有机化学英文课件chapter5.ppt
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1、Organic ChemistryUnsaturated Hydrocarbons contains one or more carbon-carbon double or triple bonds contains a carbon-carbon double bond and has the general formula CnH2n contains a carbon-carbon triple bond and has the general formula CnH2n-2 Ethylene(an alkene)H HC CC CH HH HH H Acetylene(an alkyn
2、e).H-CH-CC-HC-HUnsaturated Hydrocarbons benzene and its derivatives(Ch 21-22)we do not study arenes until Chapters 21&22 however,we show structural formulas of compounds containing the phenyl group before that time the phenyl group is not reactive under any of the conditions we describe in Ch 6-20B
3、Be en nz ze en ne eA A l lt te er rn na at ti iv ve e r re ep pr re es se en nt ta at ti io on ns s f fo or r t th he e p ph he en ny yl l g gr ro ou up pC CC CC CC CC CC CC C6 6H H5 5-P Ph h-H HH HH HH HH HH HStructure of AlkenesuA double bond consists of one sigma bond formed by the overlap of sp2
4、 hybrid orbitals and one pi bond formed by the overlap of parallel 2p orbitals the two carbon atoms of a double bond and the four atoms bonded to them lie in a plane,with bond angles of approximately 120Structure of Alkenes it takes approximately 264 kJ(63 kcal)/mol to break the pi bond in ethylene;
5、that is,to rotate one carbon by 90 with respect to the other so that there is no overlap between 2p orbitals on adjacent carbons Isomerism in Alkenes isomers that have the same connectivity but a different arrangement of their atoms in space due to the presence of either a ring or a carbon-carbon do
6、uble bondc ci is s-2 2-B Bu ut te en ne em m p p -1 13 39 9 C C,b bp p 4 4 C Ct tr ra an ns s-2 2-B Bu ut te en ne em m p p -1 10 06 6 C C,b bp p 1 1 C CC CC CC CH H3 3H HH H3 3C CH HC CC CH HC CH H3 3H H3 3C CH HIndex of Hydrogen Deficiency the sum of the number of rings and pi bonds in a moleculeu
7、To determine IHD,compare the number of hydrogens in an unknown compound with the number in a reference hydrocarbon of the same number of carbons and with no rings or pi bonds the molecular formula of the reference hydrocarbon is CnH2n+2Index of Hydrogen Deficiency for each atom of a Group 7 element(
8、F,Cl,Br,I),add one H no correction is necessary for the addition of atoms of Group 6 elements(O,S)to the reference hydrocarbon for each atom of a Group 5 element(N,P),subtract one hydrogenIDH=IDH=2 2(H(Hreferencereference -H -Hmoleculemolecule)Index of Hydrogen Deficiency isopentyl acetate has a mol
9、ecular formula of C7H14O2.Calculate its IHD reference hydrocarbon C7H16 IHD=(16-14)/2=1 calculate the IHD for niacin,molecular formula C6H6N2O reference hydrocarbon C6H16 IHD=(16-6)/2=5OOIsopentyl acetateIsopentyl acetateN NN N H H2 2O ON iacinN iacinIUPAC Nomenclature1.Number the longest chain of c
10、arbon atoms that contains the double bond in the direction that gives the carbons of the double bond the lowest numbers2.Locate the double bond by the number of its first carbon3.Name substituents4.Number the carbon,locate and name substituents,locate the double bond,and name the main chain1 1-H H e
11、 ex xe en ne e4 4-M M e et th hy yl l-1 1-h he ex xe en ne e2 2-E Et th hy yl l-4 4-m m e et th hy yl l-1 1-p pe en nt te en ne e1 12 23 34 45 56 61 12 23 34 45 56 61 12 23 34 45 5Common NamesuDespite the precision and universal acceptance of IUPAC nomenclature,some alkenes,particularly low-molecula
12、r-weight ones,are known almost exclusively by their common namesC C H H2 2=C C H H2 2C C H H3 3C C H H=C C H H2 2C C H H3 3C C=C C H H2 2P Pr ro op py yl l e en ne eE Et th hy yl l e en ne eC C o om m m m o on n:I IU UP PA A C C:2 2-M M e et th hy yl l p pr ro op pe en ne eP Pr ro op pe en ne eE Et
13、th he en ne eI Is so ob bu ut ty yl l e en ne eC C H H3 3Common Names the common names methylene,vinyl,and allyl are often used to show the presence of the following alkenyl groupsCH2=CH2=CH-CH2=CHCH2H2CCH2=CHCH2=CHCH2A A l ll ly yl lM M e et th hy yl li id de en ne ec cy yc cl lo op pe en nt ta an
14、ne e(M M e et th hy yl le en ne ec cy yc cl lo op pe en nt ta an ne e)E Et th he en ny yl l c cy yc cl lo op pe en nt ta an ne e(V V i in ny yl lc cy yc cl lo op pe en nt ta an ne e)V V i in ny yl lM M e et th hy yl le en ne eE Ex xa am m p pl le eC C o om m m m o on n N N a am m e eA A l lk ke en n
15、y yl lG G r ro ou up pM M e et th hy yl li id de en ne eE Et th he en ny yl l3 3-P Pr ro op pe en ny yl l3 3-P Pr ro op pe en ny yl lc cy yc cl lo op pe en nt ta an ne e(A A l ll ly yl lc cy yc cl lo op pe en nt ta an ne e)I IU U P PA A C C N N a am m e e(C C o om m m m o on n n na am m e e)The,Syst
16、emuConfiguration is determined by the orientation of atoms of the main chainC CH HC CC CH H3 3C CH H(C CH H3 3)2 2H H3 3C CC CH HC CC CH H2 2C CH H3 3H HC CH H3 3C CH H2 2c ci is s-3 3,4 4-D D i im m e et th hy yl l-2 2-p pe en nt te en ne e1 12 23 34 4t tr ra an ns s-3 3-H H e ex xe en ne eThe,Syst
17、em uses priority rules(Chapter 3)if groups of higher priority are on the same side,the configuration is Z(German,zusammen)if groups of higher priority are on opposite sides,the configuration is E(German,entgegen)Z Z (z zu us sa am m m m e en n)E E (e en nt tg ge eg ge en n)C Ch hi ig gh he er rC Ch
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