5.15 Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory: (i) [Fe(CN)6] 4– (ii) [FeF6] 3– (iii) [Co(C2O4)3]3– (iv) [CoF6] 3–

4.9 Hydrogen Bonding Hydrogen bond is represented by a dotted line (– – –) while a solid line represents theNitrogen, oxygen and fluorine are the highly covalent bond. Thus, hydrogen bond can beelectronegative elements. When they are attached to a hydrogen atom to form covalent defined as the attractive force which binds bond, the electrons of the covalent bond are hydrogen atom of one molecule with the shifted towards the more electronegative electronegative atom (F, O or N) of another atom. This partially positively charged molecule. hydrogen atom forms a bond with the other 4.9.1 Cause of Formation of Hydrogen more electronegative atom. This bond is Bond known as hydrogen bond and is weaker When hydrogen is bonded to stronglythan the covalent bond. For example, in HF electronegative element ‘X’, the electron pairmolecule, the hydrogen bond exists between shared between the two atoms moves farhydrogen atom of one molecule and fluorine away from hydrogen atom. As a result theatom of another molecule as depicted below : hydrogen atom becomes highly electropositive – – – Hδ+–Fδ– – – –Hδ+ – Fδ– – – – Hδ+ – Fδ– with respect to the other atom ‘X’. Since Here, hydrogen bond acts as a bridge between there is displacement of electrons towards two atoms which holds one atom by covalent X, the hydrogen acquires fractional positive bond and the other by hydrogen bond. charge (δ +) while ‘X’ attain fractional negative Reprint 2025-26 132 chemistry charge (δ–). This results in the formation of a H-bond in case of HF molecule, alcohol or polar molecule having electrostatic force of water molecules, etc. attraction which can be represented as: (2) Intramolecular hydrogen bond : It is formed when hydrogen atom is in between Hδ+ – Xδ– – – – Hδ+ – Xδ– – – – Hδ+ – Xδ– the two highly electronegative (F, O, N) The magnitude of H-bonding depends atoms present within the same molecule. For on the physical state of the compound. It is example, in o-nitrophenol the hydrogen is in maximum in the solid state and minimum in between the two oxygen atoms. the gaseous state. Thus, the hydrogen bonds have strong influence on the structure and properties of the compounds. 4.9.2 Types of H-Bonds There are two types of H-bonds (i) Intermolecular hydrogen bond (ii) Intramolecular hydrogen bond (1) Intermolecular hydrogen bond : It is formed between two different molecules of the Fig. 4.22 Intramolecular hydrogen bonding in same or different compounds. For example, o-nitrophenol molecule SUMMARY Kössel’s first insight into the mechanism of formation of electropositive and electronegative ions related the process to the attainment of noble gas configurations by the respective ions. Electrostatic attraction between ions is the cause for their stability. This gives the concept of electrovalency. The first description of covalent bonding was provided by Lewis in terms of the sharing of electron pairs between atoms and he related the process to the attainment of noble gas configurations by reacting atoms as a result of sharing of electrons. The Lewis dot symbols show the number of valence electrons of the atoms of a given element and Lewis dot structures show pictorial representations of bonding in molecules. An ionic compound is pictured as a three-dimensional aggregation of positive and negative ions in an ordered arrangement called the crystal lattice. In a crystalline solid there is a charge balance between the positive and negative ions. The crystal lattice is stabilized by the enthalpy of lattice formation. While a single covalent bond is formed by sharing of an electron pair between two atoms, multiple bonds result from the sharing of two or three electron pairs. Some bonded atoms have additional pairs of electrons not involved in bonding. These are called lone-pairs of electrons. A Lewis dot structure shows the arrangement of bonded pairs and lone pairs around each atom in a molecule. Important parameters, associated with chemical bonds, like: bond length, bond angle, bond enthalpy, bond order and bond polarity have significant effect on the properties of compounds. A number of molecules and polyatomic ions cannot be described accurately by a single Lewis structure and a number of descriptions (representations) based on the same skeletal structure are written and these taken together represent the molecule or ion. This is a very important and extremely useful concept called resonance. The contributing structures or canonical forms taken together constitute the resonance hybrid which represents the molecule or ion. Reprint 2025-26 Chemical Bonding And Molecular Structure 133 The VSEPR model used for predicting the geometrical shapes of molecules is based