Question

Determine the structure of a compound, CiH: NO), given the IR and C NMR spectra shown below. Show all of your reasoning. A correct structure with no explanation will receive no credit Transmittance -10058 3500 3000 1500 2500 2000 Warenumber(cm-1) 3H HI that the hot to stot 2H 3H

Answer 1

Tried to explain as much as possible. Hope that will help :-)

equivalent (2) Co H, NO2 : Unsaturation / Double bond Mol. wt. = 177 g/mole Rule: (10*2)+1 -11 +2 DBE - DBE 2014 +n3-n,+2 2 na = tetravatent atom number - no. of C-atoms (here) Az = trivalent IR data: 2962 cm atom number. I 2934 cm - no.7 N-otoms Chere). 2836 cm n = Monovalent atom namber. 2241 cm =no. of H-atoms (here). 1599 cm 1518 ml 1262 cm! 2962– 2934 em' = CSpz H stretches. 2836 cm =) Csp3 -H stretching. 2241 cm =) Can stretching 1599 - 1518 cms) conjugated C=C stretches 1262 cm = c-o stretching. NMR data: 'H-NMRE 86.81-6.83 ppm (3 H,) => Aromatic H's. 3-85-3.9 ppm (34,3H) =) Two -CH3 group protons attached to electrolegative atom (like 'O). 3.7 ppm (2H),s) >> -CH2- group protoors adjacent to a carbon atom having no t's adjacent telectronegative

23.1 ppm > -CH2- carbon, with o or 'N' atom attached to it. 55.9 ppm to attached -CH3 carbons 110.8-111.4 ppm 120.1 ppm ) -CH₂ carbons at o-atom. =) Aromatic & tertiary Cspz H carbons. - Another Aromatic Csp2-H carbon resonates at higher intensities). ) -CEN carbon (having low intensity ). 118.1 ppm 122.0 ppm = should be a quarternary aromatic ring Carbon , since only 3 Csp2H carbons are present in the compound. 148.7-149.3 ppm = Aromatic atoms negative quarternary c- attached to electro- o-atoms. Hence, we have : i) 3 aromatic protons. -OCH3 groups attached to the aromatic ring. in) 1 – CH2- group and one -CEN group (of Unsaturation=2) Then, the aromatic ring is benzene (Unsaluation = 4) Since the compound has total 6 unsaturations, and also also has 6- carbons left.

Now, Since, BC-NMR shows two (spłH carbons resonating at closer frequencies and the other Csp2-t carbon , resonating at higher frequency, the compound should be - H 1 H₂C-a -CEN H3C or