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Topic: HCl + CH3CH2(CH3)(OH)CCH2CH2CH3 --> alkyl halide, but why 2 enantiomer product?  (Read 2073 times)

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Offline Lo.Lee.Ta.

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HCl + CH3CH2(CH3)(OH)CCH2CH2CH3 --> alkyl halide, but why 2 enantiomer product?
« on: November 27, 2012, 11:37:58 PM »
This is the reaction: (S) CH3CH2(CH3)(OH)CCH2CH2CH3  :rarrow: 
(S) CH3CH2(CH3)(Cl)CCH2CH2CH3 + (R) CH3CH2(CH3)(Cl)CCH2CH2CH3  +  H2O

Why is the product two enantiomers? I thought it would be a pure product of
(S) CH3CH2(CH3)(Cl)CCH2CH2CH3.
I thought enantiomers were the products of only Sn1 reactions?

This is just an alcohol :rarrow: alkyl halide reaction...

Thanks! :)
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Offline orgopete

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Re: HCl + CH3CH2(CH3)(OH)CCH2CH2CH3 --> alkyl halide, but why 2 enantiomer product?
« Reply #1 on: November 28, 2012, 01:07:04 AM »
I think you need to understand the question differently. The S-alcohol is giving a racemic mixture of chlorides. What is the rate limiting step or what is the mechanism of this reaction?

We often go into the lab thinking we know what the products should be. When we analyze the products we learn something about the chemistry. You may have thought this should be an SN2 reaction, but is it?
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