PowerPoint PresentationSlide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 26Slide 27Slide 28Slide 29Slide 30Slide 31Slide 32Slide 33Slide 34Slide 35Slide 36Slide 37Slide 38Slide 39Slide 40Slide 41Slide 42Slide 43Slide 44Slide 45Slide 46Slide 47Slide 48Slide 49Slide 50Slide 51Slide 52Slide 53Slide 54Slide 55Slide 56Slide 5722.53)a)OOHOHSOCl2OClOHb)OClNHexcessONc)d)CNCH3CH2CH2MgBrH2OOCH2CH2CH3OOHOHOOe)HN OH2ONaOHNH2OOf)OOOH3O+OOOHOHg)h)BrNaCNH2O, NaOHOOOOHSOCl2CH3CH2CH2CH2NH2LiAlH4H2OHNi)j)CNH3O+COOHOOHOHOheatOOOk)l)OOONH2HNOOOH2ONaOHOO22.54)BrNaCNCNH3O+OHOSOCl2OCl(CH3)2CuLiH2OOCNLiAlH4H2ONH2(CH3CO)2ONHOOHOCH3OHH+OODIBAL-HH2OOCH3LiH2OOHPCCOOOLiAlH4H2OOHTsOHpyridineOTsNaCNCNCH3MgBrH2OOOBrLi2CO3LiBrDMF23.47)a)Ob)O OOHO OHheatOO OHc)d)OOCH2CH3LDACH3CH2IOOCH2CH3OBr(CH3)2CHNH2OHNe)f)OLDACH3CH2IOOBr2, CH3COOHLi2CO3, LiBr, DMFOOBrLi2CO3LiBrDMF23.47)a)Ob)O OOHO OHheatOO OHc)d)OOCH2CH3LDACH3CH2IOOCH2CH3OBr(CH3)2CHNH2OHNe)f)OLDACH3CH2IOOBr2, CH3COOHLi2CO3, LiBr, DMFOg)h)OI2 (exess)NaOHOOCHI3Cl CNNaHCNi)j)OBr2 excessNaOHOBr BrOClNaIOIO OOCH2CH3H3CH2CONaOCH2CH3O OOCH2CH3H3CH2COCH3CH2CH2CH2CH2BrO OOCH2CH3H3CH2COH3O+O OOHHOO OOHHOheatOHOO OOCH2CH3NaOCH2CH3O OOCH2CH3(CH3)2CHCH2CH2CH2BrO OOCH2CH3H3O+O OOHO OOHheatOPhOPhNaOCH2CH3PhOPhOPhOPhOCH3CH2OHPhOPhOOONaOHOOOOOOH2OOOO+ OHOOOOOOOOOH2OOOOH+ OHOO+ H2O1) Which of the following alkanes will give more than one monochlorination product when treated with chlorine and light?a) 2,2’-dimethylpropane, b) cyclopropnae, c) ethane, d) 2,3-dimethylbutane Cl2hvCl2hvCH3H3CCl2hvCl2hvClClH2CH3CClClCl2) Which type of halogenation is most selective? Bromination is most selective, always occurring at the site of the most stable radical.3) Which of the following absorbs at the highest frequency? a) 1,3,5-hexatriene, b) 1,3,5,7-octatetraene, c) 1,7-diphenyl-1,3,5-heptatiene, d) 1,6-diphenyl-1,3,5-heptatriene4) Draw 3,5-difluoroanisole. FFO5) Circle each of the following compounds that is aromatic.4n + 2 = 10 pi electrons4n + 2 = 2 pi electrons6) Which of the following compounds will undergo Fredal-Crafts alkylation? a) benzoic acid, b) nitrobenzene, c) aniline, d) tolueneCH37) Which of the following compounds is most acidic? OOHH2NOOHO2NOOHOOHOOHOO2N8) Which of the following gives a secondary alcohol when treated with methyl grignard?a) butyl formate, b) 3-pentanone, c)pentanal, d) methyl butanoate OOHCH3MgBrH2OOCH3MgBrH2OOHCH3MgBrH2OOHOHOHOHOOCH3MgBrH2OOHO9) Draw N-ethyl-N-propylformamide. NOH10) What does a positive iodoform test tell you is present It indicates the presence of a methyl ketone.OOClOONaOHH2O11)OOOOOO12)OOOOH3O+H2OHOOOHO13)14)OOHONaBH4, MeOHH+, H2O, heatOI2NaOHOOOO+ CHI315)16)OBr2aq. NaOHOBrBrBrOHClH2OO17)18)OONaOHheatOOO+OONaOCH2CH3H+, H2O, heatO19)20)OCH3O ONaOCH2CH3H3O+OOO+OH3COOOCH3NaOCH2CH3CH3CH2OHOOOCH3OH3COO OOCH2CH3NaOCH2CH3CHCCH2BrO OOCH2CH3OH3O+heatHOCH2CH2OHTsOHOONaOHCH3IOOOOH2Pd-CH3O+OOBr2CH3COOHOLi2CO3LiBrDMFOBrNaOCH2CH3O OOCH2CH3H2OOOOH3CH2COOOH3O+heatOOOCH2CH3+OOO1) NaOEt, EtOH2) H+, H2O, heatOOOCH2CH3HOHOOOCH2CH3OOOOCH2CH3OOH HOOOCH2CH3OOOOCH2CH3OH OH2OHOOCH2CH3OH2OOHOOCH2CH3OOHHH2OOHOOCH2CH3OOHOHOOCH2CH3OOHH3O+OHOOCH2CH3OOHHOHOOOHH2OOOOOHOHOOOOOMolecular ion peak is a single peak at 164, so no halogens or nitrogen.164/12 = 13 x 12 =156164 – 156 = 8 hydrogensC13H82(13) + 2 – 8 = 20/2 = 10C12H20 2(12) + 2 – 20 = 6/2 = 3C11H16O 2(11) + 2 – 16 = 8/2 = 4C10H12O22(10) + 2 – 12 = 10/2 = 5Looking at this you have sp2 and sp3 hydrogens as well at a carbonyl.You have 5 aromatic protons, so most likely a benzene ring is present, so that takes 4 of your five degrees of unsaturation. And the final degree is taken by the carbonyl.8 different types of carbon, 4 of those are taken up by a monosubstituted benzene ring. You know this because of the 4 peaks in the 120 to 140 range. And the fact that there are 5 aromati protons.So what is the substituents. We have 4 carbons and 2 oxygens left, plus 7 hydrogens. You know there is a carbonyl so that is one carbon plus one oxygen. What is the other oxygen?So it could either be an ether, an alcohol, or part of an ester. We know it isn’t an alcohol b/c there is no large peak around 3400 in the IR.So ester or ether?This is not an ether b/c at least one of the remaining 3 carbons would be a singlet somewhere up field and that isn’t present. So it must be an ester.OOOOOOOOOOOOSo here are all the possible esters, which one is it?For the nonaromatic protons we see a triplet a multiplet and a quartet, The multiplet tells you that the three carbons are in a group
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