on the assumption that electron pairs repel each other and, therefore, tend to remain as far apart as possible. According to this model, molecular geometry is determined by repulsions between lone pairs and lone pairs; lone pairs and bonding pairs and bonding pairs and bonding pairs. The order of these repulsions being : lp-lp > lp-bp > bp-bp The valence bond (VB) approach to covalent bonding is basically concerned with the energetics of covalent bond formation about which the Lewis and VSEPR models are silent. Basically the VB theory discusses bond formation in terms of overlap of orbitals. For example the formation of the H2 molecule from two hydrogen atoms involves the overlap of the 1s orbitals of the two H atoms which are singly occupied. It is seen that the potential energy of the system gets lowered as the two H atoms come near to each other. At the equilibrium inter-nuclear distance (bond distance) the energy touches a minimum. Any attempt to bring the nuclei still closer results in a sudden increase in energy and consequent destabilization of the molecule. Because of orbital overlap the electron density between the nuclei increases which helps in bringing them closer. It is however seen that the actual bond enthalpy and bond length values are not obtained by overlap alone and other variables have to be taken into account. For explaining the characteristic shapes of polyatomic molecules Pauling introduced the concept of hybridisation of atomic orbitals. sp, sp2, sp3 hybridizations of atomic orbitals of Be, B, C, N and O are used to explain the formation and geometrical shapes of molecules like BeCl2, BCl3, CH4, NH3 and H2O. They also explain the formation of multiple bonds in molecules like C2H2 and C2H4. The molecular orbital (MO) theory describes bonding in terms of the combination and arrangment of atomic orbitals to form molecular orbitals that are associated with the molecule as a whole. The number of molecular orbitals are always equal to the number of atomic orbitals from which they are formed. Bonding molecular orbitals increase electron density between the nuclei and are lower in energy than the individual atomic orbitals. Antibonding molecular orbitals have a region of zero electron density between the nuclei and have more energy than the individual atomic orbitals. The electronic configuration of the molecules is written by filling electrons in the molecular orbitals in the order of increasing energy levels. As in the case of atoms, the Pauli exclusion principle and Hund’s rule are applicable for the filling of molecular orbitals. Molecules are said to be stable if the number of elctrons in bonding molecular orbitals is greater than that in antibonding molecular orbitals. Hydrogen bond is formed when a hydrogen atom finds itself between two highly electronegative atoms such as F, O and N. It may be intermolecular (existing between two or more molecules of the same or different substances) or intramolecular (present within the same molecule). Hydrogen bonds have a powerful effect on the structure and properties of many compounds. EXERCISES 4.1 Explain the formation of a chemical bond. 4.2 Write Lewis dot symbols for atoms of the following elements : Mg, Na, B, O, N, Br. 4.3 Write Lewis symbols for the following atoms and ions: S and S2–; Al and Al3+; H and H– 4.4 Draw the Lewis structures for the following molecules and ions : H2S, SiCl4, BeF2, CO32−, HCOOH 4.5 Define octet rule. Write its significance and limitations. Reprint 2025-26 134 chemistry 4.6 Write the favourable factors for the formation of ionic bond. 4.7 Discuss the shape of the following molecules using the VSEPR model: BeCl2, BCl3, SiCl4, AsF5, H2S, PH3 4.8 Although geometries of NH3 and H2O molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss. 4.9 How do you express the bond strength in terms of bond order ? 4.10 Define the bond length. 4.11 Explain the important aspects of resonance with reference to the CO32− ion. 4.12 H3PO3 can be represented by structures 1 and 2 shown below. Can these two structures be taken as the canonical forms of the resonance hybrid representing H3PO3 ? If not, give reasons for the same. 4.13 Write the resonance structures for SO3, NO2 and NO3−. 4.14 Use Lewis symbols to show electron transfer between the following atoms to form cations and anions : (a) K and S (b) Ca and O (c) Al and N. 4.15 Although both CO2 and H2O are triatomic molecules, the shape of H2O molecule is bent while that of CO2 is linear. Explain this on the basis of dipole moment. 4.16 Write the significance/applications of dipole moment. 4.17 Define electronegativity. How does it differ from electron gain enthalpy ? 4.18 Explain with the help of suitable example polar covalent bond. 4.19 Arrange the bonds in order of increasing ionic character in the molecules: LiF, K2O, N2, SO2 and ClF3. 