8.7 FUNDAMENTAL CONCEPTS IN Br– as shown below. ORGANIC REACTION MECHANISM In an organic reaction, the organic molecule A species having a carbon atom possessing(also referred as a substrate) reacts with an sextext of electrons and a positive charge isappropriate attacking reagent and leads to the formation of one or more intermediate(s) called a carbocation (earlier called carbonium and finally product(s) ion). The H3 ion is known as a methyl cation or methyl carbonium ion. Carbocations areThe general reaction is depicted as follows : classified as primary, secondary or tertiary Attacking depending on whether one, two or three Reagent [Intermediate] Product(s) carbons are directly attached to the positivelyOrganic molecule charged carbon. Some other examples of Byproducts + (Substrate) carbocations are: CH3C H2 (ethyl+ cation, a primary carbocation), (CH3)2C H (isopropyl+ Substrate is that reactant which supplies cation, a secondary carbocation), and (CH3)3C carbon to the new bond and the other reactant (tert-butyl cation, a tertiary carbocation). is called reagent. If both the reactants Carbocations are highly unstable and supply carbon to the new bond then choice reactive species. Alkyl groups directly is arbitrary and in that case the molecule on attached to the positively charged carbon which attention is focused is called substrate. stabilise the carbocations due to inductive In such a reaction a covalent bond and hyperconjugation effects, which you will between two carbon atoms or a carbon and be studying in the sections 8.7.5 and 8.7.9. some other atom is broken and a new bond is The+ observed+ order of carbocation+ stability+ formed. A sequential account of each step, is: C H3 < CH3C H2 < (CH3)2CH < (CH3)3C. These describing details of electron movement, carbocations have trigonal planar shape energetics during bond cleavage and bond with positively charged carbon+ being sp2 formation, and the rates of transformation hybridised. Thus, the shape of C H3 may be of reactants into products (kinetics) is considered as being derived from the overlap referred to as reaction mechanism. The of three equivalent C(sp2) hybridised orbitals knowledge of reaction mechanism helps with 1s orbital of each of the three hydrogen Reprint 2025-26 272 chemistry atoms. Each bond may be represented as curved arrow. Such cleavage results in C(sp2)–H(1s) sigma bond. The remaining the formation of neutral species (atom or carbon orbital is perpendicular to the group) which contains an unpaired electron. molecular plane and contains no electrons. These species are called free radicals. Like [Fig. 8.3(a)]. carbocations and carbanions, free radicals are also very reactive. A homolytic cleavage can be shown as: Alkyl free radical Alkyl radicals are classified as primary, secondary, or tertiary. Alkyl radical stability increases as we proceed from primary to Fig. 8.3 (a) Shape of methyl carbocation tertiary: The heterolytic cleavage can also give a , species in which carbon gets the shared Methyl Ethyl Isopropyl Tert-butyl pair of electrons. For example, when group free free free free Z attached to the carbon leaves without radical radical radical radical Organic reactions, which proceed by homolytic fission are called free radical or electron pair, the methyl anion is homopolar or nonpolar reactions. formed. Such a carbon species carrying a 8.7.2 Substrate and Reagent negative charge on carbon atom is called Ions are generally not formed in the reactions carbanion. Carbon in carbanion is generally of organic compounds. Molecules as such sp3 hybridised and its structure is distorted participate in the reaction. It is convenient to tetrahedron as shown in Fig. 8.3(b). name one reagent as substrate and other as reagent. In general, a molecule whose carbon is involved in new bond formation is called substrate and the other one is called reagent. When carbon-carbon bond is formed, the choice of naming the reactants as substrate and reagent is arbitrary and depends on molecule under observation. Example: Fig. 8.3 (b) Shape of methyl carbanion (i) CH2 = CH2 + Br2 → CH2 Br – CH2Br Substrate Reagent Product Carbanions are also unstable and reactive species. The organic reactions which proceed through heterolytic bond cleavage are called (ii) ionic or heteropolar or just polar reactions. In homolytic cleavage, one of the electrons of the shared pair in a covalent bond goes with each of the bonded atoms. Thus, in homolytic cleavage, the movement of a Nucleophiles and Electrophilessingle electron takes place instead of an electron pair. The single electron movement Reagents attack the reactive site of the is shown by ‘half-headed’ (fish hook: ) substrate. The reactive site may be electron Reprint 2025-26 organic chemistry – some basic principles and techniques 273 deficient portion of the molecule (a positive Problem 8.11reactive site) e.g., an atom with incomplete electron shell or the positive end of the dipole Using curved-arrow notation, show the in the molecule. If the attacking species is formation of reactive intermediates when electron rich, it attacks these sites. If attacking the following covalent bonds undergo heterolytic cleavage.species is electron deficient, the reactive site for it is that part of the substrate molecule (a) CH3–SCH3, (b) CH3–CN, (c) CH3–Cu which can supply electrons, e.g., π electrons Solution in a double bond. A reagent that brings an electron pair to the reactive site is called a nucleophile (Nu:) i.e., nucleus seeking and the reaction is then called nucleophilic. A reagent that takes away an electron pair from reactive site is called electrophile (E+) i.e., electron seeking and the reaction is called electrophilic. Problem 8.12 During a polar organic reaction, a Giving justification, categorise the nucleophile attacks an electrophilic centre following molecules/ions as nucleophile of the substrate which is that specific atom or electrophile: or part of the substrate which is electron deficient. Similarly, the electrophiles attack at nucleophilic centre, which is the electron rich centre of the substrate. Thus, the electrophiles receive electron pair from the Solution substrate when the two undergo bonding Nucleophiles: HS–,C2H5O–,(CH3)3N:H2N:– interaction. A curved-arrow notation is used These species have unshared pair of to show the movement of an electron pair electrons, which can be donated and from the nucleophile to the electrophile. Some shared with an electrophile. examples of nucleophiles are the negatively + + Electrophiles: BF3,C1+ H3–C =O,N O2.charged ions with lone pair of electrons such Reactive sites have only six valence as hydroxide (HO– ), cyanide (NC–) ions and electrons; can accept electron pair from carbanions (R3C:–). Neutral molecules such a nucleophile. as etc., can also act as Problem 8.13nucleophiles due to the presence of lone Identify electrophilic centre in thepair of electrons. Examples of electrophiles + following: CH3CH=O, CH3CN, CH3I.include carbocations (C H3) and neutral molecules having functional groups like Solution carbonyl group (>C=O) or alkyl halides Among CH 3HC*=O, H 3CC*≡N, and (R3C-X, where X is a halogen atom). The H3C*–I, the starred carbon atoms arecarbon atom in carbocations has sextet electrophilic centers as they will haveconfiguration; hence, it is electron deficient partial positive charge due to polarity ofand can receive a pair of electrons from the the bond.nucleophiles. In neutral molecules such as alkyl halides, due to the polarity of the C-X 8.7.3 Electron Movement in Organicbond a partial positive charge is generated Reactionson the carbon atom and hence the carbon atom becomes an electrophilic centre at which The movement of electrons in organic a nucleophile can attack. reactions can be shown by curved-arrow Reprint 2025-26 274 chemistry notation. It shows how changes in bonding 8.7.5 Inductive Effect occur due to electronic redistribution during When a covalent bond is formed between the reaction. To show the change in position atoms of different electronegativity, the of a pair of electrons, curved arrow starts electron density is more towards the more from the point from where an electron pair is electronegative atom of the bond. Such a shift shifted and it ends at a location to which the of electron density results in a polar covalent pair of electron may move. bond. Bond polarity leads to various electronic Presentation of shifting of electron pair is effects in organic compounds. given below : Let us consider cholorethane (CH3CH2Cl) in which the C–Cl bond is a polar covalent(i) from π bond to bond. It is polarised in such a way that the adjacent bond position carbon-1 gains some positive charge (δ+) (ii) from π bond to and the chlorine some negative charge (δ–). adjacent atom The fractional electronic charges on the two (iii) from atom to adjacent atoms in a polar covalent bond are denoted bond position by symbol δ (delta) and the shift of electron density is shown by an arrow that points from Movement of single electron is indicated by δ+ to δ– end of the polar bond.a single barbed ‘fish hooks’ (i.e. half headed curved arrow). For example, in transfer of δδ+ δ+ δ− hydroxide ion giving ethanol and in the CH3→CH2→Cl dissociation of chloromethane, the movement 2 1 In turn carbon-1, which has developedof electron using curved arrows can be partial positive charge (δ+) draws somedepicted as follows: electron density towards it from the adjacent C-C bond. Consequently, some positive charge (δδ+) develops on carbon-2 also, where δδ+ symbolises relatively smaller positive charge as compared to that on carbon – 1. In other words, the polar C – Cl bond induces polarity in the adjacent bonds. Such polarisation of8.7.4 Electron Displacement Effects in σ-bond caused by the polarisation of adjacent Covalent Bonds σ-bond is referred to as the inductive effect. The electron displacement in an organic This effect is passed on to the subsequent molecule may take place either in the ground bonds also but the effect decreases rapidly state under the influence of an atom or a as the number of intervening bonds increases substituent group or in the presence of an and becomes vanishingly small after three appropriate attacking reagent. The electron bonds. The inductive effect is related to the displacements due to the influence of ability of substituent(s) to either withdraw or an atom or a substituent group present in donate electron density to the attached carbon the molecule cause permanent polarlisation atom. Based on this ability, the substitutents of the bond. Inductive effect and resonance can be classified as electron-withdrawing or effects are examples of this type of electron electron donating groups relative to hydrogen. displacements. Temporary electron Halogens and many other groups such as displacement effects are seen in a molecule nitro (- NO2), cyano (- CN), carboxy (- COOH), when a reagent approaches to attack it. ester (COOR), aryloxy (-OAr, e.g. – OC6H5), etc. This type of electron displacement is called are electron-withdrawing groups. On the other electromeric effect or polarisability effect. hand, the alkyl groups like methyl (–CH3) and In the following sections we will learn about ethyl (–CH2–CH3) are usually considered as these types of electronic displacements. electron donating groups. ­ Reprint 2025-26 organic chemistry – some basic principles and techniques 275 benzene cannot be adequately represented Problem 8.14 by any of these structures, rather it is Which bond is more polar in the following a hybrid of the two structures (I and II) pairs of molecules: (a) H3C-H, H3C-Br called resonance structures. The resonance (b) H3C-NH2, H3C-OH (c) H3C-OH, structures (canonical structures or H3C-SH contributing structures) are hypothetical and individually do not represent any Solution real molecule. They contribute to the actual (a) C–Br, since Br is more electronegative structure in proportion to their stability. than H, (b) C–O, (c) C–O Another example of resonance is provided Problem 8.15 by nitromethane (CH3NO2) which can be represented by two Lewis structures, (I and In which C–C bond of CH3CH2CH2Br, the II). There are two types of N-O bonds in these inductive effect is expected to be the least? structures. 