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.. (c )
1 B r2 � Br � Br
Br2 T his structure shows a trans alkene in a to-membered ring, just the rotated view of the structure to the right. � anti a ddition of B r 2 requires t ra ns alkene to gi ve meso product tate around C-2 .. (d) (e) CO
� o C l2 (X trans-cyclodecene C! �'" :
�2 Br "" OH B H3 - THF .. ~
OH or ( f) � V-/ or 8 -60 CO � � Hg (OA c h� N aBH _ _ _ _ _ -l.. t, C H30H A ) Unknown X, C SH9Br, h as one element of unsaturation. X r eacts with neither bromi ne nor KMn04, so the unsaturation in X c annot be an alkene; it must be a ri ng. B ) Upon treatment with strong base (t-butoxide), X l oses H and Br to give Y, CSHg, which does react with bromine and KMn0 4 ; it must have an alkene and a ring. Only one isomer is formed. C) Catalytic hydrogenation of Y gives methylcyc lobutane. This is a B IG c l ue because it gi ves the carbon skeleton of the unknown. Y must have a double bond in the methylcyclobutane skeleton, and X m ust have a B r on the methylc yc lobutane skeleton . D ) Ozonolysis of Y gi ves a dialdehyde Z, CSHg0 2' which contains all the original carbons, so the alkene c leaved in the ozonolysis had to be in the ring. Let's consider the possible answers for X a nd see if each fits the information.
I KO-t-B u i f this is X .---:f' >U o zonolysis this must be Y; only one product > DOES NOT FIT OZONOLYSIS RESULTS so this c annot be X a nd Y d' ° + O =CH2 more possi bilities on the next page
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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