Chemistry Chapter 9 Hydrocarbons Class 11 Notes PDF

Revision Notes for Class 11 Chemistry Chapter 9 – Hydrocarbons ALKANES Alkanes are the simplest saturated organic compounds or hydrocarbons. They are also called paraffins. General formula of alkanes is Cn H 2n+ 2.The carbon atoms in alkanes are sp 3...

Revision Notes for Class 11 Chemistry Chapter 9 – Hydrocarbons ALKANES Alkanes are the simplest saturated organic compounds or hydrocarbons. They are also called paraffins. General formula of alkanes is Cn H 2n+ 2.The carbon atoms in alkanes are sp 3 o hybridized and have a tetrahedral shape. The bond length between C–H is 1.12 A and between o C – C is 1.54 A. The simplest alkane is methane ( CH 4 ). Conformation Alkanes have various conformers existing in space due to free rotation of the C – C single bond. Different spatial arrangements of atoms give rise to isomers called conformers or rotamers. Conformation of ethane ( C2 H 6 ) Ethane has two types of conformation: 1. Newman Projection Newman Projection of Ethane Class XI Chemistry www.vedantu.com 1 Order of stability is, staggered > skew > eclipsed 1. SawHorse projection Sawhorse Projection of Ethane Methods of Preparation 1. By catalytic hydrogenation of unsaturated hydrocarbons: Hydrogenation is the process of adding hydrogen to the unsaturated hydrocarbon in the presence of a catalyst. When the catalyst is Nickel and the temperature is 200 − 300o C (Raney Ni) then the reaction is called Sabatier and Sanderson’s reduction. CH 2 = CH 2 + H 2 ⎯⎯⎯Ni 200o C → CH 3 − CH 3 ethylene ethane 2. CH  CH + 2H 2 ⎯⎯⎯Ni 300o C → CH3 − CH3 acetylene ethane Class XI Chemistry www.vedantu.com 2  At room temperature, hydrogenation is carried out using platinum or palladium as a catalyst in place of Ni. Methane cannot be produced by this method as an unsaturated hydrocarbon does not have a single carbon atom. 2. By the reduction of alkyl halides: This method uses alkyl halide which is reduced in presence of Zn + CH 3COOH , Zn + HCl , Zn + NaOH , Zn + Cu along with an alcohol (C2 H 5OH ) or aluminum amalgam in C2 H 5OH or LiAlH 4. R − X + 2H → RH + HX Alkyl halides can also be reduced by heating them with HI and red phosphorus in a sealed tube (under pressure). R − I + HI ⎯⎯⎯⎯⎯ P 150o , pressure → RH + I 2 The function of red phosphorus is to remove iodine. 3. By the reduction of alcohols, aldehydes, ketones and fatty acids: These compounds and their derivatives can be reduced in the presence of hot hydroiodic acid and red phosphorus at 150o C in a sealed tube to produce alkanes. The reactions are: ROH + 2HI ⎯⎯⎯ Red P → RH + H 2O + I 2 RCHO + 4HI ⎯⎯⎯ Red P → RCH3 + 2 I 2 + H 2O R − COR + 4HI ⎯⎯⎯ Red P → R − CH 2 − R + H 2O + 2I 2 R − COOH + 6HI ⎯⎯⎯ Red P → RCH3 + 2H 2O + 3I 2 Aldehydes and ketones are also reduced to alkanes in the presence of amalgamated zinc and conc. HCl. This reaction is known as Clemmensen reduction. Zn − Hg R − CHO + 2 H 2 ⎯⎯⎯⎯ conc. HCl → R − CH 3 + H 2O Class XI Chemistry www.vedantu.com 3  Aldehydes can also be reduced to alkanes using hydrazine and KOH at 150 − 200o C. This reaction is known as Wolff−Kishner reduction. CH3CHO + NH 2 NH 2 → CH3CH = NNH 2 ⎯⎯⎯ KOH → CH3CH3 + N2 4. By condensing two molecules of alkyl halides: When two molecules of alkyl halides are treated with