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. However, it is known that the two N–O bonds of nitromethane are of the same8.7.6 Resonance Structure length (intermediate between a N–O single There are many organic molecules whose bond and a N=O double bond). The actual behaviour cannot be explained by a single Lewis structure of nitromethane is therefore structure. An example is that of a resonance hybrid of the two canonical benzene. Its cyclic structure forms I and II. containing alternating C–C single The energy of actual structure of the molecule and C=C double bonds shown (the resonance hybrid) is lower than that of is inadequate for explaining its Benzene any of the canonical structures. The difference characteristic properties. in energy between the actual structure and the As per the above representation, benzene lowest energy resonance structure is called the should exhibit two different bond lengths, resonance stabilisation energy or simply the resonance energy. The more the numberdue to C–C single and C=C double bonds. of important contributing structures, theHowever, as determined experimentally more is the resonance energy. Resonance isbenzene has a uniform C–C bond distances particularly important when the contributingof 139 pm, a value intermediate between the structures are equivalent in energy. C–C single(154 pm) and C=C double (134 The following rules are applied while writingpm) bonds. Thus, the structure of benzene resonance structures:cannot be represented adequately by the The resonance structures have (i) the sameabove structure. Further, benzene can be positions of nuclei and (ii) the same number of represented equally well by the energetically unpaired electrons. Among the resonance identical structures I and II. structures, the one which has more number of covalent bonds, all the atoms with octet of electrons (except hydrogen which has a duplet), less separation of opposite charges, (a negative charge if any on more electronegative atom, a positive charge if any on more Therefore, according to the resonance electropositive atom) and more dispersal of theory (Unit 4) the actual structure of charge, is more stable than others. Reprint 2025-26 276 chemistry Problem 8.16 Solution Write resonance structures of CH3COO– The two structures are less important and show the movement of electrons by contributors as they involve charge curved arrows. separation. Additionally, structure I contains a carbon atom with an Solution incomplete octet. First, write the structure and put unshared pairs of valence electrons on 8.7.7 Resonance Effect appropriate atoms. Then draw the arrows The resonance effect is defined as ‘the polarity one at a time moving the electrons to get produced in the molecule by the interaction the other structures. 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. There are two types of resonance or mesomeric effect designated as R or M effect. Problem 8.17 (i) Positive Resonance Effect (+R effect) Write resonance structures of In this effect, the transfer of electrons is away CH2=CH–CHO. Indicate relative stability from an atom or substituent group attached of the contributing structures. to the conjugated system. This electron displacement makes certain positions in the Solution molecule of high electron densities. This effect in aniline is shown as : Stability: I > II > III (ii) Negative Resonance Effect (- R effect) [I: Most stable, more number of covalent This effect is observed when the transfer of bonds, each carbon and oxygen atom has electrons is towards the atom or substituent an octet and no separation of opposite group attached to the conjugated system. charge II: negative charge on more For example in nitrobenzene this electron electronegative atom and positive charge displacement can be depicted as : on more electropositive atom; III: does not contribute as oxygen has positive charge and carbon has negative charge, hence least stable]. Problem 8.18 Explain why the following two structures, I and II cannot be the major contributors The atoms or substituent groups, which to the real structure of CH3COOCH3. represent +R or –R electron displacement effects are as follows : +R effect: – halogen, –OH, –OR, –OCOR, –NH2, –NHR, –NR2, –NHCOR, – R effect: – COOH, –CHO, >C=O, – CN, –NO2 Reprint 2025-26 organic chemistry – some basic principles and techniques 277 The presence of alternate single and system or to an atom with an unshared double bonds in an open chain or cyclic system p orbital. The σ electrons of C—H bond of the is termed as a conjugated system. These alkyl group enter into partial conjugation with systems often show abnormal behaviour. the attached unsaturated system or with the The examples are 1,3- butadiene, aniline unshared p orbital. Hyperconjugation is a and nitrobenzene etc. In such systems, the permanent effect. π-electrons are delocalised and the system To understand hyperconjugation effect, let develops polarity. + us take an example of CH3 C H2 (ethyl cation) in 8.7.8 Electromeric Effect (E effect) which the positively charged carbon atom has an empty p orbital. One of the C-H bonds ofIt is a temporary effect. The organic compounds the methyl group can align in the plane of thishaving a multiple bond (a double or triple empty p orbital and the electrons constitutingbond) show this effect in the presence of the C-H bond in plane with this p orbital canan attacking reagent only. It is defined as then be delocalised into the empty p orbitalthe complete transfer of a shared pair of as depicted in Fig. 8.4 (a).π-electrons to one of the atoms joined by a multiple bond on the demand of an attacking reagent. The effect is annulled as soon as the attacking reagent is removed from the domain of 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: Fig. 8.4(a) Orbital diagram showing hyperconjugation in ethyl cation This type of overlap stabilises the carbocation because electron density from (ii) Negative Electromeric Effect (–E effect) In the adjacent σ bond helps in dispersing the this effect the π - electrons of the multiple positive charge. bond are transferred to that atom to which the attacking reagent does not get attached. For example: When inductive and electromeric effects operate in opposite directions, the electomeric effect predominates. In general, greater the number of alkyl 8.7.9 Hyperconjugation groups attached to a positively charged carbon Hyperconjugation is a general stabilising atom, the greater is the hyperconjugation interaction. It involves delocalisation of interaction and stabilisation of the cation. σ electrons of C—H bond of an alkyl group Thus, we have the following relative stability directly attached to an atom of unsaturated of carbocations : Reprint 2025-26 278 chemistry Problem 8.19 + Explain why (CH3)3C is more stable + + than CH3C H2 and C H3 is the least stable cation. Hyperconjugation is also possible in Solution alkenes and alkylarenes. + Hyperconjugation interaction in (CH3)3C is+ + D e l o c a l i s a t i o n o f e l e c t r o n s b y greater than in CH3C H2 as+ the (CH3)3Chyperconjugation in the case of alkene can has nine C-H bonds. In C H3, vacant pbe depicted as in Fig. 8.4(b). orbital is perpendicular to the plane in which C-H bonds lie; hence cannot + overlap with it. Thus, C H3 lacks hyperconjugative stability. 8.7.10 Types of Organic Reactions and Mechanisms Organic reactions can be classified into the Fig. 8.4(b) Orbital diagram showing following categories: hyperconjugation in propene (i) Substitution reactions (ii) Addition reactions There are various ways of looking at the (iii) Elimination reactions hyperconjugative effect. One of the way is to (iv) Rearrangement reactions regard C—H bond as possessing partial ionic You will be studying these reactions incharacter due to resonance. Unit 9 and later in class XII. 8.8 Methods of Purification of Organic Compounds Once an organic compound is extracted from a natural source or synthesised in the laboratory, it is essential to purify it. Various methods used for the purification of organic compounds are based 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 Finally, the purity of a compound is ascertained by determining its melting or boiling point. Most of the pure compounds have sharp melting points and boiling points. The hyperconjugation may also be New methods of checking the purity of an regarded as no bond resonance. organic compound are based on different Reprint 2025-26 organic chemistry – some basic principles and techniques 279 types of chromatographic and spectroscopic carefully. On boiling, the vapours of lower techniques. boiling component are formed first. The vapours are condensed by using a condenser8.8.1 Sublimation and the liquid is collected in a receiver. You have learnt earlier that on heating, some The vapours of higher boiling component solid substances change from solid to vapour form later and the liquid can be collected state without passing through liquid state. separately. The purification technique based on the above principle is known as sublimation and is used to separate sublimable compounds from non- sublimable impurities. 8.8.2 Crystallisation This is one of the most commonly used techniques 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 impure compound is dissolved in a solvent in which it is sparingly soluble at room temperature but appreciably soluble at higher temperature. The solution is concentrated to get a nearly saturated solution. 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. If the compound is highly soluble in one solvent and very little soluble in another solvent, crystallisation can be Fig.8.5 Simple distillation. The vapours of a substance formed are condensed andsatisfactorily carried out in a mixture of these the liquid is collected in conical flask. solvents. Impurities, which impart colour to the solution are removed by adsorbing over Fractional Distillation: If the differenceactivated charcoal. Repeated crystallisation in boiling points of two liquids is not much,becomes necessary for the purification of compounds containing impurities of simple distillation cannot be used to separate comparable solubilities. them. The vapours of such liquids are formed within the same temperature range and are 8.8.3 Distillation condensed simultaneously. The technique of This important method is used to separate fractional distillation is used in such cases. In (i) volatile liquids from nonvolatile impurities this technique, vapours of a liquid mixture are and (ii) the liquids having sufficient difference passed through a fractionating column before in their boiling points. Liquids having condensation. The fractionating column is different boiling points vaporise at different fitted over the mouth of the round bottom temperatures. The vapours are cooled and flask (Fig.8.6, page 280). the liquids so formed are collected separately. Vapours of the liquid with higher boiling Chloroform (b.p 334 K) and aniline (b.p. 457 point condense before the vapours of the K) are easily separated by the technique of liquid with lower boiling point. The vapours distillation (Fig 8.5). The liquid mixture is rising up in the fractionating column become taken in a round bottom flask and heated richer in more volatile component. By the Reprint 2025-26 280 chemistry unit in the fractionating column is called a theoretical plate. Commercially, columns with hundreds of plates are available. One of the technological applications of fractional distillation is to separate different fractions of crude oil in petroleum industry. 