sn1

[BrownieDDD]

Consider these halides. which reacts faster in an sn1 rxn?

source: Dat destroyer orgo odyssey ch. 8 #30 year 2012

a) a diels alder product from a 1,3 cyclopentane with a ethylene
*this prodduct has Iodine attached to a tertiary carbon (next to the bridge)
b) 2-chloride,2 methyl butane

answer is B

why isn't a an answer? book says a won't react....but a is tertidary and I is a better leaving group than Cl.

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[Tangymoose]

Diel-Alder is not a type of SN1 reaction. This reaction occurs in 1 step (concerted process similar to SN2). I would say its closer to an addition reaction than substitution.

I'm sure someone on this forum could give you a more complicated, longer answer but
I just wanted to mention that your problem should say cyclopentene instead of cyclopentane.

 

[FairytaleDMD]

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 cyclopentadiene 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.

 

[Tunaman]

Despite it being a good leaving group in the tert position, you cant form a double bond at a bridgehead carbon, think carbocation rearrangement...that is if you're explaining the problem correctly

 

[Dotoday]

The problem is asking you how the Diels Alder PRODUCT will react. The diels alder product will have Iodine at the bridge head. Yes, you are right the Iodine is a good leaving group located on the tertiary carbon, but that carbon is also a bridgehead carbon. No carbocation can form on the bridgehead. Since Sn1 requires the formation of a carbocation, that molecule cannot utilize Sn1 mechanism. However, another molecule is linear and has a halogen on the tertiary carbon, therefore it has no problem forming a carbocation. Thus is the answer. Hope this helps.

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