sodium metal in the presence of dry ether, then they are coupled to form an alkane. This reaction is known as Wurtz synthesis. R − Br + 2 Na + R − Br ⎯⎯⎯⎯ Dryether → R − R + 2 NaBr 5. By decarboxylation of carboxylic acid: when the sodium salt of carboxylic acid is strongly heated with soda lime, then an alkane is formed by elimination of CO2 as carbonate. R − COONa + NaOH ⎯⎯⎯ Heat CaO → R − H + Na2CO3 6. Kolbe’s electrolysis: This involves sodium or potassium salts of fatty acids to be electrolyzed that form higher alkanes at anode. 2CH 3COONa + 2 H 2O → CH 3 − CH 3 + 2CO2 + 2 NaOH + H 2 As the product is higher in alkanes, so methane can’t be prepared by this method. 7. By action of water on aluminium carbide or beryllium carbide: AC 4 3 + 12 H 2O → 4 Al (OH )3 + 3CH 4  aluminiumcarbide 8. Be C2 + 4 H 2O → 2 Be(OH ) 2 + CH 4  Berylliumcarbide Class XI Chemistry www.vedantu.com 4 Physical Properties 1. State: They consist of weak forces, therefore alkanes up to four carbon atoms are colourless and odourless gases, the next thirteen members are colourless and odourless liquids. Alkanes from carbon – 18 onwards are colourless and odourless solids. In alkenes, except ethene, all are odourless and follow some trend as alkanes. Ethene has a pleasant odour. All are colourless. Alkynes also follow the same trend as alkanes. 2. Density: The density of alkanes increases steadily with the rise of molecular mass and becomes constant at 0.8. 3. Solubility: They are insoluble in polar solvents such as water but soluble in non-polar solvents like ether, carbon tetrachloride CCl4 , benzene etc. 4. Boiling and melting points: As the number of carbons in straight chain alkanes increases, the boiling point also increases. But the increase in melting point is irregular with respect to increase in molecular mass. Alkenes and alkynes also show a gradual increase in boiling and melting points with the increase in molecular mass in homologous series. They are less volatile than alkanes, i.e., their boiling point and melting point are higher than corresponding alkanes. Chemical Properties Alkanes are highly stable compounds and inert substances due to the presence of non-polar C – C and C – H bonds. Alkanes are saturated compounds with strong sigma bonds which don't break under ordinary conditions. Therefore, Alkanes react at high temperature by a free radical mechanism. 1. Halogenation (free radical substitution): Alkanes react with halogens ( Cl2 , Br2 ) in the presence of light or in dark at high temperature to form corresponding substituted products. CH 4 ⎯⎯⎯ Cl2 − HCl → CH 3C1 ⎯⎯⎯ Cl2 − HCl → CH 2C12 ⎯⎯⎯ Cl2 − HCl → CHC13 ⎯⎯⎯ Cl2 − HCl → CC1 4 methene methyl chloride methylene chloride chloroform carbontetrachloride Class XI Chemistry www.vedantu.com 5  The relative reactivity of halogens has the order F2  Cl2  Br2  I 2 and alkanes follow    the order, 3  2  1  CH3 2. Nitration: Nitration is the introduction of the nitro group. It is possible for alkanes having three or more carbon atoms. Nitration of propane yields a mixture of nitro products. NO2 | CH 3CH 2CH 3 ⎯⎯⎯→ CH 3CH 2CH 2 NO2 + CH 3 CH CH 3 + CH 3CH 2 NO2 + CH 3 NO2 HNO3 400o C 3. Sulphonation: Higher alkanes (C- 6 hexane onwards) undergo sulfonation when treated with fuming H 2 SO4. n − C6 H14 + HOSO3 H → C6 H13 SO3 H + H 2O hexane hexanesulphonicacid 4. Oxidation or combustion: Alkanes burn in the presence of oxygen to form carbon dioxide and water along with evolution of heat. CH 4 + 2O2 → CO2 + 2H 2O 2C2 H 6 + 7O2 → 4CO2 + 6 H 2O Illustration 1: The order of reactivity of halogens towards halogenation of alkane is A. F2  Br2  Cl2 B. F2  Cl2  Br2 C. Cl2  F2  Br2 D. Cl2  Br2  F2 Class XI Chemistry www.vedantu.com 6 Solution: (B). In a group, electronegativity of the atom decreases. So, the reactivity of halogen also decreases in a group. Thus, the order of reactivity is F2  Cl2  Br2. Illustration 2 : Consider the following reaction: Reaction of Free Radical with Alkane Identify the structure of the major product ‘X’. Structure of the Major Product ‘X’ Solution: (B). Br is less reactive and more selective and so the most stable free radical ( 3o ) will be the major product. Exercise 1 1. The compound with the highest boiling point is A. n- hexane B. n- pentane Class XI Chemistry www.vedantu.com 7 C. 2,2- dimethylpropane D. 2- methyl butane 2. Relative reactivity of halogens on alkanes follow the order A. F2Cl2 Br2 I 2 B. Cl2 Br2 I 2 F2 C. F2 I2 Br2Cl2 D. I 2 Br2Cl2 F2 3. Propane is obtained from propene by which of the following methods? A. Wurtz reaction B. Dehydration C. Frankland reaction D. Catalytic hydrogenation ALKENES Alkenes are unsaturated hydrocarbons having a double bond between two carbon atoms. Alkenes have the general formula Cn H 2 n. Isomerism 1. Structural isomerism Example, Butene has 3structural isomers, CH 3 − CH 2 − CH = CH 2 But −1− ene Class XI Chemistry www.vedantu.com 8  CH 3 − CH = CH − CH 3 But − 2 −ene Isobutene 2. Geometrical isomers The hindered rotation around the C – C bond gives rise to stereoisomers having different spatial arrangements. Two isomers exist as follows: Cis and Trans Isomers of but-2-ene Methods of Preparation 1. By dehydration of alcohol: Dehydration of alcohol in the presence of acids forms alkene. The reaction is an elimination reaction. + R − CH 2 − CH 2 − OH ⎯⎯ H  → R − CH = CH 2 + H 2O Class XI Chemistry www.vedantu.com 9 2. By the dehydrohalogenation of alkyl halides: it involves an alkyl halide in the presence of alcoholic KOH to yield alkene. CH3CH 2CH 2 Br ⎯⎯⎯⎯ alc. KOH → CH3CH = CH 2 + HBr If dehydrogenation of alkyl halide gives two products, the major product will be according to Saytzeff’s rule that states that the most substituted alkene will be the major product. Dehydrohalogenation of Alkyl Halides The ease of dehydrohalogenation has the order: Tertiary alkyl halide > secondary alkyl halide > primary alkyl halide. Alkyl halides follow the order: alkyl iodide > alkyl bromide > alkyl chloride. 3. By the dehalogenation of vicinal dihalides: Dehydrohalogenation of vicinal dihalides in the presence of Zinc dust along with alcoholic solution yields pure alkene. Dehydrohalogenation of Vicinal Dihalides 4. Kolbe’s electrolysis method: The electrolysis of sodium or potassium salts of dicarboxylic acid produces alkene at anode. Class XI Chemistry www.vedantu.com 10  Kolbe’s Electrolysis Method If, Na / Liq.NH 3 is used, trans alkene is formed, and in presence of Ni cis alkene is formed. Physical Properties 1. Melting point: the trans isomers have a high melting point than cis isomer due to symmetry and crystal lattice. 