12.6 Isomerism

Definition box: Isomerism: The phenomenon of existence of two or more compounds possessing the same molecular formula but different properties is known as isomerism. Isomers: Compounds which exhibit isomerism are called as isomers.

The following flow chart shows different types of isomerism.

12.6.1 Structural Isomerism

Compounds having the same molecular formula but different structures (manners in which atoms are linked) are called as structural isomers.

Types of structural isomerism:

(i) Chain isomerism: When two or more compounds have similar molecular formula but different carbon skeletons, these are referred to as chain isomers and the phenomenon is termed as chain isomerism.

For example, $C_5H_{12}$ represents three compounds or three chain isomers:

(ii) Position isomerism: When two or more compounds differ in the position of substituent atom or in the position of functional group on the carbon skeleton, they are called position isomers and this phenomenon is termed as position isomerism.

For example, the molecular formula $C_3H_8O$ has two alcohols or, position isomers as shown:

(iii) Functional group isomerism: Two or more compounds having the same molecular formula but different functional groups are called functional isomers and this phenomenon is termed as functional group isomerism.

For example, the molecular formula $C_3H_6O$ represents an aldehyde and a ketone which are functional isomers as depicted:

(iv) Metamerism: It arises due to different alkyl chains on either side of the functional group in the molecule.

For example, $C_4H_{10}O$  represents methoxypropane ($CH_3OC_3H_7$) and ethoxyethane $(C_2H_5OC_2H_5)$.

Note box: Tautomerism: Compounds having same molecular formula but differing in position of hydrogen atom and in dynamic equilibrium are said to exhibit tautomerism.

12.6.2 Stereoisomerism

Definition box: Stereoisomerism: The compounds that have the same constitution and sequence of covalent bonds but differ in relative positions of their atoms or groups in space are called stereoisomers. This special type of isomerism is called as stereoisomerism.

It can be classified as: a) geometrical and b) optical isomerism.

Questions from section 12.6:

1. Define “isomerism”.

2. What is structural isomerism? Explain its types.

3. What is stereoisomerism? State its types.

12.7 Fundamental Concepts in Organic Reaction Mechanism

Some Important terms:

• Organic reaction: The reaction in which, the organic molecule called, substrate reacts with an appropriate attacking reagent and leads to the formation of one or more intermediate(s) and finally product(s).
  The general reaction is depicted as follows:

• Substrate is that reactant which supplies carbon to the new bond. The other reactant is called, reagent.
  If both the reactants supply carbon to the new bond then choice is arbitrary and in that case the molecule on which attention is focused is called substrate.
• Reaction mechanism: In organic reactions, a covalent bond between two carbon atoms or a carbon and some other atom is broken and a new bond is formed. A sequential description of such a reaction with details of electron movement, energies during bond cleavage and bond formation, and the rates of transformation of reactants into products (kinetics) is referred to as reaction mechanism.

The knowledge of reaction mechanism helps in understanding the reactivity of organic compounds and in their synthesis.

12.7.1 Fission of a Covalent Bond

A covalent bond can get cleaved either by:

(i) Homolytic cleavage (or, heterolysis),

(ii) Heterolytic cleavage (or, homolysis)

1. Homolysis: During homolytic cleavage of a covalent bond, the bond breaks symmetrically and hence, each species formed gets an odd electron. These species are called, free radicals.

Free radicals: The reactive intermediates containing odd number of electrons are called, free radicals.

Free radicals are formed by the homolytic cleavage of a covalent bond.

Note box: Free radicals are very reactive since they are one electron short of completing their outermost shell of electrons. All free radicals are electrically neutral.

Stability of free radicals: The stability of alkyl free radical is in the order: $CH_3<1^0<2^0<3^0$.

2. Heterolysis: During heterolysis cleavage of a covalent bond, the bond breaks asymmetrically and one of the two species gets both the electrons and the other loses electrons, resulting in the formation of ions.

The ions which are positively charged act as electrophiles whereas, the one with negative charge are called, nucleophiles. During heterolytic cleavage, if the carbon atom gets a positive charge, the species is called, carbocation ($^+CH_3$) and the species in which the carbon atom gets a negative charge is called, carbanion ($^-CH_3$).

Reactive Intermediate	Example	Hybridization	Structure	Stability
Free radicals	·CH₃	sp²	Planar (will become pyramidal if substituent groups are present)	3° > 2° > 1° > CH₃
Carbocations	⁺CH₃	sp²	Planar	3° > 2° > 1° > CH₃
Carbanions	⁻CH₃	sp³	Pyramidal	CH₃ > 1° > 2° > 3°

Nucleophiles:

Nucleophiles are electron rich species which attack the electron deficient centre during a chemical reaction. These are lewis bases which donate a pair of electrons during the reaction.

Examples include: $OH^-, CN^-, NH_2^-, H_2O, ROH$ and $NH_3$.

Electrophiles:

Electrophiles are electron deficient species which attack the electron rich centre during a chemical reaction.

Examples include: $ H^+, Cl^+, NO_2^+, BF_3, AlCl_3, SO_3$ and $CO_2$.

12.7.2 Electron Movement in Organic Reactions

The movement of electrons in organic reactions can be shown by curved-arrow notation. This notation represents changes in bonding which occur due to electronic redistribution during the reaction.

To show the change in position of a pair of electrons, curved arrow starts from the point from where an electron pair is shifted and it ends at a location to which the pair of electron may move. Presentation of shifting of electron pair is given below:

Movement of single electron is indicated by a single barbed ‘fish hooks’ (i.e. half headed curved arrow).

For example, in transfer of hydroxide ion giving ethanol and in the dissociation of chloromethane, the movement of electron using curved arrows can be depicted as follows:

12.7.3 Electron Displacement Effects in Covalent Bonds

During a chemical reaction, the bonding electrons of the reacting molecule are partially or completely displaced. This type of electronic displacement is called, electronic effects. Four kinds of electronic effects are observed in organic reactions as discussed below.