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. A liquid boils at a temperature at which its vapour pressure is equal to the external pressure. The pressure is reduced with the help of a water pump or vacuum pump (Fig.8.8). Glycerol can be separated from Fig.8.6 Fractional distillation. The vapours of lower boiling spent-lye in soap industry by fraction reach the top of the column first followed by using this technique. vapours of higher boiling fractions. time the vapours reach to the top of the fractionating column, these are rich in the more volatile component. Fractionating columns are available in various sizes and designs as shown in Fig.8.7. A fractionating column provides many surfaces for heat exchange between the ascending vapours and the descending condensed liquid. Some of the condensing liquid in the fractionating column obtains heat from the ascending vapours and revaporises. The vapours thus become richer in low boiling component. The vapours of low boiling component ascend to the top of the column. On reaching the top, the vapours become pure in low boiling component and pass through the condenser and the pure liquid is collected in a receiver. After a series of successive distillations, the remaining liquid in the distillation flask gets enriched in high boiling component. Each Fig.8.7 Different types of fractionating columns.successive condensation and vaporisation Reprint 2025-26 organic chemistry – some basic principles and techniques 281 Fig.8.8 Distillation under reduced pressure. A liquid boils at a temperature below its vapour pressure by reducing the pressure. Steam Distillation: This technique is 8.8.4 Differential Extraction applied to separate substances which are When an organic compound is present in an steam volatile and are immiscible with water. aqueous medium, it is separated by shaking In steam distillation, steam from a steam it with an organic solvent in which it is more generator is passed through a heated flask soluble than in water. The organic solvent and containing the liquid to be distilled. The the aqueous solution should be immiscible mixture of steam and the volatile organic with each other so that they form two distinct compound is condensed and collected. The layers which can be separated by separatory compound is later separated from water using funnel. The organic solvent is later removed a separating funnel. In steam distillation, by distillation or by evaporation to get back the liquid boils when the sum of vapour the compound. Differential extraction is pressures due to the organic liquid (p1) carried out in a separatory funnel as shown in and that due to water (p2) becomes equal to Fig. 8.10 (Page 282). If the organic compound the atmospheric pressure (p), i.e. p =p1+ p2. is less soluble in the organic solvent, a very Since p1 is lower than p, the organic liquid large quantity of solvent would be required vaporises at lower temperature than its to extract even a very small quantity of the boiling point. compound. The technique of continuous extraction is employed in such cases. In this Thus, if one of the substances in the technique same solvent is repeatedly used formixture is water and the other, a water extraction of the compound.insoluble substance, then the mixture will boil close to but below, 373K. A mixture of water 8.8.5 Chromatography and the substance is obtained which can Chromatography is an important technique be separated by using a separating funnel. extensively used to separate mixtures into Aniline is separated by this technique from their components, purify compounds and aniline – water mixture (Fig.8.9, Page 282). also to test the purity of compounds. The Reprint 2025-26 282 chemistry Fig.8.9 Steam distillation. Steam volatile component volatilizes, the vapours condense in the condenser and the liquid collects in conical flask. 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 this technique, the mixture of substances is applied onto a stationary phase, which may be a solid or a liquid. A pure solvent, a mixture of solvents, or a gas is allowed to move slowly over the stationary phase. The components of the mixture get gradually separated from one another. The moving phase is called the mobile phase. Based on the principle involved, chromatography is classified into different categories. Two of these are: (a) Adsorption chromatography, and Fig.8.10 Differential extraction. Extraction of com- (b) Partition chromatography. pound takes place based on difference in solubility a) Adsorption Chromatography: Adsor- ption chromatography is based on the fact distances over the stationary phase. Followingthat different compounds are adsorbed on are two main types of chromatographican adsorbent to different degrees. Commonly techniques based on the principle of differentialused adsorbents are silica gel and alumina. adsorption.When a mobile phase is allowed to move over a stationary phase (adsorbent), the (a) Column chromatography, and components of the mixture move by varying (b) Thin layer chromatography. Reprint 2025-26 organic chemistry – some basic principles and techniques 283 Column Chromatography: Column The glass plate is then placed in a closed jar chromatography involves separation of containing the eluant (Fig. 8.12a). As the a mixture over a column of adsorbent eluant rises up the plate, the components of (stationary phase) packed in a glass tube. the mixture move up along with the eluant to The column is fitted with a stopcock at its different distances depending on their degree lower end (Fig. 8.11). The mixture adsorbed of adsorption and separation takes place. on adsorbent is placed on the top of the The relative adsorption of each component adsorbent column packed in a glass tube. of the mixture is expressed in terms of its An appropriate eluant which is a liquid or a retardation factor i.e. Rf value (Fig.8.12 b). mixture of liquids is allowed to flow down the Rf = Distance moved by the substance from base line (x) column slowly. Depending upon the degree to Distance moved by the solvent from base line (y) 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 (Fig.8.11). Fig.8.12 (a) Thin layer chromatography. Chromatogram being developed. Fig.8.11 Column chromatography. Different stages of separation of components Fig.8.12 (b) Developed chromatogram. of a mixture. The spots of coloured compounds are Thin Layer Chromatography: Thin layer visible on TLC plate due to their original colour. chromatography (TLC) is another type of The spots of colourless compounds, which are adsorption chromatography, which involves invisible to the eye but fluoresce in ultraviolet separation of substances of a mixture over light, can be detected by putting the plate under a thin layer of an adsorbent coated on glass ultraviolet light. Another detection technique plate. A thin layer (about 0.2mm thick) of is to place the plate in a covered jar containing an adsorbent (silica gel or alumina) is spread a few crystals of iodine. Spots of compounds, over a glass plate of suitable size. The plate which adsorb iodine, will show up as brown is known as thin layer chromatography plate spots. Sometimes an appropriate reagent or chromaplate. The solution of the mixture may also be sprayed on the plate. For example, to be separated is applied as a small spot amino acids may be detected by spraying the about 2 cm above one end of the TLC plate. plate with ninhydrin solution (Fig.8.12b). Reprint 2025-26 284 chemistry Partition Chromatography: Partition spot on the chromatogram. The spots of the chromatography is based on continuous separated colourless compounds may be differential partitioning of components of observed either under ultraviolet light or by a mixture between stationary and mobile the use of an appropriate spray reagent as phases. Paper chromatography is a type discussed under thin layer chromatography. of partition chromatography. In paper 8.9 Qualitative Analysis of Organic chromatography, a special quality paper Compounds known as chromatography paper is used. The elements present in organic compoundsChromatography paper contains water trapped are carbon and hydrogen. In addition toin it, which acts as the stationary phase. these, they may also contain oxygen, nitrogen, A strip of chromatography paper spotted sulphur, halogens and phosphorus.at the base with the solution of the mixture is suspended in a suitable solvent or a mixture 8.9.1 Detection of Carbon and Hydrogen of solvents (Fig. 8.13). This solvent acts as the Carbon and hydrogen are detected by heating mobile phase. The solvent rises up the paper the compound with copper(II) oxide. Carbon by capillary action and flows over the spot. The present in the compound is oxidised to paper selectively retains different components carbon dioxide (tested with lime-water, which according to their differing partition in the develops turbidity) and hydrogen to water two phases. The paper strip so developed is (tested with anhydrous copper sulphate, known as a chromatogram. The spots of the which turns blue). separated coloured compounds are visible at C + 2CuO 2Cu + CO2different heights from the position of initial 2H + CuO Cu + H2O CO2 + Ca(OH)2 CaCO3↓ + H2O 5H2O + CuSO4 CuSO4.5H2O White Blue 8.9.2 Detection of Other Elements Nitrogen, sulphur, halogens and phosphorus present in an organic compound are detected by “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: Na + C + N NaCN 2Na + S Na2S Na + X Na X (X = Cl, Br or I) C, N, S and X come from organic compound. 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 Fig.8.13 Paper chromatography. Chromatography paper in two different The sodium fusion extract is boiled with shapes. iron(II) sulphate and then acidified with Reprint 2025-26 organic chemistry – some basic principles and techniques 285 concentrated sulphuric acid. The formation bromine and a yellow precipitate, insoluble of Prussian blue colour confirms the presence in ammonium hydroxide shows the presence of nitrogen. Sodium cyanide first reacts of iodine. with iron(II) sulphate and forms sodium X– + Ag+ → AgX hexacyanidoferrate(II). On heating with X represents a halogen – Cl, Br or I. concentrated sulphuric acid some iron(II) If nitrogen or sulphur is also present in the ions are oxidised to iron(III) ions which compound, the sodium fusion extract is react with sodium hexacyanidoferrate(II) first boiled with concentrated nitric acid to to produce iron(III) hexacyanidoferrate(II) decompose cyanide or sulphide of sodium (ferriferrocyanide) which is Prussian blue in formed during Lassaigne’s test. These ions colour. would otherwise interfere with silver nitrate 6CN– + Fe2+ → [Fe(CN)6]4– test for halogens. 