A Structural Puzzle

A Second Structural Puzzle

You have four C6H14O constitutional isomers, identified as A, B, C & D. All of these compounds react with sodium metal to liberate hydrogen gas. A & C are chiral, B & D are achiral. Compounds A, B, C & D behave differently on treatment with PCC (pyridinium chlorochromate) in methylene chloride. A gives a C6H12O chiral aldehyde M, C gives an isomeric chiral ketone K, B an isomeric achiral aldehyde N, and D fails to react at all.
When heated with phosphoric acid D forms a C6H12 mixture, predominantly compound E, which reacts rapidly with bromine. Ozonolysis of E followed by a zinc dust workup produced a mixture of 2-butanone and acetaldehyde (ethanal). Wolff-Kishner reduction of A, B & C give the same C6H14 hydrocarbon that is obtained from E by catalytic addition of hydrogen. These transformations are outlined below.


A   +   PCC
methylene chloride

M

B   +   PCC
methylene chloride

N

C   +   PCC
methylene chloride

K

D   +   PCC
methylene chloride

No Change

D
  +  
H3PO4  
heat

  E  
 

  E  
 
  (i) O3

  (ii) Zn (dust)
CH3COCH2CH3
+
CH3CHO

After thinking about these facts and relationships, you should be able to assign structural formulas to all the unknown compounds designated by a bold letter above. To do so select an unknown from the following list and draw a formula using the drawing window opened by the "Draw Formulas" button. When you are finished, "Submit" your formula and then check it by pressing the "Check Molecule" button. A help screen is available for the drawing window.

You have to enable Java and JavaScript on your machine !   Unknown A B C D E K M N

     




This script written by William Reusch, Dept. of Chemistry, Michigan State University. Please send comments and corrections to whreusch@pilot.msu.edu.
JME Molecular Editor v2002.05 by: by Peter Ertl