bond energy is the highest in carbon and it decreases in order C > Si > S > P > N > O. The stability of the bond decreases in the same order. Hence the ability to form long chains decreases in the same order.

Sources of Organic Compounds:

Natural Sources:

• Fungi and microorganisms produce alcohols, acids, vitamins and antibiotics during fermentation.
• From coal by a destructive distillation aromatic hydrocarbons, dyes, drugs, perfumes are obtained.
• Natural gas and petroleum is a major source of organic compounds like gasoline, kerosene, lubricants, machine oils, paraffin waxes, vaseline, etc.
• 10 % of organic compounds are sourced from natural sources.

Synthetic Sources:

Man-made or artificially prepared substances are called synthetic substances.  Simple organic substances are obtained from petroleum and then applying certain processes they are converted into complex organic substances in industry or laboratories. 90 % of organic compounds are sourced from synthetic sources.

The Significance of Organic Compounds in Modern Life:

The significance of organic compounds in modern life can be explained w.r.t. following industries.

• Food Industry: The food materials contain nutrients like fats, carbohydrates, proteins are organic compounds. Artificial sweetener, flavouring agent, food preservatives are organic compounds.
• Petroleum Industry: The fuels like kerosene, gasoline, petrol, diesel, compressed natural gas (C.N.G.), liquefied petroleum gas (L.P.G.) are organic compounds. Paraffins (Wax), Vaseline, Boot Polish etc. are organic compounds.
• Textile Industry: Fibres like wool, silk, cotton, jute are organic compounds. Synthetic fibres like nylon, terylene, polyester are organic compounds.
• Solvent Industry: Water is a good solvent but there are many substances which are not getting dissolved in water. Organic substances like chloroform, alcohol, benzene, acetone, carbon tetrachloride are good solvents.
• Plastic Industry: Polyvinyl chloride (PVC), polyethene, bakelite, rubber are organic compounds.
• Paints and Dyestuff Industry: Paints, varnishes, dyes, indigo, azo dyes, printing inks are organic compounds. e.g. Malachite gree, Alizarin, etc.
• Soap and Detergent industry: Detergents are organic compounds. soaps are alkali salts of higher fatty acids are organic compounds.
• Miscellaneous Industry: Perfumes, insecticides (D.D.T., Gammexane, Malathion, etc.), fertilisers like urea, drugs are carbon compounds.
• Medicines: Antibiotics Penicillin, Streptomycin, Chloromycetin, Sulphadiazine, Aspirin, Morphine, Cocaine, iodoform are organic compounds,
• Explosives: Nitroglycerine, Nitrocellulose, T.N.B., T.N.T., etc. are organic compounds.

Characteristics of Organic Compounds:

• Types of bond and linkage: Organic compounds are generally covalent compounds and do not ionize when dissolved in water.
• Solubility: Organic compounds are mostly insoluble in water. They are soluble in organic solvents like benzene, alcohol, chloroform.
• Melting and Boiling points: Organic compounds have generally low melting points and boiling points.
• Electrical Conductivity: Organic compounds are bad conductors of electricity.
• Isomerism: Two or more compounds having the same molecular formula but different structural formulae are called isomers and the phenomenon is known as isomerism. Isomerism is a unique characteristic of organic compounds.

• Polymerization: Organic compounds have a tendency to form a polymer.
• Nature: Organic compounds have a well-established structure and are complex compounds and possess high molecular weight.
• Odour: Many organic substances have a characteristic odour. e.g. esters have a sweet odour, amines have a fishy odour, phenol has a clinical smell.
• Rate of Reaction: Reactions involving organic compounds are slow as breaking and forming of covalent bonds takes place.

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

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In the last article, we have studied monosaccharides. In this article, we shall study disaccharides and polysaccharides.

Disaccharides:

Di-saccharides on hydrolysis give two molecules of monosaccharide. They on hydrolysis with dilute acids or enzymes yield two molecules of either the same or different monosaccharides.

e.g. Cane sugar (Sucrose) (C₁₂H₂₂O₁₁)on hydrolysis gives one molecule of glucose and one molecule of fructose, Maltose (C₁₂H₂₂O₁₁) on hydrolysis gives two molecules of glucose, Lactose (C₁₂H₂₂O₁₁) on hydrolysis gives one molecule of glucose and one molecule of galactose.

• Disaccharides are crystalline, water-soluble, and sweet in taste.
• They have the general formula (C₁₂H₂₂O₁₁).
• The two monosaccharides are joined together by an oxide linkage formed by the loss of a water molecule. Such a linkage between two monosaccharide units through oxygen atom is called glycosidic linkage.

Examples of Disaccharides:

Sucrose:

One of the common disaccharides is sucrose which on hydrolysis gives an equimolar mixture of α -D-Glucapyranose and β-D-Fructofuranose. These two monosaccharides are held together by a glycosidic linkage between C1 of α-glucose and C2 of β-fructose.

Since the reducing groups of glucose and fructose are involved in the glycosidic bond formation, sucrose is a non-reducing sugar.