3[Fe(CN)6]4– + 4Fe3+ Fe4[Fe(CN)6]3.xH2O (D) Test for Phosphorus Prussian blue The compound is heated with an oxidising (B) Test for Sulphur agent (sodium peroxide). The phosphorus (a) The sodium fusion extract is acidified present in the compound is oxidised to with acetic acid and lead acetate is added phosphate. The solution is boiled with nitric to it. A black precipitate of lead sulphide acid and then treated with ammonium indicates the presence of sulphur. molybdate. A yellow colouration or precipitate indicates the presence of phosphorus. S2– + Pb2+ → PbS Black Na3PO4 + 3HNO3 → H3PO4+3NaNO3 (b) On treating sodium fusion extract with H3PO4 + 12(NH4)2MoO4 + 21HNO3 → sodium nitroprusside, appearance of Ammonium a violet colour further indicates the molybdate presence of sulphur. (NH4)3PO4.12MoO3 + 21NH4NO3 + 12H2O AmmoniumS2– + [Fe(CN)5NO]2– → [Fe(CN)5NOS]4– phosphomolybdate Violet In case, nitrogen and sulphur both are 8.10 Quantitative Analysis present in an organic compound, sodium Quantitative analysis of compounds is very thiocyanate is formed. It gives blood red colour important in organic chemistry. It helps and no Prussian blue since there are no free chemists in the determination of mass per cyanide ions. cent of elements present in a compound. You Na + C + N + S → NaSCN have learnt in Unit-1 that mass per cent of Fe3+ +SCN– → [Fe(SCN)]2+ elements is required for the determination of Blood red emperical and molecular formula. If sodium fusion is carried out with excess The percentage composition of elements of sodium, the thiocyanate decomposes to present in an organic compound is determined yield cyanide and sulphide. These ions give by the following methods: their usual tests. 8.10.1 Carbon and Hydrogen NaSCN + 2Na → NaCN+Na2S (C) Test for Halogens Both carbon and hydrogen are estimated in one experiment. A known mass of an organicThe sodium fusion extract is acidified with compound is burnt in the presence of excessnitric acid and then treated with silver nitrate. of oxygen and copper(II) oxide. Carbon andA white precipitate, soluble in ammonium hydrogen in the compound are oxidised tohydroxide shows the presence of chlorine, carbon dioxide and water respectively.a yellowish precipitate, sparingly soluble in ammonium hydroxide shows the presence of CxHy + (x + y/4) O2 → x CO2 + (y/2) H2O Reprint 2025-26 286 chemistry Fig.8.14 Estimation of carbon and hydrogen. Water and carbon dioxide formed on oxidation of substance are absorbed in anhydrous calcium chloride and potassium hydroxide solutions respectively contained in U tubes. The mass of water produced is determined 8.10.2 Nitrogen by passing the mixture through a weighed There are two methods for estimation of U-tube containing anhydrous calcium chloride. nitrogen: (i) Dumas method and (ii) Kjeldahl’s Carbon dioxide is absorbed in another U-tube method. containing concentrated solution of potassium (i) Dumas method: The nitrogen containing hydroxide. These tubes are connected in series organic compound, when heated with copper (Fig. 8.14). The increase in masses of calcium oxide in an atmosphere of carbon dioxide, chloride and potassium hydroxide gives the yields free nitrogen in addition to carbon amounts of water and carbon dioxide from dioxide and water. which the percentages of carbon and hydrogen CxHyNz + (2x + y/2) CuO →are calculated. Let the mass of organic compound be x CO2 + y/2 H2O + z/2 N2 + (2x + y/2) Cu m g, mass of water and carbon dioxide Traces of nitrogen oxides formed, if produced be m1 and m2 g respectively; any, are reduced to nitrogen by passing the 12 × m2 × 100 gaseous mixture over a heated copper gauze.Percentage of carbon= 44 × m The mixture of gases so produced is collected over an aqueous solution of potassium 2 × m1 × 100Percentage of hydrogen = hydroxide which absorbs carbon dioxide. 18 × m Nitrogen is collected in the upper part of the Problem 8.20 graduated tube (Fig.8.15). On complete combustion, 0.246 g of an Let the mass of organic compound = m g organic compound gave 0.198g of carbon Volume of nitrogen collected = V1 mL dioxide and 0.1014g of water. Determine Room temperature = T1K the percentage composition of carbon and hydrogen in the compound. PV1 1 × 273 Volumeof nitrogen atSTP = Solution 760 × T1 12 × 0.198 × 100 (Letit be V mL) Percentageof carbon = 44 × 0.246 Where p1 and V1 are the pressure and = 21.95% volume of nitrogen, p1 is different from the atmospheric pressure at which nitrogen gas 2 × 0.1014 × 100 is collected. The value of p1 is obtained by Percentage of hydrogen = 18 × 0.246 the relation; = 4.58% p1= Atmospheric pressure – Aqueous tension 22400 mL N2 at STP weighs 28 g. Reprint 2025-26 organic chemistry – some basic principles and techniques 287 Fig. 8.15 Dumas method. The organic compound yields nitrogen gas on heating it with copper(II) oxide in the presence of CO2 gas. The mixture of gases is collected over potassium hydroxide solution in which CO2 is absorbed and volume of nitrogen gas is determined. 28 × V V mL N 2 atSTP weighs = g 28 × 41.9 22400 41.9mLof nitrogen weighs = g 22400 28 × V × 100 Percentage of nitrogen = 28 × 41.9 × 100 22400 × m Percentageof nitrogen = 22400 × 0.3 Problem 8.21 = 17.46% In Dumas’ method for estimation of nitrogen, 0.3g of an organic compound gave 50mL of nitrogen collected at 300K (ii) Kjeldahl’s method: The compound temperature and 715mm pressure. containing nitrogen is heated with concentrated Calculate the percentage composition sulphuric acid. Nitrogen in the compound of nitrogen in the compound. (Aqueous gets converted to ammonium sulphate tension at 300K=15 mm) (Fig. 8.16). The resulting acid mixture is then Solution heated with excess of sodium hydroxide. Volume of nitrogen collected at 300K and The liberated ammonia gas is absorbed in 715mm pressure is 50 mL an excess of standard solution of sulphuric Actual pressure = 715-15 =700 mm acid. The amount of ammonia produced is 273 × 700 × 50 determined by estimating the amount of Volumeof nitrogen atSTP sulphuric acid consumed in the reaction. It 300 × 760 is done by estimating unreacted sulphuric 41.9mL acid left after the absorption of ammonia by 22,400 mL of N2 at STP weighs = 28 g titrating it with standard alkali solution. The difference between the initial amount of acid Reprint 2025-26 288 chemistry Fig.8.16 Kjeldahl method. Nitrogen-containing compound is treated with concentrated H2SO4 to get ammonium sulphate which liberates ammonia on treating with NaOH; ammonia is absorbed in known volume of standard acid. taken and that left after the reaction gives the 14 × M × 2(V − V1 / 2) 100amount of acid reacted with ammonia. Percentageof N = × 1000 m Organic compound + H2SO4 → (NH4)2SO4 1.4 × M × 2(V − V / 2) = Na2SO4 + 2NH3 + 2H2O m 2NH3 + H2SO4 → (NH4)2SO4 Kjeldahl method is not applicable to Let the mass of organic compound taken = m g compounds containing nitrogen in nitro and Volume of H2SO4 of molarity, M, azo groups and nitrogen present in the ring taken = V mL (e.g. pyridine) as nitrogen of these compounds Volume of NaOH of molarity, M, used for does not change to ammonium sulphate titration of excess of H2SO4 = V1 mL under these conditions. V1mL of NaOH of molarity M Problem 8.22 = V1 /2 mL of H2SO4 of molarity M During estimation of nitrogen presentVolume of H2SO4 of molarity M unused in an organic compound by Kjeldahl’s= (V - V1/2) mL method, the ammonia evolved from (V- V1/2) mL of H2SO4 of molarity M 0.5 g of the compound in Kjeldahl’s = 2(V-V1/2) mL of NH3 solution of estimation of nitrogen, neutralized 10 mL molarity M. of 1 M H2SO4. Find out the percentage of 1000 mL of 1 M NH3 solution contains nitrogen in the compound. 17g NH3 or 14 g of N Solution 2(V-V1/2) mL of NH3 solution of molarity M 1 M of 10 mL H2SO4=1M of 20 mL NH3contains: 1000 mL of 1M ammonia contains 14 g 14 × M × 2(V − V1 / 2) nitrogen g N 1000 20 mL of 1M ammonia contains Reprint 2025-26 organic chemistry – some basic principles and techniques 289 14 × 20 Percentage of halogen g nitrogen 1000 atomicmassof X × m1g = 14×20×100 molecular massof AgX Percentageof nitrogen = =56.0% 1000×0.5 Problem 8.23 8.10.3 Halogens In Carius method of estimation of Carius method: A known mass of an organic halogen, 0.15 g of an organic compound compound is heated with fuming nitric acid in gave 0.12 g of AgBr. Find out the the presence of silver nitrate contained in a hard percentage of bromine in the compound. glass tube known as Carius tube, (Fig.8.17) Solution Molar mass of AgBr = 108 + 80 = 188 g mol-1 188 g AgBr contains 80 g bromine 80 × 0.12 0.12 g AgBr contains g bromine 188 80 × 0.12 × 100 Percentage of bromine = 188×0.15 = 34.04% 8.10.4 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 Fig. 8.17 Carius method. Halogen containing calculated from the mass of barium sulphate. organic compound is heated with fuming Let the mass of organic nitric acid in the presence of silver nitrate. compound taken = m g and the mass of barium in a furnace. Carbon and hydrogen present in sulphate formed = m1g the compound are oxidised to carbon dioxide 1 mol of BaSO4 = 233 g BaSO4 = 32 g sulphur and water. The halogen present forms the 32 × m1 corresponding silver halide (AgX). It is filtered, m1 g BaSO4 contains g sulphur 233 × washed, dried and weighed. 32 × m1 × 100Let the mass of organic Percentageof sulphur = compound taken = m g 233 × m Mass of AgX formed = m1 g 1 mol of AgX contains 1 mol of X Problem 8.24Mass of halogen in m1g of AgX atomicmassof X × m1g In sulphur estimation, 0.157 g of an = organic compound gave 0.4813 g of molecular massof AgX Reprint 2025-26 290 chemistry percentage composition (100) and the sum of barium sulphate. What is the percentage the percentages of all other elements. However, of sulphur in the compound? oxygen can also be estimated directly as follows: Solution A definite mass of an organic compound is Molecular mass of BaSO4 = 137+32+64 decomposed by heating in a stream of nitrogen gas. The mixture of gaseous products containing = 233 g oxygen is passed over red-hot coke when all 233 g BaSO4 contains 32 g sulphur the oxygen is converted to carbon monoxide. 32 × 0.4813 This mixture is passed through warm iodine 0.4813 g BaSO4 contains g pentoxide (I2O5) when carbon monoxide is g sulphur 233 oxidised to carbon dioxide producing iodine. 