4.20 The skeletal structure of CH3COOH as shown below is correct, but some of the bonds are shown incorrectly. Write the correct Lewis structure for acetic acid. 4.21 Apart from tetrahedral geometry, another possible geometry for CH4 is square planar with the four H atoms at the corners of the square and the C atom at its centre. Explain why CH4 is not square planar ? 4.22 Explain why BeH2 molecule has a zero dipole moment although the Be–H bonds are polar. 4.23 Which out of NH3 and NF3 has higher dipole moment and why ? 4.24 What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp2, sp3 hybrid orbitals. 4.25 Describe the change in hybridisation (if any) of the Al atom in the following reaction. AlCl 3  Cl   AlCl 4 Reprint 2025-26 Chemical Bonding And Molecular Structure 135 4.26 Is there any change in the hybridisation of B and N atoms as a result of the following reaction? 4.27 Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in C2H4 and C2H2 molecules. 4.28 What is the total number of sigma and pi bonds in the following molecules? (a) C2H2 (b) C2H4 4.29 Considering x-axis as the internuclear axis which out of the following will not form a sigma bond and why? (a) 1s and 1s (b) 1s and 2px; (c) 2py and 2py (d) 1s and 2s. 4.30 Which hybrid orbitals are used by carbon atoms in the following molecules? CH3–CH3; (b) CH3–CH=CH2; (c) CH3-CH2-OH; (d) CH3-CHO (e) CH3COOH 4.31 What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one exmaple of each type. 4.32 Distinguish between a sigma and a pi bond. 4.33 Explain the formation of H2 molecule on the basis of valence bond theory. 4.34 Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals. 4.35 Use molecular orbital theory to explain why the Be2 molecule does not exist. 4.36 Compare the relative stability of the following species and indicate their magnetic properties; (superoxide), O22− (peroxide) 4.37 Write the significance of a plus and a minus sign shown in representing the orbitals. 4.38 Describe the hybridisation in case of PCl5. Why are the axial bonds longer as compared to equatorial bonds? 4.39 Define hydrogen bond. Is it weaker or stronger than the van der Waals forces? 4.40 What is meant by the term bond order? Calculate the bond order of : N2, O2, O2+ and O2–. Reprint 2025-26 Unit 5 Thermodynamics It is the only physical theory of universal content concerning which I am convinced that, within the framework of the applicability of its basic concepts, it will never be overthrown. After studying this Unit, you will be Albert Einstein able to • explain the terms : system and surroundings; • discriminate between close, open and isolated systems; Chemical energy stored by molecules can be released as• explain internal energy, work and heat; heat during chemical reactions when a fuel like methane, • state first law of thermodynamics cooking gas or coal burns in air. The chemical energy may and express it mathematically; also be used to do mechanical work when a fuel burns • calculate energy changes as in an engine or to provide electrical energy through a work and heat contributions in galvanic cell like dry cell. Thus, various forms of energy chemical systems; are interrelated and under certain conditions, these may • explain state functions: U, H. be transformed from one form into another. The study • correlate ∆U and ∆H; of these energy transformations forms the subject matter • measure experimentally ∆U and of thermodynamics. The laws of thermodynamics deal ∆H; with energy changes of macroscopic systems involving• define standard states for ∆H; • calculate enthalpy changes for a large number of molecules rather than microscopic various types of reactions; systems containing a few molecules. Thermodynamics is • state and apply Hess’s law of not concerned about how and at what rate these energy constant heat summation; transformations are carried out, but is based on initial and • differentiate between extensive final states of a system undergoing the change. Laws of and intensive properties; thermodynamics apply only when a system is in equilibrium • define spontaneous and non- or moves from one equilibrium state to another equilibrium spontaneous processes; state. Macroscopic properties like pressure and temperature• e x p l a i n e n t r o p y a s a thermodynamic state function do not change with time for a system in equilibrium state. and apply it for spontaneity; In this unit, we would like to answer some of the important • explain Gibbs energy change (∆G); questions through thermodynamics, like: and How do we determine the energy changes involved in a • establish relationship between chemical reaction/process? Will it occur or not? ∆G and spontaneity, ∆G and equilibrium constant. What drives a chemical reaction/process? To what extent do the chemical reactions proceed? Reprint 2025-26 THERMODYNAMICS 137

