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Looking at the following reaction, I understand how the product forms, however, can somebody help me understand why two products are formed? Would the identical mechanism produce the two products?

enter image description here


Edit: This is how I understand the mechanism.

enter image description here

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    $\begingroup$ Can you draw a mechanism to show what you think happens to form the product. The mechanism contains an enol as one of the key intermediates, and drawing this intermediate should give you an indication of why you get both 'products'. The 'products' are infact two enantiomers of the same product (i.e. the reaction produces a racemic mixture). $\endgroup$ – NotEvans. Dec 27 '15 at 13:20
  • $\begingroup$ @NotNicolaou I've added it. $\endgroup$ – Sal Dec 27 '15 at 13:43
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    $\begingroup$ Based on what you've drawn, you've answered your own question! xD If you look at the enol, its planar with respect to C=C, attack can happen from above that plane, or below that plane, leading to both enantiomers of product formed via an identical mechanism. $\endgroup$ – NotEvans. Dec 27 '15 at 13:46
  • $\begingroup$ @NotNicolaou Oh... That makes sense then. I expected there to be two different mechanisms. Thank you for the help. $\endgroup$ – Sal Dec 27 '15 at 13:51
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    $\begingroup$ Glad it makes sense. If you understand now, you can write and accept your own answer so anyone searching for the answer in the future can find it. $\endgroup$ – NotEvans. Dec 27 '15 at 13:53
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The mechanism that I have submitted in the edit of my post is correct. The reason both products are formed, is due to the the enol intermediate being planar with respect to the oxygen, allowing it to be attacked from either plane. This allows for the formation of a the so-called racemic mixture.

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The first step involves the formation of enol by protonation of the carbonyl oxygen. Then the electrophile adds to the enolate, and it may approach in either direction, from the top or the bottom of the plane of the enol.And so, a racemic mixture is obtained.

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    $\begingroup$ I’m not sure why this answer is downvoted. From what I can tell, it’s correct. $\endgroup$ – Jan Dec 27 '15 at 20:13

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