12.7.4 Inductive Effect

Let us consider a chlorine atom attached to a carbon atom. Since chlorine is more electronegative than the carbon atom, chlorine attracts the shared pair of electrons towards itself and hence, carbon becomes partially positive and chlorine becomes partially negative leading to the induction of permanent polarity on the molecule.

The permanent effect whereby, the polarity is induced on the carbon atom and the substituent attached to it due to minor displacement of the bonding electron pair towards more electronegative atom is known as inductive effect.

Inductive effect involves sigma electrons of a single bond.

Depending on the electronegativity of the substituent group, carbon atom to which substituent group is attached can acquire partial negative or partial positive charge. Thus, two types of inductive effects are defined. They are:

a) –I effect (Negative inductive effect): If the carbon atom is attached to a higher electronegative atom like chlorine, as in, methyl chloride, the shared electron pair is attracted towards the more electronegative (electron withdrawing) chlorine. The carbon atom gets the partial positive charge [δ+] and the electronegative atom gets a partial negative charge [δ-]. This type of inductive effect is known as –I effect.

A substituent group shows –I effect when the electron displacement takes place towards it. Hence, the substituent group acquired a partial negative charge and the carbon atom to which the substituent group is attached, gets partial positive charge.

The groups that show –I effect (in decreasing order) when joined to carbon atom are: $-NO_2 > -CN > -F > -COOH > -Cl > -Br > -I > -OCH_3> -OH > -C_6H_5$.

b) +I effect(Positive inductive effect):

If the carbon atom is attached to a lower electronegative group [Y] (electron donating group), the shared pair of electrons is attracted towards the carbon atom. Hence, a polarity is developed. The carbon atom gets a slight negative (δ-) and the lower electronegative atom gets a slight positive charge (δ+). This type of inductive effect is called, +I effect.

A substituent group shows +I effect when the electron displacement takes place away from it. Hence, the substituent group acquires a partial positive charge and carbon atom to which the substituent group is attached gets a partial negative charge.

The groups that show +I effect (in the decreasing order) when joined to carbon atom are:

$$ (CH_3)_3C- > (CH_3)_2CH- > CH_3-CH_2- > -CH_3$$

Example 12.11: Which bond is more polar in the following pairs of molecules: (a) H$_3$C-H, H$_3$C-Br (b) H$_3$C-NH$_2$, H$_3$C-OH (c) H$_3$C-OH, H$_3$C-SH

Solution:

(a) C–Br, since Br is more electronegative than H,

(b) C–O, (c) C–O

Example 12.12: In which C–C bond of CH$_3$CH$_2$CH$_2$Br, the inductive effect is expected to be the least?

Solution:

Magnitude of inductive effect diminishes as the number of intervening bonds increases. Hence, the effect is least in the bond between carbon-3 and hydrogen.

12.7.5 Resonance Structure

There are many organic molecules whose behaviour cannot be explained by a single Lewis structure.

For example,

1. For benzene, a structure containing alternating C–C single and C=C double bonds shown below is inadequate for explaining its characteristic properties.

As per the above representation, benzene should exhibit two different bond lengths, due to C–C single and C=C double bonds. However, as determined experimentally benzene has a uniform C–C bond distances of 139 pm, which is a value intermediate between the C–C single (154 pm) and C=C double (134 pm) bonds. Thus, the structure of benzene cannot be represented adequately by the above structure. However, benzene can be represented equally well by the energetically identical structures I and II.

Thus, the actual structure of benzene is a hybrid of the two structures (I and II) called resonance structures.

2. Nitromethane ($CH_3NO_2$) can be represented by two Lewis structures, (I and II). There are two types of N-O bonds in these structures.

However, it is known that the two N–O bonds of nitromethane are of the same length (intermediate between a N–O single bond and a N=O double bond). The actual structure of nitromethane is therefore a resonance hybrid of the two canonical forms, I and II.

Note box: The resonance structures (canonical structures or contributing structures) are hypothetical and individually do not represent any real molecule. But they do contribute to the actual structure and account for their stability.

Resonance Stabilisation Energy:

The energy of actual structure of the molecule (the resonance hybrid) is lower than that of any of the canonical structures. The difference in energy between the actual structure and the lowest energy resonance structure is called the resonance stabilisation energy or simply the resonance energy. The more the number of contributing structures, the more is the resonance energy. Resonance is particularly important when the contributing structures are equivalent in energy.

The following rules are applied while writing resonance structures:

1. The resonance structures have:

(i) the same positions of nuclei and,

(ii) the same number of unpaired electrons.

2. Among the resonance structures, the one which has more number of covalent bonds, all the atoms with octet of electrons, less separation between opposite charges, a negative charge if any on more electronegative atom, a positive charge if any on more electropositive atom and; more dispersal of charge, is more stable than others.

Example 12.13:  Write resonance structures of $CH_3COO^{–}$ and show the movement of electrons by curved arrows.

Solution:

First, write the structure and put unshared pairs of valence electrons on appropriate atoms. Then draw the arrows one at a time moving the electrons to get the other structures.

Example 12.14: Write resonance structures of CH2=CH–CHO. Indicate relative stability of the contributing structures.

Solution:

[I: Most stable, more number of covalent bonds, each carbon and oxygen atom has an octet and no separation of opposite charge II: negative charge on more electronegative atom and positive charge on more electropositive atom; III: does not contribute as oxygen has positive charge and carbon has negative charge, hence least stable].

Example 12.15: Explain why the following two structures, I and II cannot be the major contributors to the real structure of $CH_3COOCH_3$.

Solution:

The two structures are less important contributors as they involve charge separation. Additionally, structure I contains a carbon atom with an incomplete octet.

12.7.6 Resonance Effect

Definition box: Resonance Effect: The resonance effect is defined as the polarity produced in the molecule by the interaction of two π-bonds or between a π-bond and lone pair of electrons present on an adjacent atom.

The effect is transmitted through the chain.

Concept box: Conjugated System: The presence of alternate single and double bonds in an open chain or cyclic system is termed as a conjugated system. These systems often show abnormal behaviour. Examples include: 1,3- butadiene, aniline and nitrobenzene. In such systems, the π-electrons are delocalised and the system develops polarity.

