13-29 Since allyl bromide was the starting material, it is reasonable to expect the allyl group to be present in the impurity: the triplet at 1 1 5 is a =CH2 ' the doublet at 1 38 is =CH- , and the triplet at 63 is a deshielded aliphatic CH2 ; assembling the pieces forms an allyl group. The formula has changed from C3HSBr to C3H60, so OH has replaced the Br. H2C=CH-CH20H 8 1 1 5 ---' 8 Its "-8 63 tnplet doublet triplet Allyl bromide is easily hydrolyzed by water, probably an SN 1 process. 1 3-30 1 3-3 1 H2C=CH-CH2Br ° II c /C, C a ° \ / C-C d b b a /C, c C C II I C C a 'C/ c b 13-32 Compound 2 H20 H2C=CH -CH20H + HBr a = 8 180, singlet b = 8 70, triplet c = 8 28, triplet d = 8 22, triplet a = 8 1 28, doublet b = 8 25, triplet c = 8 23, triplet Two elements of un saturation in C4�02' one of which is a carbonyl, and no evidence of a C=C, prove that a ring must be present. Two elements of unsaturation in C6HIO must be a C=C and a ring. Only three peaks indicates symmetry.
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