2. Boiling point: Cis isomer has a more dipole moment as it is more polar and therefore has a high boiling point than trans isomer of alkene. Chemical Properties Alkenes are reactive due to the presence of double bonds. Due to THE presence of  bonds alkenes are able to react towards electrophilic addition reaction. They also give free radical addition reactions. 1. Addition reactions: (i) Addition of hydrogen (catalytic hydrogenation) CH 2 = CH 2 + H 2 ⎯⎯⎯⎯ Ni 200 −300o C → CH 3 − CH 3 (ii)Addition of halogens (Chlorine and bromine) CH 2 = CH 2 + Br2 → BrCH 2 − CH 2 Br ethylenedibromide ( colorless ) Addition of bromine is used as a test for detecting the presence of unsaturation (C – C double bond or triple bond). (ii) Addition of hydrogen halides Class XI Chemistry www.vedantu.com 11  CH 2 = CH 2 + HX → CH 3CH 2 X The order of reactivity among hydrogen halides is HI > HBr > HCl > HF In case of unsymmetrical alkenes, addition occurs according to Markonikov’s rule. This reaction takes place through an ionic mechanism. Electrophilic addition to a carbon–carbon double bond involves the formation of an intermediate that is the more stable carbocation. Deviation from Markonikov’s rule: It has been observed that addition of HBr to unsymmetrical alkenes like propene in presence of air, peroxide or light yields n-propyl bromide by anti-Markovnikov's rule. The effect is also called the peroxide effect or Kharasch effect. Markonikov’s and Anti-Markovnikov's Rule for Halogenation of Propene i. Addition of hypochlorous acid Reaction of Hypochlorous Acid of Ethylene ii. Addition of sulphuric acid Class XI Chemistry www.vedantu.com 12 Reaction of Sulphuric Acid of Propene Alkyl hydrogen sulphates are water soluble, when heated at about 160o C , they give olefins. In reaction with water they give alcohol. Formation of Alcohol from Alkyl Hydrogen Sulphates iii. Addition of water It is also in accordance with Markovnikov’s rule. Addition of Water to Alkene iv. Addition of alkanes (alkylation) Alkylation of Alkanes in Alkene Class XI Chemistry www.vedantu.com 13  v. Addition of diborane (hydroboration) Hydroboration of Alkanes in Alkene In case of unsymmetrical alkenes, addition follows Anti Markovnikov's rule. 6CH 3CH 2CH = CH 2 + B2 H 6 → 2 ( CH 3CH 2CH 2CH 2 )3 B tributyl borane Trialkyl borane on oxidation ( H 2O2 / OH − ) gives alcohol and on reduction ( LiAlH 4 ) gives alkane. vi. Oxymercuration – demercuration Oxymercuration – Demercuration of Alkene vii. Addition of oxygen Addition of Oxygen in Alkene Class XI Chemistry www.vedantu.com 14  1. Oxidation: i. Oxidation by cold alkaline KMnO4 (Bayer’s reagent) Oxidation of Alkene It is a test for detecting double bonds in alkene. Hydroxylation by KMnO4 is always a syn addition. The cis alkene on hydroxylation gives meso compound and trans alkene gives racemic mixture. Like Bayer’s reagent OsO4 also gives glycol and the hydroxylation is a syn addition. Hydroxylation of Alkene Using OsO4 ii. Oxidation by per acids ( RCO3 H ) Oxidation of Alkene by per Acids ( RCO3 H ) Class XI Chemistry www.vedantu.com 15 Reaction of Acetic Acid with Alkene This addition occurs in trans manners. The cis alkene gives racemic mixture and trans alkenes give meso compound. iii. Ozonolysis