32 × 0.4813 × 100 Compound heat → O2 + other gaseous Percentageof sulphur = products 233 × 0.157 1373 K 2C + O2 → 2CO]× 5 (A) = 42.10% I2O5 + 5CO → I2 + 5CO2]× 2 (B) 8.10.5 Phosphorus On making the amount of CO produced in equation (A) equal to the amount of CO usedA known mass of an organic compound is in equation (B) by multiplying the equationsheated with fuming nitric acid whereupon (A) and (B) by 5 and 2 respectively; we findphosphorus present in the compound is that each mole of oxygen liberated fromoxidised to phosphoric acid. It is precipitated the compound will produce two moles ofas ammonium phosphomolybdate, (NH4)3 PO4.12MoO3, by adding ammonia and carbondioxide. ammonium molybdate. Alternatively, Thus 88 g carbon dioxide is obtained if 32 g phosphoric acid may be precipitated as oxygen is liberated. MgNH4PO4 by adding magnesia mixture which Let the mass of organic compound taken be m g on ignition yields Mg2P2O7. Mass of carbon dioxide produced be m1 g Let the mass of organic compound taken ∴ m1 g carbon dioxide is obtained from = m g and mass of ammonium phospho 32 × m1 g O2molydate = m1g 88 Molar mass of (NH4)3PO4.12MoO3 = 1877g 32 × m1 × 100 ∴Percentage of oxygen = % 31 × m1 × 100 88 × mPercentage of phosphorus = % 1877 × m The percentage of oxygen can be derived from the amount of iodine produced also. If phosphorus is estimated as Mg2P2O7, Presently, the estimation of elements in 62 × m1 × 100 an organic compound is carried out by usingPercentage of phosphorus = × 222 microquantities of substances and automatic experimental techniques. The elements,where, 222 u is the molar mass of Mg2P2O7, carbon, hydrogen and nitrogen present in am, the mass of organic compound taken, compound are determined by an apparatusm1, the mass of Mg2P2O7 formed and 62, the known as CHN elemental analyser. Themass of two phosphorus atoms present in the analyser requires only a very small amountcompound Mg2P2O7. of the substance (1-3 mg) and displays the 8.10.6 Oxygen values on a screen within a short time. A The percentage of oxygen in an organic compound detailed discussion of such methods is beyond is usually found by difference between the total the scope of this book. Reprint 2025-26 organic chemistry – some basic principles and techniques 291 Summary In this unit, we have learnt some basic concepts in structure and reactivity of organic compounds, which are formed due to covalent bonding. The nature of the covalent bonding in organic compounds can be described in terms of orbitals hybridisation concept, according to which carbon can have sp3, sp2 and sp hybridised orbitals. The sp3, sp2 and sp hybridised carbons are found in compounds like methane, ethene and ethyne respectively. The tetrahedral shape of methane, planar shape of ethene and linear shape of ethyne can be understood on the basis of this concept. A sp3 hybrid orbital can overlap with 1s orbital of hydrogen to give a carbon - hydrogen (C–H) single bond (sigma, σ bond). Overlap of a sp2 orbital of one carbon with sp2 orbital of another results in the formation of a carbon–carbon σ bond. The unhybridised p orbitals on two adjacent carbons can undergo lateral (side- by-side) overlap to give a pi (π) bond. Organic compounds can be represented by various structural formulas. The three dimensional representation of organic compounds on paper can be drawn by wedge and dash formula. Organic compounds can be classified on the basis of their structure or the functional groups they contain. A functional group is an atom or group of atoms bonded together in a unique fashion and which determines the physical and chemical properties of the compounds. The naming of the organic compounds is carried out by following a set of rules laid down by the International Union of Pure and Applied Chemistry (IUPAC). In IUPAC nomenclature, the names are correlated with the structure in such a way that the reader can deduce the structure from the name. Organic reaction mechanism concepts are based on the structure of the substrate molecule, fission of a covalent bond, the attacking reagents, the electron displacement effects and the conditions of the reaction. These organic reactions involve breaking and making of covalent bonds. A covalent bond may be cleaved in heterolytic or homolytic fashion. A heterolytic cleavage yields carbocations or carbanions, while a homolytic cleavage gives free radicals as reactive intermediate. Reactions proceeding through heterolytic cleavage involve the complimentary pairs of reactive species. These are electron pair donor known as nucleophile and an electron pair acceptor known as electrophile. The inductive, resonance, electromeric and hyperconjugation effects may help in the polarisation of a bond making certain carbon atom or other atom positions as places of low or high electron densities. Organic reactions can be broadly classified into following types; substitution, addition, elimination and rearrangement reactions. Purification, qualitative and quantitative analysis of organic compounds are carried out for determining their structures. The methods of purification namely : sublimation, distillation and differential extraction are based on the difference in one or more physical properties. Chromatography is a useful technique of separation, identification and purification of compounds. It is classified into two categories : adsorption and partition chromatography. Adsorption chromatography is based on differential adsorption of various components of a mixture on an adsorbent. Partition chromatography involves continuous partitioning of the components of a mixture between stationary and mobile phases. After getting the compound in a pure form, its qualitative analysis is carried out for detection of elements present in it. Nitrogen, sulphur, halogens and phosphorus are detected by Lassaigne’s test. Carbon and hydrogen are estimated by determining the amounts of carbon dioxide and water produced. Nitrogen is estimated by Dumas or Kjeldahl’s method and halogens by Carius method. Sulphur and phosphorus are estimated by oxidising them to sulphuric and phosphoric acids respectively. The percentage of oxygen is usually determined by difference between the total percentage (100) and the sum of percentages of all other elements present. Reprint 2025-26 292 chemistry Exercises 8.1 What are hybridisation states of each carbon atom in the following compounds ? CH2=C=O, CH3CH=CH2, (CH3)2CO, CH2=CHCN, C6H6 8.2 Indicate the σ and π bonds in the following molecules : C6H6, C6H12, CH2Cl2, CH2=C=CH2, CH3NO2, HCONHCH3 8.3 Write bond line formulas for : Isopropyl alcohol, 2,3-Dimethylbutanal, Heptan-4-one. 8.4 Give the IUPAC names of the following compounds : (a) (b) (c) (d) (e) (f) Cl2CHCH2OH 8.5 Which of the following represents the correct IUPAC name for the compounds concer ned ? (a) 2,2-Dimethylpentane or 2-Dimethylpentane (b) 2,4,7-Trimethyloctane or 2,5,7-Trimethyloctane (c) 2-Chloro-4-methylpentane or 4-Chloro-2-methylpentane (d) But-3-yn-1-ol or But-4-ol-1-yne. 8.6 Draw formulas for the first five members of each homologous series beginning with the following compounds. (a) H–COOH (b) CH3COCH3 (c) H–CH=CH2 8.7 Give condensed and bond line structural formulas and identify the functional group(s) present, if any, for : (a) 2,2,4-Trimethylpentane (b) 2-Hydroxy-1,2,3-propanetricarboxylic acid (c) Hexanedial 8.8 Identify the functional groups in the following compounds (a) (b) (c) 8.9 Which of the two: O2NCH2CH2O– or CH3CH2O– is expected to be more stable and why? 8.10 Explain why alkyl groups act as electron donors when attached to a π system. 8.11 Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation. + (a) C6H5OH (b) C6H5NO2 (c) CH3CH=CHCHO (d) C6H5–CHO (e) C6H5–CH2 + (f) CH3CH=CH C H2 8.12 What are electrophiles and nucleophiles ? Explain with examples. 8.13 Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles: (a) CH3COOH + HO– → CH3COO–+H2O Reprint 2025-26 organic chemistry – some basic principles and techniques 293 – (b) CH3COCH3+ C N → (CH3)2C(CN)(OH) + (c) C6H6 + CH3C O → C6H5COCH3 8.14 Classify the following reactions in one of the reaction type studied in this unit. (a) CH3CH2Br + HS– → CH3CH2SH + Br– (b) (CH3)2C = CH2 + HCI → (CH3)2CIC – CH3 (c) CH3CH2Br + HO– → CH2 = CH2 + H2O + Br– (d) (CH3)3C– CH2OH + HBr → (CH3)2CBrCH2CH2CH3 + H2O 8.15 What is the relationship between the members of following pairs of structures ? Are they structural or geometrical isomers or resonance contributors ? (a) (b) (c) 8.16 For the following bond cleavages, use curved-arrows to show the electron flow and classify each as homolysis or heterolysis. Identify reactive intermediate produced as free radical, carbocation and carbanion. (a) (b) (c) (d) 8.17 Explain the terms Inductive and Electromeric effects. Which electron displacement effect explains the following correct orders of acidity of the carboxylic acids? (a) Cl3CCOOH > Cl2CHCOOH > ClCH2COOH (b) CH3CH2COOH > (CH3)2CHCOOH > (CH3)3C.COOH 8.18 Give a brief description of the principles of the following techniques taking an example in each case. (a) Crystallisation (b) Distillation (c) Chromatography 8.19 Describe the method, which can be used to separate two compounds with different solubilities in a solvent S. 8.20 What is the difference between distillation, distillation under reduced pressure and steam distillation ? Reprint 2025-26 294 chemistry 8.21 Discuss the chemistry of Lassaigne’s test. 8.22 Differentiate between the principle of estimation of nitrogen in an organic compound by (i) Dumas method and (ii) Kjeldahl’s method. 8.23 Discuss the principle of estimation of halogens, sulphur and phosphorus present in an organic compound. 8.24 Explain the principle of paper chromatography. 8.25 Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens? 8.26 Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens. 8.27 Name a suitable technique of separation of the components from a mixture of calcium sulphate and camphor. 8.28 Explain, why an organic liquid vaporises at a temperature below its boiling point in its steam distillation ? 8.29 Will CCl4 give white precipitate of AgCl on heating it with silver nitrate? Give reason for your answer. 8.30 Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound? 8.31 Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test? 8.32 An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this substance is subjected to complete combustion. 8.33 A sample of 0.50 g of an organic compound was treated according to Kjeldahl’s method. The ammonia evolved was absorbed in 50 ml of 0.5 M H2SO4. The residual acid required 60 mL of 0.5 M solution of NaOH for neutralisation. Find the percentage composition of nitrogen in the compound. 8.34 0.3780 g of an organic chloro compound gave 0.5740 g of silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound. 8.35 In the estimation of sulphur by Carius method, 0.468 g of an organic sulphur compound afforded 0.668 g of barium sulphate. Find out the percentage of sulphur in the given compound. 