Organic Chemistry Past Paper PDF - Hydrocarbons

ORGANIC CHEMISTRY PRE-MEDICAL NURTURE COURSE Study Material Hydrocarbons English Medium All rights including trademark and copyrights and rights of translation etc. reserved and vested exclusively with ALLEN Career Institute Private Limited. (ALLEN) No part of this work may be copied, reprodu...

ORGANIC CHEMISTRY PRE-MEDICAL NURTURE COURSE Study Material Hydrocarbons English Medium All rights including trademark and copyrights and rights of translation etc. reserved and vested exclusively with ALLEN Career Institute Private Limited. (ALLEN) No part of this work may be copied, reproduced, adapted, abridged or translated, transcribed, transmitted, stored or distributed in any form retrieval system, computer system, photographic or other system or transmitted in any form or by any means whether electronic, magnetic, chemical or manual, mechanical, digital, optical, photocopying, recording or otherwise, or stood in any retrieval system of any nature without the written permission of the Allen Career Institute Private Limited. Any breach will entail legal action and prosecution without further notice. This work is sold/distributed by Allen Career Institute Private Limited subject to the condition and undertaking given by the student that all proprietary rights (under the Trademark Act, 1999 and Copyright Act, 1957) of the work shall be exclusively belong to ALLEN Career Institute Private Limited. Neither the Study Materials and/or Test Series and/or the contents nor any part thereof i.e. work shall be reproduced, modify, re-publish, sub-license, upload on website, broadcast, post, transmit, disseminate, distribute, sell in market, stored in a retrieval system or transmitted in any form or by any means for reproducing or making multiple copies of it. Any person who does any unauthorised act in relation to this work may be liable to criminal prosecution and civil claims for damages. Any violation or infringement of the propriety rights of Allen shall be punishable under Section- 29 & 52 of the Trademark Act, 1999 and under Section- 51, 58 & 63 of the Copyright Act, 1957 and any other Act applicable in India. All disputes are subjected to the exclusive jurisdiction of courts, tribunals and forums at Kota, Rajasthan only. Note:- This publication is meant for educational and learning purposes. All reasonable care and diligence have been taken while editing and printing this publication. ALLEN Career Institute Private Limited shall not hold any responsibility for any error that may have inadvertently crept in. ALLEN Career Institute Private Limited is not responsible for the consequences of any action taken on the basis of this publication.  Chemistry : Hydrocarbons ® Pre-Medical HYDROCARBONS 4.0 ALKANES 4.1.1 Introduction of Alkanes (a) Branched and unbranched aliphatic saturated open chain hydrocarbons are called member of alkanes. (b) CH4 is also known as Marsh gas (fire damp). (c) Calore gas : Mixture of n–butane and isobutane. (d) LPG (Liquefied petroleum gas) : liquid propane, isobutane. (e) Natural gas : 80% methane + 10% ethane + 10% propane + small amounts of H2, N2, CO2 etc. (f) Water gas : CO + H2 (1:1) ® (g) Synthesis gas : CO + 3H2 (1:3) 4.1.2 General Methods of Preparation (1) From alkenes and alkynes (Sabatier and Senderens reaction) or (By hydrogenation of alkenes and alkynes) : Alkenes and alkynes on catalytic hydrogenation give alkanes. Catalyst R—CH=CH—R + H2  → R—CH2—CH2—R Alkene Alkane Catalyst R—C≡C—R + 2H2  → R—CH2—CH2—R Alkyne Catalyst : (a) Pd/Pt at ordinary temperature and pressure (b) Ni, 200–300° C (Sabatier and Senderens reaction) (c) Raney Nickel at room temp. (d) Methane can not be prepared by this method (2) From alkyl Halides (By reduction) : 2H R—X  → R—H + HX (Nascent Hydrogen) Catalyst : (i) Zn + HCl (ii) Zn + CH3 COOH (iii) Zn—Cu couple in C2H5OH (iv) Red P + HI (v) Al – Hg + ethanol GOLDEN KEY POINTS Alkyl halides can also be reduced to alkane by H2/Pd or LiAlH4 or H2/Ni. Halogen atom of alkyl halide is replaced by hydrogen atom to obtain an alkane. 87 ®  Chemistry : Hydrocarbons Pre-Medical (3) From alkyl halide (By Wurtz reaction) : A solution of alkyl halide in ether on heating with sodium gives alkane. Dry R — X + 2Na + X — R  → R — R + 2NaX ether (a) Two moles of alkyl halide are treated with Na in presence of dry ether. If ether is wet then we obtain alcohol. 2Na + H2O → 2NaOH + H2 CH3I + NaOH → CH3OH + NaI Methanol (b) Methane can not be prepared by this method. The alkane produced is higher and symmetrical i.e. it contains double the number of carbon atoms present in the alkyl halide taken. (c) Two different alkyl halides, on wurtz reaction give all possible alkanes. (d) The separation of mixture into individual members is not easy because their B.P. are near to each other and thus wurtz reaction is not suitable for the synthesis of alkanes containing odd ® number of carbon atom. (4) From Frankland Reagent: R—X + 2Zn +RX → R2Zn + ZnX2 Frankland reagent R2Zn + R—X → R—R + RZnX (5) From Carboxylic Acid (By decarboxylation) : Sodium salt of saturated monocarboxylic acid on dry distillation with sodalime give alkane. ∆ RCOONa + NaOH  Cao → R—H + Na2CO3 Note :- Sodalime ⇒ NaOH + CaO BEGINNER'S BOX-1 1. If two moles of Isopropyl chloride reacts with Na in presence of dry ether. Which alkane is obtained ? (1) Hexane (2) 2, 3-Dimethyl butane (3) Isopentane (4) Neopentane 2. If isopropyl chloride and ethyl chloride both react with Na in presence of dry ether which alkanes are obtained ? (1) n-Butane (2) 2-Methyl butane (3) 2, 3-Dimethyl butane (4) All of them 3. Which of the following compound can not be obtained from single alkyl halide by wurtz reaction ? (1) Ethane (2) Butane (3) Isobutane (4) Hexane 4. Give reactivity order for decarboxylation :- (I) CH3—CH2—COOH (II) CH2=CH—COOH (III) CH≡C —COOH (1) I > II > III (2) III > II > I (3) III > I > II (4) None is correct 5. Which of the following reaction can not be used to obtained propane in good yield ? (1) Wurtz reaction (2) Corey-house reaction (3) Decarboxylation of acid salt (4) All of them 6. Decarboxylation of isobutyric acid yields :- (1) Isobutane (2) Propane (3) 2-Methyl propane (4) None of the above 7. When alkyl halide is made to react with magnesium in presence of dry ether, alkyl magnesium halide RMgX is formed. To generate R—H the use is made of the reagent :- (1) RNH2 (2) ROH (3) NH3 (4) Any one of the above 88  Chemistry : Hydrocarbons ® Pre-Medical GOLDEN KEY POINTS The process of elimination of Carbon-di-oxide from Carboxylic acid called decarboxylation. The alkane formed by decarboxylation contains one carbon atom less than the original acid. Decarboxylation of sodium formate gives H2 ∆  HCOONa + NaOH (CaO) → H2 +Na2 CO3     CH COONa + NaOH (CaO) → CH + Na CO  ∆  3 4 2 3 If in a compound two carboxylic groups are present and they are attached to same carbon atom then decarboxylation of one of the carboxylic groups takes place simply on heating. COOH ∆ CH2 → CH3COOH + CO2 COOH ® CH3 —CH2 —CH3 can be prepared by Butanoic acid and 2–Methyl propanoic acid. β-Keto acids are decarboxylated readily simply on heating (soda lime is not required) ∆ R–C–CH2COOH → R–C–CH3 O O (6) From carboxylic acid (By Kolbe's electrolysis process) : Alkanes are formed on electrolysis of concentrated aqueous solution of sodium or potassium salt of saturated monocarboxylic acids. Electrolysis 2RCOONa + 2H2O  → R – R + 2CO2 + 2NaOH + H2      At Anode At Cathode electro. 2C2H5 − COONa  2H O → C2H5 − C2H5 + 2CO2 + 2NaOH + H2 2 Electrolysis of potassium or sodium salt of saturated dicarboxylic acid gives alkene. CH2COONa electrolysis CH2 + 2CO2 + 2NaOH + H2 → CH2COONa CH2 At Cathode Aqueous At Anode CH3–CHCOONa electrolysis CH3–CH + 2CO2 + 2NaOH + H2 → CH3–CHCOONa CH3–CH At Cathode Aqueous At Anode By the electrolysis of aqueous Solution. of sodium or potassium fumarate or maleate, acetylene is formed at anode. CH–COOK Electrolysis CH → + CO2 CH–COOK CH (Aqueous) 89 ®  Chemistry : Hydrocarbons Pre-Medical GOLDEN KEY POINTS Electrolysis of an acid salt gives symmetrical alkane, However in case of a mixture of Carboxylic acid salts, all probable alkanes are formed. Electrolysis R'COOK + R"COOK  → (R'—R" + R'—R' + R"—R") +2CO2 + H2 +2NaOH At anode alkane and CO2 gas is formed while at cathode NaOH and H2 gas is formed. The concentration of NaOH in solution is increased with time so pH of solution is also increased. (7) From alkanol, alkanals, Alkanone and alkanoic acid (By reduction) : The reduction of either of the above compound in presence of red P and HI gives corresponding alkane. Re d P R—OH + 2HI  150° C  → R—H + H2O + I2 ® Re d P R—CHO + 4HI  150° C  → RCH3 + H2O + 2I2 Re d P R—CO—R + 4 HI  150° C  → R—CH2—R + H2O + 2I2 Re d P RCOOH + 6HI  150° C  → R—CH3 + H2O + 3I2 In the above reaction I2 is formed which may react with alkane to form alkyl halide. So red P is added in the reaction to remove I2 formed in the reaction.   R—CH3 + I2  R—CH2—I + HI 2P + 3I2 → 2PI3 (8) From alkanones (By Clemmensen's method) : Carbonyl compound may also be reduced with Zinc amalgam and concentrated HCl (Zn–Hg/HCl), this reaction is called Clemmensen reduction. Zn − Hg R—CO—R'+4H  → R—CH2—R'+H2O con.HCl (9) From alkanals and alkanones (By Wolf Kishner reaction) : (1) NH2NH2 >= C O  − → > CH2 (2) OH /heat (10) From G.R. : (a) Formation of alkanes with same number of C atoms : Grignard reagent reacts with the compounds having active hydrogen to form alkane. R Mg X+H O H → R H + Mg (OH) X +R O H → R H + Mg (OR) X +R NH H → R H + Mg (NHR) X This reaction is used to determine the number of active H-atoms in the compound this is known as Zerewitnoff's method. (b) G.R. react with alkyl halide to give higher alkanes : RMgX + R'—X → R—R' + MgX2 90  Chemistry : Hydrocarbons ® Pre-Medical (11) Corey-House Synthesis : This method is suitable for the preparation of unsymmetrical alkanes i.e. those of type R—R' (i) Li R—X  (ii)CuX → R − R '+ RCu + LiX (iii) R ' − X Note: In Corey-house reaction symmetrical and unsymmetrical alkane both can be formed. (12) From metal carbide (By hydrolysis) : Only CH4 can be obtained by the hydrolysis of Be or Al carbides ∆ Al 4 C3 + 12H2O  → 4Al(OH)3 + 3CH 4 ∆ Be2C + 4H2O  → 2Be(OH)2 + CH4 BEGINNER'S BOX-2 ® 1. Which of the following does not give alkane with R—Mg—X ? (1) Ph—OH (2) C6H6 (3) CH3COOH (4) HCl 2. The reagent