There are two types of resonance or mesomeric effect designated as R or M effect. They are:

(i) Positive Resonance Effect (+R effect)

In this effect, the transfer of electrons is away from an atom or substituent group attached to the conjugated system. This electron displacement makes certain positions in the molecule high in electron densities.

This effect in aniline is shown as:

(ii) Negative Resonance Effect (- R effect)

This effect is observed when the transfer of electrons is towards the atom or substituent group attached to the conjugated system.

For example, in nitrobenzene this electron displacement can be depicted as:

The atoms or substituent groups, which show +R or –R electron displacement effects are as follows:

a) +R effect: – halogen, –OH, –OR, –OCOR, –$NH_2, –NHR, –NR_2$, –NHCOR,

b) – R effect: – COOH, –CHO, >C=O, – CN, –$NO_2$.

Differences between Inductive and Mesomeric Effect

Inductive effect	Mesomeric effect
1. Involves the displacement of σ electrons	1. Involves the displacement of π electrons
2. Electrons get partially displaced	2. Electrons get completely displaced
3. Inductive effect mainly operates in saturated compounds	3. Mesomeric effect operates in the conjugated unsaturated system
4. Does not involve delocalisation of electrons	4. Involves delocalisation of electrons
5. Inductive effect weakens as the distance from the substituent group increases	5. Mesomeric effect does not decrease with the increase in the distance from the substituent group

12.7.7 Electrometric Effect (E effect)

Definition Box: Electromeric Effect (E) is defined as the complete transfer of a shared pair of π-electrons to one of the atoms joined by a multiple bond on the demand of an attacking reagent.

• It is a temporary effect.
• The organic compounds having a multiple bond (a double or triple bond) show this effect in the presence of an attacking reagent only.
• The effect is ceased as soon as the attacking reagent is removed from the reaction.
• It is represented by E and the shifting of the electrons is shown by a curved arrow: .

• There are two distinct types of electromeric effect:

(i) Positive Eelctromeric Effect (+E effect): In this effect, the π−electrons of the multiple bond are transferred to that atom to which the reagent gets attached.

For example,

(ii) Negative Electromeric Effect (–E effect): In this effect, the π – electrons of the multiple bond are transferred to that atom to which the attacking reagent does not get attached.

For example,

Note box: When inductive and electromeric effects operate in opposite directions, the electomeric effect predominates.

Difference Between inductive and electromeric effect

Inductive effect	Electromeric effect
1. Involves σ electrons	1. Involves π electrons
2. Permanent effect	2. Temporary effect: The effect will remain only till the attacking reagent is present.
3. Displacement of the σ electron takes place	3. Transfer of the π electron takes place
4. Electrons involved do not leave the original orbitals	4. Electrons involved occupy the new orbitals
5. Develops partial charge in the molecule	5. Develops complete charge separation in the molecule

12.7.8 Hyperconjugation

Definition box: Hyperconjugation is the phenomenon in which electron displacement takes place resulting in the conjugation between π electrons and σ electrons.

This effect is also called Baker Nathan effect (or) No Bond Resonance. Only the electrons of C—H bond give rise to this type of effect.

Hyperconjugative effect is used to explain the following phenomenon:

a) Orienting influence of –$CH_3$ group

b) Stability of carbocation

c) Stability of free radicals

Orienting influence of –$CH_3$ group:

The carbon atom of methyl group which is directly attached to the benzene ring does not contain lone pair of electrons. The ortho-para directing nature of –$CH_3$  group can be explained by the hyperconjugative effect or, no bond resonance.

The C—H bonds of the –$CH_3$   group interacts with the π electrons of the benzene ring. This gives rise to the following resonance structures:

The actual structure is represented by the resonance hybrid of these structures. In this resonance hybrid, the ortho and para positions have higher electron density than meta positions. Therefore, the incoming substituent group which is an electrophile will prefer ortho-para position over meta position. Hence, –$CH_3$   acts as an ortho-para directing group.

Stability of Carbocations:

The stability of carbocations is in the order: 3⁰>2⁰>1⁰

This can be explained by the number of hyperconjugative structures possible for a given carbocation.

Therefore, there is 9 C–H bonds which involve in hyperconjugation and hence, 9 resonance structures are possible by distribution of positive charge.

On the other hand, secondary carbocation has only 6 similar contributing structures and primary has only 3 resonance structures. Therefore, maximum distribution of charge takes place in tertiary carbocation. Hence, maximum will be the stability.

Stability of free radicals:

The stability of free radicals is in the order: 3⁰>2⁰>1⁰.

This can also be explained by the hyperconjugative effect. The contributing structures of tertiary free radicals due to hyperconjugation are:

Hence, similar to the distribution of charge that takes place in case of carbocation, the distribution of unpaired electron takes place in the case of free radical. Greater the distribution, higher is the stability.

Example 12.16:  Explain why $(CH_3)_3 \overset{+}{C}$ is more stable than $CH_3\overset{+}{C}H_2$ and  $\overset{+}{C}H_3$ is the least stable cation.

Solution:   

Hyperconjugation interaction in $(CH_3)_3\overset{+}{C}$ is greater than in $CH_3\overset{+}{C}H_2$ as the  $(CH_3)_3 \overset{+}{C}$ has nine C-H bonds. In $\overset{+}{C}H_3$ vacant p orbital is perpendicular to the plane in which C-H bonds lie; hence cannot overlap with it. Thus, $\overset{+}{C}H_3$ lacks hyperconjugative stability. 

12.7.9 Types of Organic Reactions and Mechanisms

Organic reactions can be classified into the following categories:

(i) Substitution reactions

(ii) Addition reactions

(iii) Elimination reactions

(iv) Rearrangement reactions

We will be studying these reactions in the next unit.

Questions from section 12.7:

1. Describe the terms:

a) Substrate

b) Reagent

c) Reaction mechanism

d) Homolytic cleavage

e) Heterolytic cleavage

f) Nucleophile

g) Electrophile.

