Solutions_Manual_for_Organic_Chemistry_6th_Ed 157

Solutions_Manual_for_Organic_Chemistry_6th_Ed 157 - 7 -29...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 7 -29 (d) continued E1 o n rearranged carbocation � � <:;LJ H20 :. + 7 -30 Please refer to solution 1 -20, page 1 2 of this Solutions Manual. 7 -3 1 (a) 00 (c) + H,o' Br Br (d) � (g) � ( h) (f) Br � r; 0 1 (") ( e) Z E � o 7 -32 ( a) 2-ethylpent-l-ene (number the longest chain containing the double bond) ( b) 3-ethylpent-2-en e ( c ) (3E,6E)-octa-l,3,6-triene ( d) (E)-4-ethylhept-3 -ene ( e) l-cyc lohexy lcyclohexa-l,3-diene (f) (3Z, 5E)-6-ch loro-3-(chloromethyl)octa-l ,3,5 -triene 7 -33 7-34 (a) � (a) E (b) nei ther-two methyl groups on one carbon (c) Z (d) Z ( Z)-l-fluoro­ prop-l-ene F �F ( E)-l-fluoroprop-l-ene 2 -fluoroprop- l -ene A F 3 -fluoroprop-l-ene 0 F fluorocyclopropane [>-F (b) C4H 7 B r has one element of unsaturation, but no rings are permitted in the problem, so all the isomers must have one double bond. Only four i somers are possible with four c arbons and one double bond, so Br Y y Br /'o.... B r - � '- � ��� H Br Br �" � � H Br �Br (b) C holesterol, four must be rings" C27H460, Br � Br � � Br Br h as fi ve elements of unsaturation . If only one of those is a pi bond, the other 150 ...
View Full Document

Ask a homework question - tutors are online