in the Clemmensen's reduction is :- (1) LiAlH4 (2) Zn-Hg/HCl (3) Zn/HCl (4) Na/EtOH 4.1.3 Physical & Chemical Properties of alkane Physical properties (i) Solubility : Alkanes being non polar and thus insoluble in water but soluble in nonpolar solvents Ex. C6H6, CCl4 ,ether etc. (ii) Boiling point - ∝ molecular weight (for n-alkanes)  Vander Waals force of attraction ∝ surface area of molecule ∝ molecular weight. i.e. boiling point Pentane < hexane < heptane 1 Also boiling point ∝ number of side chain because the shape approaches to spherical which results in decrease in van der Waal's forces (as surface area decreases) Thus boiling point n–Pentane > Isopentane > neopentane (iii) Melting Point : M.P. of alkanes do not show regular trend. Alkanes with even number of carbon atoms have higher M.P. than their alkanes of odd number of carbon atoms. The abnormal trend in M.P. is due to the fact that alkanes with odd carbon atoms have their carbon atom on the same side of the molecule and in even carbon atom alkane the end Carbon atom on opposite side. Thus alkanes with even carbon atoms are packed closely in crystal lattice to permit greater intermolecular attractions. C C C C C < C C C C C C Odd number of carbon Even number of carbon 91 ®  Chemistry : Hydrocarbons Pre-Medical Chemical Properties (1) Free radical substitution reactions : Substitution reaction in alkanes show free radical mechanism. They give following substitution reaction. (a) Halogenation : Replacement of H-atom by halogen atom R—H + X2 → R—X + HX Halogenation is made on exposure to (halogen + alkane) mixture to UV or at elevated temp. The reactivity order for halogens shows the order. F2 > Cl 2 > Br2 > I2 Reactivity order of hydrogen atom in alkane is Tertiary C – H > Sec. C – H > primary C – H ® (i) Fluorination : Reacts explosively even in dark. Fluorination can be achieved without violence when alkane is treated with F2 diluted with an inert gas like N2. (ii) Chlorination : Cl2 Cl2 Cl2 Cl2 CH4  → CH3 Cl  → CH2 Cl2  → CHCl 3  → CCl 4 The monochloro derivative of alkane is obtained as major product by taking alkane in large excess. When chlorine is in excess then perchloro derivative is obtained as major product. At 12 noon explosively CH 4 + 2Cl 2 → C + 4HCl UV Mechanism for CH4 + Cl2  → CH3Cl + HCl UV Step I Chain initiation step : Cl :Cl  → Cl + Cl or ∆ Step II Chain propagation step : Cl + H CH3 → H Cl + CH3 Methyl radical Methane CH3 + Cl Cl → CH3Cl + Cl Step III Chain termination step : Cl + Cl → Cl2 , CH3 + Cl → CH3 Cl , CH3 + CH3 → CH3 CH3 (iii) Bromination : Br2 reacts with alkanes in a similar manner but less vigorously. (iv) Iodination : Iodine reacts with alkanes reversibly. HI formed as the by product is a powerful reducing agent and is capable of reducing the CH3I to CH4. Iodination may be carried out in the presence of an oxidising agent such as HIO3, HNO3, HgO etc. which decompose HI,  CH4 + I2    CH3 I + HI 5HI + HIO3 → 3I2 + 3H2O Iodination is very slow because energy of activation of the reaction is very large CH 4 + I → HI + CH3 92  Chemistry : Hydrocarbons ® Pre-Medical (b) Nitration : (Vapour phase nitration) This involves the substitution of a hydrogen atom of alkane with –NO2 group. At ordinary temperature, alkanes do not react with HNO3. But reacts with vapours of Conc. HNO3 at ° 450 C and in pressure. 400 − 500° C R – H + HO – NO2  pressure → R – NO2 + H2 O Since the reaction is carried at high temperature and in pressure, so the C—C bonds of alkanes also break during the reaction and a mixture of nitroalkanes is formed. 450 C 0 Ex. CH3—CH3 + HNO3  → CH3CH2NO2 + CH3NO2 + H2O 450°C CH3CH2CH3 + HNO3 1–Nitro propane 2–Nitro propane (major) Nitro ethane Nitromethane (c) Sulphonation : Replacement of H atom of alkane by –SO3H is known sulphonation. ® Alkane react with fuming H2SO4 or oleum (H2S2O7). CH3 CH3 Ex. CH3 C H + HO SO3H → CH3 C SO3H + H2O CH3 CH3 2–Methyl propane 2–Methyl propane–2–sulphonic acid The reactivity order for sulphonation is tert. H > Sec. H > prim. H Note : The reaction is observed in higher alkanes and the alkanes having 3° H. BEGINNER'S BOX-3 1. In the following reaction, the major product is :- CH3 Br → hν 2 CH3 CH3 CH2Br CH3 Br Br (1) (2) (3) (4) Br 2. The bond dissociation energy of C–H bond for the compound :- (I) CH3 − H (II) CH3 – CH2 – H (III) CH2 = CH – CH2 – H (IV) C6 H5 – H (1) I > II > III > IV (2) IV > III > II > I (3) IV > I > II > III (4) II > I > IV > III 3. Arrange the following in correct order of reactivity towards Cl2/hυ – CH3 (A) CH 4 (B) CH3C H3 (C) CH3CH2 CH3 (D) CH3–CH–CH3 (1) A > B > C > D (2) D > C > B > A (3) B > C > A > D (4) C > B > D > A 93 ®  Chemistry : Hydrocarbons Pre-Medical (2) Oxidation : (a) Complete oxidation or combustion : Burn readily with non-luminous flame in presence of air or oxygen to give CO2 and water with evolution of heat. Therefore, alkanes are used as fuels.  