8.36 In the organic compound CH2 = CH – CH2 – CH2 – C ≡ CH, the pair of hydridised orbitals involved in the formation of: C2 – C3 bond is: (a) sp – sp2 (b) sp – sp3 (c) sp2 – sp3 (d) sp3 – sp3 8.37 In the Lassaigne’s test for nitrogen in an organic compound, the Prussian blue colour is obtained due to the formation of: (a) Na4[Fe(CN)6] (b) Fe4[Fe(CN)6]3 (c) Fe2[Fe(CN)6] (d) Fe3[Fe(CN)6]4 8.38 Which of the following carbocation is most stable ? + + + + (a) (CH3)3C. C­H2 (b) (CH3)3C­ (c) CH3CH2C­ H2 (d) CH3C­ H CH2CH3 8.39 The best and latest technique for isolation, purification and separation of organic compounds is: (a) Crystallisation (b) Distillation (c) Sublimation (d) Chromatography 8.40 The reaction: CH3CH2I + KOH(aq) → CH3CH2OH + KI is classified as : (a) electrophilic substitution (b) nucleophilic substitution (c) elimination (d) addition Reprint 2025-26 Hydrocarbons 295 Unit 9 Hydrocarbons Hydrocarbons are the important sources of energy. After studying this unit, you will be able to • name hydrocarbons according to The term ‘hydrocarbon’ is self-explanatory which means IUPAC system of nomenclature; compounds of carbon and hydrogen only. Hydrocarbons • recognise and write structures play a key role in our daily life. You must be familiar o f i s o m e r s o f a l k a n e s , with the terms ‘LPG’ and ‘CNG’ used as fuels. LPG is the alkenes, alkynes and aromatic abbreviated form of liquified petroleum gas whereas CNG hydrocarbons; stands for compressed natural gas. Another term ‘LNG’ • learn about various methods of (liquified natural gas) is also in news these days. This is preparation of hydrocarbons; • distinguish between alkanes, also a fuel and is obtained by liquifaction of natural gas. alkenes, alkynes and aromatic Petrol, diesel and kerosene oil are obtained by the fractional hydrocarbons on the basis of distillation of petroleum found under the earth’s crust. physical and chemical properties; Coal gas is obtained by the destructive distillation of • draw and differentiate between coal. Natural gas is found in upper strata during drilling various conformations of ethane; of oil wells. The gas after compression is known as • a p p r e c i a t e t h e r o l e o f compressed natural gas. LPG is used as a domestic fuel hydrocarbons as sources of with the least pollution. Kerosene oil is also used as a energy and for other industrial domestic fuel but it causes some pollution. Automobiles applications; need fuels like petrol, diesel and CNG. Petrol and CNG • pr edict the for mation of the addition products of operated automobiles cause less pollution. All these fuels unsymmetrical alkenes and contain mixture of hydrocarbons, which are sources of alkynes on the basis of electronic energy. Hydrocarbons are also used for the manufacture mechanism; of polymers like polythene, polypropene, polystyrene etc. • comprehend the structure of Higher hydrocarbons are used as solvents for paints. They benzene, explain aromaticity are also used as the starting materials for manufacture and understand mechanism of many dyes and drugs. Thus, you can well understand of electrophilic substitution the importance of hydrocarbons in your daily life. In this reactions of benzene; unit, you will learn more about hydrocarbons. • predict the directive influence of substituents in monosubstituted 9.1 CLASSIFICATION benzene ring; • learn about carcinogenicity and Hydrocarbons are of different types. Depending upon toxicity. the types of carbon-carbon bonds present, they can be classified into three main categories – (i) saturated Reprint 2025-26 296 chemistry (ii) unsaturated and (iii) aromatic of the general formula for alkane family hydrocarbons. Saturated hydrocarbons or homologous series? If we examine the contain carbon-carbon and carbon-hydrogen formula of different alkanes we find that single bonds. If different carbon atoms are the general formula for alkanes is CnH2n+2. It joined together to form open chain of carbon represents any particular homologue when n atoms with single bonds, they are termed is given appropriate value. Can you recall the as alkanes as you have already studied in structure of methane? According to VSEPR Unit 8. On the other hand, if carbon atoms theory (Unit 4), methane has a tetrahedral form a closed chain or a ring, they are termed structure (Fig. 9.1), in which carbon atom lies as cycloalkanes. Unsaturated hydrocarbons at the centre and the four hydrogen atoms lie contain carbon-carbon multiple bonds – at the four corners of a regular tetrahedron. double bonds, triple bonds or both. Aromatic All H-C-H bond angles are of 109.5°. hydrocarbons are a special type of cyclic compounds. You can construct a large number of models of such molecules of both types (open chain and close chain) keeping in mind that carbon is tetravalent and hydrogen is monovalent. For making models of alkanes, you can use toothpicks for bonds and plasticine balls for atoms. For alkenes, alkynes and aromatic hydrocarbons, spring models can be constructed. Fig. 9.1 Structure of methane In alkanes, tetrahedra are joined together9.2 ALKANES in which C-C and C-H bond lengths are As already mentioned, alkanes are saturated 154 pm and 112 pm respectively (Unit 8). open chain hydrocarbons containing You have already read that C–C and C–H σ carbon - carbon single bonds. Methane (CH4) bonds are formed by head-on overlapping of is the first member of this family. Methane is 3 sp hybrid orbitals of carbon and 1s orbitals a gas found in coal mines and marshy places. of hydrogen atoms. If you replace one hydrogen atom of methane by carbon and join the required number of 9.2.1 Nomenclature and Isomerism hydrogens to satisfy the tetravalence of the You have already read about nomenclature other carbon atom, what do you get? You of different classes of organic compounds get C2H6. This hydrocarbon with molecular in Unit 8. Nomenclature and isomerism formula C2H6 is known as ethane. Thus you in alkanes can further be understood with can consider C2H6 as derived from CH4 by the help of a few more examples. Common replacing one hydrogen atom by -CH3 group. names are given in parenthesis. First three Go on constructing alkanes by doing this alkanes – methane, ethane and propane have theoretical exercise i.e., replacing hydrogen only one structure but higher alkanes can atom by –CH3 group. The next molecules will have more than one structure. Let us write be C3H8, C4H10 … structures for C4H10. Four carbon atoms of H H H C4H10 can be joined either in a continuous replace any H by - CH3 chain or with a branched chain in the H—C—H H—C—C—H or C2H6 following two ways : H H H I These hydrocarbons are inert under normal conditions as they do not react with acids, bases and other reagents. Hence, they were earlier known as paraffins (latin : Butane (n- butane), (b.p. 273 K)parum, little; affinis, affinity). Can you think Reprint 2025-26 Hydrocarbons 297 II isomers. It is also clear that structures I and III have continuous chain of carbon atoms but structures II, IV and V have a branched chain. Such structural isomers which differ in chain of carbon atoms are known as chain isomers. Thus, you have seen that C4H10 2-Methylpropane (isobutane) and C5H12 have two and three chain isomers (b.p.261 K) respectively. In how many ways, you can join five Problem 9.1 carbon atoms and twelve hydrogen atoms of Write structures of different chain C5H12? They can be arranged in three ways isomers of alkanes corresponding to the as shown in structures III–V molecular formula C6H14. Also write their IUPAC names.III Solution (i) CH3 – CH2 – CH2 – CH2– CH2– CH3 n-Hexane Pentane (n-pentane) (b.p. 309 K) IV 2-Methylpentane 3-Methylpentane 2-Methylbutane (isopentane) (b.p. 301 K) 2,3-Dimethylbutane V 2,2 - Dimethylbutane Based upon the number of carbon atoms attached to a carbon atom, the carbon atom is 2,2-Dimethylpropane (neopentane) termed as primary (1°), secondary (2°), tertiary (b.p. 282.5 K) (3°) or quaternary (4°). Carbon atom attached Structures I and II possess same molecular to no other carbon atom as in methane or to formula but differ in their boiling points and only one carbon atom as in ethane is called other properties. Similarly structures III, IV primary carbon atom. Terminal carbon and V possess the same molecular formula atoms are always primary. Carbon atom but have different properties. Structures I and attached to two carbon atoms is known as II are isomers of butane, whereas structures secondary. Tertiary carbon is attached to III, IV and V are isomers of pentane. Since three carbon atoms and neo or quaternary difference in properties is due to difference in carbon is attached to four carbon atoms. Can their structures, they are known as structural you identify 1°, 2°, 3° and 4° carbon atoms in Reprint 2025-26 298 chemistry structures I to V ? If you go on constructing compounds. These groups or substituents structures for higher alkanes, you will be are known as alkyl groups as they are derived getting still larger number of isomers. C6H14 from alkanes by removal of one hydrogen has got five isomers and C7H16 has nine. As atom. General formula for alkyl groups is many as 75 isomers are possible for C10H22. CnH2n+1 (Unit 8). In structures II, IV and V, you observed Let us recall the general rules for that –CH3 group is attached to carbon atom nomenclature already discussed in Unit 8. numbered as 2. You will come across groups Nomenclature of substituted alkanes can like –CH3, –C2H5, –C3H7 etc. attached to further be understood by considering the carbon atoms in alkanes or other classes of following problem: Problem 9.2 Write structures of different isomeric alkyl groups corresponding to the molecular formula C5H11. Write IUPAC names of alcohols obtained by attachment of –OH groups at different carbons of the chain. Solution Structures of – C5H11 group Corresponding alcohols Name of alcohol (i) CH3 – CH2 – CH2 – CH2– CH2 – CH3 – CH2 – CH2 – CH2– CH2 – OH Pentan-1-ol (ii) CH3 – CH – CH2 – CH2 – CH3 CH3 – CH – CH2 – CH2– CH3 Pentan-2-ol | | OH (iii) CH3 – CH2 – CH – CH2 – CH3 CH3 – CH2 – CH – CH2– CH3 Pentan-3-ol | | OH CH3 CH3 3-Methyl- | | butan-1-ol (iv) CH3 – CH – CH2 – CH2 – CH3 – CH – CH2 – CH2– OH CH3 CH3 2-Methyl- | | butan-1-ol (v) CH3 – CH2 – CH – CH2 – CH3 – CH2 – CH – CH2– OH CH3 CH3 2-Methyl- | | butan-2-ol (vi) CH3 – C – CH2 – CH3 CH3 – C – CH2 – CH3 | | OH CH3 CH3 2,2- Dimethyl- | | propan-1-ol (vii) CH3 – C – CH2 – CH3 – C – CH2OH | | CH3 CH3 CH3 CH3 OH 3-Methyl- | | | | b u t a n - 2 - o l (viii) CH3 – CH – CH –CH3 CH3 – CH – CH –CH3 Reprint 2025-26 Hydrocarbons 299 Table 9.1 Nomenclature of a Few Organic Compounds Structure and IUPAC Name Remarks Lowest sum and alphabetical (a) 1CH3–2CH – 3CH2 – 4CH – 5CH2 – 6CH3 (4 – Ethyl – 2 – methylhexane) arrangement Lowest sum and (b) 8CH3 – 7CH2 – 6CH2 – 5CH – 4CH – 3C – 2CH2 – 1CH3 alphabetical arrangement (3,3-Diethyl-5-isopropyl-4-methyloctane) sec is not considered (c) 1CH3–2CH2–3CH2–4CH–5CH–6CH2–7CH2–8CH2–9CH2–10CH3 while arranging alphabetically; isopropyl is taken 5-sec– Butyl-4-isopropyldecane as one word (d) 1CH3–2CH2–3CH2–4CH2–5CH–6CH2–7CH2–8CH2–9CH3 Further numbering to the substituents of the side chain 5-(2,2– Dimethylpropyl)nonane (e) 1CH3 – 2CH2 – 3CH – 4CH2 – 5CH – 6CH2 – 7CH3 Alphabetical priority order 3–Ethyl–5–methylheptane important to write the correct structure Problem 9.3 from the given IUPAC name. To do this, first Write IUPAC names of the following of all, the longest chain of carbon atoms compounds : corresponding to the parent alkane is written. (i) (CH3)3 C CH2C(CH3)3 Then after numbering it, the substituents are (ii) (CH3)2 C(C2H5)2 attached to the correct carbon atoms and (iii) tetra – tert-butylmethane finally valence of each carbon atom is satisfied by putting the correct number of hydrogen Solution atoms. This can be clarified by writing the (i) 2, 2, 4, 4-Tetramethylpentane structure of 3-ethyl-2, 2–dimethylpentane in (ii) 3, 3-Dimethylpentane the following steps : (iii) 3,3-Di-tert-butyl -2, 2, 4, 4 - i) Draw the chain of five carbon atoms: tetramethylpentane C – C – C – C – C If it is important to write the correct ii) Give number to carbon atoms: IUPAC name for a given structure, it is equally C 1– C2– C3– C4– C5 Reprint 2025-26 300 chemistry iii) Attach ethyl group at carbon 3 and two Longest chain is of six carbon atoms and methyl groups at carbon 2 not that of five. Hence, correct name is 3-Methylhexane. 