Answer to Question #178671 in Organic Chemistry for sahra

Question

Cyclopentadiene (5.000 mL) was added to a premade solution of 2,5-cyclohexadiene-1,4-dione (5.000 g) in ether (20.00 mL) in a 100 mL round bottom flask.. The resulting solution was purged with nitrogen and heated to reflux for 2 h. Reaction progress was monitored by TLC. Once the reaction reached completion, the mixture was cooled to room temperature and then extracted with water. The product was purified by recrystallization and analyzed by IR spectroscopy. 1,4,4a,8a-tetrahydro-1,4-methanonaphthalene-5-8-dione was obtained as a white precipitate (2.00 g).

1. Calculate the moles of each reagent and the % yield of your product.

2. Draw the HOMO and LUMO on the compounds during the transition state for the endo isomer of this product. Why is this the more favoured transition state?

3. Write the mechanism for this reaction.

Answer 1

1. Volume of cyclopentadiene C5H6= 5.00ml

Mass of 2,5-cyclohexadiene-1,4-dione, C6H4O2= 5.000g

Molarass of C6H4O2= 108g/mol

Mole= mass/molar mass= 5.000g/108g/mol = 0.046mol

Mass = 2.00g

molar mass of methanonaphthalene-5,8-dione, C11H10O22= 174.1959g/mol

Mole= mass/molar mass= 2.00/174.1959= 0.011mol

since 1 mol of C6H4O2 react with 1 mol of C5H6,

0.062mol of C6H4O2 will react with 0.062mol of C5H6

mole oc cyclopentadiene= 0.062mol

% yield of product= 0.011/0.062 x 100% = 17.74%

II) HOMO and LUMO which mean highest occupied molecular orbital and lowest unoccupied molecular orbital are molecular orbitals with high energy and low stabilities. since the molecular orbital is unstable at the transition state, it tends to quickly transit to the product with lower energy and higher stability.

III) The mechanism follows an electrophilic addition reaction in which the action of C5H6 on C6H4O2 yields C11H10O22.