The question isn't asking whether a Diels-Alder proceeds via Sn1 (in which case, Tangymoose is correct, it's a cycloaddition and not a substitution reaction). You are comparing the halide product of a Diels-Alder with option (b) to see which would undergo an Sn1 faster.
This question is testing:
- Your knowledge of a Diels-Alder
- Your knowledge of Sn1 reaction conditions
We know Sn1's proceed via carbocations, so the more stable the carbocation and the better the leaving group, the faster the Sn1.
Actually, for the Diels-Alder there needs to be a
diene and a dienophile, so I suspect you are starting with a cyclopenta
diene and an iodoethylene. (OP -- I'm not sure if you posted the question correctly from the DAT Destroyer, but that would be helpful lol
...it makes more sense that the iodine would be on the dienophile, but in order for the product to have "Iodine attached to a tertiary carbon," the halide must have been on the pentadiene.) Anyway, draw out the D-A product and you should have a 6-membered ring fused with a 5-membered ring. Let's just say the iodine really is attached to the bridgehead C. If the iodine leaves, there will be a carbocation on the bridgehead C of a fused 5- and 6- membered ring. What is the hybridization of a tertiary carbocation? It's looking to be sp2, which is trigonal planar and 120 degrees. Think of how crazy that C would have to contort to accomodate the 2 rings and the carbocation...it's not gonna happen.
For the halobutane, yes -- Cl is generally a bad leaving group, but it *can* still leave to form a
tertiary carbocation, just at a pretty slow Sn1 rate.
Learning orgo through words is the worst -- I wish I could draw you a picture, but hopefully this helps! Feel free to ask for further clarification.