Part 8: STRUCTURE DETERMINATION

The question is based on a REAL environmental analysis carried out during the summer of 1998 by undergraduate students working for Dr. Dixon.

The important pieces of information from the question are that:

• A - F were hydrocarbons
• their molecular weights
• 13C nmr data suggested Ar-C and some alkane C
• D,E and F are isomers

The MW and 13C is enough to determine that A is benzene and get you moving in the right direction. The number of types of Ar-C is enough to determine the substitution pattern of D,E and F . Since C has two types of alkane C is must be ethylbenzene.

• A = benzene (MW=78, 1 type Ar-C)
• B = methylbenzene (MW=92, 1 type alkane C and 4 types Ar-C)
• C = ethylbenzene (MW=106, 2 types alkane C and 4 types Ar-C)
• D = 1,4-dimethylbenzene (MW=106, 1 type alkane C and 2 types Ar-C)
• E = 1,3-dimethylbenzene (MW=106, 1 type alkane C and 4 types Ar-C)
• F = 1,2-dimethylbenzene (MW=106, 1 type alkane C and 3 types Ar-C)

Gas chromatography (Ch4, and lab. manual) separates based on the boiling point, with lower boiling materials having short retention times. Peak 1 = A, peak 2 = B, and peak 3 = C since the increasing molecular weight and therefore size, will increase the Van der Waals interactions between these systems thus raising their boiling points.