7 6 5 4 3 2 1 C1 – 2C – 3C – 4C – 5C (ii) CH3 – CH2 – CH – CH2 – CH – CH2 – CH3 iv) Satisfy the valence of each carbon atom by putting requisite number of hydrogen Numbering is to be started from the atoms : end which gives lower number to ethyl group. Hence, correct name is 3-ethyl-5- methylheptane. CH3 – C – CH – CH2 – CH3 9.2.2 Preparation Petroleum and natural gas are the main Thus we arrive at the correct structure. sources of alkanes. However, alkanes can be If you have understood writing of structure prepared by following methods : from the given name, attempt the following 1. From unsaturated hydrocarbonsproblems. Dihydrogen gas adds to alkenes and alkynes Problem 9.4 in the presence of finely divided catalysts like Write structural formulas of the following platinum, palladium or nickel to form alkanes. compounds : This process is called hydrogenation. These metals adsorb dihydrogen gas on their (i) 3, 4, 4, 5–Tetramethylheptane surfaces and activate the hydrogen – hydrogen (ii) 2,5-Dimethyhexane bond. Platinum and palladium catalyse the reaction at room temperature but relatively Solution higher temperature and pressure are required with nickel catalysts. (i) CH3 – CH2 – CH – C – CH– CH – CH2 =CH2 + H2 Pt/Pd/Ni CH3−CH3 CH3 Ethene Propane (9.1) CH2–CH=CH2 + H2 Pt/Pd/Ni CH3−CH2CH3 Propane Propane (9.2) (ii) CH3 – CH – CH2 – CH2 – CH – CH3 Problem 9.5 CH3–C≡ C–H + 2H Pt/Pd/Ni CH3−CH2CH3 Write structures for each of the Propyne Propane following compounds. Why are the given (9.3) names incorrect? Write correct IUPAC names. 2. From alkyl halides (i) 2-Ethylpentane i) Alkyl halides (except fluorides) on reduction (ii) 5-Ethyl – 3-methylheptane with zinc and dilute hydrochloric acid give Solution alkanes. (i) CH3 – CH – CH2– CH2 – CH3 Zn,H CH–C1+H2 + CH4+HC1 (9.4) Chloromethane Methane Reprint 2025-26 Hydrocarbons 301 C2H5–C1+H2 Zn,H+ C2H6+HC1 alkane containing even number of Chloroethane Ethane (9.5) carbon atoms at the anode. − + Zn,H+ 2CH3 COO Na + 2H2 OCH3CH2CH2C1 + H2 CH3CH2CH3+CH1 Sodium acetate1-Chloropropane Propane (9.6) ↓ Electrolysts ii) Alkyl halides on treatment with sodium CH3 − CH3 + 2CO2 + H2 + 2NaOH (9.9) metal in dry ethereal (free from moisture) solution give higher alkanes. This reaction The reaction is supposed to follow the is known as Wurtz reaction and is used following path : O for the preparation of higher alkanes containing even number of carbon i) 2CH3COO–Na+ 2CH3 – C – O –+2Na+ atoms. ii) At anode: CH3Br+2Na+BrCH3 dry ether CH3+2Na O O Bromomenthane Ethane – –02e– 2C.H3+2CO2↑ (9.7) 2CH3 –C–O 2CH3 – C – dry ether Acetate ion Acetate Methyl freeC2H5Br+2Na+BrC2H5 C2H5–C2H free radical radical Bromoethane n–Butane iii) H3C + CH3 H3C–CH3↑ (9.8) iv) At cathode : What will happen if two different alkyl halides are taken? H2O+e–→–OH+ 2 →H2↑3. From carboxylic acids i) Sodium salts of carboxylic acids on Methane cannot be prepared by this heating with soda lime (mixture of sodium method. Why? hydroxide and calcium oxide) give alkanes 9.2.3 Properties containing one carbon atom less than the carboxylic acid. This process of elimination Physical properties of carbon dioxide from a carboxylic acid is Alkanes are almost non-polar molecules known as decarboxylation. because of the covalent nature of C-C and CaO C-H bonds and due to very little difference CH3COO– Na++NaOH ∆ CH4+Na2CO3 of electronegativity between carbon and Sodium ethanoate hydrogen atoms. They possess weak van der Waals forces. Due to the weak forces, the first Problem 9.6 four members, C1 to C4 are gases, C5 to C17 are liquids and those containing 18 carbon Sodium salt of which acid will be needed atoms or more are solids at 298 K. They are for the preparation of propane ? Write colourless and odourless. What do you think chemical equation for the reaction. about solubility of alkanes in water based Solution upon non-polar nature of alkanes? Petrol Butanoic acid, is a mixture of hydrocarbons and is used CH3CH2CH2COO–Na++ NaOH CaO as a fuel for automobiles. Petrol and lower fractions of petroleum are also used for dry CH3CH2CH3+Na2CO3 cleaning of clothes to remove grease stains. On the basis of this observation, what do you think about the nature of the greasy ii) Kolbe’s electrolytic method: An aqueous substance? You are correct if you say that solution of sodium or potassium salt of grease (mixture of higher alkanes) is non- a carboxylic acid on electrolysis gives Reprint 2025-26 302 chemistry polar and, hence, hydrophobic in nature. It is reducing agents. However, they undergo generally observed that in relation to solubility the following reactions under certain of substances in solvents, polar substances conditions. are soluble in polar solvents, whereas the 1. Substitution reactions non-polar ones in non-polar solvents i.e., like One or more hydrogen atoms of alkanesdissolves like. can be replaced by halogens, nitro group Boiling point (b.p.) of different alkanes are and sulphonic acid group. Halogenationgiven in Table 9.2 from which it is clear that takes place either at higher temperaturethere is a steady increase in boiling point with (573-773 K) or in the presence of diffusedincrease in molecular mass. This is due to the sunlight or ultraviolet light. Lower alkanesfact that the intermolecular van der Waals do not undergo nitration and sulphonationforces increase with increase of the molecular reactions. These reactions in which hydrogensize or the surface area of the molecule. atoms of alkanes are substituted are known You can make an interesting observation as substitution reactions. As an example,by having a look on the boiling points of chlorination of methane is given below:three isomeric pentanes viz., (pentane, 2-methylbutane and 2,2-dimethylpropane). It Halogenation is observed (Table 9.2) that pentane having a continuous chain of five carbon atoms has CH2 + C1 hv CH3C1 + HC1 the highest boiling point (309.1K) whereas Chloromethane (9.10) 2,2 – dimethylpropane boils at 282.5K. With increase in number of branched chains, hv CH3C1 + CH2 C12 + HC1 the molecule attains the shape of a sphere. Dichloromethane (9.11)This results in smaller area of contact and therefore weak intermolecular forces between CH2C12 hv CHC13 + HC1 spherical molecules, which are overcome at Trichloromethane (9.12) relatively lower temperatures. Chemical properties CHC13 + C12 hv CC14 + HC1 As already mentioned, alkanes are generally Tetrachloromethane (9.13) inert towards acids, bases, oxidising and Table 9.2 Variation of Melting Point and Boiling Point in Alkanes Molecular Name Molecular b.p./(K) m.p./(K) formula mass/u CH4 Methane 16 111.0 90.5 C2H6 Ethane 30 184.4 101.0 C3H8 Propane 44 230.9 85.3 C4H10 Butane 58 272.4 134.6 C4H10 2-Methylpropane 58 261.0 114.7 C5H12 Pentane 72 309.1 143.3 C5H12 2-Methylbutane 72 300.9 113.1 C5H12 2,2-Dimethylpropane 72 282.5 256.4 C6H14 Hexane 86 341.9 178.5 C7H16 Heptane 100 371.4 182.4 C8H18 Octane 114 398.7 216.2 C9H20 Nonane 128 423.8 222.0 C10H22 Decane 142 447.1 243.3 C20H42 Eicosane 282 615.0 309.7 Reprint 2025-26 Hydrocarbons 303 CH3–CH3 + C12 hv CH3–CH2C1 + HC1 steps are possible and may occur. Two such steps given below explain how more highly Chloroethane (9.14) haloginated products are formed. It is found that the rate of reaction of C. 1 → C. H2C1 + HC1alkanes with halogens is F2 > Cl2 > Br2 > I2. CH3C1 + Rate of replacement of hydrogens of alkanes is : C. H2C1 + C1– C1 → CH2C12 + C. 1 3° > 2° > 1°. Fluorination is too violent to (iii) Termination: The reaction stops afterbe controlled. Iodination is very slow and a some time due to consumption of reactantsreversible reaction. It can be carried out in the and / or due to the following side reactions :presence of oxidizing agents like HIO3 or HNO3. The possible chain terminating steps are: CH4+I2 CH3I+HI (9.15) (a) C. 1 + C. 1 → C1–C1 HIO3+5HI→312+3H2O (9.16) (b) H3 C. + C. H3 → H3C– CH3 Halogenation is supposed to proceed via (c) H3 C. 1 + C. 1 → H3C–C1 free radical chain mechanism involving three Though in (c), CH3 – Cl, the one of thesteps namely initiation, propagation and products is formed but free radicals aretermination as given below: consumed and the chain is terminated. The Mechanism above mechanism helps us to understand (i) Initiation : The reaction is initiated the reason for the formation of ethane as a by homolysis of chlorine molecule in the byproduct during chlorination of methane. presence of light or heat. The Cl–Cl bond 2. Combustion is weaker than the C–C and C–H bond and Alkanes on heating in the presence of air or hence, is easiest to break. dioxygen are completely oxidized to carbon dioxide and water with the evolution of large C1–C1 hv C. H3 + C1 amount of heat. homolysis Chlorine free radicals CH4 (g) + 20 2 (g) → CO2 (g) + 2H2 O(1); (ii) Propagation : Chlorine free radical Äc Hè − 890kJmol -1 attacks the methane molecule and takes the (9.17) reaction in the forward direction by breaking C 4 H10 (g) + 13/2O2 (g) → 4CO2 (g) + 5H2 O(1)the C-H bond to generate methyl free radical with the formation of H-Cl. Äc Hè = −2875.84kJmol -1 + hv + (9.18)(a) CH4 + C1 C H3 + H–C1 The general combustion equation for any The methyl radical thus obtained attacks alkane is : the second molecule of chlorine to form  3n + 1 O2 → nCO2 + (n + 1)H2 O C n H2n+2 +  CH3 – Cl with the liberation of another chlorine   2 free radical by homolysis of chlorine molecule. (9.19) hv Due to the evolution of large amount of(b) CH3 + C1–C1 CH3 – C1 + C1 heat during combustion, alkanes are used as The chlorine and methyl free radicals fuels. generated above repeat steps (a) and (b) During incomplete combustion of alkanes respectively and thereby setup a chain of with insufficient amount of air or dioxygen, reactions. The propagation steps (a) and carbon black is formed which is used in (b) are those which directly give principal the manufacture of ink, printer ink, black products, but many other propagation pigments and as filters. Reprint 2025-26 304 chemistry incomplete pressure in the presence of oxides of vanadium,CH4(g) + O2(g) combustion C(s)+2H2 O(1) molybdenum or chromium supported over (9.20) alumina get dehydrogenated and cyclised to 3. Controlled oxidation benzene and its homologues. This reaction is Alkanes on heating with a regulated supply known as aromatization or reforming. of dioxygen or air at high pressure and in the presence of suitable catalysts give a variety of oxidation products. (i) 2CH4 + O2 Cu/523K/100atm 2CH3 OH Methanol (9.21) Mo2O3 (9.26)(ii) CH4 + O2 HCHO + H2O ∆ Toluene (C7H8) is methyl derivative of Methanal benzene. Which alkane do you suggest for preparation of toluene ? (9.22) 6. Reaction with steam (iii) 2CH3CH3 + 3O2 (CH3COO)Mn 2CH3COOH ∆ Methane reacts with steam at 1273 K in the Ethanoic acid presence of nickel catalyst to form carbon + 2H2O monoxide and dihydrogen. This method is (9.23) used for industrial preparation of dihydrogen (iv) Ordinarily alkanes resist oxidation but gas alkanes having tertiary H atom can be Ni oxidized to corresponding alcohols by CH4 + H2IO ∆ CO + 3H2 (9.27) potassium permanganate. 