1559T.eh17.304-32211/3/0s10:48Page319 Solutions to Problems319 CH CHs- 上9MCH,Co Oxidation of the alcohol to a ketone,followed by Wolff-Kishner reduction works as well. (c)The"target"molecule isa hemiacetal Rewrite it in"open"form(that is,as its acyclic isomer CICH.CHCHOH .CH.COH CICH2CH-CH2OC(CHa)3 CHCHOCHO OH 0 CH,CHCH,CH.CH,OC(CH)CH,CCH.CH.CH,OC(CH, 0 CH CCH2CH2CH2OH -product 51.The answer is actually not that hard:Cyclopropanone has more bond angle strain than does the corresponding hemiacetal.Why?The carbonyl carbon wants to be sphybridized and have 12angles.Being The strain it experi nces is therefore less, onding toa compression of10g°-60°=49°relative to this halocyclopropanes(Chapter 6,Problem 53). is mostly free, te s more electrophilic by protonation:C=OH.Above pH 7.no carbonyl groups are protonated,so the rate is reduced to that of free NHOH attacking unactivated C=O groups.Oxidation of the alcohol to a ketone, followed by Wolff-Kishner reduction works as well. (c) The “target” molecule is a hemiacetal. Rewrite it in “open” form (that is, as its acyclic isomer). Synthesis of the open form will automatically result in formation of the desired product. So, with care in protecting the alcohol, O B CH3CCH2CH2CH2OH 88zy product 51. The answer is actually not that hard: Cyclopropanone has more bond angle strain than does the corresponding hemiacetal. Why? The carbonyl carbon wants to be sp2 hybridized and have 120° angles. Being stuck in the triangular ring with an angle near 60°, it experiences the strain of a bond angle compression of 120° 60° 60°. In contrast, the hemiacetal carbon only wants to be tetrahedral, with 109° bond angles. The strain it experiences is therefore less, corresponding to a compression of 109° 60° 49° relative to the desired angle. We’ve seen situations like this before, in the relative unreactivity of cyclopropane in radical halogenation (Chapter 4, Problem 20), and in the sluggishness of SN2 displacements on halocyclopropanes (Chapter 6, Problem 53). 52. Below pH 2, the N in NH2OH is protonated (
NH3OH), so the nucleophilic atom is effectively gone. At pH 4, the N is mostly free, but the solution is still acidic enough for some carbonyl groups to be made more electrophilic by protonation: Above pH 7, no carbonyl groups are protonated, so the rate is reduced to that of free NH2OH attacking unactivated C O groups. C OH.
CH3CHCH2CH2CH2OC(CH3)3 OH PCC, CH2Cl2 CH3CCH2CH2CH2OC(CH3)3 O H
, H2O ClCH2CH2CH2OH H
, (CH3)3COH ClCH2CH2CH2OC(CH3)3 1. Mg, (CH3CH2)2O 2. CH3CHO O H O CH3 CH3 HO O C O CH3 CH3 C H OH 2. H
, H2O 1. CH3MgI, (CH3CH2)2O CrO3, H2SO4, H2O Solutions to Problems • 319 1559T_ch17_304-322 11/3/05 10:48 Page 319