Chapter 11: Hydrocarbons

Hydrocarbons Definition: Hydrocarbons are organic compounds made up of only carbon (C) and hydrogen (H) atoms. Natural Occurrence: Found abundantly in petroleum and natural gas. Uses of Hydrocarbons • Used as domestic and industrial fuels. • Have many synthetic applications. Serve as raw materials for: Medicines, plastics, perfumes, polymers, unsaturated hydrocarbons especially alkenes are starting materials in industrial synthesis. • Used in the production of various commercial products. Classification of Hydrocarbons: Hydrocarbons are broadly divided into: (i) Aliphatic Hydrocarbons (ii) Aromatic Hydrocarbons tree Explanation • Hydrocarbons contain only carbon and hydrogen. • They are major components of fossil fuels. • Play a vital role in the chemical industry: • Serve as precursors to numerous synthetic materials. ALIPHATIC AND AROMATIC HYDROCARBONS (i) Aliphatic Hydrocarbons Definition: Aliphatic hydrocarbons are organic compounds that do not contain aromatic rings. Examples • Open-chain (acyclic) or closed-chain (cyclic) • Saturated (alkanes) or unsaturated (alkenes and alkynes) • If they contain only carbon and hydrogen atoms, they are called aliphatic hydrocarbons. Types (a) Acyclic Aliphatic Hydrocarbons: They are open chain hydrocarbons. • Alkanes: Saturated hydrocarbons (methane, ethane) CH4, CH3 - CH3 etc. • Alkenes: Unsaturated hydrocarbons with at least one double bond (ethene) CH2 = CH2, (propene) CH2 = CH- CH3 etc. • Alkynes: Unsaturated hydrocarbons with at least one triple bond (ethyne) CH = CH, (propyne) CH = C - CH3 etc. (b) Cyclic aliphatic hydrocarbons (cycloalkanes): cyclopropane, cyclobutane, cyclopentane and cyclohexane. H H H H H -H H H H Ethane Methane H2 H,C- CH, H,C CH, H,C- CH2p Cyclopropatte Cyclobutane Fig: Aliphatic hydrocarbons (ii) Aromatic Hydrocarbons Definition: Aromatic hydrocarbons are a special class of cyclic compounds characterized by a high carbon-to-hydrogen ratio and delocalized electrons (resonance) based on benzene (C.H6) and its derivatives Examples: Benzene, methylbenzene (toluene), ethylbenzene, phenol and naphthalene etc. Interesting Information! S.Q. Give use of aroma in organic chemistry. Ans. The term aromatic was derived from the Greek word 'aroma' meaning "fragrance" and is used in organic chemistry for a special class of compounds (aromatic compounds) having characteristic odour. H CH HC HC HC CH HC H Benzene Methylbenzene Fig: Some aromatic hydrocarbons H H H-CEC-H H H Ethyne Ethene H2 H,C CH2 H,C CH, CH2 H,C H,C CH2 H2 Cyclopentane Cyclohexane CH3 CH, - CH3 CH CH HC HC CH CH C H H Ethylbenzene NOMENCLATURE

Alkanes or Paraffins are saturated Alkanes Definition: hydrocarbons containing only single bonds between carbon and hydrogen atoms. • General formula: CnH2n+2 • Simplest alkane: Methane (CH4) • They are present organic compounds. • The homologous series of alkanes with condensed formula and molecular formula are shown in table 11.1. Table 11.1: Condensed structural formulae and molecular formulae of alkanes Molecular IUPAC Name Formula CH4 CH4 Methane CH: CH: CzH6 Ethane CH: CH2 CH3 СзН& 519 Propane CH3 CH2 CH2 CH3 C4H1O Butane CH: CH2 CH2 CH2 CH3 CsH12 Pentane CH: CH2 CH2 CH2 CH2 CH3 C.H14 Hexane CH: CH2 CH2 CH2 CH2 CH2 CH3 C-H16 Heptane CH3 CH2 CH2 CH2 CH2 CH2 CH2 CH3 C&H18 Octane CH CH2 CH2 CH2 CH2 CH2 CH. CH2 CH Nonane CH: CH≥ CH≥ CH› CH› CH› CH2 CH› CH› CH: C1oH22 Decane ALKYL GROUPS Definition: When one hydrogen atom is removed from an alkane, the remaining part is called an alkyl group. Named by replacing the "-ane" suffix of alkanes with "-yl" Table 11.2: Alkyl groups their names and abbreviations Alkyl Group Alkane Methyl (CH3-) Methane (CH4) Ethyl (CH3-CH2-) Ethane (CH3-CH3.) For alkanes with more than two atoms, more than one alkyl group can be derived. Example: Propyl can give n-propyl and iso-propyl. Did You Know? S. Q. Write uses of isomers of butane. Ans. Butane is a fuel used is lighters. Isobutene is used as a propellant in products such as shaving gel. Condensed Structure Abbreviation Me- Et- Table 11.3: Alkyl groups, Straight-chain and branched and their names Alkane CH3-CH2-CH2- CH:CH2CH3 n-Propyl group (Propyl). Propane CH:CH.CH2CH- CHCH_CHCH: CH,CHCH, - CH:CH CH2CH: Butane sec-Butyl group iso-Butyl group tert-Butyl group n-Butyl group (1-Methylpropyl) (2-Methylpropyl) (1,1-Dimethylpropyl) (Butyl) NOMENCLATURE OF ALKANES

IUPAC Naming Rules To systematically name branched alkanes, follow these eight IUPAC rules: Rule 1: Identify the longest continuous chain Locate the longest continuous chain of carbon atoms; this chain determines the parent name for the alkane. We designate the following compound as a hexane because the longest continuous chain contains six carbon atoms, that is a bent chain. Ch, -CH, The longest continuous chain may not always be obvious from the way the formula is written Example: That the following alkane is designated as a heptane because the longest chain contains seven carbon atoms. • If the longest chain has 7 carbon atoms → heptanes #CH CH, Rule 2: Number the chain from the end nearest a substituent Number the longest chain beginning from the end nearest to the substituent. Applying this rule, we number the two alkanes shown below in the following way. CHL Possible Alkyl Groups CH3-CH-CH3 Iso-Propyl group (1-Methylethyl) CH,CCH: CH, CH, CH, CH- CH CH CH CH CHA CH, CH Substituent CH CH CH, CH, 4 6. -CH, -CH, H,C Substituent Rule 3: Indicate the position of the substituents Use the numbers obtained by applying rule 2 to designate the location of the substituent group. The parent name is placed last, and the substituent group, preceded by the number designating its location on the chain, is placed first. Example: 3-Methylheptane 7 6 CH, CH, H,C 3-Methylheptane 4 6 CH, CH, H,C 2-Methylhexane Rule 4: Alphabetical order for multiple substituents When two or more substituents are present, give each substituent a number corresponding to its location on the longest chain. Example: We designate the following compound as 4-Ethyl-2-methylhexane. 2~ H,C CH, 4-Ethy1-2-methylhexane The substituent groups should be listed alphabetically (i.e. ethyl before methyl). While deciding on alphabetical order ignore multiplying prefixes such as "di" and "tri" Rule 5: Two substituents on the same carbon When two substituents are present on the same carbon atoms, use that number twice. Example: 3-Ethyl-3-miethylhexane CH H,C-CH,-C-CH,-CH,-CH, CH 3 - Ethyl - 3 - methylhexane 1 3 CH, CH, CH. 3 4 CH CH, CH, 2 CH, ,CH CH, Зн, CH, SH, CH, CH. CH, Rule 6: Identical substituents When two or more substituents are identical, indicate this by the prefixes di, tri, tetra, and so on. Then make certain that each and every substituent has a number. Examples: 2,3-Dimethylbutane, 2,3,4-Trimethylpentane, 2,2,3,3-Tetramethylpentane. CH3 Н.с-CH-CH-CH, НС-СН-СН-СН-СН, НіС-С-С-СН,-СН, CH3 CH, CH 2,3 - Dimethylbutane 2,3,4-Trimethylpentane Applying these six rules allows us to name most of the alkanes that we shall encounter. Two other rules, however, may be required occasionally. Rule 7: Choosing the Parent Chain When two chains of equal length compete for selection as the parent chain, choose the chain with the greater number of substituents. Example: 2,3,5-Trimethyl-4-iso-propylheptane н,с-CH,-Сн-Сн-Сн-Сн-СН, CH: cH, cH, cH, m CH, CH3 2,3,5-Trimethyl-4-iso - propylheptane Rule 8: First point of difference rule When branching first occurs at an equal distance from either end of the longest chain, choose the name that give the lower number at the first point of difference. Example: 2,3,5-Trimethylhexane CH3 CH3 СHз 2, 3, 5- Trimethylhexane (not - 2,4,5 - Trimethylhexane)

