September 07, 2024, 11:10:59 PM
Forum Rules: Read This Before Posting


« previous next »
Pages: [1]   Go Down

Topic: HCl + CH3CH2(CH3)(OH)CCH2CH2CH3 --> alkyl halide, but why 2 enantiomer product?  (Read 2050 times)

0 Members and 1 Guest are viewing this topic.

Offline Lo.Lee.Ta.

  • Regular Member
  • ***
  • Posts: 51
  • Mole Snacks: +3/-1
  • Gender: Female
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! :)
Logged

Offline orgopete

  • Chemist
  • Sr. Member
  • *
  • Posts: 2636
  • Mole Snacks: +213/-71
    • Curved Arrow Press
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?
Logged
Author of a multi-tiered example based workbook for learning organic chemistry mechanisms.
Pages: [1]   Go Up
« previous next »
Sponsored Links