{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

0243Pb44-09.cwk (WP)

0243Pb44-09.cwk (WP) - We strongly suggest that you begin...

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

View Full Document Right Arrow Icon
Problem 44, Chemistry 0243 - 2009 A mixture of two isomers of the formula C 5 H 9 IO is treated with a strong base, sodium hydride Na + – :H. (what does NaH always do? Two new compounds, each C 5 H 8 IONa are formed. Only one of these is converted into a new compound of the formula C 5 H 8 O; the other survives. What is the structure of the C 5 H 8 O product, and why is only one stereoisomer reactive under these conditions?
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: We strongly suggest that you begin with a determination of the structures of the two stereoisomers. Next consider what are the products of the reactions with sodium hydride. Finally, explain why only one of the initial products reacts to give C 5 H 8 O. OH I H NaH C 5 H 8 IO – Na + C 5 H 8 O squiggley line means stereochemistry not specified two compounds two compounds one compound...
View Full Document

{[ snackBarMessage ]}