Question

4. (E)-1-methoxyprop-1-ene has two vinylic protons. Does the proton at C1 or C2 have a chemical shift of 4.6 ppm? Explain using multiplicity and resonance. IH 1Н 09 3 PPM 4. The proton with this chemical shift is off (circle one) C1 C2 Show the splitting diagram for this proton on the back AND the resonance structure of this molecule on the back. The multiplicity of this proton is and the resonance structure of the molecule indicates this proton is (circle one) electron rich electron deficient

Answer 1

H3CY CH3 C1 carbon is attached to electronegative atom Oxygen. Whereas, C2 is attached to methyl group. Hence due to electron withdrawing inductive effect of oxygen, the vinylic proton at C1 will be most downfield as compared to the vinylic proton attached to C2 C2 is electron rich. Thus the proton at 4.6 ppm will be proton attached to C2. As the C2 is attached to CH3 the respective vinylic proton will show pentet (n+1, where n=4). The proton with this chemical shift is off C2. H The pentet: Resoance:- Hzc Vad CH3 H2C1 The multiplicity of this proton is pentet and the resonance structure of the molecule indicates this proton is electron rich.