2. Explain the terms:

a) Inductive effect (with +I and –I effects)

b) Resonance Stabilisation Energy

c) Resonance effect (with +R and –R effects)

d) Electromeric effect

3. Explain the phenomena as a consequence of hyperconjugation:

a) Orienting influence of –$CH_3$ group

b) Stability of carbocation

c) Stability of free radicals

12.8 Methods of Purification of Organic Compounds

Purification techniques of organic compounds rely on the nature of the compound and the impurity present in it.

The common techniques used for purification are as follows:

(i) Sublimation

(ii) Crystallisation

(iii) Distillation

(iv) Differential extraction and

(v) Chromatography

12.8.1 Sublimation

• Sublimation is the process of transformation directly from the solid phase into the gaseous phase without passing through the intermediate liquid phase. When the vapours are cooled, the original substance in pure form is obtained.
• This technique is used to separate sublimable compounds from nonsublimable impurities.
• In this method, the impure substance is placed in a dish covered with an inverted funnel. The dish is slowly heated. The sublimate solid vapourises and condenses on the walls of the funnel, leaving the sublimate impurities in the dish.
• Naphthalene, Camphor, Benzoic acid and iodine are purified by sublimation.

12.8.2 Crystallisation

• This method is used for the purification of solid organic compounds.
• It is based on the difference in the solubilities of the compound and the impurities in a suitable solvent.
• The solvent chosen for crystallization should have the following properties:
  • The compound should be highly soluble in solvent at high temperature and only sparingly soluble at room temperature.
  • The impurities should either be highly soluble in solvent at room temperature or insoluble even in the hot solvent which can then be filtered off.
  • The solvent should not react with the compound being purified.
• On cooling the solution, pure compound crystallises out and is removed by filtration.
• The filtrate (mother liquor) contains impurities and small quantity of the compound.
• Impurities which impart colour to the solution are removed by adsorbing over activated charcoal.
• If the compound is highly soluble in one solvent and very little soluble in another solvent, crystallisation can be satisfactorily carried out in a mixture of these solvents.
• Repeated crystallisation becomes necessary for the purification of compounds containing impurities with near solubilities.
• Benzoic acid can be purified using water as solvent.

12.8.3 Distillation

Definition box: Distillation is a process of separating the components of the mixture by their boiling points. Different chemicals distil at different temperatures and hence, the mixture can be separated and purified.

Distillation can be carried out in the following ways:

a) Simple distillation

b) Fractional distillation

c) Distillation under reduced pressure

d) Steam distillation

e) Vacuum distillation

Simple distillation

This method is used to separate:

(i) volatile liquids from nonvolatile impurities and,

(ii) liquids having sufficient difference in their boiling points.

• Liquids having different boiling points vaporise at different temperatures.
• The vapours are cooled and the liquids so formed are collected separately.
• For example, Chloroform (b.p 334 K) and aniline (b.p. 457 K) are easily separated by simple distillation (Figure 12.2).
• In this process, the liquid mixture is taken in a round bottom flask and heated carefully. On boiling, the vapours of lower boiling component are formed first. The vapours are condensed by using a condenser and the liquid is collected in a receiver. The vapours of higher boiling component form later and this liquid can be collected separately.

Fractional Distillation

• If the difference in boiling points of two liquids is not much, simple distillation cannot be used to separate them. The vapours of such liquids are formed at the same temperature range and are condensed together. The technique of fractional distillation is used in such cases.
• In this technique, vapours of a liquid mixture are passed through a fractionating column before condensation.
• The fractionating column is fitted over the mouth of the round bottom flask (Figure 12.3).

• Vapours of the liquid with higher boiling point condense before the vapours of the liquid with lower boiling point.
• In the fractionating column, the plates are arranged such that, the vapours of those components with higher boiling point gets condensed in fractionating column and fall back to the bottom of the flask.
• The compound with least boiling point vaporises first, is condensed and collected.
• When the highly volatile component is completely removed, on further heating, the less volatile component also vaporises. The less volatile component is condensed and collected.
• Similarly, a series of compounds are separated simultaneously one after the other.
• Separation of components of petroleum is carried out by this process.

Distillation under reduced pressure:

• This method is used to purify liquids having very high boiling points and those, which decompose at or below their boiling points.
• Such liquids are made to boil at a temperature lower than their normal boiling points by reducing the pressure on their surface by using a water pump or vacuum pump (Figure 12.4).

• The liquid boils at a temperature at which its vapour pressure is equal to the external pressure.
• Glycerol can be separated from spent-lye in soap industry by using this technique.

Steam Distillation:

• Steam Distillation technique is used for the purification of mixtures in which, the components are heat or temperature sensitive.
• In steam distillation, steam from a steam generator is passed through a heated flask containing the liquid to be distilled.
• The mixture of steam and the volatile organic compound is condensed and collected.
• The compound is later separated from water using a separating funnel.
• In steam distillation, the liquid boils when the sum of vapour pressures due to the organic liquid ($p_1$) and that due to water ($p_2$) becomes equal to the atmospheric pressure (p), that is, when, $p =p_1+ p_2$.
• This technique allows for evaporation of components with high boiling points at much lower temperatures merely by allowing them to form a mixture with water.
• Aniline is separated by this technique from aniline – water mixture (Figure 12.5).

12.8.4 Differential Extraction

• Differential Extraction is a method used to separate compounds based on their relative solubilities in two different immiscible liquids usually, water and an organic solvent.
• The organic solution containing the organic solute is taken in a separating funnel. It is a glass apparatus in which, the line of demarcation between the two liquids is clearly visible (As shown in figure below):

• Since the solute dissolves in the organic solvent more than that of water during extraction, the solute dissolves in the organic solvent which is separated.
• The solvent is removed by evaporation.
• If the solute is less soluble in organic solvent, then the same solvent is continuously recycled and used for extraction. This process is thus called, Continuous extraction.

12.8.5 Chromatography

Chromatography is a technique extensively used to separate mixtures into their components, purify compounds and also to test the purity of compounds.

The name chromatography is based on the Greek word “chroma”, for colour since the method was first used for the separation of coloured substances found in plants.