3n + 1  C n H2n +2 +   O2 → nCO2 + (n+1) H2O+Q; (∆H =–ve)  2  (b) Incomplete oxidation : In limited supply of air gives carbon black and CO. 2CH4 + 3O2 → 2CO + 4H2O CH4 + O2 → C + 2H2O C–black (used in printing ink) (c) Catalytic oxidation : (i) Alkanes are easily converted to alcohols and aldehydes under controlled catalytic oxidation. Red hot Cu or Fe tube 2CH4 + O2 → 2CH3OH High P and T ® Mo2O3 CH4 + O2  → HCHO + H2O 2300 C,100atm (ii) Alkanes on oxidation in presence of maganese acetate give fatty acids. (CH COO) Mn 2 CH3 (CH2 )n CH3  3 high Temp → CH3 (CH2 )n COOH (iii) Tertiary alkanes are oxidized to give tertiary alcohols by KMnO4. CH3 CH3 [O] CH3–C–H  → CH3–C–OH KMnO4 CH3 CH3 (3) Isomerization: Unbranched chain alkanes on heating with AlCl3 + HCl / 200 C are converted in to 0 branched chain alkanes CH3 AlCl 3 + HCl CH3 − CH2 − CH2 − C H3 → CH3–CH–CH3 n-butane Isobutane Branched chain alkanes converted to more branched alkane. CH3 CH3 CH3 AlCl 3 + HCl CH3–CH–CH2–CH2–CH3 → CH3–CH–CH–CH3 BEGINNER'S BOX-4 1. Which of the following reactions of alkanes involve free radical intermediates ? (1) Halogenation (2) Pyrolysis (3) Nitration (4) All of the above 2. (CH3)3CMgCl on reaction with D2O produces (1) (CH3)3CD (2) (CH3)3OD (3) (CD3)3CD (4) (CD3)3OD 3. The compound with the highest boiling point is– (1) n-hexane (2) n-pentane (3) 2,2-dimethyl propane (4) 2-methylbutane 94  Chemistry : Hydrocarbons ® Pre-Medical 4. Photochemical chlorination of alkane is initiated by a process of :- (1) Pyrolysis (2) Substitution (3) Homolysis (4) Peroxidation 5. Isomerization in alkane may be brought about by using :- (1) Al2O3 (2) Fe2O3 (3) AlCl3 and HCl (4) Concentrated H2SO4 6. Bromination of an alkane as compared to chlorination proceeds :- (1) At a slower rate (2) At a faster rate (3) With equal rates (4) With equal or different rate depends upon the temperature ® (4) Pyrolysis or Cracking or thermal decomposition : When alkanes are heated to 500-700°C they are decomposed in to lower hydrocarbon. This decomposition is called pyrolysis. 1000° C Ex. CH 4  → C + H2 500° C CH3 − CH3 → absence of air CH2 = CH2 + H2 CH2=CH2 + CH4 CH3CH2CH3 CH3–CH=CH2 + H2 Cracking n-Butane  → 1–Butene + 2–Butene + Ethane + Ethene + Propene + CH4 + H2 (5) Aromatization: Unbranched higher alkanes (from 6 to 10 carbon atoms) when heated in presence of oxides of Cr, Mo, V on Al2O3 support at 5000 C aromatic hydrocarbons are formed. Cr O / Al O n − hexane  2 3 → 2 3 + 4H2 500° C CH3 Cr O / Al O CH3 (CH2 )5 − CH3  2 3 → 2 3 + 4H2 500° C n-heptane Toluene 4.2 ALKENES (OLEFINS) 4.2.1 Introduction of Alkenes Alkene are also called olefins (oil forming) since the first member ethylene (C2H4) was found to form an only liquid on reaction with chlorine. CH2=CH2 + Cl2 → Cl—CH2—CH2—Cl 95 ®  Chemistry : Hydrocarbons Pre-Medical 4.2.2 General Methods of Preparation (1) Elimination reactions : These reactions are brought about elimination of small molecule from the substrate Elimination α–elimination β–elimination or 1, 1–elimination or 1, 2–elimination. α-Elimination (1, 1-Elimination) : Removal of H and X from one C-atom KOH Example : CHCl3  → : CCl2 (dichloro carbene) Mechanism : Cl OH   Cl ® H C Cl → –H2O C Cl (acidic H) Cl Cl.. Cl.. –Cl C..Cl → :CCl2  Cl α, β Elimination (β-elimination) : Removal of H and X from adjacent C-atoms E1 unimolecular elimination β E2 bimolecular elimination (a) Unimolecular elimination (E1) :- CH3 − CH2 − OH  443° K  95%H2SO4 → CH2 = CH2 Mechanism of Reaction: The acidic dehydration of alcohol proceeds through the formation of a carbocation intermediate and is explained as follows : Step I : Alcohol being a Lewis base accepts a proton (H+) from the acid in a reversible step as follows: ⊕ ⊕ CH3 CH2 O   H + H  CH3 CH2 O H H Ethanol (From acid) Protonated ethanol Step II : Due to presence of positive charge on electronegative oxygen, its electron accepting tendency increases. As a result C – O bond becomes weak and cleaves as follows : ⊕ ⊕ Slow CH3 CH2 O H → RDS CH3 CH2 + H2O H Ethyl carbocation This is a slow and is regarded as rate determining step in E1 reaction. Step III : Base removes αH (proton) from carbocation and changes it into ethene in a fast step as follows: ⊕ Base H—CH2— CH2  fast → CH2=CH2 Ethene 96  Chemistry : Hydrocarbons ® Pre-Medical Saytzeff rule : When two possible alkenes are obtained by the elimination reaction then that alkene containing maximum number of alkyl group on double bonded C − atoms is called Saytzeffs product and formed as major product. Note : The alkene having less number of alkyl groups on double bonded C-atoms is called Hofmann's product. OH H2 SO4 Example : (i) CH3 CH2 CH CH3  → CH3—CH=CH—CH3 + CH3—CH2—CH=CH2 ∆ 2–butanol main product 1–butene 2–butene 80% 20% (Saytzeff's product) (Hoffmann's product) H2 SO4 (ii) CH3—CH2—CH2—CH2—OH  → CH3—CH=CH—CH3 + CH3CH2CH=CH2 ∆ ® 1–butanol 2–butene80% 1–butene 20% Main product Mechanism : Acid catalyzed dehydration of alkanols proceeds via the formation of more stable carbonium ion. ⊕ ⊕ CH3CH2CH2—CH2— O —H + H → CH3CH2CH2CH2 OH 2 ⊕ ⊕ CH3 CH2 CH2 CH2 OH2 → CH3 CH2 CH2 — CH2 + H2 O Primary Carbonium ion H H H H ⊕ Re arrangement by ⊕ CH3 C C C H → CH3 C C C H 1, 2 hydride ion shift H H H H H H 10 Carbonium ion 0 2 Carbonium ion more stable H H ⊕ CH3 CH CH CH3 Elimination of a proton CH3 C C C H → 2-butene (major product) H H H CH3 CH2 CH CH2 1–butene (minor product) OH (iii) H PO /heat → 3 4 + H2O Cyclohexanol Cyclohexene Reactivity order of acidic dehydration of alcohols is : 3° > 2° > 1° R–OH  Rate of reaction ∝ [substrate]  Molecularity of reaction = 1 (So reaction is called as E1)  In reaction intermediate carbocation is formed, so carbocation rearrangement is possible. 