9.7 (c)]. This can be represented in the form This is often referred to as Hückel Rule.of two doughtnuts (rings) of electron clouds [Fig. 9.7 (d)], one above and one below the Some examples of aromatic compounds are plane of the hexagonal ring as shown below: given below: (electron cloud) Fig. 9.7 (c) Fig. 9.7 (d) The six π electrons are thus delocalised and can move freely about the six carbon nuclei, instead of any two as shown in Fig. 9.6 (a) or (b). The delocalised π electron cloud is attracted more strongly by the nuclei of the carbon atoms than the electron cloud localised between two carbon atoms. Therefore, presence of delocalised π electrons in benzene makes it more stable than the hypothetical cyclohexatriene. X-Ray diffraction data reveals that benzene is a planar molecule. Had any one of the above structures of benzene (A or B) been correct, 9.5.4 Preparation of Benzene two types of C—C bond lengths were expected. Benzene is commercially isolated from coalHowever, X-ray data indicates that all the tar. However, it may be prepared in thesix C—C bond lengths are of the same order laboratory by the following methods.(139 pm) which is intermediate between C— C single bond (154 pm) and C—C double (i) Cyclic polymerisation of ethyne: bond (133 pm). Thus the absence of pure (Section 9.4.4) double bond in benzene accounts for the (ii) Decarboxylation of aromatic acids: reluctance of benzene to show addition Sodium salt of benzoic acid on heating reactions under normal conditions, thus with sodalime gives benzene. explaining the unusual behaviour of benzene. 9.5.3 Aromaticity Benzene was considered as parent ‘aromatic’ compound. Now, the name is applied to all the ring systems whether or not having benzene (9.70) ring, possessing following characteristics. Reprint 2025-26 322 chemistry (iii) Reduction of phenol: Phenol is reduced (ii) Halogenation: Arenes react with halogens to benzene by passing its vapours over in the presence of a Lewis acid like anhydrous heated zinc dust FeCl3, FeBr3 or AlCl3 to yield haloarenes. (9.71) Chlorobenzene 9.5.5 Properties (9.73) (iii) Sulphonation: The replacement of aPhysical properties hydrogen atom by a sulphonic acid group inAromatic hydrocarbons are non- polar a ring is called sulphonation. It is carried outmolecules and are usually colourless liquids by heating benzene with fuming sulphuricor solids with a characteristic aroma. You are acid (oleum).also familiar with naphthalene balls which are used in toilets and for preservation of clothes because of unique smell of the compound and the moth repellent property. Aromatic hydrocarbons are immiscible with water but are readily miscible with organic solvents. They burn with sooty flame. Chemical properties (9.74) Arenes are characterised by electrophilic (iv) Friedel-Crafts alkylation reaction:substitution reactions. However, under When benzene is treated with an alkyl halidespecial conditions they can also undergo in the presence of anhydrous aluminiumaddition and oxidation reactions. chloride, alkylbenene is formed. Electrophilic substitution reactions The common electrophilic substitution reactions of arenes are nitration, halogenation, sulphonation, Friedel Craft’s alkylation and acylation reactions in which attacking reagent is an electrophile (E +) (i) Nitration: A nitro group is introduced (9.75) into benzene ring when benzene is heated with a mixture of concentrated nitric acid and concentrated sulphuric acid (nitrating mixture). (9.76) Why do we get isopropyl benzene on treating benzene with 1-chloropropane instead of n-propyl benzene? (v) Friedel-Crafts acylation reaction: The reaction of benzene with an acyl halide or (9.72) acid anhydride in the presence of Lewis acids (AlCl3) yields acyl benzene. Nitrobenzene Reprint 2025-26 Hydrocarbons 323 (9.77) In the case of nitration, the electrophile, nitronium ion, is produced by transfer of a proton (from sulphuric acid) to nitric acid in the following manner: (9.78) Step I If excess of electrophilic reagent is used, further substitution reaction may take place in which other hydrogen atoms of benzene Step II ring may also be successively replaced by the electrophile. For example, benzene on treatment with excess of chlorine in the presence of anhydrous AlCl3 can be Protonated Nitronium chlorinated to hexachlorobenzene (C6Cl6) nitric acid ion It is interesting to note that in the process of generation of nitronium ion, sulphuric acid serves as an acid and nitric acid as a base. Thus, it is a simple acid-base equilibrium. (b) F o r m a t i o n o f C a r b o c a t i o n (arenium ion): Attack of electrophile results in the formation of σ-complex or (9.79) 3 arenium ion in which one of the carbon is sp Mechanism of electrophilic substitution hybridised. reactions: According to experimental evidences, SE (S = substitution; E = electrophilic) reactions are supposed to proceed via the following three steps: (a) Generation of the eletrophile sigma complex (arenium ion) (b) Formation of carbocation intermediate The arenium ion gets stabilised by resonance:(c) Removal of proton from the carbocation intermediate (a) Generation of electrophile E ⊕: During chlorination, alkylation and acylation of benzene, anhydrous AlCl3, being a Lewis acid helps in generation of the elctrophile Cl⊕, R ⊕, RC⊕O (acylium ion) respectively by combining with the attacking reagent. Reprint 2025-26 324 chemistry Sigma complex or arenium ion loses its chemical equation: aromatic character because delocalisation of 3 CxHy + (x + ) O2 → x CO2 + H2O n (9.83)electrons stops at sp hybridised carbon. (c) Removal of proton: To restore the 9.5.6 Directive influence of a functional aromatic character, σ -complex releases group in monosubstituted benzene proton from sp3 hybridised carbon on attack – When monosubstituted benzene is subjected by [AlCl4] (in case of halogenation, alkylation – to further substitution, three possible and acylation) and [HSO4] (in case of disubstituted products are not formed in nitration). equal amounts. Two types of behaviour are observed. Either ortho and para products or meta product is predominantly formed. It has also been observed that this behaviour depends on the nature of the substituent already present in the benzene ring and not on the nature of the entering group. This is known as directive influence of substituents. Reasons for ortho/para or meta directive nature of groups are discussed below: Addition reactions Ortho and para directing groups: The Under vigorous conditions, i.e., at high groups which direct the incoming group to temperature and/ or pressure in the presence ortho and para positions are called ortho and of nickel catalyst, hydrogenation of benzene para directing groups. As an example, let us gives cyclohexane. discuss the directive influence of phenolic (–OH) group. Phenol is resonance hybrid of following structures: Cyclohexane (9.80) Under ultra-violet light, three chlorine molecules add to benzene to produce benzene hexachloride, C6H6Cl6 which is also called gammaxane. Benzene hexachloride, It is clear from the above resonating (BHC) structures that the electron density is more on (9.81) o – and p – positions. Hence, the substitution Combustion: When heated in air, benzene takes place mainly at these positions. However, burns with sooty flame producing CO2 and it may be noted that –I effect of – OH group also H2O operates due to which the electron density on 15 ortho and para positions of the benzene ring C H6 + O2 → 6CO2 +3H2 O is slightly reduced. But the overall electron 6 2 (9.82) density increases at these positions of the General combustion reaction for any ring due to resonance. Therefore, –OH group hydrocarbon may be given by the following activates the benzene ring for the attack by Reprint 2025-26 Hydrocarbons 325 an electrophile. Other examples of activating In this case, the overall electron density groups are –NH2, –NHR, –NHCOCH3, –OCH3, on benzene ring decreases making further –CH3, –C2H5, etc. substitution difficult, therefore these groups are also called ‘deactivating groups’. TheIn the case of aryl halides, halogens are moderately deactivating. Because of their electron density on o – and p – position strong – I effect, overall electron density on is comparatively less than that at meta benzene ring decreases. It makes further position. Hence, the electrophile attacks on substitution difficult. However, due to comparatively electron rich meta position resonance the electron density on o– and resulting in meta substitution. p– positions is greater than that at the 9.6 Carcinogenicity and Toxicitym-position. Hence, they are also o– and p – directing groups. Resonance structures of Benzene and polynuclear hydrocarbons chlorobenzene are given below: containing more than two benzene rings fused together are toxic and said to possess cancer producing (carcinogenic) property. Such polynuclear hydrocarbons are formed on incomplete combustion of organic materials like tobacco, coal and petroleum. They enter into human body and undergo various biochemical reactions and finally damage DNA and cause cancer. Some of the carcinogenic hydrocarbons are given below (see box). Meta directing group: The groups which direct the incoming group to meta position are called meta directing groups. Some examples of meta directing groups are –NO2, –CN, –CHO, –COR, –COOH, –COOR, –SO3H, etc. Let us take the example of nitro group. Nitro group reduces the electron density in the benzene ring due to its strong–I effect. Nitrobenzene is a resonance hybrid of the following structures. Reprint 2025-26 326 