• In the chromatography technique, the mixture of substances is applied onto a stationary phase, which may be a solid or a liquid.
• The other is the mobile phase which moves over the stationary phase in a definite direction. The mobile phase is either a liquid or a gas.
• Distribution of solute between stationary and mobile phase occur either by adsorption or by partition based on which, chromatography is classified into different categories. Two of these are:

(a) Adsorption chromatography, and

(b) Partition chromatography.

a) Adsorption Chromatography:

• Adsorption chromatography is based on the fact that different compounds are adsorbed on an adsorbent to different degrees.
• Commonly used adsorbents are silica gel and alumina.
• When a mobile phase constituted by the mixture is allowed to move over a stationary phase (adsorbent), the components of the mixture move by varying distances over the stationary phase.
• Following are two main types of chromatographic techniques based on the principle of differential adsorption:

(a) Column chromatography, and

(b) Thin layer chromatography.

(a) Column Chromatography:

• Column chromatography involves separation of a mixture in a column of adsorbent (stationary phase) packed in a glass tube.
• The column is fitted with a stopcock at its lower end (as shown in Figure 12.7).

• The mixture adsorbed on the adsorbent is placed on the top of the adsorbent column packed in the glass tube.
• An appropriate eluant which is a liquid or a mixture of liquids is allowed to flow down the column slowly.
• Depending upon the degree to which the compounds are adsorbed, complete separation takes place.
• The most readily adsorbed substances are retained near the top and others come down to various distances in the column due to the flow of liquid (as depicted by Figure 12.7).

(b) Thin Layer Chromatography:

• Thin layer chromatography (TLC) is another type of adsorption chromatography, which involves separation of substances of a mixture over a thin layer of an adsorbent coated on glass plate.
• The layer of the adsorbent is about 0.2mm thick, made of silica gel or alumina.
• The adsorbent is spread over a glass plate of suitable size. The glass plate is known as thin layer chromatography plate or chromaplate.
• The solution of the mixture to be separated is applied as a small spot about 2 cm above one end of the TLC plate.
• The glass plate is then placed in a closed jar containing the eluant (Figure 12.8a).

• As the eluant rises up the plate, the components of the mixture move up along with the eluant to different distances depending on their degree of adsorption and thus, separation takes place.
• The relative adsorption of each component of the mixture is expressed in terms of its retardation factor (or) $ R_f$ value (Figure 12.12 b).

$$ R_f = \frac {Distance \space moved \space by \space the \space substance \space from \space base \space line \space (x)}{ Distance \space moved \space by \space the \space solvent \space from \space base \space line \space (y)} $$

• The spots of coloured compounds are visible on TLC plate due to their original colour.
• The spots of colourless compounds, which are invisible to the eye but fluoresce in ultraviolet light, can be detected by putting the plate under ultraviolet light. Another detection technique is to place the plate in a covered jar containing a few crystals of iodine.
• Spots of compounds, which adsorb iodine, will show up as brown spots. Sometimes an appropriate reagent may also be sprayed on the plate.
• For example, amino acids may be detected by spraying the plate with ninhydrin solution (as obtained in Figure 12.8b).

(b) Partition Chromatography:

• Partition chromatography is a technique in which, immobilized liquid acts as the stationary phase and another liquid as the mobile phase.
• Paper chromatography is a type of partition chromatography.
• In paper chromatography, a special quality paper known as chromatography paper is used. Chromatography paper contains water trapped in it, which acts as the stationary phase.
• A strip of chromatography paper spotted at the base with the solution of the mixture is suspended in a suitable solvent (or a mixture of solvents) as shown in Figure 12.9. This solvent acts as the mobile phase.

• The solvent rises up the paper by capillary action and flows over the spot. The paper selectively retains different components according to their differing adsorption capacities in the two phases.
• The paper strip so developed is known as a chromatogram. The spots of the separated colourless compounds may be observed either under ultraviolet light or by the use of an appropriate spray reagent.

Questions from section 12.8:

1. Explain how organic molecules are purified by the method of:

a) Sublimation

b) Crystallization

2. What is distillation?

3. Explain with relevant figures, purification by:

a) Simple distillation

b) Fractional distillation

c) Distillation under reduced pressure

d) Steam distillation

e) Vacuum distillation

4. Explain the technique of differential extraction

5. Explain:

a) Column Chromatography

b) Thin layer Chromatography

c) Partition Chromatography

12.9 Analysis of Organic Compounds

Through elemental analysis, the molecular formula of unknown compounds can be determined. Elemental analysis is performed by following two steps:

a) Qualitative analysis: Qualitative analysis is the process of detecting different elements present in a given compound. All the organic compounds essentially contain carbon and hydrogen. In addition, they may also contain oxygen, nitrogen and sulphur.

b) Quantitative analysis: Quantitative analysis is the process of estimating the amount of each of the elements present in an organic compound. The results are usually expressed in percentages.

12.9.1 Detection of Carbon and Hydrogen

Carbon and hydrogen are detected by heating the compound with copper(II) oxide.

During the reaction, carbon present in the compound is oxidised to carbon dioxide (which can be detected by lime-water, as it turns turbid) and hydrogen is oxidized to water (since anhydrous copper sulphate turns blue).

$ \ce { C + 2CuO ->[{\Delta}] 2Cu + CO2 } $

$ \ce { 2H + CuO ->[{\Delta}]Cu + H2O } $

$ \ce { CO2 + Ca(OH)2 ⎯→ CaCO3↓ + H2O } $

$ \ce {5H2O + CuSO4 → CuSO4.5H2O }$
              White        Blue

12.9.2 Detection of Other Elements

• Nitrogen, sulphur, halogens and phosphorus present in an organic compound are detected by Lassaigne’s test.
• During the Lassaigne’s test, the elements present in the compound are converted from covalent form into the ionic form by fusing the compound with sodium metal. Following reactions take place:

$$ \ce { Na + C + N ->[{\Delta}] NaCN }$$

$$ \ce { 2Na + S ->[{\Delta}] Na2S } $$

$$ \ce { Na + X ->[{\Delta}] Na X }$$ (X = Cl, Br or I)

• Cyanide, sulphide and halide of sodium so formed on sodium fusion are extracted from the fused mass by boiling it with distilled water. This extract is known as sodium fusion extract.