Nomenclature of cycloalkanes Cycloalkanes are named by attaching the prefix 'cyclo' to the names of the alkanes possessing the same number of carbon atoms. Examples: For three carbons shape = cyclopropane for four carbons shape = cyclobutane etc. HC CH H,C CH CH 3 2,2,3,3-Tetramethylpentane CH: - CH: H,C-CH-CH, Common name: Isopropylcyclohexane TUPAC name: I-Methylethylcyclohexane 1.1,4-Trimethylcyclohexane Ethylcyclohexane Fig: Naming cycloalkanes Quick Check 11.1 (a) Write down the structural formulas Dimethylpropane. Ans. Structural formulas: 2-Methylbutane CH3-CH(CH3)-CH2-CH3(OR) 2,2-Dimethylpropane Fren (OR) (CH3)зC-CH3 (b) Write down displayed formula of 2,3,3-Trimethylhexane and Methylcyclohexane. Ans. Displayed formulas 2,3,3-Trimethylhexane H H-C-HH-C-H H H H C C - H C H- 1 H H H H-C-H H H H (2,3,3-Trimethyl hexane) Methylcyclohexane Cyclohexane ring with one CH: group attached to any carbon. (c) Give two differences between molecules of cyclopentane and pentane. Ans. Two ditterences between cyclopentane and pentane: Cyclopentane is cyclic, pentane is straight-chain (acyclic). Cyclopentane has fewer hydrogen atoms than pentane (CHio and CsH1z) is formula of pentane. CH: CH, - CH: -CH. H,C Common name: n-Propylcyclohexane IUPAC name: Propyclcyclohexane for 2-Methylbutane and 2,2- CH, CH, - CH-CH, - CH CH, - C-CH, 1 CH3 H- C-H H #>c- C H H (Methylcyclohexane) (d) Eicosane is a straight chain alkane whose molecules contain 20 carbon atoms. What is the molecular formula of eicosane? Ans. Molecular formula of eicosane is C2oH42. Structure of Alkanes • Each carbon atom in an alkane is sp hybridized. • It forms four sigma (o) bonds arranged tetrahedrally and bond angle 109.5°. C-C bond length is 1.54. H H 109.5° H Methane Fig: Sp hybridized methane and ethane Bonding • Alkanes are often shown using ball-and-stick models to illustrate 3D geometry. • Carbon atoms in alkanes are surrounded by four pairs of bonding electrons which repel equally to each other to form an angle of 109.5. The shapes of some straight chain alkanes are shown in table 11.4 Table 11.4: Shapes of some straight chain alkanes IUPAC Name & Structural Formula Molecular Formula H H - H Methane CH4 H H -C H Ethane C2H6 H H C H- C Propane C3H8 H H H H Butane C4H10 H (H H H Ethane 3D Structure H -H C -H H H H •C -H H H Structure of cycloalkanes: Cycloalkanes are cyclic meaning that the carbon atoms of the molecules are arranged in the form of a ring. Definition: Cycloalkanes are saturated cyclic hydrocarbons. • All C-C bonds are single bonds forming a closed ring. • Each carbon is sp' hybridized and forms 4 sigma bonds. • The structural shapes of cyclopropane, cyclobutane and cyclopentane are shown in table 11.5. Table 11.5: Shapes of few cycloalkanes IUPAC name & Structural Formula Molecular Formula H, Cyclopropane 3Н6 H,C H,C- Cyclobutane C4H8 H,C H20 Cyclopentane C4H10 H,C Physical properties of alkanes (i) Physical state and appearance • Ci to C4 (methane to butane) are colourless, odourless, gases at room temperature. • Cs to C17 (pentane to heptadecane) are colourless, odourless liquids. The higher members from Cis to onwards are above waxy solids. (ii) Solubility and polarity Alkanes are non-polar or very weakly polar. • Insoluble in polar solvents like water. • Soluble in non-polar solvents such as hexane, benzene and diethyl ether etc. (iii) Variation of physical constants with size • The physical constants like boiling point, melting point, and density increase with the increase in number of carbon atoms. • Solubility decreases as molecular size increases. (iv) Effect of branching on boiling point • Branched-chain alkanes have lower boiling points than their straight-chain isomers. Example: n-Butane has low boiling point (-0.5 °C) than 2-Methylpropane (isobutane) boiling point (-11.7 °C). Reason: Branched molecules have less surface area, leading to weaker intermolecular forces (Van der Waals interactions). Quick Check 11.2 (a) Why do branched alkanes have lower boiling points than the straight-chain alkanes? Ans. Branched alkanes have lower boiling points than straight-chain alkanes because they have less surface area for intermolecular van der Waals forces, so weaker attractions. (b) Explain why alkanes have a tetrahedral shape. Ans. Alkanes have a tetrahedral shape because each carbon forms sp' hybridization and four single bonds arranged to minimize electron repulsion, giving bond angles of about 109.5° 3D Structure -CH2 -CH, CH2 H2 -CH2 (c) Draw the shapes of cyclopropane and cyclopentane. Ans. Cyclopropane: Triangle-shaped ring (3 carbons forming a triangle). CH2 H,C - CH, Cyclopentane: Pentagonal ring (5 carbons forming a pentagon). H,C CH, Cyclopentane H,C - CH2 REACTION MECHANISM AND MODES OF BOND BREAKING

