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Alkanes ALKANES General Methods of Preparation

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Alkanes ALKANES General Methods of Preparation Organic Chemistry : Alkanes ALKANES General Methods of Preparation: 1. Wurtz Reaction: * It is reaction of alkyl halides with sodium metal in anhydrous ether(Et2O) solvent to form alkane with double the number fo carbon atoms. * It is coupling between two R–(alkyl groups) from two R–X molecules. Hence it is best suitable to prepare symmetrical alkanes with even number of carbon atoms (R–R). * Unsymmetrical alkanes can be prepared by taking two different alkyl halides RX and R′X(Cross Wurtz Reaction), but there will be three products R–R′(cross product), R–R and R′–R′ (self coupling products) in the ratio 2 : 1 : 1 and the separation of the components will be uneconomical. For unsymmetrical alkanes, the best method available now is Corey-House synthesis(to be discussed later). * Order of reactivity among alkyl halides: 10 >> 20 >>>> 30 Tert-halides are not used to prepare alkane by this method, as we shall see in sometime that the major product in that case will be alkene containing same number of carbon atoms, not alkane with double the number of carbon atoms. * Methane cannot be prepared by this method. ether(anhy.) RX + 2Na + XR R R + 2 NaX alkane ether(anhy.) CH3 I + 2Na + ICH3 CH3 CH3 + 2 NaI methyl iodide ethane CH CH3 CH3 CH3 3 Et2O CH3 CH Br + 2Na + Br CH CH3 CH3 CH CH CH3 2-bromopropane 2,3-dimethylbutane (isopropyl bromide) Mechanism: 0 Wurtz coupling is SN2 reaction and as you know for 3 halide, E2 elimination completely 0 0 takes over SN2. So 3 halides are not used for Wurtz reaction. In fact 1 halides are best candidates for Wurtz coupling, as 20 halides behave closer to 30 halides in reactions. Step-I: + - RX+ Na R + Na Cl Step-II : R + Na R + Na carbanion Step-III(S 2): N R + RX R R + X Dr. S. S. Tripathy 1 Organic Chemistry : Alkanes In the first step, alkyl free radical is formed, and in the second step alkyl carbanion is formed. That is why 2 moles of sodium are needed. In the final step, the carbanion acts as a nucleophile to bring about SN2 reaction with another alkyl halide molecule. Side Reaction(Elimination) You know that elimination and substituion go hand in hand. For 10 halide, the elimination is the minor product. Hence some unwanted side products are always formed in Wurtz reaction along with alkane(R–R) in small quantities. For 30 halide, the elimination is the exclusive product, 0 hence not used. For 2 halide, we get both SN2 and E2 products with appreciable quantities. 2Na 2CH3 CH2 Br CH3 CH2 CH2 CH3 + CH3 CH3 + CH2 CH2 ethyl bromide ether butane (major product) elimination products (minor) When ethyl bromide is subjected to Wurtz coupling, we get butane as the major product(>90%), and also we get a mixture of ethane and ethene as byeproducts. The latter mixture is due to the elimination reaction which always go together with substitution reaction. H E2 CH3 CH2 + CH2 CH2 CH3 CH3 + CH2 CH2 ethane ethene X For 20 halide, the elimination products will be quite appreciable and for 30 halide, there is only the elimination product. Moreover, the carbanion formation in a 30 halide is very less probable as it is the least stable carbanion. Precaution: Why anydrous ether is used ? If a trace of water would be there then all the carbanions will be destroyed by accepting protons from H2O.The SN2 mechanism will not occur to give alkane. Wurtz reaction of dihalides: 3,4,5,6-membered cycloalkanes can be prepared by the reaction of terminal dihalides with Na metal in ether. 1,3-dibromopropane reacts with Na metal in ether to form cyclopropane. ether + 2 NaBr Br Br 2Na 2 Na cyclopropane 1,3-dibrompropane Cross-Wurtz Reaction: If two different alkyl halides are used(RX and R′X), then we can get cross-product of R– R′ along with homo coupling products R–R and R′–R′ in the ration 50 : 25 : 25. Hence is not a convenient method to prepare R-R′ Ether CH3 Cl + 2 Na + Cl CH2 CH3 CH3 CH2 CH3 propane(50%) cross product Ether CH3 CH2 Cl + 2 Na + Cl CH2 CH3 CH3 CH2 CH2 CH3 butane(25%) 2 Dr. S. S. Tripathy Organic Chemistry : Alkanes Ether CH3 Cl + 2 Na + Cl CH3 CH3 CH3 ethane(25%) So for getting propane, if we take mixture of methyl chloride and ethyl chloride, you will get the cross-product, propane(R-R′) to the extent of 50% statistically and homo coupling products R-R and R′-R′ i.e n-butane and ethane to the extents of 25% each. The separation of these alkanes from each other is