Solutions manual for part b
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Solutions manual for part b
Solutions to the ProblemsChapter 11.1.ITicsc questions can be answered by comparing the electron-accepting capacity and relative location of the subst Solutions manual for part b tituents groups. 1T1C most acidic compounds arc those with the most stabilized anions.a.In (a) the most difficult choice is between nitroethane and dicyanomethane, fable 1.1 indicates that nilroclhanc (pK — 8.6) is more acidic in hydroxylic solvents, hut that the order might be reversed in DMSO. jud Solutions manual for part b ging from the high PROVISO (17.2) for nitromethane. For hydroxyl ic solvents, the order should be CHjCHjNO, > CH2(CN)2 > (CH3)2CHC-O(Ph) > CHjCHjCN.b.Solutions manual for part b
The comparison in (b) is between N II. o II. and c II bonds. This order is dominated by the electronegativity difference, which is o > N > c. Of the tSolutions to the ProblemsChapter 11.1.ITicsc questions can be answered by comparing the electron-accepting capacity and relative location of the subst Solutions manual for part b (CH3)2CHPh > CH3CH2CH3.c.In (c) the two P-dicarhonyl compounds are more acidic, with the diketone being a bit more acidic than the (3-ketocstcr. Of the two monocslcrs. the phenyl conjugation will enhance the acidity of methyl phcnylacclalc. whereas the nonconjugated phenyl group in benzyl acetate h Solutions manual for part b as little effect on the pA'.ỌỌỌỌ(CH3C)2CH2 > CH3CCH2CO2CH3 > CH3OCCH2Ph > CH3COCH2Phd.In (d) the extra stabilization provided by the phenyl ring makesSolutions manual for part b
benzyl phenyl ketone the most acidic compound of the group. The cross-conjugation in I-phenylbutanone has a smaller effect, but makes it more acidic Solutions to the ProblemsChapter 11.1.ITicsc questions can be answered by comparing the electron-accepting capacity and relative location of the subst Solutions manual for part b abilization of the enolate of the latter.ỌỌỌ0_ H_____ ĩ___________________ ĩ ___________L.PhCCH?Ph > PhCCH2CH?CH3 > (CH^CCHi > (CH3)3CCH(CH3)212Solution! lo the Problem!1.2.a. This is a monosubstituted cyclohexanone where (he less-substiuited enolate is (he kinetic enolate and the more-substituted e Solutions manual for part b nolate is the thermodynamic enolate.C(CH3)3C(CH3)3kinetic thermodynamich. The conjugated dienolate should he preferred under both kinetic and thermodySolutions manual for part b
namic conditions.kinetic and thermodynamicc.This presents a comparison between a trisubstituted and disubstituted enolate. The steric destabilization Solutions to the ProblemsChapter 11.1.ITicsc questions can be answered by comparing the electron-accepting capacity and relative location of the subst Solutions manual for part b s on the base that is used, ranging from 60:40 favoring z with I DA to 2:98 favoring z with I.ÍỈỈMDS or I.i 2.4.6-trichloroanilide (see Section 1.1.2 for a discussion).p-(CH3)?CHSolutions manual for part b
strain present in the alternate enolate would also make this the more stable.kinetic and thermodynamice.The kinetic enolate is the less-substituted oSolutions to the ProblemsChapter 11.1.ITicsc questions can be answered by comparing the electron-accepting capacity and relative location of the subst Solutions manual for part b s the cross-conjugated enolate arising from ct'-raihcr than 7-deprotonation. No information was found on the conjugated a.-y-isomer, which, while conjugated, may suffer from steric destabilization.kinetic<1,7 -isomerg.The kinetic enolate is the cross-conjugated enolate arising from a'-rather than 7- Solutions manual for part b deprotonation. The conjugated 7-isomer would he expected to he the morestable enolate.Solutions to the ProblemsChapter 11.1.ITicsc questions can be answered by comparing the electron-accepting capacity and relative location of the substSolutions to the ProblemsChapter 11.1.ITicsc questions can be answered by comparing the electron-accepting capacity and relative location of the substGọi ngay
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