Reaction Mechanism Definition: A reaction mechanism is the step-by-step sequence of elementary steps or reactions by which an overall chemical change occurs. Every organic reaction involves bond breaking and bond formation. Covalent bonds in organic molecules can break in two main ways: (i) Homolytic fission In homolytic fission, the covalent bond breaks evenly to produce free radicals and each atom retains one A electron from the shared pair • A free radical is a species with electron and is an unpaired highly reactive. • Reason: Free radicals are highly reactive due to the tendency of unpaired electrons to pair up. Fig: Homolytic fission and free radical formation Example (ii) Heterolytic Fission In heterolytic fission, the covalent bond breaks unevenly, and one atom takes both electrons from the bond • Produces oppositely charged ions (a cation and an anion). • Shown in mechanisms using full curly arrows to indicate electron pair movement. Example Heterolytic cleavage B A H H Fig: Heterolytic fission and formation of radical ions Homolytic Cleavage Homolysis + B - B Free radicals hy ci + ci Free radicals + i° H → H +:8 H UNREACTIVE NATURE OF ALKANES TOWARDS POLAR REAGENT The alkanes or paraffins: The term "paraffins" comes from Latin: Parum means little, Affinis means affinity This name reflects their low chemical reactivity, especially towards polar reagents. Did You Know? S.Q. Give source and effects of oxides of carbon. Ans. Hydrocarbon combustion produces CO2, a greenhouse gas contributing to climate change. Incomplete combustion can release carbon monoxide (CO) and particulate matter, which have serious health and environmental consequences. Inertness of Alkanes Under ordinary conditions, alkanes are inert or unreactive towards acids, alkalis, oxidizing agents and reducing agents. Reasons for Unreactivity (i) Non-Polarity of Bonds Alkanes are composed of C-C and C-H bonds. • Electronegativity of carbon is 2.5 and hydrogen is 2.1 • The small difference in electronegativity makes these bonds non-polar. • Polar reagents do not interact easily with alkanes. (ii) Stability of sigma (o) bonds • The bonds in alkanes are sigma (o) bonds. • In o-bond, electrons are tightly held between two nuclei making it very stable bond. • Breaking a o-bond requires a large amount of energy, contributing to alkane inertness. REACTIONS OF ALKANES

Reactions of alkanes Under high temperature or suitable conditions, alkanes undergo two main types of reactions. Thermal and catalytic reactions: These reactions typically occur at high temperatures or in the presence of catalysts. Examples: Combustion and cracking Free radical substitution reactions: These reactions involve the replacement of one (ii) atom or group of atoms by another proceed via a free radical mechanism. Example: Halogenation of alkanes. FREE RADICAL SUBSTITUTION REACTIONS Free radical reactions Halogenation: The substitution of a halogen atom with hydrogen atom of an alkane called halogenation. • These reactions occur in the presence of light energy (UV light), forming highly reactive free radicals. • Alkanes react with chlorine (Clz) or bromine (Brz) in sunlight or UV light. Hydrogen atoms are successively replaced by halogens. • The extent of halogenation depends on the amount of halogen used Reactivity of halogens F2 > Cl > Br2 >12 Order of reactivity: • Fluorine (F2): Reacts violently to produces carbon, fluorinated alkanes, and hydrofluoric acid (HF). • Chlorine (Clz) and Bromine (Brz): Undergo substitution in presence of UV light. • Iodine (12): Reaction is too slow and reversible. Mechanism of halogenation (Three steps) (i) Initiation Step • It is the first step in the mechanism of free radical substitution of alkanes by halogen. • Halogen molecule breaks into radicals using UV light: hv CI- CI (ii) Propagation step • It is the second step in free radical substitution. Radicals formed in initiation step attack the alkanes, generating more free radicals and continuing the chain. CH4 + •Cl → •CH3 + HCl • This step repeats to produce various chloroalkanes. (iii) Termination Step: It is the final step in the free radical substitution in which two free radicals combine to form a product. The termination may take place in different ways: • The chloride and alkyl radicals may combine to form chloroalkanes. : Two chloride may react to form chlorine molecule. Alkane with double number of carbons may formed by merging two alkyl radicals, such products formed in traces and it is rare in this mechanism. H H - → H-C-Cl (1) H-C. + H H H H (ii) > H - C-C-H •C-H H-C® H H → cI + CI H > H-c• + H--Cl -CI + CI > H-C H H H (iii) CI + CI → CI- CI By repetition of step two, a mixture of choloroalkanes (halogen substituted alkanes) are produced. The chlorine gas in the reaction mixture to start with, the greater proportions of chlorine atoms in the chloroalkanes molecules formed. Final Products • A mixture of hałogenated alkanes. • More chlorine is higher degree of substitution. Trichloromethane (Chloroform) Tetrachloromethane (Carbon tetrachloride) Interesting Information! S.Q. What are precautions of haloalkanes as solvent? Ans. Haloalkanes are anaesthetic in nature. The anesthetic power of a haloalkane increase with the number of halogen atoms. Use of these as organic solvents in lab is strictly monitored due to their toxic nature. Further substitution Further substitution with Cl radical with Cl radical CI CI CH,CI CH C12 Dichloromethane Chloromethane Quick Check 11.3 (a) Explain why alkanes have high stability? Ans. Alkanes have high stability because they have strong C-C and C-H sigma bonds and are nonpolar, making them less reactive. (b) What are major types of reactions that alkanes undergo? Ans. Major types of reactions alkanes undergo combustion catalytic oxidation and free radical substitution (halogenation). (c) How termination step occurs in the halogenation of alkanes? Ans. Termination step in halogenation: Two radicals combine to form a stable molecule, ending the chain reaction, due to removal of sunlight. Example: CI + CI → Cl2 CH3+CI® → CHCI CH' + CH3 → C2H6 (d) State the conditions under which a mixture of halogenoalkanes is obtained from the halogenation of alkanes. Ans. A mixture of halogenoalkanes is obtained when excess alkane and UV light are used during halogenation, leading to substitution at different positions. Further substitution with Cl radical CHCh CCL4 Tetrachloromethane Trichloromethane (e) Predict the products of the free radical chlorination of ethane. Ans. Products of free radical chlorination of ethane: Chloroethane (CHsCI) (major) Minor amounts of 1,1-dichloroethane, 1,2-dichloroethane, etc. from further substitution. ALKENES

