Unformatted text preview: 19-19 ¢ continued OCHl + S02CI 6
(b) N H2 � 9'-'::
h- HCI H20
!1 • 19-20 (a)� lll (d) 1 9-2 1 Orientation of the Cope elimination is similar to Hofmann elimination: the less substituted alkene is the major product. HO' � N (b) H2C=CH2 (a)� � minor major (CH3hNOH + (CH3CH2)2NOH CXJ 0
+ H3C, ,CH3 (e) V \H1 t? 6 S02 NH 9
, NH2 sulfapyridine H3C, +
,CH3 o a CH3 H1C-N�
,..CH3 N CH3 (c)
I 6 S02 NH CH3 N + H2C=CH2 (I) �N-CH] + + � +0 N
OH (c) (CH,hNOH (d) H2C=CH2 0 +rN� OH nunor 1 9-22 The key to this problem is to understand that Hofmann elimination occurs via an E2 mechanism requiring anti coplanar stereochemistry, whereas Cope elimination requires syn coplanar stereochemistry. (a) H(CH3h H CH3 , .:- ...; Hofmann orientation loses a hydrogen from the CH3 and the N(CH3h C � / group to make the less substituted double bond � \ H 2C H F 444 ...
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This note was uploaded on 02/27/2010 for the course CHEM 140 taught by Professor Wade during the Spring '10 term at Whitman.
- Spring '10