# Why do non-polar solvents favour Sn2 reactions and why do polar solvents favour Sn1 reactions? Please simple ways. Thanks a lot

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#### Explanation

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#### Explanation:

I want someone to double check my answer

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anor277 Share
Feb 17, 2018

Well, ${S}_{N} 1$ reactions feature a reactive intermediate...a carbocation....

#### Explanation:

In an ${S}_{N} 1$ reaction, say of isopropyl halide, the reaction is presumed to proceed thru a carbocationic intermediate....i.e. a bond-breaking rate determining step.

${H}_{3} C - C H X C {H}_{3} \stackrel{\text{polar solvent, slow step}}{\rightarrow} {H}_{3} C - \stackrel{+}{C} H C {H}_{3} + {X}^{-}$

And the presumed intermediate ${H}_{3} C - \stackrel{+}{C} H C {H}_{3}$ is certainly stabilized by interaction with the solvent....

On the other hand, should isopropyl react with a NUCLEOPHILE, the reaction is presumed to occur by a bond-making mechanism that involves reaction of (say) hydroxide anion at the ipso carbon...

${H}_{3} C - C H X C {H}_{3} + H {O}^{-} \stackrel{\text{fast step}}{\rightarrow} {H}_{3} C - C H \left(O H\right) C {H}_{3} + {X}^{-}$

And this is a concerted mechanism, that involves no reactive intermediates. Less polar solvents magnify the reactivity of the intermediate...hydroxide, for instance is a more powerful nucleophile in ethanol or ether than it is in water, and the nucleophile is more reactive in the alcoholic/ethereal solvent....of course there is a trade-off in solubility versus reactivity...

I would look in your text for bond making versus bond-breaking reactivity....

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