(A) Test for Nitrogen

• The sodium fusion extract is boiled with iron(II) sulphate and then acidified with concentrated sulphuric acid.
• The formation of Prussian blue colour confirms the presence of nitrogen.
• Sodium cyanide first reacts with iron(II) sulphate and forms sodium hexacyanoferrate(II). On heating with concentrated sulphuric acid, some iron(II) ions are oxidised to iron(III) ions which react with sodium hexacyanoferrate(II) to produce iron(III) hexacyanoferrate(II) (ferriferrocyanide) which is Prussian blue in colour.

$$ \ce {6CN– + Fe2+ → [Fe(CN)6]4– }$$

$$ \ce { 3[Fe(CN)6]^{4–} + 4Fe^{3+} ->[{xH_2O}]Fe4[Fe(CN)6]3.xH2O }$$

                                                                                        Prussian blue

(B) Test for Sulphur

(a) The sodium fusion extract is acidified with acetic acid and lead acetate is added to it. A black precipitate of lead sulphide indicates the presence of sulphur.

$$ \ce { S^{2-} + Pb^{2+} -> PbS }$$

                                                 Black

(b) Also, on treating sodium fusion extract with sodium nitroprusside, appearance of a violet colour confirms the presence of sulphur.

$$ \ce { S^{2-} + [Fe(CN)_5NO]^{2-} -> [Fe(CN)_5NOS]^{4-} }$$

                                                                          Violet

In case, nitrogen and sulphur both are present in an organic compound, sodium thiocyanate is formed. It gives blood red colour and no Prussian blue since there are no free cyanide ions.

$$ \ce { Na + C + N + S -> NaSCN }$$

$$ \ce { Fe^{3+} + SCN^- -> [Fe(SCN)]^{2+}}$$

                                                Blood red

Note box: If sodium fusion is carried out with excess of sodium, the thiocyanate decomposes to yield cyanide and sulphide. These ions give their usual test results. $$ \ce { NaSCN + 2Na -> NaCN + Na_2S } $$

(C) Test for Halogens

• The sodium fusion extract is acidified with nitric acid and then treated with silver nitrate.
• A white precipitate, soluble in ammonium hydroxide indicates the presence of chlorine, a yellowish precipitate, sparingly soluble in ammonium hydroxide shows the presence of bromine and a yellow precipitate, insoluble in ammonium hydroxide shows the presence of iodine.

$$ \ce { X^- + Ag^+ -> AgX}$$

Note box: Detection of nitrogen, sulphur and halogens by Lassaigne’s test: Nitrogen, sulphur and halogens in the organic compound can be detected by Lassaigne’s test. In this method, sodium fusion extract of the organic compound is prepared. During the fusion reaction, nitrogen is converted to cyanide, sulphur to sulphide and halogens to sodium halides. By testing for radicals, the presence of these elements can be established.

(D) Detection of phosphorus:

Organic compound is fused with sodium peroxide, extracted with water + boiled with concentrated nitric acid + ammonium molybdate. The appearance of yellow colour indicates the presence of phosphorus.

$$ \ce { Na_3PO_4 + 3HNO_3 -> H_3PO_4 + 3NaNO_3 } $$

$$ \ce { H_3PO_4 + 12(NH_4)_2MoO_4 + 21HNO_3 ->

                                         Ammonium molybdate

 (NH_4)_3PO_4.12MoO_3 + 21 NH_4NO_3 + 12H_2O } $$

Ammonium phosphomolybdate

Questions from section 12.9:

1. Define:

a) Qualitative analysis

b) Quantitative analysis

2. Explain how carbon and hydrogen can be detected in an organic compound

3. What is sodium fusion extract?

4. Explain the tests for the presence of:

a) Nitrogen

b) Sulphur

c) Halogen

d) Phosphorus

12.10 Qualitative Analysis

The percentage of elements in organic compounds is determined by the methods based on following compounds:

Estimation of Carbon and Hydrogen:

Principle: A known mass of organic compound is oxidized with cupric oxide in $CO_2$-free-atmosphere. Carbon gets oxidized to carbon dioxide and hydrogen to water. Water formed is absorbed in weighed potassium hydroxide bulbs. The increase in mass of calcium chloride and potassium hydroxide tubes are determined. These correspond to mass of water and carbon dioxide respectively. Knowing the increase in mass, the percentage of carbon and hydrogen in the given organic compound can be determined.

$$ \ce { C_xH_y + (x + y/4)O_2 -> x CO_2 + (y/2)H_2O}$$

Tabulations and calculations

Mass of organic compound taken = W g

Increase in the mass of calcium chloride U tube = mass of water formed = m₁ g

Increase in the mass of potassium hydroxide U tube = mass of carbon dioxide formed = m₂ g

Estimation of hydrogen:

Mass of hydrogen in 18g of water = 2g

Mass of hydrogen in m₁ grams of water = $ \frac {2}{18} \times m_1 =a \space g$

Therefore, percentage of hydrogen in the organic compound =  $ \frac {a}{w} \times 100 $

Estimation of carbon:

Mass of carbon in 44g of carbon dioxide = 12g

Mass of carbon in m₂ g of carbon dioxide = $\frac {12}{44} \times m_2$ =b g

Therefore, percentage of carbon in the organic compound =  $ \frac {b}{w} \times 100 $

Example 12.17: On complete combustion, 0.246 g of an organic compound gave 0.198g of carbon dioxide and 0.1014g of water. Determine the percentage composition of carbon and hydrogen in the compound.

