SN1 Forced RXN

[StraightShooter]

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Question: Which rxn occurs faster and why? Reaction A: reaction of chlorocyclopent-2-ene with sodium meth-oxide or Reaction B: reaction of chlorocyclopenta-2,4-diene with sodium meth-oxide. Assume SN1 forced rxn.

My original answer: Reaction A is faster since its non aromatic(more stable) than that of B(antiaromatic).

Curious as to what the right answer would be, or if these reactions would be very close in speed?


Reaction A starting material is on left, B is on the right.

 

[yoyohomieg5432]

reaction B should be faster. the rate limiting step in SN1 is dissociation of the leaving group and carbocation formation. This step occurs the fastest in the reactant that best stabilizes the carbocation intermediate. reactant B has 2 pi bonds that can form form resonance structures with the carbocation. reactant a only has 1. more resonance structures=greater stability=the reaction will occur faster.

 

[PostBaccHopeful]

Because this is a forced Sn1 reaction, I think dissociation of the leaving group should not be the main concern. I think the question focuses more on what's left after the L.G. cleaves. And, what we're left with concerns both thermodynamics and kinetics. Because the question asks about kinetics (aka how fast reaction goes), we should consider which product would form the fastest. If you draw out the carbocation for B (the di-ene), you will see that you have more sites available for the nucleophile (the methoxide) than A. Thus, you will form product B faster because of this reasoning. Further, it happens to be that B is more thermodynamically favorable (stability of charges), so you will also form product B first if the question had asked about thermodynamics.

So, yes product B will form faster like yoyo had mentioned. 🙂

 

[StraightShooter]

Actually based on huckels rule wouldn't the diene be more difficult to create since if you look at the energy diagram its forming a di-radical vs the nonaromatic.

So it would be harder to create the SN1 on the diene leading to faster at A

 

[HotHamH2O]

I would say B because pi bonds donate electrons to stabalize the carbocation intermediate. Two of those bonds would offer more stabality to the carbocation intermediate.

 

[mehc012]

Actually based on huckels rule wouldn't the diene be more difficult to create since if you look at the energy diagram its forming a di-radical vs the nonaromatic.

So it would be harder to create the SN1 on the diene leading to faster at A

The cyclopentadienyl cation would be antiaromatic for sure (sorry, I was unsure if that's what you were getting at with your phrasing).

So product B is more stable, but product A has a lower energy intermediate. They're asking for the kinetic product...