Question

5. The following spectra are from the isomers of C.H.O. Identify the different compounds based on the NMR spectra. 10 9 8 7 R 5 4 10 9 8 7 6 5 s 4 2 1 0

Answer 1

Firstly, we will calculate double bond equivalent DBE = No. of Carbon atom + - + - monovalentatem A trivalent atom 2 = = 8+1-8 - 0 + 0 0 8 ON - 9-4 [DBE = 5 so, I DBE may be equal to one double bond or one ring. so in all cases, there will be five double bond equivalents. From NMR Data b; (9) A singlet at $ 3.8 ppm (b) Two doublets in the aromatic region one at - 7ppm and other around 7.8 ppm. (C) One singlet at 9.8 ppm from the point (b), there are two doublet in aromatice region which indicates para substituted phenyl ring possible structure; O-CH3 - - H-coo Singlet at 3-8 ppm corresponds to - och comes in b/w 3.5-4.0 ppm. as it

linght at 9.8 plang correspords to oldshydic potton -CHOLA A from NMR Data 2; - 19) A singlet at 2.4 ppm. in a broad singlet at 5.1-5.4 ppm jes Tuso doublut in aromatic region These points to be concluded that Tion doublits in aromatic region corresponds to para disubstituted Phenyl ring... . A broad singlet corresponds to the presence of -oh • A singlet at 2.4 ppm is for - CH3 group. group. should be The structure 01 osechs from Nm R Data 37 lettend (01 A singlet at 2.2 bpm . (4 A multiplit in aromatic region at Fit ppm.

t there e from the multiplet it can be concluded that a ! es presence of mono substituted phenyl ring." • A singlet at 2.2 ppm corresponding to a Chash so, there are two possibilities of structure 1 Ethock ® ETO'Creta structure t can be ruled out because according to that - CH3 group i present next to oxygen atom if this would be the case, s value must be deshielded in between 3.5-4.0 ppm. So, the correct structure ® & from NMR Data 4; (a) A singlet at 2:3 ppm (h) two doublets in aromatic region at 7.3 ppm and at 7.g plems (c) A singht at long bom It is para disubstituted phenyl sing COOH