Alkenes • Alkenes are unsaturated hydrocarbons. • They consist only of carbon (C) and hydrogen (H). • Contain at least one carbon-to-carbon double bond (C=C): • Alkenes have two fewer hydrogen atoms than the corresponding alkanes and their general formula is C.H2n. IUPAC rules for naming Alkenes The IUPAC nomenclature for alkenes is largely similar to that of alkanes, with a few key modifications: (i) Select the longest carbon chain containing the double bond Select the longest continuous chain that contains the C = C as the parent chain. Change the ending of the name of the alkane of identical length from - ane to - ene. Example H,C-CH,-CH, -C=CH-CH, 3- n- Propyl - 2 - hexene (ii) Number the carbon chain Number the chain so as to include both carbon atoms of the double bond are included in the chain. Numbering begins from the end nearer to the double bond. CH 3 н,°-ČH-CH,-CH=CH-CH, 5 - Methyl - 2 - hexene (iii) Indicate the position of the double bond Designate the location of the double bond by using the number of the first atom of the double bond as a prefix. Example: 2-Butene (double bond starts at carbon 2) H,C=CH-CH,-CH, 2 - Butene (iv) Identify and number substituent groups Indicate the locations of the substituent groups by the numbers of the carbon atoms to which they are attached. H,C-CH, - CH H,C=ĆH-ČH,-CH,-ČH, 1-Pentene Example: 2-Methyl-2-butene and 2,5,5-Trimethyl-2-hexene. CH3 H,C-C=CH-CH3 2 - Methyl - 2 - butene (v) Naming alkenes with more than one double bonds "diene" for two double bonds (But-1,3-diene) "triene" for three double bonds (Hept-1,3,5-triene) • Indicate the positions of all double bonds in the name. Example Names • CH=CH-CH=CH› (But-1,3-diene) • CH3-CH=CH-CH=CH-CH=CH2 (Hept-1,3,5-triene) PREPARATION OF ALKENES

Preparation of Alkenes Alkenes are commonly prepared through elimination reactions. The chemical reactions in which small atoms or groups (like H, OH, or X) are removed from adjacent carbon atoms of a saturated compound, resulting in the formation of a carbon-carbon double bond (C=C). Elimination reactions • Small molecules such as H2O or HX (Hcl, HBr) are eliminated. • These reactions convert saturated compounds (alcohols, or alkyl halides etc) into unsaturated alkenes. Methods of preparation (i) Dehydration of Alcohols • Ethanol is dehydrated using concentrated sulfuric acid (H2SO4) at 180°C. • A water molecule is removed from ethanol to form ethene. Reaction CH,CH2OH H,SO. heat Ethanol Mechanism: Ethanol reacts with H2SO4 to form ethyl hydrogen sulfate. Upon heating, this intermediate loses H2SO4, forming ethene. 180°C → CH,-CH, -OSO,H+H2O CH, CH2OH+H,SOA Heat → H,C=CH,†H,SO4 CH, CH, OSO,H (ii) Dehydrohalogenation of Alkyl Halides • Ethene is formed by heating ethyl halide (like ethyl bromide) with alcoholic KOH. • Elimination of hydrogen and halogen from adjacent carbon atoms occurs to form a double bond. CH, CH3 2,5,5 - Trimethyl - 2 - hexene _CH,= CH_+H2O Ethene water Reaction H3C-CH, - Br + KOH (alcoholic) PHYSICAL PROPERTIES OF ALKENES Alkenes display certain distinct physical properties due to the presence of the carbon-carbon double bond (C= C) and the nature of their molecular structure. (i) Physical state • The first three alkenes like ethene, propene, and butane are gases at room temperature. • Alkenes with 5 to 15 carbon atoms (Cs-C1s) are liquids. Higher alkenes (Cro and above) are solids. (ii) Solubility • Insoluble in water (due to non-polar nature). • Soluble in organic solvents like alcohol, ether, and benzene. (iii) Odour and combustion • Alkenes have a characteristic smell. • They burn with a luminous (yellow, smoky) flame due to incomplete combustion and higher carbon content. (iv) Polarity • Unlike alkanes, alkenes exhibit weak polarity: • This is due to the sp' hybridization of carbon atoms in the double bond, which slightly affects electron distribution Quick Check 11.4 ----- (a) Give the reactions of propanol with H2SO4, write down all the reaction involved. Ans. Reactions of propanol with H2SO4: Propanol + HSO• (acid catalyst) → CHCH2CH2OH > CHCH = CH2 + H20 Propanol + H2SO4 → Propyl hydrogen sulfate intermediate (at lower temp) On hydrolysis, propyl hydrogen sulfate gives back propanol or propene depending on conditions. (b) Name the following alkenes: (i) CH3C(CH3) = CH-CH3 Ans. Name the following alkenes: (i) CH3-C(CH3)=CH-CH3 Name: 2-Methyl-2-butene (ii) CH2=CH-CH(CH3)2 Name: 3-Methyl-1-butene (c) Draw the structural formulas of the following alkenes: (i) 3-Methyl-2butene Ans. Structural formulas of alkenes: (i) 3-Methyl-2-butene(incorrect) & 2-Methyl-2-Butene (is correct). CH3-C(CH3)=CH-CH3 heat, H,C=CH, + KBr +H2O Propene + H2O (Dehydration) (ii) CH2=CHCH(CH3)2 (ii) 2-Methyl-4-ethyl-2pentene (il) 2-Methyl-4-ethyl-2-pentene (Incorrect) H.C-C= CH-CH - CH (2,4-Dimethyl-2-hexene) (correct) (d) Give the dehydrohalogenation reaction of bromopropane. Ans. Dehydrohalogenation of bromopropane: CHCH›CH›Br + KOH (alc) → CH3CH=CH2 + KBr + H2O (e) Can propanol undergo dehydration? If yes give reactions involving dehydration. Ans. Yes. Reaction: CH3CH2CH2OH → CHCH = CH2 + H2O (in presence of conc. H2SO4, heat) STRUCTURE AND REACTIVITY OF ALKENES

Structure of Ethene (CzH4) • The simplest alkene is ethane. • Each carbon atom in ethene is sp' hybridized • Three equivalent sp' orbitals are formed by mixing one 2s orbital, two 2p orbitals which are present in the same plane. • One unhybridized p-orbital remains on each carbon atom lies perpendicular (90°) to the plane of hybrid orbitals. Geometry and Bonding • In ethene each carbon used sp' orbitals to form two C-H bonds each. • The remaining sp' orbitals forms C-C o bonds by linear overlap of sp' orbitals. • The unhybridized p-orbitals which are parallel to each other (one on each carbon) overlap sideways to form a r bond. • A double bond (C=C) consists of one o bond and one a bond. • The two carbon atoms forming a double bond and four H - atoms attach directly to them lie in the same plane. Molecular Shape • The molecule has a trigonal planar geometry around each carbon. *• All atoms connected to the C=C bond. • Bond angles are approximately 120°. H CH, 1s Is r-bond H 2p, SP: o-bond o-bond o-bond SPZ SP? o-bond o-bond SP2 H H 1s 1s Fig: Bond formation of carbons in ethene Unhybrid p orbital -x Hybrid sp2 The set of orbitals sp? & p Fig: Shows sp? hybrid and unhybrid p-orbitals. Interesting Information! S.Q. How ethane helps in ripening of fruits & vegetables? Ans. Plants produce ethene as a natural ripening hormone. For efficient harvesting and transportation to market, fruits and vegetables are often picked unripe and exposed to ethene so they could ripen in short time. Illustrations H H 120° H H Fig: Structure of ethane Reactivity of the a bond (i) Nature of the a bond Alkenes are more reactive than alkanes due to presence of the double bond • The r-electrons are located above and below the plane of the bonded atoms, not directly between the nuclei. • The probability of finding electrons is away from the line joining the two nuclei. • This makes the it bond weaker and more easily broken than a o bond. (ii) Electrophilic attack • Because the i electrons are more exposed, they readily attacked by electrophilic (electron-seeking) reagents. • This leads to electrophilic addition reactions, which are characteristic of alkenes. • As a result, electrophiles (electron-seeking species) can easily attack the a bond Did you know? S.Q. Differentiate between electrophiles and nucleophile. Ans. Electrophile is a specie that can act as electron pair acceptor, it is represented by Et Alkenes act as nucleophile because they contain at electrons. Nucleophile is a specie that can act as a donor of a pair of electrons, it is represented by Nü. Quick Check 11.5 (a) How does an alkane differ from an alkene in stability? Ans. Alkanes are more stable than alkenes because they contain only single bonds (o- bonds), while alkenes have a reactive double bond (r-bond) which is less stable and more reactive (b) Which of the following species is likely to act as a nucleophile? Why? (i) H2 (ii) H+ (ії) ОН Ans. OH (hydroxide ion), OH has lone pairs of electrons and a negative charge, so it can donate electrons to an electron-deficient species. That's the characteristic of a nucleophile. H2, and H* cannot behave as nucleophile due to absence of lone pairs. (c) Which of the following species is likely to act as an electrophile? Why? (i) H2 (ii) H+ (іїї) ОН- Ans. Only Ht can act as a electrophile from these species. Ht (proton), H* has no electrons and a positive charge, so it can accept a pair of electrons, which makes it an electrophile. CARBOCATION STABILITY AND INDUCTIVE EFFECT OF ALKYL GROUP

