Question

Label HMNR peaks and indenify structure

(a) HN (2) CH N (6) CHAN

a) C8H11N

b) C8H11N

c) C7H9N

Answer 1

Questions :- 1-HNMRs :- @) Co M, N :-As per molecular formula, calculate DBE DBE( Double Bond Equivalence) = C+1+H+X-N Where C= No. of Carbons - 0 +1-/11 +0- H=) No. of Hydrogens X = No. of halogen or oxygen M= No. of Nitrogen = 9-5 DBE = 4 2 It means that there are 4 double bonds or 3 double bonds and I ring in the structure. NMR spectral Analysis S. No. Chemical shift Integration splitting Inference (5ppm (No. of H a 1.3 ppm ЗН heighbours CH2CH- Doublet 1Н Quartet 4 ppm 7.3 ppm the CCH 5H Multiplet HN- CH3 Overall Structure of (a)

DBE by Molecular frarossa, DBE = 8+1- (6) COMIN &- Calculating c+ 2 H+X-N) 4-69 70-4 - 9-5 IDBE = 4 It meary that there are 4 double bonds or 1 3 double bonds in the structure NMR Spectral Analysis : :- Chemical süft call lutegration Splitting sufressure sught to 2:6–3ppm Sextet I sing and 0.9 ppm 2H 4H |(2 tiplets mer CHHOH 7.3 ppm 5H Multiplet CH₂-CH₂-NIL Overall structure of (b).