Sucrose is dextrorotatory but after hydrolysis gives dextrorotatory glucose and laevorotatory fructose. Since the laevorotation of fructose (–92.4°) is more than dextrorotation of glucose (+ 52.5°), the mixture is laevorotatory. Thus, hydrolysis of sucrose brings about a change in the sign of rotation, from dextro (+) to laevo in the sign of rotation, from Dextro (+) to Laevo (–) and the product is named as invert sugar.

Maltose:

Another disaccharides, maltose is composed of two α-D-glucose units in which C1 of one glucose (I) is linked to C4 of another glucose unit (II).

The free aldehyde group can be produced at C1 of second glucose in solution and it shows  reducing properties so it is a reducing sugar

Cellobiose:

Cellobiose is obtained by partial hydrolysis of cellulose, C1 of one β-D-Glucapyranose is linked to C4 of another β-D-Glucapyranose by glucosidic linkage. Thus cellobiose contains 1→ 4 β- glucosic bond.

Cellobiose is reducing sugar because free aldehyde group can be produced at C1 in second glucose molecule

Lactose:

It is more commonly known as milk sugar since this disaccharide is found in milk. It is composed of β-D-galactose (β-D-Galactopyranose) and β-D-glucose (β-D-Glucopyranose). The glucosidic linkage is between C1 of β-D-galactose (b-D-Galactopyranose) and C4 of β-D-glucose (β-D-Glucopyranose). Hence it is also a reducing sugar.

Reducing Sugars:

The saccharides in which aldehydic and ketonic groups are free are called reducing sugars. They reduce Fehling’s solution and Tollen’s reagent. They contain either α-hydroxy aldehyde or α -hydroxy ketone group or contain cyclic hemiacetal or cyclic hemiketal structures.

All monosaccharides are reducing sugars. Diasaccharides like maltose, lactose and cellobiose are reducing sugars.

Nonreducing sugars:

The saccharides in which aldehydic and ketonic groups are not free are called non-reducing sugars. They do not reduce Fehling’s solution and Tollen’s reagent. They contain stable acetal or ketal structures which cannot be opened into the free carbonyl group.

Sucrose, Starch, Cellulose, Glycogen, Dextrin are reducing sugars.

Hemiacetal and Hemiketal Structures:

They are formed when an alcohol oxygen atom adds to the carbonyl carbon of an aldehyde or a ketone. When alcohol adds to an aldehyde, the result is called a hemiacetal; when alcohol adds to a ketone the resulting product is a hemiketal.

This happens through the nucleophilic attack of the hydroxyl group at the electrophilic carbonyl group. Since alcohols are weak nucleophiles, the attack on the carbonyl carbon is usually promoted by protonation of the carbonyl oxygen.

PolySaccharides:

Carbohydrates which on hydrolysis give indefinite or large no. of monosaccharides (more than 10)  are called polysaccharides. They contain a large number of monosaccharide units joined together by glycosidic linkages. These are the most commonly encountered carbohydrates in nature. They mainly act as food storage or structural materials.

• They are natural polymeric carbohydrates. They are insoluble in water, amorphous, and tasteless.
• They are nonsugars.
• They have the general formula (C6H10O5)n.  e.g. Starch, Cellulose, Inulin, Dextrin, etc
• Cellulose is a linear polymer of β-Glucose units while starch is a branched polymer of a-glucose units.

Examples of Polysaccharides:

Starch:

Starch is the main storage polysaccharide of plants. It is the most important dietary source for human beings. High content of starch is found in cereals, roots, tubers, and some vegetables. It is a polymer of α-glucose (α-D-Glucopyranose) and consists of two components Amylose and Amylopectin.

Amylose is a water-soluble component which constitutes about 15-20% of starch. Chemically amylose is a long unbranched chain with 200-1000 α-D-(+)-glucose units held by C1– C4 glycosidic linkage.

Amylopectin is insoluble in water and constitutes about 80-85% of starch. It is a branched chain polymer of α-D-glucose units in which chain is formed by C1–C4 glycosidic linkage whereas branching occurs by C1–C6 glycosidic linkage

Glycogen:

The carbohydrates are stored in the animal body as glycogen. It is also known as animal starch because its structure is similar to amylopectin and is rather more highly branched. It is present in the liver, muscles, and brain. When the body needs glucose, enzymes break the glycogen down to glucose (hydrolysis). Glycogen is also found in yeast and fungi.

Importance of Carbohydrates:

• Carbohydrates are essential for life in both plants and animals.
• They form a major portion of our food. Honey has been used for a long time as an instant source of energy by ‘Vaids’ in Ayurvedic system of medicine.
• Carbohydrates are used as storage molecules as starch in plants and glycogen in animals.
• The cell wall of bacteria and plants is made up of cellulose. We build furniture, etc. from cellulose in the form of wood and clothe ourselves with cellulose in the form of cotton fibre.
• They provide raw materials for many important industries like textiles, paper, lacquers and breweries.
• Two aldopentoses viz. D-ribose and 2-deoxy D-ribose (Section are present in nucleic acids. Carbohydrates are found in biosystem in the combination with many proteins and lipids.

Science > Chemistry > Biomolecules > Disaccharides and Polysaccharides

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