Ozonolysis of Alkene iv. Oxidation by hot concentrated alkaline KMnO4 RCH = CH 2 + KMnO4 (conc.) → RCOOH + CO2 + H 2O 1. Substitution reaction: Chlorination is done by treating the alkene with carbon tetrachloride in liquid phase or with chlorine gas: Class XI Chemistry www.vedantu.com 16 Chlorination of Alkene Allylic bromination (bromination at allylic carbon atom) is very easily achieved by treating the alkene having hydrogen atom at the allylic carbon atom with N-bromosuccinimde (NBS). Allylic Bromination of Alkene Illustration 3 : Which of the following alkene has the lowest heat of hydrogenation? Alkene with Lowest Heat of Hydrogenation Class XI Chemistry www.vedantu.com 17 Solution: (B). Higher the stability of alkene, lower the heat of hydrogenation. Illustration 4 : Reaction of Alkene with Alkaline KMnO4 Which is true about this reaction? (A) A is meso 1, 2-butan-di-ol formed by syn addition. (B) A is meso 1, 2-butan-di-ol formed by anti addition. (C) A is a racemic mixture of d and l, 1, 2-butan-di-ol formed by anti addition. (D) A is a racemic mixture of d and l, 1, 2-butan-di-ol formed by syn addition. Solution: (A). On cis alkene there is syn addition of two –OH groups forming meso compound Exercise 2 Halogenation of Alkene Class XI Chemistry www.vedantu.com 18 Predominant A is: Predominant Product A Product X and Y from Alkene X and Y are: A. CH3 − CH 2 − CH = CH 2 , ( CH 3CH 2COOH + CO 2 ) B. CH3 − CH = CH − CH3 , CH3COOH (2 moles) C. CH3 − CH = CH − CH3 , CH3CHO (2 moles) D. CH3 − CH 2 − CH = CH 2 , ( CH3CH 2CHO + HCHO ) The reaction of propene with HOCl proceeds via the addition of: E. H + in the first step E. Cl + the first step E. OH − in the first step Class XI Chemistry www.vedantu.com 19  E. Cl + and OH − in a single step ALKYNES Alkynes are characterized by the presence of a triple bond between two carbon atoms. The general formula of alkyne is Cn H 2 n −2. Methods of Preparation 1. By the dehydrohalogenation of vicinal dihalides: CH 2 = CH 2 ⎯⎯ Br2 → Br − CH 2 − CH 2 − Br + KOH (alc.) → Br − CH = CH 2 ⎯⎯⎯ NaNH 2 → CH  CH CH3 − CHBr2 + KOH (alc.) → CH 2 = CH − Br ⎯⎯⎯ NaNH 2 → CH  CH 2. By dehalogenation of vicinal tetrahalides: Reaction with active metals like Zinc, Mg etc. gives acetylene. Dehalogenation of Vicinal Tetrahalides 3. By Kolbe electrolysis method: Kolbe Electrolysis Method Class XI Chemistry www.vedantu.com 20 4. By heating iodoform or chloroform with silver powder or zinc: This method can be used for the preparation of only acetylene. CHI 3 + 6 Ag + CHI 3 → CH  CH + 6 AgI 5. From acetylene: Higher alkynes can be prepared from acetylene when treated with sodium metal in liquid ammonia. CH  CH + Na ⎯⎯⎯ ⎯ Liq. NH3 → CH  C − Na +1 / 2 H 2 CH  CNa + CH 3 Br → CH  C − CH 3 + NaBr sodiumacetylide propyne Similarly, CH  CH ⎯⎯⎯2 Na Liq. NH ⎯ → NaC = CNa ⎯⎯⎯→ 3 2CH Br CH 3C = CCH 3 3 Chemical Properties Alkyne gives electrophilic addition reactions due to the presence of loosely held  electrons, but electrophilic addition reactions in alkynes are slower than that of alkenes. Terminal hydrogen present in alkynes is acidic in nature. Since s electrons are closer to the nucleus than p electrons, the electrons present in the bond having more s character will be closer to the nucleus. The amount of s