Part
7: LABORATORY
- 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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