Solution:

Percentage of carbon = $\frac{12 \times 0.198 \times 100}{44 \times 0.246}$

$= 21.95\%$

Estimation of Nitrogen

Estimation of Nitrogen is carried out by either of these methods:

a) Duma’s method

b) Kjeldahl method

a) Duma’s method:

Principle: A known mass of organic compound containing nitrogen is heated with cupric oxide in carbon dioxide. Organic nitrogen gets converted to free nitrogen gas along with water and carbon dioxide.

$$ \ce { C_xH_yN_z + (2x + y/2 )CuO -> x CO_2 + y/2 H_2O + z/2 N_2 + (2x + y/2) Cu}$$

If nitrogen gets converted to oxides of nitrogen, it is reduced to nitrogen gas by passing over heated- copper-gauze. The gaseous mixture formed is collected in KOH which absorbs all gases except nitrogen. The volume of nitrogen obtained is measured and is reduced to STP. From this, mass of nitrogen present in organic compound is calculated.

Procedure: A hard glass combustion tube is filled to two thirds with coarse cupric oxide. On both ends of the tube, oxidized copper gauze is placed. About 0.2g of organic compound is mixed with cupric oxide and taken in the tube. One end of combustion tube is connected to Schiff’s nitrometer containing 50% KOH and little mercury. Dry carbon dioxide is passed through the combustion tube from the other end. When air present in combustion tube is completely displaced, bubbles will stop forming in KOH solution. The tap present on top of nitro-meter is opened and the side tube is raised till nitrometer is completely with KOH solution and then the tap is closed.

Now the combustion tube is heated. Organic nitrogen gets converted into free nitrogen gas which displaces KOH solution and gets collected in nitrometer. The level of KOH in the reservoir and in the nitrometer is equalized. The volume of nitrogen gas collected is thus noted.

Calculations:

Mass of organic compound = ‘w’ g

Volume of nitrogen = $’V_1’cm^3$

To reduce this volume to STP

$ \frac {P_oV_o}{T_o} = \frac {(P_1 – f)V_1}{T_1}$

$\quad$ (standard) $\quad$ (lab)

We know, 28 g of (1 mole) of nitrogen occupies 22,710 $cm^3$ at STP.

$ \therefore $ Mass of nitrogen which occupies $V_o cm^3$ of nitrogen at STP = $ \frac {28 \times V_o}{22710} = a \space g $

$ \therefore $ Percentage of Nitrogen = $ \frac {a \times 100}{w}$

Example 12.18: In Dumas’ method for estimation of nitrogen, 0.3g of an organic compound gave 50mL of nitrogen collected at 300K temperature and 715mm pressure.

Calculate the percentage composition of nitrogen in the compound. (Aqueous tension at 300K=15 mm)

Solution:

Volume of nitrogen collected at 300K and 715mm pressure is 50 mL

Actual pressure = 715 – 15 = 700 mm

Volume of nitrogen at STP = $\frac{273 \times 700 \times  50}{300 \times  760} \space = \space 41.9mL$

22,400 mL of N$_2$ at STP weighs = 28 g

41.9 mL of nitrogen weigh = $\frac{28 \times 41.9 \times 100}{22400 \times 0.3} \space = \space 17.46\%$

b) Kjeldahl’s method:

Principle:

• The compound containing nitrogen is heated with concentrated sulphuric acid. Nitrogen in the compound gets converted to ammonium sulphate (Figure 12.12).

• The resulting acid mixture is then heated with excess of sodium hydroxide. The liberated ammonia gas is absorbed in an excess of standard solution of sulphuric acid.
• The amount of ammonia produced is determined by estimating the amount of sulphuric acid consumed in the reaction. The estimation is done by determining the unreacted sulphuric acid left after the absorption of ammonia by titrating it with standard alkali solution.
• The difference between the initial amount of acid taken and that left after the reaction gives the amount of acid reacted with ammonia.

Calculations:

$$ \ce { Organic compound + H2SO4 → (NH4)2SO4

->[{NaOH}] Na_2SO_4 + 2NH_3 + 2H_2O }$$

 $$ \ce { 2NH3 + H2SO4 → (NH4)2SO4 } $$

 Let the mass of organic compound taken = m g

Volume of $H_2SO_4$ of molarity, M, taken = V mL

Volume of NaOH of molarity, M, used for titration of excess of $H_2SO_4$ = V1 mL

$V_1$ mL of NaOH of molarity M = $V_1$ /2 mL of $H_2SO_4$ of molarity M

Volume of $H_2SO_4$ of molarity M unused = $(V – V_1/2)$ mL

$(V- V_1/2$) mL of $H_2SO_4$ of molarity M = 2($V-V_1/2$) mL of $NH_3$ solution of molarity M.

1000 mL of 1 M $NH_3$ solution contains 17g $NH_3$ or 14 g of N

2($V-V_1/2$) mL of $NH_3$ solution of molarity M contains:

$$ \frac { 14 \times M \times 2 (V -V_1/2)}{1000} g \space N $$

$$ Percentage \space of \space N= \frac {14 \times M \times 2 (V -V_1/2)}{1000} \times \frac {100}{m} $$

$$ = \frac {1.4 \times M \times 2 (V -V_1/2)}{m}$$

Note box: Kjeldahl method is not applicable to compounds containing nitrogen in nitro and azo groups and nitrogen present in the ring (like, pyridine) as nitrogen of these compounds does not change to ammonium sulphate under these conditions.

Example 12.19: During estimation of nitrogen present in an organic compound by Kjeldahl’s method, the ammonia evolved from 0.5g of the compound in Kjeldahl’s estimation of nitrogen, neutralized 10 mL of 1 M H$_2$SO$_4$. Find out the percentage of nitrogen in the compound.

Solution:

1 M of 10 mL H$_2$SO$_4$ = 1M of 20 mL NH$_3$ 1000 mL of 1M ammonia contains 14 g nitrogen 20 mL of 1M ammonia contains $\frac{14 \times 20}{1000}$g nitrogen

Percentage of nitrogen = $\frac{14 \times 20 \times 100}{1000 \times 0.5} = 56.0\%$

12.10.1 Halogens

Carius method:

• A known mass of an organic compound is heated with fuming nitric acid in the presence of silver nitrate contained in a hard glass tube known as Carius tube, (Figure 12.12) in a furnace.

