1559T_ch09_148-17410/30/0518:03Pa9e157 EQA Solutions to Problems.157 Line formulas will be used for most of the cyclic struct below.Note that methyl groups are understood to be present at the ends of lines even whenCHis not written in. CH. 代七女☆ 心心88 Each product arises from loss of a proton from a carbon adjacent to the positively charged carbon of a structure in the previous problem 29.Rea nditions:The is eaker (HO'rather hero muclie Nome of h pmh (b)CH,CHBrCH3 (e)CH,CH2CH.CH2Br (d)(CH)CHCH,Br (e)(CH)CCH,CH,Br e likely.Products will result fron sent.See answers to Problem 27(f)through 27(h). Na 30.(a) HO: Br Solutions to Problems • 157 Line formulas will be used for most of the cyclic structures below. Note that methyl groups are understood to be present at the ends of lines even when “CH3” is not written in. (f ) (g) (h) (h) Each product arises from loss of a proton from a carbon adjacent to the positively charged carbon of a structure in the previous problem. 29. Rearrangements are much less likely under these conditions: The acid is much weaker (H3O rather than H2SO4), and there is a good nucleophile around. None of the primary alcohols rearrange. (a) CH3CH2CH2Br (b) CH3CHBrCH3 (c) CH3CH2CH2CH2Br (d) (CH3)2CHCH2Br (e) (CH3)3CCH2CH2Br With secondary or tertiary alcohols, rearrangements become more likely. Products will result from attachment of Br
to any positively charged carbon in the carbocations present. See answers to Problem 27(f ) through 27(h). 30. (a) (b) HO H HOH H Br
Br Rearrangement from secondary to tertiary O H Na H
O
Na H2 CH3 CH3 CH3 CH3 C(CH3)3 CH3 CH CH3 2 CH3 CH3 CH3 CH2 (major product) 1559T_ch09_148-174 10/30/05 18:03 Page 157