Question

Organic chemistry, trans-9-(2phenylethenyl)anthracene synthesis via Wittig Reaction.

"Does the H-NMR data suggest this a symmetrical molecule? Does the C-13 NMR data suggest it is symmetrical?"

Please be detailed, thank you. Below is the data provided:

Compound 21c: (67%); mp 131 °C; IR Vmax (KBr) 3024 cm'' (=CH), 1621 cm'(C=C); UV Nmax (CH3CN) 270 (€ 19,800), 300 (€ 7,900), 350 (€ 1,100), 390 (€ 60,000); 'H NMR (CDC13) 8 6.95 (1H, d, Jax = 16.56 Hz, vinylic), 7.25-8.41 (14H, m, aromatic), 7.92 (1H, d, Jax= 16.56 Hz, vinylic); C NMR (CDC13) 124.87, 125.12, 125.41, 125.98, 126.43, 126.56, 127.96, 128.65, 128.78, 129.71, 131.48, 132.72, 137.26. Anal. Calcd for C22H16: C, 94.25; H, 5.75. Found: C, 94.47; H, 5.42.

Answer 1

trans-9-(2phenylethenyl)anthracene synthesis via Wittig Reaction.

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1H-NMR data : two peaks for trans protons on -C=C-, as suggested by coupling constant : J = 16.56 Hz ; other 14 H appear as multiplet in a range. As both alkenic protons appear at different ppm values with large coupling constant , thus 1H-NMR data doesn't suggest it to be symmetrical. But, we cannot conclude based on it completely as , detailed 1H-NMR data in aromatic range is lacking.

13C-NMR data : Total Carbon in molecule : 22 , # peaks in 13C-NMR is 13. Thus there are 13 types of non-equivalent Carbons, against total # carbon : 22 ==> it sugests that there is presence of symmetry elements in thus molecule.