97 ®  Chemistry : Hydrocarbons Pre-Medical (b) Bimolecular elimination (E2) : Example : (i) Dehydrohalogenation of halides by alcoholic KOH/NaNH2 : CH3–CH2–Cl + KOH(alc.) → CH2 = CH2 + KBr + H2O Mechanism : H H HO + H – C – C – H H2C = CH2 H Cl  Rate of reaction α [substrate] [base] ®  Order of reaction = 2 (So reaction is E2)  In E2 reaction intermediate (carbocation) is not formed. So there will be no carbocation rearrangement. Alc. CH3–CH–CH3 → CH2=CH–CH3 KOH Cl Alc. CH3–CH2–CH2–CH2–Cl  → CH3–CH2–CH=CH2 KOH Alc. CH3–CH–CH2–CH3 → CH3–CH=CH–CH3 + CH2=CH–CH2–CH3 KOH Cl (major) (minor) (Saytzeff's product) (Hoffmann's product) (ii) Pyrolysis of tetra alkyl ammonium ion : CH3 H CH3  ⊕ OH CH3–N–CH2–CH2 → CH3–N + CH2=CH2 + H2O CH3 CH3 CH–CH2–CH3 CH3 ⊕ α β  CH2 Ex. CH3–N—–CH–CH OH 2–CH3 → –H2O major product CH3 CH3 β CH=CH–CH3 CH3 minor product 98  Chemistry : Hydrocarbons ® Pre-Medical BEGINNER'S BOX-5 1. Acidic dehydration of alcohol involves :- (1) E1 elimination (2) Carbocation rearrangement if possible (3) Saytzeff's product is formed as major product (4) All 2. Arrange the following in order of their reactivity toward dehydrohalogenation :- Cl Cl Cl Cl I II III IV (1) II > I > III > IV (2) III > II > I > IV ® (3) IV > III > I > II (4) I > II > III > IV 3. Kolbe electrolytic synthesis can be used to obtain all hydrocarbons as a major product except :- (1) 2-Butene (2) Ethyne (3) Methane (4) Ethene (iii) From Alkyl dihalide (By dehalogenation of Vicinal or Gem dihalide) : Removal of X2 from a substrate by Zn dust or Zn—Cu in alcoholic Solution.. (a) From Vicinal dihalide : same number carbon alkene is obtained. X X Zn H–C–C–H → ∆ CH2=CH2 + ZnX2 H H (b) From Gem dihalide : Higher alkene obtained ∆ CH3 CH X 2 + 2Zn + X 2 CHCH3  → CH3 − CH = CH − CH3 + ZnX 2 (2) By the controlled hydrogenation of alkynes : Alkynes can be converted into alkenes as a result of controlled reduction in two ways: (a) By the use of Lindlar's catalyst : Lindlar's catalyst is a mixture of palladium catalyst deposited over barium sulphate or calcium carbonate. The catalytic mixture is slightly poisoned by quinoline or sulphur and allows the reduction or hydrogenation of alkyne with hydrogen only upto the alkene stage. CH3 CH3 Ex. CH3–C≡C–CH3 + H2  Lindlar ' s catalyst → C=C Pd/CaCO3 H H But–2–yne cis–but–2–ene In place of Lindlar's catalyst Nickel-boride (Ni–B also called P-2 catalyst) can also be used. 99 ®  Chemistry : Hydrocarbons Pre-Medical (b) By the action of sodium in liquid ammonia : This is known as Birch reduction and the major product is a trans alkene i.e., the two hydrogen atoms get attached on the opposite side of the double bond. For example, Na/Liquid NH3 CH3 H CH3–C≡C–CH3  → C=C H CH3 But–2–yne trans–but–2–ene 4.2.3 Physical & Chemical Properties of Alkenes Physical Properties (1) All are colourless and have no characteristic odour. Ethene has pleasant smell. (2) The B.P., M.P. and specific gravity show a regular increase with increase in molecular weight ® (3) The increase in branching in carbon chain decreases the B.P. among isomeric alkenes. (4) The B.P. and M.P. of alkenes are slightly higher than the corresponding alkanes because the intermolecular forces of attraction are stronger due to the presence of easily polarizable π bond. (5) Insoluble in water because they can not form H-bond with water molecule, they dissolve freely in organic solvent like benzene, CHCl3, CCl4 etc. Chemical Properties : Alkenes are more reactive than alkane this is because - (a) The π electrons of double bond are located much far from the carbon nuclei and are thus less firmly bound to them. (b) π bond is weaker than σ bond and more easily broken. The reactivity order for alkenes - CH2 =CH2 > R—CH=CH2 > R2C=CH2 ≈ RCH=CHR > R2C=CHR > R2C=CR2 (Trans < Cis) The reactivity order of alkenes has been delt in terms of heat of hydrogenation of alkene, more is the heat of hydrogenation (∆H = –ve), more is the reactivity, the reactivity of alkene is however also related to (i) Steric hinderance (ii) Hyperconjugation Alkenes give the following type of reactions : (A) Addition reaction. (B) Oxidation reaction. (C) Substitution reaction. (D) Polymerization Reaction. (A) Addition Reaction : I. Electrophilic addition reaction :- Because of the presence of >C=C< bond in molecules, alkenes generally take part in the addition reactions. C=C + AB → –C–C– A B Alkene Attacking molecule Addition product (Adduct) 100  Chemistry : Hydrocarbons ® Pre-Medical From mechanism point of view, the addition in alkenes is generally electrophilic in nature which means that attacking reagent which carries the initial attack is an electrophile (E+). This is quite expected also as there is high electron density in the double bond. The mechanism proceeds in two steps. Step I : The π–electron cloud of the double bond causes the polarisation of the attacking molecule (E–Nu) which cleaves to release the electrophile (E+) for the attack. The double bond simultaneously undergoes electromeric