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I was reading Solomon's Organic Chemistry and came across the Dissolving Metal Reduction of 4-Octyne to (E)-4-Octene. (The same as can be found this question)

But the book states that the (E)-4-Octene is only produced in a 52% yield. This doesn't seem very impressive to me. What is the other 48%? If it's (Z)-4-Octene, can we really say that the reduction favors anti addition? If it's octane, I'd understand, but still would think you're losing a lot of product.

So does anyone know what the other products in the reaction are?

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    $\begingroup$ Without detail of the experimental conditions it is difficult to tell where the remaining mass balance is. Typical yields I've seen are much higher (e.g. here books.google.co.uk/…) It could just be an inefficient work-up and isolation procedure on the part of the experimenters. $\endgroup$ – Waylander Nov 16 '19 at 16:09
  • $\begingroup$ @Waylander Those yields are better. I noticed the resulting alkenes are drawn with wavy bonds, does that imply that cis and trans products are produced? Because this link suggests that the mechanism also produces some cis product. $\endgroup$ – user137 Nov 16 '19 at 16:14
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    $\begingroup$ They are indeed drawn wavy, but the text refers to "reductions in good yields, and in a remarkable state of isomeric purity" . The link you highlighted seems confused as they offer the cis product as the major product in one diagram, yet (correctly) show the mechanism as giving trans. I'm inclined to blame an inefficient isolation procedure for the low yield, possibly losses on distillation. $\endgroup$ – Waylander Nov 16 '19 at 16:31

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