really inconvenient and uneconomical. Hence corss Wurtz reaction is not popular to get an unsymmetrical alkane with odd number of carbon atoms. Frankland Reaction: If Zn metal is used in stead of Na metal, it is called Frankaland reaction of preparing alkane. However the yield of alkane in this reaction is poor, though the bye products are minimum. Ether R X + Zn + X R R R + ZnX2 The preparation of R-R is similar to preparation of alkane from Grignard reagent(to be discussed later). (2) Corey-House Synthesis: * This is the most convenient method of preparing an symmetrical as well as unsymmetrical alkane in three steps. * Two different alkyl halides R–X and R′–X can be used in this method and the cross product R–R′ is obtained in pure condition unlike Wurtz Reaction. * It involves three steps. (a) Formation of Alkyl Lithium from 1st Alkyl Halide: The first alkyl halide reacts with Li metal in presence of ether solvent(anydrous) to form alkyl litium which is an organomatellac compound. Ether R X + 2 Li R Li + LiX The first alkyl hilde(R–X) can be 10, 20, 30 . (b) Preparation of Gilman Regent (Lithium dilalkyl cuprate) Alkyl lithium formed in the first step reacts with cuprous halide to form Lithium dialkyl cuprate, called Gilman reagent. 2RLi + CuX R2Cu Li lithium dialkyl cuprate(Gilman Reagen (c) Gilman reagent is then treated with the same(R-X) or different alkyl halide(R′- X) to form a pure homo coupling product or a cross-coupling product respectively. R2Cu Li + R' X R R' + R'Cu + LiX The alkane R–R′ is distilled out leaving behind the organo copper residue(RCu) along with LiX. * The 2nd halide can be 10 and 20 alkyl, alkenyl(liky allyl, vinyl etc), alkynyl, benzyl, aryl(phenyl), cycloalkyl halides. Only 30 halides cannot be used as 2nd halide, although it can be used as the first halide. Example: Devise a Corey-House synthetic scheme for the prepaation of 2,2,3-trimethylbutane. CH3 CH3 CH3 C CH CH3 CH3 Dr. S. S. Tripathy 3 Organic Chemistry : Alkanes We have to take tert-butyl halide as first halide and isopropyl halide as second halide, as we cannot take the former as 2nd halide. CH3 CH3 CH3 (1) Li/ether CH3 CH Cl 2 CH3 C Cl CH3 C Cu Li (2) CuCl CH3 CH3 2 lithium di-tert-butyl cuprate CH3 CH3 CH3 CH C Cu LiCl CH3 C CH + 3 + CH 3 CH CH3 3 (We have not shown in three steps and also not balanced the equation) (2) Br CH3 CH3 CH3 (1) Li/ether CC 2 CH3 C Cl CH3 C Cu Li H H (2) CuCl Z-alkene CH3 CH3 2 lithium di-tert-butyl cuprate C(CH3)3 CH3 CC + (CH3)3CCu + LiCl H H Z-alkene In the 2nd example, you find that the stereochemistry of the 2nd halide is retained in the product. Note that although, we get an alkene here(not alkane), but Correy House synthesis is verstatile to the extent of coupling an alkyl group with alkenyl group(or with alkynyl, benzyl, cycloalkyl group etc.) Mechanism: R R - LiX Cu Li + R' X Cu R' R R' + RCu R R Gilman Reagent Dialkyl cuprate part is anionic(with R–Cu–R covalent bonds). SN2 reaction takes place between – ′ ′ R2Cu with R –X to form R2CuR (trialky copper). This intermediate rearranges with a shift of R′ from Cu to R to form R-R′ and leaving behind the residue R-Cu. Some authors use fish hook arrow to show one-electron shift with radical transfer. Since there is no other intermediate formed in this reaction to corroborate the process, i chose a two-electron shift mechanism for the process. (3) Hydrogenation of Alkenes and Alkynes: * Alkenes react with one mole of H2 in presence of cataysts like Ni(Raney Ni) at high 0 temperature (200 - 300 C)to form alkanes. Pd (Pd-C), Pt (PtO2) catalysts can be used at lower temeprature and at lower pressure of H2 gas and hence are more efficient. Similarly alkynes react with one mole of H2 under similar catalytic conditions to give alkene and two moles of H2 to form alkane. * It is a case of heterogenous catalysis, in which both alkene/alkyne and H2 get adsorbed on the surface of the catalyst to form the product. 4 Dr. S. S. Tripathy Organic Chemistry : Alkanes * The mechanism of addition is SYN ,i.e both the H atoms join on the same side of the multiple bond. Refer the detailed mechanism in GOC-part-III. * It is called Sabatier and Senderen’s reduction. Ni/ 3000C R CH CH R' + H2 RCH2 CH2 R' alkene alkane H Me H Et Me Et H Me Et H H H Me Me Et Me Et Et H H Z(cis)-alkene Mes o product H Me H Et Me Et H Me Et H H + H Et Et Me Et Me Me H H E(trans)-alkene d/l pair In the above example, where stereochemistry comes into picture, hydrogenation produces two chiral centres.
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