effect and the attack by the electrophile is accomplished in slow step (also called rate determining step) to form a carbocation intermediate. E δ+ δ– ⊕ C C E Nu  ( Slow ) → C–C + :Nu–  fast → C–C Rate determining step (RDS) E Nu Addition product Step II : The nucleophile (:Nu ) released in the slow step combines with the carbocation to give the desired – addition product in the fast step. Reactivity for Electrophilic addition reaction ∝ stability of carbocation formed in RDS ® (1) Addition of Halogen : It is a electrophilic addition reaction. X R–CH=CH2 + X2 → R–CH–CH2 X (Vicinal halides) (a) The addition of Br2 on alkenes provides a useful test for unsaturation in molecule. The brown colour of the bromine being rapidly discharged. Thus decolarization of 5% Br2 in CCl4 by a compound suggest unsaturation in it. Colourless dibromo compound is formed. (b) I2 reacts with alkenes to form Vicinal di-iodides which are unstable and I2 gets eliminated to give original alkene. I CH3–CH=CH2 + I2  CH3–CH–CH2 I Unstable δ+ δ– ⊕ Mechanism : Br–Br + CH2=CH2 → (Slow) CH2–CH2 → CH2–CH2 Br ⊕ Br (Halonium ion) – The attack of Br ion on the cyclic cation takes place from the side opposite to side where bromine atom is present in order to minimise steric hindrance. H2C ⊕ Fast Br + Br → Br–H2C H2C CH2–Br 1,2-Dibromoethane (Anti) Br Br2 Eg. CH3–CH–CH=CH2 → CH3–CH–CH–CH2 CH3 CH3 Br Anti addition No carbocation rearrangement takes place, anti addition product is formed. 101 ®  Chemistry : Hydrocarbons Pre-Medical (2) Addition of halogen acid : X R–CH=CH–R + HX → R–CH2–CH–R X R–CH=CH2 + HX → R–CH–CH3 GOLDEN KEY POINTS The order of reactivity of hydrogen halide is : HI > HBr > HCl > HF Addition on alkene proceeds via the formation of more stable carbonium ion. Addition of HX on unsymmetrical alkenes (R—CHCH2) takes place according to Markovnikov's rule. Carbocation rearrangement is observed in the reaction. ® Markovnikov's Rule States : (a) First Rule : When molecule of HX add up on unsymmetrical unsaturated hydrocarbon, the electrophile (H+) goes to the unsaturated carbon atom bearing more number of hydrogen atoms. X CH3–CH=CH2 + HX → CH3–CH–CH2 H Mechanism : It is electrophilic addition and is illustrated by the action of HCl to propene. δ+ Slow ⊕ CH3–CH=CH2 + H–Cl → CH3–CH–CH3 + Cl– Secondary carbocation ⊕ Fast Cl– + CH3–CH–CH3 → CH3–CH–CH3 Cl 2-Chloropropane ⊕ Primary carbocation (CH3—CH2— CH2 ) is also formed but only in very small proportion since it is less stable than the secondary carbocation. Note ; The electrophilic addition to unsymmetrical alkenes always occurs through the formation of a more stable carbocation intermediate. (b) Second Rule : In the addition of HX to vinyl halide and analogous compounds, the halogen attaches itself to the carbon atom, on which the halogen atom is already present. CH2=CH—Cl + HCl → CH3–CH–Cl Cl Ethylidene chloride Cl ⊕ ⊕ Mechanism : CH2=CH–Cl → H Cl CH3–CH–Cl → CH3–CH–Cl 102  Chemistry : Hydrocarbons ® Pre-Medical ⊕ Θ ⊕ Θ ⊕ Θ ⊕ Θ ⊕ Θ All polar reagents of the general structure Y Z (such as H− X, H− OH H− SO3 , X − OH ) add on unsymmetrical unsaturated compound in accordance with Markovnikov's rules. Such additions are called normal Markovnikov's rule, where as additions in the opposite manner are referred to as abnormal or anti Markovnikov's additions. BEGINNER'S BOX-6 1. The intermediate in the Electrophilic addition–reaction is :- (1) Carbocation (2) Carbanion (3) Free radical (4) Carbene CH3 2. +HI —→ major product is ® I CH3 CH3 CH2–I CH3 (1) I (2) (3) (4) I 3. Give reactivity order towards EAR. (i) (ii) OCH3 (iii) Cl (iv) CH3 (1) (i) > (ii) > (iii) > (iv) (2) (iv) > (iii) > (ii) > (i) (3) (ii) > (iv) > (i) > (iii) (4) (iii) > (ii) > (iv) > (i) (3) Addition of Hypohalous acid (or X2/H2O, or HOX) : It is a electrophilic addition and follows Markovnikov's rule, and anti addition. δ− δ+ Slow ⊕ Cl Cl + H2C CH2 → CH2 CH2 → H2C – CH2 ⊕ Cl Cl Carbocation H ⊕ H O OH ⊕ (Fast) –H CH2 CH2 + H O H → CH2 CH2 → CH2 CH2 ⊕ Cl Cl Cl Ethylene chlorohydrin R—C≡CH + HOCl → R—C—CHCl2 O (4) Addition of water (Hydration of alkenes) : Alkene react with water in the presence of acid to form alcohol. This reaction is known as acidic hydration reaction. Intermediate in this reaction is carbocation, so rearrangement may take place. + H (i) CH3–CH=CH2 + H2O → CH3–CH–CH3 OH Propene Propan-2-ol 103 ®  Chemistry : Hydrocarbons Pre-Medical OH + H (ii) CH3–C=CH2 + H2O → CH3–C–CH3 CH3 CH3 2–Methylpropene 2-Methylpropan-2-ol Mechanism : (Slow) ⊕ CH3—CH=CH2 + H+ → CH3–CH–CH3 Carbocation (2°) ⊕ (Fast) + –H CH3–CH–CH3 + H–O–H → CH3–CH–CH3 → CH3–CH–CH3 H—O–H O–H ⊕ Propan-2-ol ® (5) Hydroboration : Borane readily reacts with alkenes giving trialkyl boranes. The reaction is called hydroboration. δ+ δ– T.H.F. R CH CH2 + BH3 → (R CH2 CH2)BH2  R CH CH2 ↓ R–CH=CH2 (R CH2 CH2)3B ← (R CH2 CH2) BH2 Trialkylborane BH3 does not exist or stable as monomer so a solvent THF (tetra hydro furane) is used. δ− δ+ δ− H δ+ THF Ex. 3CH3 CH CH2 + B H  → (CH3—CH2—CH2)3B H BHR2 also can be taken. δ+ δ− Ex. CH3 CH CH2 + BHR2 → CH3—CH2—CH2–BR2 [(Hydroboration reduction (HBR)] + H2O/H 3CH3–CH2–CH3+H3BO3 [(Hydroboration Propane reduction (HBR)] (CH3–CH2–CH2)3B