• The molecular formula of C₉H₉NO₂ implies and IHD = 6.
• The solubility in HCl suggests that we have a basic functional group.
  

From the chemical tests we get:

• 2,4-DNP test is positive and therefore indicates the presence of an aldehyde or ketone.
• The Tollen's test is postive and indicates the presence of an aldehyde.
• The Schiff's test is postive and indicates the presence of an aldehyde.
• The iodoform test is postive and indicates the presence of a methyl ketone.

The H NMR data tells us:

• 10 ppm is due the aldehyde H (-CHO)
• 7 to 7.8 ppm suggest 3 aromatic H and therefore probably a trisubstituted benzene system.
• 4.2 ppm represents two H on an electronegative atom (so here either N or O)
• 2.5 ppm shows a 3H singlet for a methyl substitutent.

The pieces then are:

O=CH, O=C-CH₃ and C₆H₃ (trisubs. benzene), this leaves the -NH₂.

Looking at the two doublets and a singlet for the 3 ArH (7 to 8 ppm) this implies that the aromatic unit is 1,2,4 substituted. At this level, it's not possible to tell the relative position of the three substituents.

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