2
$\begingroup$

I have a lot of troubles with what I circled in red with the big question mark near. I am not speaking about stereochemistry but about conservation of number of hydrogen atoms, and also there is actually no reason for the electrons on the carbon near $R_1$ to attack the other carbon. In my lesson, the teacher didn't show how the extra hydrogen was removed.

Can you explain what my teacher didn't show or forgot to include?

Ramberg-Bäcklund rearrangement

Thank you in advance!

Improve this question
$\endgroup$
9
  • 1
    $\begingroup$ Yes there is a problem with the left-hand side. There is no leaving group and you would end up with a pentavalent carbon. $\endgroup$
    – orthocresol
    Commented Mar 27, 2016 at 0:22
  • 1
    $\begingroup$ You're right. That one on the left is definitely not right. I've never seen anything like that as a mechanism. Maybe he left a step out where a carbene forms and inserts into one of the C-H bonds? $\endgroup$
    – SendersReagent
    Commented Mar 27, 2016 at 0:26
  • $\begingroup$ @DGS he didn't speak about carbene at this time of the lesson. :/ $\endgroup$
    – ParaH2
    Commented Mar 27, 2016 at 0:35
  • 1
    $\begingroup$ It's also the Ramberg-Bäcklun>d< reaction. $\endgroup$
    – pH13 - Yet another Philipp
    Commented Mar 27, 2016 at 1:08
  • $\begingroup$ I wouldn't expect him to do so, since that isn't how the reaction is thought to work, but that's the only way I could see that anion reacting. $\endgroup$
    – SendersReagent
    Commented Mar 27, 2016 at 1:13

1 Answer 1

Reset to default
4
$\begingroup$

As discussed above, the pathway shown on the left is incorrect because it essentially shows insertion into a $\ce{C-H}$ bond by an anion with loss of $\ce{X^-}$, which is not the easiest path for this reaction to take by a long shot.

The actual mechanism for this reaction is as shown below.

  1. First, the non-halogenated alpha-carbon is deprotonated.

  2. The carbanion formed arranges itself such that the it can attack the halogenated carbon and in such a way that the two R-groups are cis to each other.

  3. Attack occurs, with loss of $\ce{X^-}$. Product is cis.

  4. In water, due to lack of solubility, decomposition is faster than redeprotonation and the (Z)-alkene forms.

  5. In organic solvent, base deprotonates an alpha-carbon, which can then scramble.

  6. The main reprotonation product is the trans product.

  7. The trans product decomposes, and the (E)-alkene is the major product.

Ramberg-Bäcklund mechanism

A few notes:

  • The red path makes almost completely (Z) product. The blue path makes a mix of the two with the (E) product as the main product.

  • Purple star: the reason the anion attacks to form the cis ring is unknown, but there are several theories, apparently.

  • Green stars: cheletropic ring decomposition is symmetry forbidden, so this is theorized to be a two-step decomposition through either diradical bond fissions or a dipolar mechanism.

Improve this answer
$\endgroup$
2
  • $\begingroup$ DGS, could you try to rebuild your picture wirh some program like ChemDraw, Chemsketch or Marvin Sketch? Though white board self drawn graphics are very fancy, they are not as easy to read as a proper digital version. $\endgroup$
    – pH13 - Yet another Philipp
    Commented Mar 28, 2016 at 13:09
  • 1
    $\begingroup$ @pH13 I can and will, but at this point I haven't had the chance. $\endgroup$
    – SendersReagent
    Commented Mar 28, 2016 at 13:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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