Question

Help with problems #3 and #4

3 Draw the arrow-pushing mechanism of the reaction, including all proton transfer steps. eOH 4. Below is the structure of a chalcone, with protons labelled 1-8. The H NMR spectrum is provided on the next page and lettered A-H for clarity. Describe the multiplicity and integration you would expect to observe for each of the protons. Which peak would you expect to have a higher chemical shift, 4 or 5? Draw resonance structures to explain your reasoning. 6 8 1 5

Answer 1

3. The mentioned reaction is known as Aldol condensation, the base catalyzed aldol formed in the undergo elimination reaction under basic condition to give the corresponding chalcone product (alpha, beta unsaturated ketone).

The reaction mechanism is attached as below image,

Lollo carbamion cho enolato 1 Std Dehydeations chalcone the

4. The 1H NMR of the labelled protons have been described with the multiplicity in the attached below image.

Based on the rhe resonance structure, the proton 4 will have higher chemical shift (down field region/desheilding region) and proton shall have lower chemical shift (up field region/sheilding region) in the 1HNMR spectrum .

IHNUR chemical shifts Resonance strome e 2= H = dd, A-H 3 4 - or the double doletinichigher chemical stuff 5-H -d, olefinie - lower chemical shult 7=64 the AH-upsild, region proton 8=-M3 Stupet, oHz, uppeldflower regions olefinic proton 4 has higher chemical Thi H than proton because of the desonance ducere as above the poster is a benzylie and more alky chander