Part 7: LABORATORY

• The insolubility data suggests that we have a neutral functional group.
  

From the chemical tests we get:

• 2,4-DNP test (orange solution) is negative and therefore indicates the absence of an aldehyde or ketone.
• The iodoform test (yellow solution) is negative indicating that it is not a methyl ketone.
• The Tollen's test is negative confirms it's not an aldehyde.
• The ferric chloride test is negative indicating that it's not a phenol.

The IR absorption at about 1720 cm^-1 indicates the presence of a C=O

The H NMR data tells us:

• 7.25 to 8.0 ppm suggest 4 aromatic H and therefore probably a disubstituted benzene system.
• 4.3 ppm shows a 2H quartet, i.e. a -CH₂- coupled to 3H deshielded (by O ?)
• 2.3 ppm shows a 3H singlet, i.e. a -CH₃ with no neighbouring H.
• 1.35 ppm shows a 3H triplet, i.e. a -CH₃ coupled to 2H.

Summary....
Functional group tests and IR data suggest a carbonyl C=O from an acid derivative (but not a carboxylic acid).
H NMR 2 x doublets in the aromatic region suggests a para disubstituted benzene, C₆H₄, and two alkyl groups -CH₂CH₃ and -CH₃
Given the H NMR chemical shift for the -CH₂- group and the fact that we know we have a carboxylic acid derivative, it looks like the ethyl group is actually ethoxy: -OCH₂CH₃

Altogether...

The final step should always be to check what you have drawn. The easiest thing to check is usually the coupling patterns you would expect to see, and the chemical shifts of each unit.  You should be asking yourself : "Does my answer give me what the H-nmr shows ?"

Common general errors:

Poor interpretation of FG tests, could not interpret the simple coupling patterns from the H NMR.