第11题 Ha is an aliphatic proton; its pKa > 50 Hb's carbon is alpha to a carbonyl group; its pKa ~20 Hc's carbon is alpha to 2 carbonyl group; its pKa ~10 Hd and He are aromatic proton, 40 > pKa >30 第12题 The leaving group is ethoxide (-OEt). 第15题 It is a Friedel-Craft's acylation reaction. The E+ (acyl cation) would NOT rearrange. (textbook p.290 paragraph 1) Therefore Friedel-Craft's acylation followed by Wolff-Kishner reduction is a useful rxn sequence to synthesize n-alkyl benzene derivatives. 第16题_ The first two reaction conditions would give double methylated product. From F, hydrolysis followed by decarboxylation would occur to give product G. 第20题 EWG would increase electron deficiency of carbonyl's carbon and make it a better electrophile. 第22题 For the last reaction, a tertiary chloride would undergo solvolysis reaction (SN1-like) and generate a tertiary carbocation (and also a benzylic carbocation) intermediate. Therefore, a EDG at para position would stabilize the intermediate and give negative rho value. 第24题 You should note that the nitrogen atom on pyridinium ring is also a chirality center. You may replace either Ha or Hb with D. You would find that the two deuterated NADHs are diastereomers. 有其他问题欢迎继续讨论 助教 杨云钦 --

※ 发信站: 批踢踢实业坊(ptt.cc) ◆ From: 140.112.54.79