Tripropyl Borane H2O2/OH CH3–CH2–CH2–OH [(Hydroboration Propanol oxidation (HBO)] (1° alcohol) BH3 3R—C≡C—R  THF → (R C C )3B H R δ– δ+ δ– H δ+ δ– R C C R+B H → R CH C BH2 – Hδ R R CH C BH2 + 2R C C R → (R CH C ) 3B R R 104  Chemistry : Hydrocarbons ® Pre-Medical H2O2/(OH ) oxidation Basic R C C OH + H3BO3 → R CH2 C R H R O (enol form) (Ketones) (R C C )3B H R R R + H / H2O C C + H3BO3 H H (cis-Alkene) Note : 1) The overall process of HBO appears to be addition of water according to anti Markovnikov's rule Θ and involves syn. addition. In this form BH3 and OH come from H2O2 / OH. 2) The overall process of HBR appears to be addition of two H on C=C. In this H come from BH3 + link with C having fewer number of H and H come from H2O/H link with C having greater ® number of H. It is also syn addition. (6) Oxymercuration – demercuration : Mercuric acetate in water is treated with an alkene. The addition product on reduction with sodium Boro hydride in aqueous NaOH Solution gives alcohol. It follows the Markovnikov's rule. CH3—CH=CH2 → CH3–CH–CH3 OH (i) (AcO)2 Hg/H2O (Mercuric acetate) or (CH3COO)2 Hg/H2O (ii) NaBH4 Mechanism : CH3–COO Hg H2O  – + → CH3–COO + CH3–COOHg (Electrophile) CH3–COO + + CH3 CH CH2 + HgOOCCH3 → CH3 CH CH2 HgOOCCH3 ⊕ CH3 CH CH2 H2O H O H + OH —H ⊕ → → HgOOCCH3 CH3 CH CH2 CH3 CH CH2 (cyclic cation) HgOOCCH3 HgOOCCH3 (Oxymercuration) OH NaBH4 CH3 CH CH2 + CH3COO + Hg ← H (Product) Note : Intermediate is cyclic cation so there is no rearrangement. 105 ®  Chemistry : Hydrocarbons Pre-Medical OH + H /H2O CH3 C CH2 CH3 with rearrangement markownikoff's rule CH3 OH CH3 CH CH CH2 → (i) BH3/THF CH3 CH CH CH2 without rearrangement CH3 (ii) H2O2/OH anti-markownikoff's rule CH3 H Cl HCl CH3 C CH2 CH3 with rearrangememt markonikoff's rule CH3 OH (i) (AcO)2Hg/H2O CH3 CH CH CH2 without rearrangement (ii) NaBH4 markownikoff's rule ® CH3 H HBr Peroxide CH3 CH CH CH2 without rearrangement anti - markownikoff's rule CH3 H Br BEGINNER'S BOX-7 1. What is the product formed when acetylene reacts with excess hypochlorous acid ? (1) CH3COCl (2) ClCH2CHO (3) Cl2CHCHO (4) ClCH2COOH 2. Primary alcohol can be formed as major product by CH3 CH3 (1) BH3 ,THF dil. H2 SO4 (1) CH3–C=CH2  Θ  → (2) CH3–C=CH2  → (2) H2 O2 / OH CH3 (1) (CH3 COO)2 Hg, H2 O (3) CH3–C=CH2 (2) NaBH4 → (4) 2 & 3 both II. Free radical addition reactions :- Addition of HBr on alkene or alkyne in presence of peroxide. HBr ( A ) CH3–CH=CH2  ROOR → CH3–CH–CH2 H Br Anti Markovnikov's rule or peroxide effect or Kharasch rule (i) In the presence of peroxides the addition of HBr on unsaturated unsymmetrical compound takes place contrary to Markovnikov's rule. This is called peroxide effect and is due to the difference in the mechanism of the addition. (ii) In the normal Markovnikov's addition the mechanism is ionic. (iii) In the presence of peroxide the addition of HBr takes place via free radicals. (iv) Peroxide effect is not observed in case of H–F, HCl and HI. Reactions follows electrophilic addition mechanism. 106  Chemistry : Hydrocarbons ® Pre-Medical HBr CH3 CH CH3 Markownikoff's addition. Br CH3 CH CH2 Isopropyl bromide HBr CH3 CH2—CH2—Br Anti Markownikoff's addition R O O R n–Propyl bromide Mechanism : (i) Chain initiation - (a) R—O—O—R → 2RO (b) HBr + RO → ROH + Br (ii) Chain propagation ®   HBr CH3 CH CH2Br → CH3CH2CH2Br + Br  2° free radical more stable (major) CH3 CH CH2 + Br Br Br   HBr CH3 CH CH2 → CH3CHCH3 + Br 1° free radical less stable (iii) Chain termination : CH3 − CH− CH2 − Br + Br → CH3 − CH(Br) − CH2 (Br) CH3 – CH – CH2 – Br + R – CH – CH2 – Br → CH3 – CH – CH2 – Br CH3 – CH – CH2 – Br Br + Br → Br—Br Question : CH3–CH=CH2  HCl ROOR → CH3–CH–CH3 Cl Ans. no effect simple EAR GOLDEN KEY POINTS Addition of dil. H2SO4 (Hydration in alkyne) : The addition of water takes place in the presence of Hg+2 and H2SO4. OH O 2+ Hg Tautomerism HC≡CH → CH2=CH → CH3–C–H dil H2SO4 OH O Hg2+ Tautomerism CH3–C≡CH → CH2=C–CH3 → CH3–C–CH3 dil H2SO4 III. Addition of H2 : Ni,Pt or Pd R—CH=CH2+H2  → R—CH2—CH3 + Heat of Hydrogenation. (a) Reaction is exothermic, Heat released in reaction is called heat of hydrogenation. 1 1 (b) Stability of alkene ∝ ∝ heat of hydrogenation reactivity of alkene with H2 (c) The process is used to obtain vegetable (saturated fats) ghee from hydrogenation of oil. 107 ®  Chemistry : Hydrocarbons Pre-Medical (B) Oxidation Reaction : Alkenes are easily oxidised by oxidising agents. Oxidising agents attack on double bond and product formed during oxidation depends on oxidising agents. (1) Combustion: 3n CnH2n + O2 → nCO2 + nH2O 2 3n One mole of alkene requires moles of O2 for complete combustion. 