Bicyclo[3.3.0] Octane

Chemistry Level 3

B i c y c l o [ 3 . 3. 0 ] o c t a n e \ce{Bicyclo[3}.\ce{3}.\ce{0] octane} is reacted in the following order:

  • B r 2 \ce{Br}_2 in l i g h t \ce{light}

  • K O H \ce{KOH} in a l c o h o l \ce{alcohol}

  • O s O 4 \ce{OsO}_4 ( O s m i u m T e t r o x i d e \ce{Osmium Tetroxide} )

  • N a I O 4 \ce{NaIO}_4 ( S o d i u m P e r i o d a t e \ce{Sodium Periodate} )

What will be the d e g r e e s o f u n s a t u r a t i o n \ce{degrees~of~unsaturation} for the m a j o r e n d p r o d u c t \ce{major~end~product} ?

David's Organic Chemistry Set

David's Physical Chemistry Set

4 5 2 1 3

David Hontz by David Hontz , 26, USA

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

David Hontz
Jul 9, 2016

  • B r Br first adds to one of the preferred 3 ∘ 3^{\circ} carbons

  • K O H KOH will remove the B r Br to form a double bond. The most substituted double bond between the two 3 ∘ 3^{\circ} carbons is preferred.

  • O s O 4 OsO_4 forms a diol among the double bonded carbons

  • Lastly, I O 4 IO_4 breaks the bond between the diol and forms ketones with the alcohols

The final product is 1 , 5 c y c l o o c t a d i o n e \ce{1,5~cyclooctadione} A n s w e r = 1 D . U . ( r i n g ) + 2 D . U . ( c a r b o n y l ) = 3 D . U . Answer = 1 ~ D.U. ~ (ring) + 2 ~ D.U. ~ (carbonyl) = \boxed{3 ~ D.U. } D . U . = D e g r e e s o f U n s a t u r a t i o n D.U. = \ce{Degrees ~of~ Unsaturation}

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Same solution

Vignesh S - 4 years, 11 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...