character in various types of C – H bond are as follows Hybridisation of Alkene, Percentage of S-Character Class XI Chemistry www.vedantu.com 21  Relative acidities:.. HOH = HOR  CH  CR  NH3  CH 2 = CH 2  CH3 − CH3 Relative basicities: OH− = OR −  C−  C − R  NH−2  CH− = CH2  CH−2 − CH3 1. Addition of hydrogen: CH  CH + H 2 ⎯⎯ Ni → CH 2 = CH 2 ⎯⎯ H2 → CH 3 − CH 3 acetylene ethylene ethane In case of alkynes where triple bond is not present at the end of the chain, on reduction gives cis or trans alkene, which depends upon the choice of reducing agent. With sodium in liquid ammonia the alkene is trans form and on catalytic reduction the alkene is cis form. Addition of Hydrogen to Alkyne 2. Electrophilic addition: i.Addition of halogens Electrophilic Addition of Chlorine in Alkyne Class XI Chemistry www.vedantu.com 22  The order of reactivity of halogens is Cl2  Br2  I 2 ii. Addition of halogen acid Electrophilic Addition Halogen Acid of Alkyne The order of reactivity of halogen acids is HI > HBr > HCl. In the presence of peroxide, anti Markonikov’s product is obtained. Class XI Chemistry www.vedantu.com 23  Halogenation of Alkyne iii. Addition of hypohalous acids Addition of Hypohalous Acids of Alkyne 3. Nucleophilic addition reaction: In these reactions, the addition is initiated by a nucleophile and are generally catalysed by salt of heavy metals (e.g. Hg 2+ , Pb 2+ , Ba 2+ )which are found to form  compound with multiple bonds. HC  CH + Hg 2+ → CH = CH  Hg 2+ i.Addition of water Class XI Chemistry www.vedantu.com 24  Addition of Water to Alkyne ii. Addition of hydrogen cyanide HC  CH + HCN ⎯⎯⎯⎯ Ba ( CN )2 → CH 2 = CHCN vinyl cyanide iii. Addition of acetic acid Addition of Acetic Acid to Alkyne iv. Addition of alcohol HC  CH + C2 H 5OH ⎯⎯⎯ H 2 SO4 130o C → CH 2 = CHOC2 H 5 ⎯⎯⎯ H 2O → CH 3CHO + C2 H 3OH v. Addition of ozone and ozonolysis Ozonolysis of alkyne 4. Oxidation: i.Oxidation with alkaline KMnO4 HC  CH + 4[O] ⎯⎯⎯⎯ alk. KMnO4 → HOOC − COOH acetylene oxalicacid Class XI Chemistry www.vedantu.com 25  Oxidation of Alkyne ii. Oxidation with acidic K 2Cr2O7 KMnO4 HC  CH + ⎯⎯⎯⎯ K 2Cr2O7 H+ → HOOC − COOH → CH 3COOH acetylene aceticacid Oxidation of Acidic K 2Cr2O7 / KMnO4 5. Formation of metallic derivatives: The group –C≡C–H in alkynes is slightly acidic in nature and hence its hydrogen atom can be easily replaced by certain metals to give metallic derivatives called acetylides or alkynides. Acidic Hydrogen in Alkenes Class XI Chemistry www.vedantu.com 26  i.Formation of sodium acetylides HC  CH + Na ⎯⎯⎯ ⎯ Liq. NH3 → HC  C − Na + Na ⎯⎯⎯ NaNH 2 120o C → NaC = CNa monosodiumacetylide disodiumacetylide Formation of Sodium Propynide ii. Formation of copper and silver acetylides HC  CH + 2CuCl2 + 2 NH 4OH → Cu − C  C − Cu + 2 NH 4Cl + 2 H 2O ammonical cuprouschloride copper acetylide ( red ppt.) HC  CH + 2 AgNO3 + 2 NH 4OH → AgC  CAg  + 2 NH 4 NO3 + 2 H 2O ammonical silver nitrate silver acetylide ( white ppt ) These reactions are used for detecting the presence of acetylenic hydrogen atoms. Illustration 4 : The products obtained via oxymercuration ( HgSO4 + H 2 SO4 ) of 1–butyne would be Oxymercuration Product of 1–Butyne Class XI Chemistry www.vedantu.com 27 Solution: (A) Oxymercuration reaction of 1–butyne Class XI Chemistry www.vedantu.com 28