1559r.eh17_304-3211/3/0510:48Page309 Soutionsto Problems39 (b)C NMR:One C-O(=193.2)and two C-C groups (=129.0.135.2.146.7.and 152.5)this time HNMR:The carbonyl group is an aldehyde9.56 for-C-H.At the other end.we have CHy-CH2-CH2- 094482 一groups. CH,CH.CH.CH-CHCH-CHCH 23.Each is a conjugated carbonyl compound.giving an intense UV absorption with200nm.The first spectrum matches 人CH with r→◆abs0 at 308 om Th absorption of the more extended conjugated system. 24.(a)MS:Mof 128 confirms that CH is the mol 16+18:degrees of unsaturation I bond or ring present 0 CH3-CH2-.CHy C-CH2-CH2- 0.90 2.0s) 2.240 MS:Base peak (m/43)is CHc:next largest is m/58.consistent with McLafferty rearrangement as follows: H CH,CH2CH2-CH OH 1+ CH:CH2CH-CH-CH2+ CH2=CCH: CH2 m/收58 This answer seems quite reasonable
(b) 13C NMR: One CPO ( 193.2) and two CPC groups ( 129.0, 135.2, 146.7, and 152.5) this time. O B 1 H NMR: The carbonyl group is an aldehyde 9.56 for OCOH. At the other end, we have CH3OCH2OCH2O h hh 0.94 1.48 2.21 This adds up to C4H8O, leaving C4H4 to account for. All four of these H’s are alkene hydrogens ( 5.8–7.1), so these could most simply be two OCHPCHO groups. The result: O B CH3CH2CH2CHPCHCHPCHCH 23. Each is a conjugated carbonyl compound, giving an intense UV absorption with max 200 nm. The first spectrum matches with n * absorption at 232 nm, and carbonyl n n * absorption at 308 nm. The second spectrum matches the diene-aldehyde, with the longer wavelength 272 nm band corresponding to the n * absorption of the more extended conjugated system. 24. (a) MS: M• of 128 confirms that C8H16O is the molecular formula as well Hsat 16 2 18; degrees of unsaturation (18 2 16) 1 bond or ring present IR, UV: A ketone CPO appears to be present NMR: are likely pieces, adding up to C6H12O; only C2H4 are left to add in. Is 2-octanone a reasonable answer? MS: Base peak (m/z 43) is ; next largest is m/z 58, consistent with McLafferty rearrangement as follows: This answer seems quite reasonable. CH3CH2CH2 CH3CH2CH2CH CH2 CH2 m/z 58 CCH3 OH CH2 C CH2 2-Octanone CH3 CH O H CH3C O CH3 0.9(t) 2.0(s) 2.2(t) CH2 CH3 C O , CH2 CH2 O CH3 Solutions to Problems • 309 1559T_ch17_304-322 11/3/05 10:48 Page 309
1559T_ch17-304-32211/3/0510:48Page310 EQA 310.Chapter 17 ALDEHYDES AND KETONES:THE CARBONYL GROUF 25.(a)CrO3.HSO acetone,or MnO2.CH2Cla (better): (b)PCC.CH2Clz;(e)1.O3.CH2Cl2,2.Zn,CHaCOOH,H2O: (d)HgSO.HaO.HSO:(e)same as (d) mO8aa.2ra 26.QLGLGLG+Gi2○d 27.(a)Ask"How electrophilic is the carbon in question?" (CH3)C-H(CHa)C-O(CH3)C-NH Onder of ketone and imine detemined by clectronegativity 000 0 (b)CH.CCCCH,>CH.CCCH,>CH,CCH, 二 29.a01 (c) HO OCHs 31.a〔 The starting material cquilibrates with this product
25. (a) CrO3, H2SO4, acetone, or MnO2, CH2Cl2 (better); (b) PCC, CH2Cl2; (c) 1. O3, CH2Cl2, 2. Zn, CH3COOH, H2O; (d) HgSO4, H2O, H2SO4; (e) same as (d); (f) 1. 2. H H2O O O B B 26. (a) CH3CH2CH2CH HCH (b) O O B B (c) HCCH2CH2CH2CH2CH (d) 27. (a) Ask “How electrophilic is the carbon in question?”: (CH3)2CPO H (CH3)2CPO (CH3)2CPNH Order of ketone and imine determined by electronegativity. OOO OO O BBB BB B (b) CH3CCCCH3 CH3CCCH3 CH3CCH3 Adjacent carbonyl groups enhance each other’s reactivity by reinforcing the character of their respective carbons. (c) BrCH2CHO CH3CHO BrCH2COCH3 CH3COCH3 Aldehydes are more reactive than ketones; halogen substituents increase reactivity. 28. (a) (b) (c) 29. (a) (b) (c) 30. (a) (b) (c) 31. (a) The starting material equilibrates with this product. HO OCH3 OH O O OH O O OH OH H O O OH OH H O O 2 O C Cl, AlCl3, O 310 • Chapter 17 ALDEHYDES AND KETONES: THE CARBONYL GROUP 1559T_ch17_304-322 11/3/05 10:48 Page 310