A carbocation is an alkyl group where a carbon atom carries a single positive charge. It is highly unstable and reactive, typically formed during electrophilic addition reactions. Types of carbocation (i) Methyl carbocation (CH3*) Positive charge on carbon bonded only to hydrogen atoms named as methyl carbocation (CH3*) (ii) Primary (1°) carbocation Positive carbon is bonded to one other carbon atom (or alkyl group) named as primary (1°) carbocation. (ili) Secondary (2°) carbocation Positive carbon is bonded to two carbon atoms (alkyl groups) directly is named as secondary (2º) carbocation. (iv) Tertiary (3°) carbocation Positive carbon is bonded to three carbons (alkyl substituents) is named as tertiary (3°) carbocation. More alkyl groups means greater stability. Inductive effect Definition: The polarization of a o bonds due to electron-donating or withdrawing effect of adjacent groups or atoms is referred to as inductive effect. The inductive effect of alkyl group play a vital role in carbocation stability. Electron donating inductive effect (-I effect) Definition: The electron donating groups, such as alkyl groups (1°, 2°, 3° carbocation), exhibit a positive inductive effect denoted as (+I). • Alkyl groups exhibit +I or positive effect. Electron donating species pushing electron density toward the positively charged carbon. • It reduces the intensity of the positive charge, making the carbocation more stable. The more alkyl groups attached to the carbocation, the more stable it becomes. • The alkyl group attached to the positively charged carbon atoms are electron donating groups. • Due to the positive charge on the carbon atom, carbocation is electrophile. • The inductive effects is shown by the arrowheads on the bonds to show the alkyl groups pushing electron towards the positively charged carbon. • Forming carbocation energetically more stable by making it less positively charged • By increase in number of alkyl groups the stability of carbocation increases due to ti effect. The presence of three electron donating alkyl groups make tertiary carbocation most stable energetically. Order of stability H H H R→C® < H H Fig: Inductive effect and stability of carbocation Type of Carbocation Attached Groups Methyl No alky! groups 1 alkyl group Primary (1) 2 alkyl groups Secondary (2°) 3 alkyl groups Tertiary (3°) ELECTRON WITHDRAWING INDUCTIVE EFFECT Electron withdrawing inductive effect (-I effect) Definition: Electron withdrawing groups, such as halogen atoms (CI, Br, F), exhibit a negative inductive effect denoted as (-I). • Halogens are highly electronegative. • When bonded to carbon, they pull electron density through the sigma (o) bonds display negative inductive effect. • It results in a shift of electron density away from the carbon chain, creating a permanent dipole. Formation of permanent dipole • The halogen atom becomes partially negative (8) due to higher electron density. : The carbon atom directly bonded to the halogen (Ci) becomes electron deficient (8).. The inductive effect decreases as you move away from the halogen along the carbon chain. • Cı is the most electron-deficient because it is directly bonded to Cl. • Electrons are pulled from the Ci-C2 bond toward Cl, due to its high electronegativity. Less positive charge than C, ›Less electronegative St+ C C→ C C 3 1 2 4 Electron deficient C, pulling electrons from C, -C_ covalent bond Fig: Inductive effect and polarization of sigma bond R R→Co < < R->E0 R H Inductive Effect Stability Least stable No +I effect Low Weak +I effect Moderate +I effect Medium Most stable Strong +I effect > CI • More electronegative REACTIONS OF ALKENES

Reactions of Alkenes Alkenes are unsaturated hydrocarbons (contain at least one carbon-carbon double bond C=C). • Due to the presence of this a-bond, alkenes are more reactive than alkanes. • The r-electrons are loosely held and can be easily attacked on electrophiles. Electrophilic addition reactions Definition: Electrophilic addition reactions are those in which an electrophile (**) is added across the double bond (C=C) of an alkene. • The C=C double bond in alkenes consists of a strong sigma (o) bond and a weaker pi (T) bond and more exposed • The n electrons are electron-rich, making the double bond a site of high electron density. • This attracts electrophiles (electron-deficient species), initiating the reaction. General mechanism of electrophilic addition Step 1: Attack on electrophile The electrophile (Xt) is attracted to the -electrons of the double bond • The r-bond breaks, and Xt adds to one of the carbons, forming a carbocation intermediate. Step 2: Attack by nucleophile • The nucleophile (Y-) quickly attacks the carbocation, resulting in the final addition product and the completion of addition reaction. electrophile x® H H H Y nucleophile H H H H Fig: Generalized mechanism of electrophilic addition • Xt is the electrophile (Ht, Brt) • Y is the nucleophile (Br, CT, OH) Electrophile: Electron-deficient species (Xt) that initiates the reaction. Nucleophile: Electron-rich species (Y-) that completes the reaction. The intermediate is a carbocation. Quick Check 11.6 (a) How do alkenes react with an electrophile? Ans. Alkenes react with electrophiles by the electrophilic addition mechanism. The n- electrons of the double bond attack the electrophile, forming a carbocation intermediate, which is then attacked by a nucleophile to complete the reaction. H H X carbocation H H - C > X- -Y H (b) Why the order of stability of carbocation is 3° > 2º > 1°? Ans. Because in a 3° (tertiary) carbocation, the positive charge is stabilized by inductive effect and hyper-conjugation from three alkyl groups. More alkyl groups donate electron density to stabilize the carbocation. Halogenation Definition: Halogenation is the addition of halogen molecules (like Cl2 or Brz) across the double bond of an alkene, forming vicinal dihalides (halogens on adjacent carbon atoms). Test for unsaturation • Bromine water (reddish-brown in color) is used to test for the presence of a C=C double bond • When ethene is bubbled through bromine water in the presence of an inert solvent like carbon tetrachloride, CCla, the reddish-brown color disappears, indicating addition across the double bond and identifying unsaturation of an organic compound. Reaction CH, = CH, + Br, -C+ → Br-CH,-CH_-Br 1,2-Dibromoethane Mechanism of Halogenation: The above reaction proceeds through a three step mechanism. Step 1: Induced Dipole Formation • Brz is a non-polar molecule, but as it approaches the electron-rich C=C double bond, the high electron density repels the electrons in Br-Br bond, creating a temporary dipole. • One Br atom becomes partially positive (St) which is closes to the double bond acts as an electrophile. Electrophile H H H High area of electron density Non-polar molecule Step 2: Electrophilic Attack • The r electrons from the alkene attack the electrophilic Brt, leading to the formation of a highly reactive cyclic bromonium ion or a carbocation intermediate. Step 3: Nucleophilic Attack • The Br ion attacks the carbocation or bromonium ion formed in step 2. • Final product is a vicinal dibromide, 1,2-dibromoethane. Mechanism: H H б - Room Temp. Br Br Donates electrons pair to Br Fig: Electrophilic addition mechanism of ethene and Brz б.+ Br Br Induced Dipole The double bond repels the electrons away from the closest Br atom Primary carbocation H H H H C •C -H- > H - H Br Br Br -:Br 1,2 - Dibromoethane

