Question

a) Give the curved arrow-pushing and the allowed product for the following electrocyclic ring closure reaction, indicate whether the product would be achiral, a meso compound or a racemic mixture heat НОМО b) ON TOP OF THE STRUCTURE, draw the HOMO of the reactant cation c) To form YOUR ALLOWED product, did the reaction proceed via a conrotatory or a disrotatory ring closure?

Please note: This is 1 practice homework organic chemistry question with 4 parts in A-C to answer!

1. Give full curved arrow pushing for the electrocyclic ring closure reaction.

2. Indicate if product is achiral, meso compound or racemic.

3. Draw the HOMO of the reactant cation on top of the structure.

4. To form your ALLOWED product, did the reaction proceed conrotatary or disrotatory?

If you are unable to answer any of the parts PLEASE leave question alone for the next expert!

Your answer should be in this format below please!!! AND Please be clear!!!!

Answer 1

Sol: For the following electrocyclic ring closure reaction:

1. 4 pi electron system, ring closure proceed via heating medium,thus conrotatory ring closure leading to a product which is optically inactive, i.e, a racemic mixture.
2. HOMO of the reactant cation have one node.
3. The reaction proceed via conrotatory process for ${\color{Blue} }$ ${\color{Blue} 4Pi }$ electron system in heat medium ${\color{Blue} \Delta }$ .

Sol: For the following electrocyclic ring closure reaction ном о р ма ? heat HOMO HOMO 1 node heat connotatorny 4 electrons in a up cationic A.systery connotatony ring a closing O. HOMO of reactant cation Not optically active, racemic mixture. 88 8 8 8 12 & Energy level noole .. (na) node n: 2 here 27

Sol: ©. To force the allowed product the reaction proceed via a connotatory ring closure. Moleculcer orbital diagnary for the given pentadiengil cation: 1 818181818 4s, node: 4 - 888 88 Yy, node: - Y , node:2 — Lowest un ocupied 88 8 , node. the Highest occupied 88888 4node: o H CHOMO) node:2 Lowest anoccupied coleculay orbital