2 (2) Ozonolysis : (A test for unsaturation in molecule) (i) The addition of ozone on the double bonds and subsequent a reductive hydrolysis of the ozonide formed is termed as ozonolysis. (ii) Ozonides are explosive compound. (iii) On warming with Zn and H2O, ozonides cleave at the site of the double bond, the products ® are carbonyl compound (aldehyde or ketone) depending on the nature of the alkene. Ex. CH3–C=CH–CH3 → Ozonolysis CH3–C=O + CH3CHO CH3 CH3 (iv) Ozonolysis of alkenes helps in locating the position of double bond in an alkene. It can be achieved by joining together the carbon atoms of the two carbonyl compounds formed as the products of ozonolysis with double bond. H H 1 2 3 4 Ex. CH3–C=O + O=C–CH3 → CH3–CH=CH–CH3 Ethanal But-2-ene H H 1 2 3 4 Ex. H–C=O + O=C–CH2–CH3 → CH2=CH–CH2–CH3 Methanal Propanal But-1-ene CH CH3 CH3 CH3 Ex. CH3–C=O + O=C–CH3 → CH3–C = C–CH3 Propanone 2,3-Dimethyl but-2-ene It may be noted that reaction with bromine water or Baeyer's reagent detects the presence of double bond (or unsaturation) in an alkene while ozonolysis helps in locating the position of the double bond. 108  Chemistry : Hydrocarbons ® Pre-Medical (3) Hydroxylation : Oxidation of carbon-carbon double bond to –C—C– is known as hydroxylation. OH OH (a) Oxidation by Baeyer's reagent (A test for unsaturation) : Alkenes on passing through dilute alkaline 1% cold KMnO4 (i.e., Baeyer's reagent) decolourise the pink colour of KMnO4 and gives brown ppt of MnO2. The reaction involves syn addition. – OH C=C + H2O + [O] → KMnO C—C Glycol 4 (syn-addition) OH OH (b) By OsO4 : R–CH R–CH–O O R–CH–OH + OsO4 → Os HO → 2 ® R–CH R–CH–O O R–CH–OH syn-addition O OH ⊕ –HCOOH H2O/H (c) By peracid : > C=C < + H–C–O–O–H → > C–C < → > C–C < O HO (Anti-addition)(glycol) (d) By Ag2O/∆ : (a) Alkenes reacts with oxygen in the presence of Ag catalyst at 250°–400° C to form epoxide. O OH ∆ H2O/H⊕ CH2=CH2 + Ag2O → CH2–CH2 → CH2–CH2 (anti addition) OH R–CH RCH=CH2 + C6H5COOOH → O + C6H5COOH CH2 Epoxide (4) Oxidation by strong oxidising agent (Oxidative cleavage): The alkenes themselves are readily oxidised to acid or ketone by means of acid permagnate. If HCOOH is formed, it further oxidized to CO2 and H2O. Keep it in mind that no further oxidation of ketones will takes place. 2[O] CH2=CH2+4[O] → 2HCOOH  → 2CO2 + H2O 5[O] CH3CH=CH2  → CH3COOH+CO2+H2O 4[O] CH3CH=CHCH3  → 2CH3COOH CH3 CH3 4[O] C=CH2 → C=O + CO2 + H2O CH3 CH3 109 ®  Chemistry : Hydrocarbons Pre-Medical BEGINNER'S BOX-8 1. The treatment of (CH3)2 C=CHCH3 with boiling KMnO4 produces (1) CH3COCH3 + CH3COOH (2) CH3COCH3 + CH3CHO (3) CH3CHO + CO2 (4) CH3COCH3 only 2. An alkene on treating with hot acidified KMnO4 gives 4 - oxopentanoic acid. The alkene is (1) Pentene (2) 2–Pentene (3) 1–Methyl cyclobutene (4) 1, 2– Dimethyl cyclopropene 3. The addition of HCl to 3, 3, 3–trichloropropene gives (1) Cl3CCH2CH2Cl (2) Cl3CCH(Cl)CH3 (3) Cl2CHCH(Cl)CH2Cl (4) Cl2CHCH2CHCl2 (C) Substitution Reaction (Allylic Substitution) : When alkenes are treated with low concentration of Cl2 or Br2 at high temperature or with ® NBS/hν one of their allylic hydrogen is replaced by halogen atom. Allylic position is the carbon adjacent to one of the unsaturated carbon atoms. It is free radical substitution. 0 500 C CH3—CH=CH2 + Cl2  → ClCH2—CH=CH2 + HCl Allyl chloride (3-Chloro-1-propene) N-Bromosuccinimide (NBS) is an important reagent used for allylic bromination and benzylic substitution. O O CH3–CH=CH2 + CH2–C hv CH2–C NBr → NH + Br–CH2–CH=CH2 CH2–C CH2–C (NBS) O O Substitution reaction is not given by ethene. CH3 CH2–Br Ex. NBS → hv (D) Polymerization: (i) Two or more than two molecules of same compound unit with each other to form a long chain molecule with same empirical formula. This long chain molecule having repeating structural units called polymer, and the starting simple molecule as monomer and process is called addition polymerization. (ii) Molecular weight of polymer is simple multiple of monomer. (iii) Polymerization can be carried out by free radical or ionic mechanism. (iv) The presence of oxygen initiates free radical mechanism. (v) Addition polymerization can also be carried out by ionic mechanism by using Ziegler - Natta Catalysts (R3Al+TiCl4) Ex. nCH2=CH2 → (–CH2–CH2–)n ethene Poly ethene used in the manufacture of insulating Coating, sheeting and moulded products. nCH3–CH=CH2  R Al + TiCl 3 → (–CH–CH2–)n 4 CH3 Polypropene or Koylene (Plastic) 110  Chemistry : Hydrocarbons ® Pre-Medical 4.3 ALKYNES 4.3.1 Introduction of Alkynes Alkynes are unsaturated hydrocarbons and characterised by the presence of a triple bond between the two carbon atoms (C≡C). The carbon-carbon triple bond is also called acetylenic bond. It consists of a strong σ and two weak π bonds. Alkynes are isomers of alkadienes and cycloalkenes. 4.3.2 General Methods of Preparation (1) From Gem dihalides (by dehydrohalogenation) : Dehydrohalogenation agents are : NaNH2 (Sodamide) or Alc. KOH or ROH + RONa. H X X alc. KOH NaNH R–C–CH → –HX R–C=CH → –HX R–C≡CH + NaX + NH3 2 H X H ® (Vinyl halide) (2) From Vicinal dihalides (by dehydrohalogenation) : H H H alc. KOH NaNH R–C–C–H →–HX R–C=C–H → –HX R–C≡CH 2 X X X (a) Elimination of Vic. dihalides gives also alkadiene (1, 2 and 1, 3 alkadienes) but the major product is alkyne. H H HH Ex. CH–C–C–CH → NaNH CH2=CH–CH=CH2 + CH2=C=CH–CH3 + CH3 − C ≡ C – CH3 2 (Major) H X X H (3) Dehalogenation of tetrahalo alkane : By heating 1, 1, 2, 2 - tetra halo alkane with Zn dust. X X 2Zn R–C–C–H → R–C≡CH + 2ZnX2 X X (4) Preparation of higher alkynes by Grignard reagent : By this method lower alkyne is converted in to higher alkyne δ– δ+ δ– δ+ C–MgBr Br R–I CH≡C–H + CH3–Mg–Br → + CH4 → R–C≡CH + Mg CH I C–MgBr Br R–C≡CH + CH3Mg–Br → + CH4  R' I → R’–C≡C–R + Mg