Hydrohalogenation Definition: Hydrohalogenation is the addition of a dry gaseous hydrogen halide (HX) at room temperature to an alkene, resulting in the formation of a halogenoalkane (alkyl halide). Example CH, = CH, + HBr → CH, - CH, -Br Bromoethane Ethene Properties of hydrogen halides (HX) • HX are the simplest example of polar molecules with a permanent dipole. • The H atom is electrophilic (St), and X- (halide ion) is nucleophilic. • All HX (HF, Hcl, HBr, HI) react with alkenes, exactly the same way. • Reaction rate increases from HF to HI (HI gives the fastest reaction). The actual product formed depends upon whether the alkene is symmetrical or unsymmetrical. 0- Br H Electrophile Has lower electronegativity than Br so is electron deficient Fig: HBr molecule polarity (i) Addition to symmetrical alkenes: Only one product is feasible, when alkene. is symmetrical. Mechanism steps The mechanism of addition of HBr to ethene is: Step 1: Nucleophilic a electrons of the C=C bond attack the hydrogen of HBr. Br ion is formed by HBr. A carbocation intermediate is formed by adding hydrogen to one of the double bonded carbon of ethene. Step 2: Br (nucleophile) attacks the carbocation to gives the final product: H H 8+ б- Br › H- H H H Fig: Electrophilic addition mechanism of HBr to ethene (ii) Addition to symmetrical alkenes Two products are possible, but one is major, due to Markovnikov's rule, when alkenes are unsymmetrical. Pulls the electrons more strongly towards itself H H H H - H > H H H Br H 1-Bromoethane Alkyl halide Example: Two products are formed by the addition of HBr to propene. Experimentally 2- Bromoporpane is the major product and the minor product is 1-Bromopropane. Markovnikov's rule "When a polar reagent adds to an unsymmetrical alkene, the negative part of the reagent (X) attaches to the double bonded carbon which bears lesser number of hydrogen atoms," Mechanism for propene and HBr Propene reacts with HBr in two ways. The mechanism for this reaction involves the following steps. Step 1 • * bond attacks H* (Hydrogen ions) adds to the carbon with more H atoms. • Hydrogen bromide forming bromide (Br) ion by breaking heterolytically. • A more stable secondary carbocation is formed, by adding hydrogen to double bonded carbon with more number of hydrogens. Step 2 • Br attacks the carbocation to give 2-Bromopropane (major product). • Minor path, less stable primary carbocation forms 1-Bromopropane. 8+ б. -Br H H H Room Temp H •H- H \:Br Propene Secondary Carbocation Room Temp H - C H- H Primary Carbocation Fig: Electrophilic addition mechanism of HBr to propene Quick Check 11.7 (a) Compare and explain the relative rates of addition to alkenes (reactivities) of Hcl, HBr, and HI: Ans. The rate of addition follows the order: HI > HBr > Hcl This is because the H-I bond is weakest and breaks more easily, making HI more reactive. Bond strength decreases from HCl to HI, so the ease of addition increases. (b) Explain the difference between an addition reaction and an elimination reaction: Ans. Addition reaction: Two atoms or groups add to a molecule across a double bond, converting it to a single bond. (Common with alkenes) Elimination reaction: Atoms or groups are removed, usually forming a double bond from a single bond. (Common with alkanes/haloalkanes) (c) How does an alkene react with an electrophile? Ans. Alkenes have a n bond that is rich in electrons. The electrophile attacked by electron-rich double bond, forming a carbocation intermediate, which then reacts with a nucleophile to complete the electrophilic addition reaction. H H H - H-C -н - Br H H H 2-Bromopropane (Major Product) H H H H H - H-C --H -н H H H Br 1-Bromopropane (Minor Product) (d) Explain how Markovnikov's rule is applied in addition of HBr to 2-pentene: Ans. Markovnikov's rule states: "The hydrogen (Ht) adds to the carbon with more hydrogen atoms already attached." In 2-pentene, when HBr is added: Ht adds to C-2, which has more H atoms. (less substituted) Br adds to C-3, forming 3-bromopentane as the major product. CH3 - CH, - CH-CH, - CH Br Hydrogenation Definition: Addition of molecular hydrogen (H2) across the double bond of an alkene in the presence of a metal catalyst (Ni, Pt, Pd) at 250-300°C, forming a saturated alkane is called hydrogenation. General Reaction: H,C=CH, +H, 250-300•C> H,C-CH, Industrial use: This reaction is used to convert vegetable oil into margarine (Banaspati ghee) (Hydrogenation turns unsaturated fats into saturated fats). Margarine (Banaspati Ghee)