Fig. 12.13 Carius method. Halogen containing organic compound is heated with fuming nitric acid in the presence of silver nitrate

• Carbon and hydrogen present in the compound are oxidised to carbon dioxide and water.
• The halogen present forms the corresponding silver halide (AgX).
• The silver halide is filtered, washed, dried and weighed.

Calculations:

Let the mass of organic compound taken = m g

Mass of AgX formed = $m_1$ g

1 mol of AgX contains 1 mol of X

Mass of halogen in $m_1$ g of AgX

$$ = \frac {atomic \space mass \space of \space X \times m_1 g}{molecular \space mass\space of \space AgX}$$

Percentage of halogen

$$ = \frac {atomic \space mass \space of \space X \times m_1 \times 100 }{molecular \space mass  \space of \space AgX \times m }$$    

Example 12.20: In Carius method of estimation of halogen, 0.15 g of an organic compound gave 0.12 g of AgBr. Find out the percentage of bromine in the compound.

Solution:

Molar mass of AgBr = 108 + 80 = 188 g mol$^{-1}$

188 g AgBr contains 80 g bromine

0.12 g AgBr contains $\frac{80 \times 0.12}{188}$ g bromine

Percentage of bromine = $\frac{80 \times 0.12 \times 100}{188 \times 0.15} = 34.04\%$

12.10.2 Sulphur

• A known mass of an organic compound is heated in a Carius tube with sodium peroxide or fuming nitric acid.
• Sulphur present in the compound is oxidised to sulphuric acid.
• It is precipitated as barium sulphate by adding excess of barium chloride solution in water.
• The precipitate is filtered, washed, dried and weighed.
• The percentage of sulphur can be calculated from the mass of barium sulphate.

Calculations:

Let,

The mass of organic compound taken = m g

The mass of barium sulphate formed = m₁ g

1 mol of $BaSO_4$ = 233 g $BaSO_4$ = 32 g sulphur

$m_1$ g $BaSO_4$ contains $ \frac {32 \times m_1}{233} $ g sulphur

Percentage of sulphur= $ \frac {32 \times m_1 \times 100}{233 \times m}$

Example 12.21: In sulphur estimation, 0.157 g of an organic compound gave 0.4813 g of barium sulphate. What is the percentage of sulphur in the compound?

Solution:

Molecular mass of BaSO$_4$ = 137 + 32 + 64 = 233 g

233 g BaSO$_4$ contains 32 g sulphur 0.4813 g BaSO$_4$ contains $\frac{32 \times 0.4813}{233}$ g sulphur

Percentage of sulphur = $\frac{32 \times 0.4813 \times 100}{233 \times 0.157} = 42.10\%$

12.10.3 Phosphorus

• A known mass of an organic compound is heated with fuming nitric acid.
• After this, phosphorus present in the compound is oxidised to phosphoric acid.
• Phosphoric acid is precipitated as ammonium phosphomolybdate, $(NH_4)_3PO_4.12MoO_3_$, by adding ammonia and ammonium molybdate.
• Phosphoric acid may also be precipitated as $MgNH_4PO_4$ by adding magnesia mixture.
• $MgNH_4PO_4$ on ignition yields $Mg_2P_2O_7$.

Calculation:

Let the mass of organic compound taken = m g and mass of ammonium phosphomolydate = $m_1$ g

Molar mass of ($NH_4)_3PO_4.12MoO_3$ = 1877 g

Percentage of phosphorus = $ \frac {31 \times m_1 \times 100}{1877 \times m} \% $

If phosphorus is estimated as $Mg_2P_2O_7$,

Percentage of phosphorus =$ \frac {62 \times m_1 \times 100}{222 \times m} \% $

Where,

“222u” is the molar mass of $Mg_2P_2O_7$,

“m” is the mass of organic compound taken,

“m₁”, is the mass of $Mg_2P_2O_7$ formed and

“62”, is the mass of two phosphorus atoms present in the compound $Mg_2P_2O_7$.

12.10.4 Oxygen

The percentage of oxygen in an organic compound can be found by 3 methods, as described:

a) difference between the total percentage composition (which is 100%) and the sum of the percentages of all other elements.

b) Direct estimation:

• A definite mass of an organic compound is decomposed by heating in a stream of nitrogen gas.
• The mixture of gaseous products containing oxygen is passed over red-hot coke. This step converts all the oxygen to carbon monoxide.
• The mixture containing carbon monoxide is passed through warm iodine pentoxide (I₂O₅); thus oxidizing carbon monoxide to carbon dioxide, producing iodine.

$$ \ce {Compound ->[{heat}] O2 + other gaseous products }$$

$$ \ce { 2C + O2 ->[{1373K}] 2CO]× 5 …(A) } $$

$$ \ce {I2O5 + 5CO → I2 + 5CO2]× 2}…. (B)$$

• On making the amount of CO produced in equation (A) equal to the amount of CO used in equation (B) by multiplying the equations (A) and (B) by 5 and 2 respectively; we find that each mole of oxygen liberated from the compound will produce two moles of carbon dioxide.
• Thus, 88 g carbon dioxide is obtained if 32 g oxygen is liberated.

Calculation:

Let,

The mass of organic compound taken be = m g

The mass of carbon dioxide produced be = $m__1$ g

∴ m₁ g carbon dioxide is obtained from  $ \frac {32m_1}{88}$g of $O_2$

∴ Percentage of oxygen = $ \frac {32m_1 \times 100 }{88}$ %.

c) The percentage of oxygen produced can also be determined from the amount of iodine produced.

Note box: The elements, carbon, hydrogen and nitrogen present in a compound are determined by an apparatus known as CHN elemental analyser. The analyser requires only a very small amount of the substance (1-3 mg) and displays the values on a screen within a short time.

Questions from section 12.10:

1. Describe the experiment to estimate the amount of Carbon and hydrogen in a given organic compound.

2. Explain the estimation of nitrogen by:

a) Duma’s method

b) Kjeldahl method

3. Explain the estimation of:

a) halogens

b) Sulphur by Cariusmethod.

4. Describe the experiment to estimate the amount of phosphorus in a given organic compound.

5. Explain the methods by which percentage of oxygen can be determined.