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


B   +   PCC
methylene chloride


C   +   PCC
methylene chloride


D   +   PCC
methylene chloride

No Change



  (i) O3

  (ii) Zn (dust)

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