(i) Hydration Definition: Addition of steam (HO) to alkenes to form alcohols, using concentrated sulfuric acid (H2SO4) as a catalyst • Similar to hydrohalogenation, carbocation intermediate is formed, OH (from water) attacks the carbocation. Example H,50A > C,H2OH H,C= CH, + H2O - Ethanol (alcohol) (ii) Halohydration Definition: Addition of halogen (Cl2 or Br) and water to an alkene to form a halohydrin (contains -OH and -X on adjacent carbons). Example: The reaction specifically produces bromohydrin and chlorohydrin for ethene, depending upon halogen used. H2C=CH2 + Br2 + H20 → > l (iii)Epoxidation Definition: The epoxidation of alkenes gives an oxygen containing three-membered cyclic ether, called an epoxide. Explanation • The reaction can be completed using various oxidizing agents in case of ethene (C2H4). • Ethene can also be converted to ethylene oxide using molecular oxygen and metal catalysts. (3-bromopentane) • (iii) Epoxidation (iv) Ozonolysis CH, - CH2 + HBr OH Br • A transition metal catalyst facilitates the reaction, forming an epoxide directly from the alkene. The epoxides can be converted in to diols by acid hydrolysis. H H Ag + / H H H Uses: Intermediate in the production of glycols, plastics, detergents. (iv) Ozonolysis Definition: Oxidative cleavage of the C=C double bond using ozone (O3) a reactive allotrope of oxygen to form ozonides, which break down into carbonyl compounds (aldehydes, ketones), then alcohols upon reduction. Identification of alkenes: It is done by breaking alkenes into smaller and more identifiable pieces, by using this reaction. The location of double bond on carbon change can be identified by this reaction. Mechanism: Alkene reaction with ozone to form ozonide which can be further reduced to carbonyl compounds, and ultimately to alcohols. Example C 2Н +03 → C,H2O3 H3C- CH=CH, HC H,C 1-Propene Quick Check 11.8 (a) Write the equation for the ozonolysis of ethene: Ans. Ozonolysis breaks the double bond and forms carbonyl compounds. For ethene (CH2=CH2): > 2HCHO(formaldehyde) Zn/HO CH, =CH, +03 (b) Give the epoxidation reaction of 2-butene: Ans. 2-Butene reacts with oxygen in presence of Ag using high temperature to give epoxide which is called epoxidation. CH, CH-CH-CH, +½0, - N > CH, -CH-CH-C, epoxideof 2-Butene) OH H Ht/H2O H,C •CH, OH Ethane- 1,2-diol Epoxide • 2CH2O Zn/H2O CH2 H H H H,C Methanal Ethanal reduction H,C - CH, - OH CH, - OH Ethanol Methanol / (c) How can ozonolysis indicate the position of a double bond in an alkene? Explain with methanal as a product. Ans. Ozonolysis breaks the double bond and gives carbonyl compounds. If methanal (HCHO) is a product, it means the double bond was at the end of the chain (terminal alkene). Example CH, -CH-CH: > HCHO + CH, -CHO (Propene gives methanal and ethanal) So, if methanal is formed, the double bond must have included a CH2= group at the end. POLYMERIZATION

Polymerization Definition: Polymerization is the formation of extremely long molecules (polymers) from small reactive molecules that join together (monomers). • Addition polymerization: The basis of plastic industry is the addition polymerization reaction that are the most important reactions of alkenes. • Addition polymerization is a chemical reaction where many monomer units, each containing a carbon-carbon double bond (C=C), join together to form a polymer a large molecule with repeating units, without producing any by-products. • a-bond in each C = C bond breaks and then monomers link together to form a new C-C single bonds, just like other addition reactions of alkenes. • Polymer: A polymer is a long change molecule that is made up of many repeated units. Examples of Polymerization (i) Ethene to Poly-ethene Interesting Information! S.Q. How can plastic polymer is formed? Ans. Alkenes, like ethene and propene, are used to create plastics through polymerization. Ethene, for instance, is polymerized to make polyethylene, one of the world's most common plastics used in bags, bottles, and packaging. Monomer H H Polymerization n C H H Ethene Monomer Used in: Plastic bags, containers, packaging. H H →• H H n Poly (ethene) Polymer (ii) Chloroethene into polyvinylchloride (PVC) H. Polymerization H H H Chloroethene Poly(chloroethene) (PVC) Polymer Monomer Used in: Pipes, electrical insulation, pens. Deducing repeating units Definition: A repeating unit is the smallest group of atoms that when connected one after the another makeup the polymer chain that repeats over and over to form the polymer. It is represented by square in the displayed and general formula. Identification of repeating unit: In polyethene and substituted polyalkenes (such as PVC) made of one type of monomers, the repeating unit is the same as the monomer except that the C=C double bond is changed to a C-C single bond. H H n H Ethene Poly (ethene) Polymer Monomer Quick Check 11.9 (a) Ethene is a monomer used in the polymerization process to create polyethylene. (i) Discuss how the chemical structure of ethene makes it suitable for polymerization. (ii) Discuss the change in hybridization and bond angle during this reaction. Ans. (i) Ethene (CH2=CH2) contains a carbon carbon double bond (C=C), which is reactive. The a-bond in the double bond breaks easily, allowing ethene molecules to link together through addition reactions to form This makes ethene an ideal monomer for polymerization. (ii) Change in hybridization and bond angle during this reaction: Ans. In ethene, carbon atoms are sp' hybridized, and the bond angle is about 120° (trigonal planar). After polymerization, in poly(ethene), carbon atoms become sp' hybridized, and the bond angle changes to about 109.5° (tetrahedral). (b) Explain why an addition reaction increases the saturation of a molecule: Ans. Addition reactions break the double (unsaturated) bonds and add atoms (like H or halogens), forming only single bonds. This increases the number of single (o) bonds, making the molecule more saturated CI • C This ball pen is H n made of PVC H H H H •C H H H Repeating unit long chains - polyethylene. CONJUGATED DIENES Conjugated dienes Definition: A diene is a hydrocarbon containing two carbon-carbon double bonds (C=C), in a carbon chain and are separated by single bond. Explanation • The adjacent double bonds in the conjugated dienes allowed for the overlap of p-orbitals on three or more adjacent atoms. • The overlap leads to a delocalization of electron which can stabilize the molecule and affects its reactivity. • i bond overlap extends over the entire system. • The unique property of arrangement of their double bonds make conjugated alkenes the important class of organic chemistry. Example: 1,3-Butadiene CH: CH,=C-CH=CH, CH,-CH-CH=CH, 2-Methylbut-1,3-diene (Isoprene) But-1,3-diene Applications • 1,3-Butadiene is a key raw material in the manufacture of synthetic rubber. • Polymerization of conjugated dienes results in long chains with rubber-like elasticity. Example: 1,3-Butadiene CH, = CH-CH = CH, But - 1,3 - diene Used in: Tires, elastic materials and latex products Quick Check 11.10 (a) Draw the structure of simple conjugated dienes such as penta-1,3-diene. Ans. Structure of Penta-1,3-diene (a simple conjugated diene): Structure: CH2 = CH-CH= CH -CH3 It has two double bonds separated by a single bond. This makes it a conjugated diene. (b) Illustrate the delocalization electrons in conjugated dienes. Ans. Delocalization of electrons in conjugated dienes: In conjugated dienes, like buta-1,3-diene, the n-electrons of the double bonds are not isolated. They interact with each other through the single bond in between, leading to delocalization over the 4 carbon atoms. This delocalization creates a more stable system, often represented using resonance structures or molecular orbital diagrams. Resonance illustration CH, = CH-CH= CH, + Or, more correctly, we say the a electrons are shared across Ci to C4, forming a conjugated i system. C-C-C-C Delocalization of pie electrons of the conjugated bonds →CH,-CH-CH-ČH, ISOMERISM

