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Topic: Resonance on Sn1 and Sn2 reactions  (Read 7990 times)

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Offline looza

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Resonance on Sn1 and Sn2 reactions
« on: April 22, 2011, 07:04:24 PM »
There are two molecules: 1-chloro-2-propanone, and a 1,1-dimethyl-1-chloro-2-propanone.

The first molecule reacts by Sn2 and the second by Sn1.

It says that the first molecule is faster compared to molecules without the carbonyl group (in Sn2 reaction) because the resonance provides stabilization to the transition state.

However, for the second molecule, it says that the reaction (Sn1) is actually slowed because the carbonyl group destabilizes the carbocation.

I do not understand. How can it stabilize in one case and destabilize in another?

Thanks.

Offline Schrödinger

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Re: Resonance on Sn1 and Sn2 reactions
« Reply #1 on: April 23, 2011, 01:18:33 AM »
In the case of SN1, a carbocation is formed. C=O being an electron withdrawing group, destabilizes the carbocation. Note that this happens only because the carbocation intermediate has already been formed.

In the case of SN2, the transition state is stabilized by the interaction of the C=O pi bond with the C-X sigma bond (X being the nucleophile). This delocalization of electron cloud ('bonding', if I may say so) stabilizes the transition state. Note that there is no carbocation formed here.
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Offline looza

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Re: Resonance on Sn1 and Sn2 reactions
« Reply #2 on: April 23, 2011, 06:24:55 PM »
So basically, can we think of the electrophile carbon (in Sn2) to be kind of overworked during the transition state, and thus the C=O group helps ease the pressure by greater circulation of electrons?

I am always confused about e-withdrawing, donating groups. I know that OCH3 is a e- donating group, but why would the oxygen want to donate when its electronegative?

In this case, why is C=O electron withdrawing?

Offline Schrödinger

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Re: Resonance on Sn1 and Sn2 reactions
« Reply #3 on: April 24, 2011, 05:28:15 AM »
So basically, can we think of the electrophile carbon (in Sn2) to be kind of overworked during the transition state, and thus the C=O group helps ease the pressure by greater circulation of electrons?
More or less, yes

I am always confused about e-withdrawing, donating groups. I know that OCH3 is a e- donating group, but why would the oxygen want to donate when its electronegative?
In this case, why is C=O electron withdrawing?
When O donates electrons it can create a more stable resonance structure by completing the octet of a nearby carbon atom (example). This stability overrules any electronegativity barrier that the oxygen has to overcome in order to donate the electrons.

In this case, why is C=O electron withdrawing?
In the SN1 case where a carbocation is formed, the oxygen is not linked to the carbon via a sigma bond. Hence, it cannot donate its electrons to stabilize the carbocation. Rather, all it does is destabilize the carbocation further by pulling electrons towards itself, due to its high electronegativity.
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