chemistry SUMMARY Hydrocarbons are the compounds of carbon and hydrogen only. Hydrocarbons are mainly obtained from coal and petroleum, which are the major sources of energy. Petrochemicals are the prominent starting materials used for the manufacture of a large number of commercially important products. LPG (liquefied petroleum gas) and CNG (compressed natural gas), the main sources of energy for domestic fuels and the automobile industry, are obtained from petroleum. Hydrocarbons are classified as open chain saturated (alkanes) and unsaturated (alkenes and alkynes), cyclic (alicyclic) and aromatic, according to their structure. The important reactions of alkanes are free radical substitution, combustion, oxidation and aromatization. Alkenes and alkynes undergo addition reactions, which are mainly electrophilic additions. Aromatic hydrocarbons, despite having unsaturation, undergo mainly electrophilic substitution reactions. These undergo addition reactions only under special conditions. Alkanes show conformational isomerism due to free rotation along the C–C sigma bonds. Out of staggered and the eclipsed conformations of ethane, staggered conformation is more stable as hydrogen atoms are farthest apart. Alkenes exhibit geometrical (cis-trans) isomerism due to restricted rotation around the carbon–carbon double bond. Benzene and benzenoid compounds show aromatic character. Aromaticity, the property of being aromatic is possessed by compounds having specific electronic structure characterised by Hückel (4n+2)π electron rule. The nature of groups or substituents attached to benzene ring is responsible for activation or deactivation of the benzene ring towards further electrophilic substitution and also for orientation of the incoming group. Some of the polynuclear hydrocarbons having fused benzene ring system have carcinogenic property. EXERCISES 9.1 How do you account for the formation of ethane during chlorination of methane ? 9.2 Write IUPAC names of the following compounds : (a) CH3CH=C(CH3)2 (b) CH2=CH-C≡C-CH3 (c) (d) –CH2–CH2–CH=CH2 (f) CH3(CH2)4 CH (CH2)3 CH3 (e) CH2 –CH (CH3)2 (g) CH3 – CH = CH – CH2 – CH = CH – CH – CH2 – CH = CH2 | C2H5 9.3 For the following compounds, write structural formulas and IUPAC names for all possible isomers having the number of double or triple bond as indicated : (a) C4H8 (one double bond) (b) C5H8 (one triple bond) 9.4 Write IUPAC names of the products obtained by the ozonolysis of the following compounds : (i) Pent-2-ene (ii) 3,4-Dimethylhept-3-ene (iii) 2-Ethylbut-1-ene (iv) 1-Phenylbut-1-ene Reprint 2025-26 Hydrocarbons 327 9.5 An alkene ‘A’ on ozonolysis gives a mixture of ethanal and pentan-3-one. Write structure and IUPAC name of ‘A’. 9.6 An alkene ‘A’ contains three C – C, eight C – H σ bonds and one C – C π bond. ‘A’ on ozonolysis gives two moles of an aldehyde of molar mass 44 u. Write IUPAC name of ‘A’. 9.7 Propanal and pentan-3-one are the ozonolysis products of an alkene? What is the structural formula of the alkene? 9.8 Write chemical equations for combustion reaction of the following hydrocarbons: (i) Butane (ii) Pentene (iii) Hexyne (iv) Toluene 9.9 Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and why? 9.10 Why is benzene extra ordinarily stable though it contains three double bonds? 9.11 What are the necessary conditions for any system to be aromatic? 9.12 Explain why the following systems are not aromatic? (i) (ii) (iii) 9.13 How will you convert benzene into (i) p-nitrobromobenzene (ii) m- nitrochlorobenzene (iii) p - nitrotoluene (iv) acetophenone? 9.14 In the alkane H3C – CH2 – C(CH3)2 – CH2 – CH(CH3)2, identify 1°,2°,3° carbon atoms and give the number of H atoms bonded to each one of these. 9.15 What effect does branching of an alkane chain has on its boiling point? 9.16 Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism. 9.17 Write down the products of ozonolysis of 1,2-dimethylbenzene (o-xylene). How does the result support Kekulé structure for benzene? 9.18 Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also give reason for this behaviour. 9.19 Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty? 9.20 How would you convert the following compounds into benzene? (i) Ethyne (ii) Ethene (iii) Hexane 9.21 Write structures of all the alkenes which on hydrogenation give 2-methylbutane. 9.22 Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, E+ (a) Chlorobenzene, 2,4-dinitrochlorobenzene, p-nitrochlorobenzene (b) Toluene, p-H3C – C6H4 – NO2, p-O2N – C6H4 – NO2. 9.23 Out of benzene, m–dinitrobenzene and toluene which will undergo nitration most easily and why? 9.24 Suggest the name of a Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene. 9.25 Why is Wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms? Illustrate your answer by taking one example. Reprint 2025-26