2
$\begingroup$

enter image description here

Somehow I am told that my product is incorrect. Why is my product incorrect?

As far as I can see this is a Wittig reaction resulting in a two carbon increase with removal of the oxygen.

The only thing that strikes me as odd is that the Wittig is conjugated to a bromide salt. The only way this can happen is if the phosphorous part has an overall positive formal charge. Does this mean that the n-butyllithium will abstract a proton and leave the carbon backbone of the Wittig as $\ce{CH3CH}$?

Even so this doesn't seem to affect my final product ...

$\endgroup$
  • $\begingroup$ The Z isomer is also formed. $\endgroup$ – ron Nov 4 '14 at 0:58
  • $\begingroup$ @Ron does temperature have anything to do with this? $\endgroup$ – Dissenter Nov 4 '14 at 2:53
  • $\begingroup$ no there's no general trend bth are formed $\endgroup$ – RE60K Nov 4 '14 at 4:20
  • $\begingroup$ Choice of solvents will also influence the E/Z ratio $\endgroup$ – user32431 Aug 1 '16 at 20:50
5
$\begingroup$

Below is the preparation of the Wittig reagent you are given. Yes, the intermediate is a phosphonium salt, with a formal positive charge on phosphorus. Treatment with base (BuLi) deprotonates at the adjacent carbon giving a phosphorus ylide.

enter image description here

As for the Wittig reaction itself, one postulated mechanism (depicted in Organic Chemistry by Maitland Jones) is shown below. Nucleophilic attack of the ylide onto the aldehyde gives an open betaine intermediate, which closes to form an oxaphosphatane. The strained 4-membered ring decomposes readily to give an alkene and triphenylphosphine oxide.

enter image description here

The stereochemistry of the Wittig is related to the nature of the ylide used. Unstabilized ylides, such as the one here, predominantly give the Z-configured alkene, while stabilized ylides (usually through conjugation to a carbonyl) give an E-configured alkene as the major product. With substrates similar to the ones here, I obtained a 4:1 ratio of Z:E products, so the selectivity is not perfect, as K_P points out.

The stereochemical outcome is intimately related to the mechanism, and according to my copy of Jerry March's Advanced Organic Chemistry (4th ed), there is no evidence for the betaine intermediate. Instead, the oxaphosphatane is formed directly through a [2+2] cycloaddition mechanism. This mechanism is attractive because the stereochemical demands of the cycloaddition account for the preference of the (less stable) Z-product. However, this does not explain why different ylides give different selectivities.

$\endgroup$
  • $\begingroup$ there's no proof if betaine or oxaphophatane until now as Solomons Fryhle says $\endgroup$ – RE60K Nov 4 '14 at 4:07
  • $\begingroup$ For many years it was assumed that a diionic compound, called a betaine, is an intermediate on the pathway from the starting compoundsto the oxaphosphetane, and in fact it may be so, but there is little evidence for it. <sup>to the oxaphosphetane, and in fact it may be so, but there is little evidence for it.</sup> $\endgroup$ – RE60K Nov 4 '14 at 4:09
1
$\begingroup$

I suspect they want you to specify the stereochemistry too. The expected product should be the cis (predominantly at least). That's the textbook answer, reality is not that ideal.

EDIT: If I remember correctly the Z- isomer is favoured in higher temperatures and in lower temperatures you will get worse selectivity, or even predominantly trans. In any case your answer is not "wrong" maybe incomplete.

$\endgroup$
  • $\begingroup$ Could it be that the Z-isomer is favored at low temperatures? Since it's the less stable product, it must be kinetically formed. That implies that low temperature would favor the Z-product. $\endgroup$ – jerepierre Nov 4 '14 at 1:41
  • $\begingroup$ @jerepierre This (low temp E/Z selectivity) was in the back of my mind so I'll have to look it up. From a quick search, the wiki article seems to agree but I'll try to look further - at home now and missing my copy of March too. (as you mentioned it contains an extensive review of the Wittig ) $\endgroup$ – K_P Nov 4 '14 at 8:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.