Question

what are functional groups

2.1 3,6-Bis(dimethylamino)-2,7-difluoro-10-methylacridin-9(10H)-one 4. IR (ATR, cm-?): 3050, 2982, 2938, 2872, 2846, 2796, 1621, 1538, 1558, 1500, 1456, 1380, 1330, 1266, 1178, 1139, 1064, 1024, 975, 900, 819, 785; H NMR (90 MHz, 10 % CD,OD in CDCI3) 8 7.72 (d, 2H), 6.27 (d, 2H), 3.51 (s, 3H), 2.93 (s, 12H); UV-vis (CH2Cl2, 1 E-5 M, nm) 278, 377; PL (CH2Cl2, 1 E-5 M, nm) hexcitation 377, hemission 441. 0.8 05- Absorbance (a.u.) 0.4 02- 4000 3800 3000 1600 1000 500 Figure 1. (4) 2000 Frequency (wavenumbers) IR spectrum of the final product, 3,6-Bis(dimethylamino)-2,7-difluoro-10-methylacridin-9(101)-one

Answer 1

The molecular structure of 3,6-Bis(dimethylamino)-2,7-difluoro-10-methyl acridine-9(10H)-one is

4 2- V ㅂ 2.

The functional froups are as follows: -NHCH₃ groups, fluoro group, ketone group.

The given IR spectrum is Absorbance vs wave number (instead of usual transmittance vs wave number). We know that absorbance and transmittance are inversely proportional to eachother,

$Absorbance \; \alpha \frac{1}{Transmittance}$

Hence, the higher the peaks in absorbance spectrum the weaker or lesser intense bands they represent in the trasmittance spectrum.

Therefore, short peaks in the given spectrum are intense bands in transmittance spectrum and high peaks in the given spectrum are weak bands in transmittance spectrum.

1621 cm^-1 is intense strong band representing carbonyl C=O group.

3050 cm^-1 is intense strong band representing aromatic C=C

1024 cm^-1 is a medium band representing C-N

1266 cm^-1 is intense strong band representing C-F

.

.

Now, there are four non-equivalent protons (labelled as a, b, c and d in the figure below)

d u 2- . OS c 6 u 2_2

So, NMR spectrum has 4 peaks in total

$\delta$ 7.72 (d, 2H) representing protons a

$\delta$ 6.27 (d, 2H) representing protons b

$\delta$ 3.51 (s, 3H) representing protons c

$\delta$ 2.93 (s, 12H) representing protons d