7. Pyrolysis KMnO4 (iCH3)3 CH Oxidation (CH3)3 COH Higher alkanes on heating to higher temperature decompose into lower alkanes, 2-Methylpropane 2-Methylpropane-2-01 alkenes etc. Such a decomposition reaction (9.24) into smaller fragments by the application of heat is called pyrolysis or cracking.4. Isomerisation n-Alkanes on heating in the presence of anhydrous aluminium chloride and hydrogen chloride gas isomerise to branched chain alkanes. Major products are given below. Some minor products are also possible which you can think over. Minor products are (9.28) generally not reported in organic reactions. Pyrolysis of alkanes is believed to be a Anhy, AICI3/ HCI free radical reaction. Preparation of oil gasCH3(CH)2)4CH3 or petrol gas from kerosene oil or petrol n-Hexane involves the principle of pyrolysis. For CH3CH–(CH2)2–CH3+CH3CH2–CH–CH2–CH3 example, dodecane, a constituent of kerosene oil on heating to 973K in the presence of CH3 CH3 platinum, palladium or nickel gives a mixture 2-Methylpentane 3-Methylpenatone of heptane and pentene. (9.25) Pt/Pd/Ni Other 5. Aromatization C12H26 973K C7H16 + C5H10 + Products n-Alkanes having six or more carbon atoms Dodecane Heptane Pentene on heating to 773K at 10-20 atmospheric (9.29) Reprint 2025-26 Hydrocarbons 305 9.2.4 Conformations 1. Sawhorse projections Alkanes contain carbon-carbon sigma (σ) In this projection, the molecule is viewed bonds. Electron distribution of the sigma along the molecular axis. It is then projected molecular orbital is symmetrical around the on paper by drawing the central C–C bond internuclear axis of the C–C bond which is as a somewhat longer straight line. Upper not disturbed due to rotation about its axis. end of the line is slightly tilted towards This permits free rotation about C–C single right or left hand side. The front carbon is bond. This rotation results into different shown at the lower end of the line, whereas spatial arrangements of atoms in space the rear carbon is shown at the upper end. which can change into one another. Such Each carbon has three lines attached to it spatial arrangements of atoms which can corresponding to three hydrogen atoms. be converted into one another by rotation The lines are inclined at an angle of 120° to around a C-C single bond are called each other. Sawhorse projections of eclipsed conformations or conformers or rotamers. and staggered conformations of ethane are Alkanes can thus have infinite number of depicted in Fig. 9.2. conformations by rotation around C-C single bonds. However, it may be remembered that rotation around a C-C single bond is not completely free. It is hindered by a small energy barrier of 1-20 kJ mol–1 due to weak repulsive interaction between the adjacent bonds. Such a type of repulsive interaction is called torsional strain. Conformations of ethane : Ethane Fig. 9.2 Sawhorse projections of ethane molecule (C2H6) contains a carbon – carbon 2. Newman projectionssingle bond with each carbon atom attached In this projection, the molecule is viewed at theto three hydrogen atoms. Considering the C–C bond head on. The carbon atom nearerball and stick model of ethane, keep one to the eye is represented by a point. Threecarbon atom stationary and rotate the other carbon atom around the C-C axis. This hydrogen atoms attached to the front carbon rotation results into infinite number of spatial atom are shown by three lines drawn at an arrangements of hydrogen atoms attached to angle of 120° to each other. The rear carbon one carbon atom with respect to the hydrogen atom (the carbon atom away from the eye) is atoms attached to the other carbon atom. represented by a circle and the three hydrogen These are called conformational isomers atoms are shown attached to it by the shorter (conformers). Thus there are infinite number lines drawn at an angle of 120° to each other. of conformations of ethane. However, there are The Newman’s projections are depicted in two extreme cases. One such conformation Fig. 9.3. in which hydrogen atoms attached to two carbons are as closed together as possible is called eclipsed conformation and the other in which hydrogens are as far apart as possible is known as the staggered conformation. Any other intermediate conformation is called a skew conformation.It may be remembered that in all the conformations, the bond angles and the bond lengths remain the same. Eclipsed and the staggered conformations can be represented by Sawhorse and Newman projections. Fig. 9.3 Newman’s projections of ethane Reprint 2025-26 306 chemistry Relative stability of conformations: As the first member, ethylene or ethene (C2H4) mentioned earlier, in staggered form of ethane, was found to form an oily liquid on reaction the electron clouds of carbon-hydrogen bonds with chlorine. are as far apart as possible. Thus, there are 9.3.1 Structure of Double Bond minimum repulsive forces, minimum energy Carbon-carbon double bond in alkenesand maximum stability of the molecule. On the consists of one strong sigma (σ) bond (bondother hand, when the staggered form changes enthalpy about 397 kJ mol –1) due to head-oninto the eclipsed form, the electron clouds of the 2 overlapping of sp hybridised orbitals andcarbon – hydrogen bonds come closer to each one weak pi (π) bond (bond enthalpy aboutother resulting in increase in electron cloud 284 kJ mol –1) obtained by lateral or sidewaysrepulsions. To check the increased repulsive overlapping of the two 2p orbitals of the twoforces, molecule will have to possess more carbon atoms. The double bond is shorter inenergy and thus has lesser stability. As already bond length (134 pm) than the C–C singlementioned, the repulsive interaction between bond (154 pm). You have already read thatthe electron clouds, which affects stability of the pi (π) bond is a weaker bond due to poora conformation, is called torsional strain. sideways overlapping between the two 2pMagnitude of torsional strain depends upon orbitals. Thus, the presence of the pi (π) bondthe angle of rotation about C–C bond. This makes alkenes behave as sources of looselyangle is also called dihedral angle or torsional held mobile electrons. Therefore, alkenes areangle. Of all the conformations of ethane, the easily attacked by reagents or compoundsstaggered form has the least torsional strain which are in search of electrons. Suchand the eclipsed form, the maximum torsional reagents are called electrophilic reagents.strain. Therefore, staggered conformation is The presence of weaker π-bond makes alkenesmore stable than the eclipsed conformation. unstable molecules in comparison to alkanesHence, molecule largely remains in staggered and thus, alkenes can be changed into singleconformation or we can say that it is preferred bond compounds by combining with theconformation. Thus it may be inferred that electrophilic reagents. Strength of the doublerotation around C–C bond in ethane is not bond (bond enthalpy, 681 kJ mol–1) is greatercompletely free. The energy difference between than that of a carbon-carbon single bond inthe two extreme forms is of the order of 12.5 ethane (bond enthalpy, 348 kJ mol –1). OrbitalkJ mol–1, which is very small. Even at ordinary diagrams of ethene molecule are shown in temperatures, the ethane molecule gains Figs. 9.4 and 9.5. thermal or kinetic energy sufficient enough to overcome this energy barrier of 12.5 kJ mol–1 through intermolecular collisions. Thus, it can be said that rotation about carbon-carbon single bond in ethane is almost free for all practical purposes. It has not been possible to separate and isolate different conformational isomers of ethane.

7.23 Give IUPAC names of the following ethers: 7.24 Write the names of reagents and equations for the preparation of the following ethers by Williamson’s synthesis: (i) 1-Propoxypropane (ii) Ethoxybenzene (iii) 2-Methoxy-2-methylpropane (iv) 1-Methoxyethane 7.25 Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers. 7.26 How is 1-propoxypropane synthesised from propan-1-ol? Write mechanism of this reaction. 7.27 Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason. 7.28 Write the equation of the reaction of hydrogen iodide with: (i) 1-propoxypropane (ii) methoxybenzene and (iii) benzyl ethyl ether. 7.29 Explain the fact that in aryl alkyl ethers (i) the alkoxy group activates the benzene ring towards electrophilic substitution and (ii) it directs the incoming substituents to ortho and para positions in benzene ring. 7.30 Write the mechanism of the reaction of HI with methoxymethane. 7.31 Write equations of the following reactions: (i) Friedel-Crafts reaction – alkylation of anisole. (ii) Nitration of anisole. (iii) Bromination of anisole in ethanoic acid medium. (iv) Friedel-Craft’s acetylation of anisole. 7.32 Show how would you synthesise the following alcohols from appropriate alkenes? CH3 OH (i) OH (ii) OH (iii) (iv) OH 7.33 When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place: Give a mechanism for this reaction. (Hint : The secondary carbocation formed in step II rearranges to a more stable tertiary carbocation by a hydride ion shift from 3rd carbon atom. Chemistry 224 Reprint 2025-26 Answers to Some Intext Questions 7.1 Primary alcohols (i), (ii), (iii) Secondary alcohols (iv) and (v) Tertiary alcohols (vi) 7.2 Allylic alcohols (ii) and (vi) 7.3 (i) 4-Chloro-3-ethyl-2-(1-methylethyl)-butan-1-ol (ii) 2, 5-Dimethylhexane-1,3-diol (iii) 3-Bromocyclohexanol (iv) Hex-1-en-3-ol (v) 2-Bromo-3-methylbut-2-en-1-ol 7.4 OH CH2 C OCH3 7.5 (i) CH3 CH CH3 (ii) O OH (iii) CH3 CH2 CH CH2 OH CH3 7.7 (i) 1-Methylcyclohexene (ii) A Mixture of but-1-ene and but-2-ene. But-2-ene is the major product formed due to rearrangement to give secondary carbocation.

6.9 Which compound in each of the following pairs will react faster in SN2 reaction with –OH? (i) CH3Br or CH3I (ii) (CH3)3CCl or CH3Cl 6.10 Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene: (i) 1-Bromo-1-methylcyclohexane (ii) 2-Chloro-2-methylbutane (iii) 2,2,3-Trimethyl-3-bromopentane. 6.11 How will you bring about the following conversions? (i) Ethanol to but-1-yne (ii) Ethane to bromoethene (iii) Propene to 1-nitropropane (iv) Toluene to benzyl alcohol (v) Propene to propyne (vi) Ethanol to ethyl fluoride (vii) Bromomethane to propanone (viii) But-1-ene to but-2-ene (ix) 1-Chlorobutane to n-octane (x) Benzene to biphenyl. 6.12 Explain why (i) the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride? (ii) alkyl halides, though polar, are immiscible with water? (iii) Grignard reagents should be prepared under anhydrous conditions? 6.13 Give the uses of freon 12, DDT, carbon tetrachloride and iodoform. 6.14 Write the structure of the major organic product in each of the following reactions: (i) CH3CH2CH2Cl + NaI (ii) (CH3)3CBr + KOH (iii) CH3CH(Br)CH2CH3 + NaOH (iv) CH3CH2Br + KCN (v) C6H5ONa + C2H5Cl (vi) CH3CH2CH2OH + SOCl2 (vii) CH3CH2CH = CH2 + HBr (viii) CH3CH = C(CH3)2 + HBr 6.15 Write the mechanism of the following reaction: nBuBr + KCN nBuCN 6.16 Arrange the compounds of each set in order of reactivity towards SN2 displacement: (i) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane (ii) 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 2-Bromo-3-methylbutane (iii) 1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methylbutane, 1-Bromo-3-methylbutane. 6.17 Out of C6H5CH2Cl and C6H5CHClC6H5, which is more easily hydrolysed by aqueous KOH. 6.18 p-Dichlorobenzene has higher m.p. than those of o- and m-isomers. Discuss. 6.19 How the following conversions can be carried out? (i) Propene to propan-1-ol (ii) Ethanol to but-1-yne (iii) 1-Bromopropane to 2-bromopropane Chemistry 190 Reprint 2025-26 (iv) Toluene to benzyl alcohol (v) Benzene to 4-bromonitrobenzene (vi) Benzyl alcohol to 2-phenylethanoic acid (vii) Ethanol to propanenitrile (viii) Aniline to chlorobenzene (ix) 2-Chlorobutane to 3, 4-dimethylhexane (x) 2-Methyl-1-propene to 2-chloro-2-methylpropane (xi) Ethyl chloride to propanoic acid (xii) But-1-ene to n-butyliodide (xiii) 2-Chloropropane to 1-propanol (xiv) Isopropyl alcohol to iodoform (xv) Chlorobenzene to p-nitrophenol (xvi) 2-Bromopropane to 1-bromopropane (xvii) Chloroethane to butane (xviii) Benzene to diphenyl (xix) tert-Butyl bromide to isobutyl bromide (xx) Aniline to phenylisocyanide