Isomerism Definition: Two or more compounds having the same molecular formula but different structural formulas and properties are said to be isomers and the phenomenon is called isomerism. Explanation: Isomerism is the important property of organic compounds. The structural formulas show the arrangement of atoms and bonds in it. Example of Alkanes The simplest hydrocarbon to have structural isomers in butane (C4H10). Butane (C4H10) show two isomers which are: • n-Butane CHz-CH2-CH2-CH3 • Isobutane (2-Methylpropane) CH-СН(СНз)-СНз. • By the study the structural formula of butane and other higher hydrocarbons of alkanes, it is possible to arrange atoms present in the molecule in more than one way to satisfy all the valencies. • It is possible to have two or more different arrangements for the same molecular formula. Example of Chlorobutane Isomers • 1-Chlorobutane: CHz-CHz-CHz-CHz-Cl • 2-Chlorobutane: CH3-CH(Cl)-CH2-CH3 • Both have same molecular formula C4H,Cl but different structures and properties, verified by experimental evidence. • Isomerism is not only possible but common if the compound contain more than three carbon atoms. With increase in number of carbon atoms in hydrocarbons, the number of possible isomers increase very rapidly. • The five carbon compound pentane, has three isomers. Types of Isomerism A. Structural Isomerism Definition: The structural isomerism arises due to the difference in the arrangement of atoms within the molecule. The structural isomerism can be explained in five different ways: (i) Chain isomerism This type of isomerism arises due to the difference in the nature of the carbon chain. Example: The pentane (CsH12) and butane (C4H10), have the following possible arrangements: CH3 - CH- CH CH. - CH, - CH, - CH n - Butane 2 - Methylpropane (Isobutane) CH-CH, - CH, - CH, - CH3 CH3- -CH - CH3 n - Pentane (il) Position isomerism Definition: Functional group remains the same, but its position changes on the carbon chain. So, this type of isomerism is called position isomerism. The arrangement of carbon atoms remains the same. Example: chloropropane and butene (C4H8) can have two positional isomers. CH-CH, - CH, - Cl 1- Chloropropane free charpropame CH, -CH, - CH = CH, But - 1- ene (iii) Functional group isomerism Definition: The type of isomerism arises due to the same molecular formula but different functional groups are present is called functional group isomerism. Example: There are two compounds having the same molecular formula C2H6O, but different functional groups like ether and alcohols are present Carboxylic acids and esters are also the examples of this category. H.C-O-CH3 Methoxy methane (iv) Metamerism Definition: This type of isomerism arises due to the unequal distribution of carbon atoms on either side of the functional group. Example: Such compounds belong to the same homologous series, like diethyl ether and methoxy propane belongs the ether series. H,C- H,C-O-CH, - CH Ethoxy ethane (v) Tautomerism Definition: This type of isomerism arises due to shifting of proton from one atom to another within the same molecule. CH3 CH. - C-CH, 2- Methylpropane 2,2 - Dimethylpropane (Neopentane) (Isopentane) CH3 - CH - CH3 CH-CH = CH - CH But - 2- ene H,C- CH, - OH Ethanol H3C-O-CH, - CH, - CH Methoxy propane Example: This is commonly seen in amino acids. They exist as zwitter ions which are formed when a proton is shifted from the carboxylic acid end of amino acid molecule to the amine group. Zwitter ion is also known as internal salt. COOH C R H Amino acid B. Stereoisomerism Definition: Stereoisomers have the same structural formula but differ in spatial arrangement of the identical groups in space. Types: There are two types of stereoisomerism. (i) Geometrical or cis-trans Isomerism (ii) Optical Isomerism. ORGANIC REDOX REACTIONS

Organic redox reactions Definition: Oxidation-reduction reactions that involve the organic compounds are called organic redox reactions. • The redox reactions are the transfer of electrons through addition or removal of atoms or bonds, specifically oxygen and hydrogen atoms. • These reactions are characterized by the addition or removal of atoms or bonds. • Oxygen is added or hydrogen is removed during organic oxidation. • Hydrogen is added or oxygen is removed showing organic reduction. Some example of oxidation and reduction reactions are given below. 1. Oxidation (i) Combustion Reaction: The reaction of excess of oxygen or air with organic compounds is usually called combustion or oxidation of organic compounds. 2C,H, +70, -4CO, + 6H,0 Complete oxidation of ethane and ethene produces carbon dioxide and water. (ii) Ozonolysis: Definition: Oxidative cleavage of the alkenes (C=C double bond) using ozone (O3) a réactive allotrope of oxygen. Breaks double bonds, forming carbonyl compounds (aldehydes/ketones). COO + R C NH: H Zwitterion CH, H,C - CH CH, H,C- (iii) Epoxidation of alkenes Definition: The epoxidation of alkenes gives an oxygen containing three-membered cyclic ether, called an epoxide. • The epoxidation of alkene is also the example of oxidation reaction. • Formation of epoxides from alkenes using oxidizing agents like Oz, peracids, or metal catalysts. H H H C • 02. H H H Alkene Epoxide (iv) Oxidation of alcohols Definition: The primary alcohol (ethanol) oxidized to carboxylic acid using acidified K2C120т. с,H2OH +2[0]-K,C. >CH,COOH+ H2O 2. Reduction (i) Hydrogenation of alkenes Definition: The addition of hydrogen to carbon-carbon double or triple bond in the presence of a catalyst and temperature is called hydrogenation. • It converts alkenes into alkanes. • Involves addition of hydrogen across a double bond. H H / + H-H C=C H H Zn/H2O + H H,C Formaldehyde Acetaldehyde H,C-CH, ОН Сн, Он Methanol Ethanol OH H/H2O > H,C- CH, H Ethane - 1,2 - diol H H Pt H- C-C-H H (i Reducing agents There are large number of reducing agents are used in organic synthesis. Some of them are lithium aluminum hydride (LiAlH4), sodium borohydride (NaBH4), tin and hydrochloric acid (Sn + HCl). Common reductions • Nitriles to amines • Amides to amines • Carboxylic acids to primary alcohols • Nitrobenzene to phenylamine Explanation: Reduction of acetic acid (carboxylic acid) to ethanol (alcohol) using LiAlH4. And aldehyde to alcohol using LiAlH4 as well. H,C-C-H+ 2[H]-LAH, ›CH, - CH, Quick Check 11.11 (a) Draw the displayed formula and name an isomer of CHa that shows positional isomerism: Ans. One example is But-1-ene: CH2 = CH - CH2- CH3 ale H TM C=C-C-C- H / H (Displayed formula) Positional isomer: But-2-ene CH3 - CH = CH - CH3 H H H-C-C=C-C-H H (Displayed formula) Both are positional isomers because the position of the double bond differs. (b) What type of isomers are ethoxy methane (CHOCzHs) and propanol CH›ОН? Ans. They are functional group isomers. Ethoxy methane: an ether Propanol: an alcohol (c) What are the different types of structural isomerism in alkenes? Ans. Chain isomerism - different carbon chain arrangement Position isomerism - double bond at different positions Functional group isomerism (in some special cases) Geometrical (cis-trans) isomerism - due to restricted rotation around the double bond OH H H H H H (d) Write down different isomers of CaH,NH› (butylamine): Ans. These are amine isomers of butylamine: Butan-1-amine: CH3-CH2-CH2-CH2-NH2 Butan-2-amine: CH3-CH2-CH(NH2)-CH3 2-Methylpropan-1-amine: (CH3)2CH-CH2-NH2. 2-Methylpropan-2-amine: (CH3)3C-NH2 .