[synthesis87]

So I am starting out with...

a 1,3,5-cycloheptatriene with O double bonded to the 7th carbon...

(1) pentadiene + heat
I found that this involves 10 electrons, which is 4n+2, and therefore reaction proceeds under thermal conditions. And I was able to come up with a product and support that with the frontier molecular orbital analysis. But what I'm confused about is the second part (and let me know if (1) was wrong...)

(2) ethene (C2H4) + heat
This involves 8 electrons, which is just 4n, and therefore reaction does not proceed? I couldn't draw out the molecules to support this though... because drawing the molecules.... cant you draw it like:

look at the HOMO of ethene (no node), LUMO (3 nodes) of the cycloheptatriene:
let the plus orbitals of the HOMO overlap with the plus orbitals of Carbon 1 and Carbon 4? and therefore produce:

1,4-cycloheptadiene with double bonded O at 6th carbon + 2-carbon bridge connecting C3 and C7 on the heptadiene.



I'm really bad at nomenclature, I hope that made sense....
Thanks in advance to any *delete me*

[Mitch]

Isn't the sterochemistry for (2) just changed somehow but the reaction still goes? Like something about the sym-addition is not possible, but you can still twist the reactant to bond somehow? Mayve movies can give the better answer.

[AWK]

C2H4 at 400 C => CH3-CH2-CH=CH2
 at 500 C =>CH3-CH=CHCH3
at 700 C => CH2=CH-CH=CH2 + H2
at 850 C
butadiene + ethene => cyclohexene => C6H6 + 2H2

[movies]

I think that you just get a 4+2 cycloaddition which is thermally allowed by Woodward-Hoffman rules.

Do you want the whole orbital explanation?  It's pretty much just your standard Diels-Alder in this case.  You ignore the other double bond because the symmetry is wrong.

So your product would be this:

[synthesis87]

alright! thank you so much!
yeah, i had that either it attaches the way i drew it or there's no reaction... my friend said that when she asked the professor, he said that diels alder will more likely happen.