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I have a difficult question in my problem sheet that I cannot solve. I have been asked to explain the following observation:

enter image description here

I have a number of issues with this:

1) The fast reaction must be a syn-elimination. How can syn E2 elimination occur? If an E2 elimination is stereospecific because it must go through an anti-periplanar transition state surely syn-peri-planar elimination cannot happen.

2) How does the slow reaction even happen? The H and the Cl are never going to be in the same plane (syn or anti).

As a side question: does the boat conformation (which the bicyclic molecule must adopt) have equatorial and axial sites like the chair conformation does?

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    $\begingroup$ You might find this link helpful. Do you know how to draw a Newman projection? If so, draw the Newman projections, sighting down the C2-C3 bond (C2 and C3 are the two chlorine containing carbon atoms), for both of your dichloro compounds. This should make it easy to understand the arguments presented in the first link above. Hope this helps. $\endgroup$ – ron Jan 21 '15 at 1:42
  • $\begingroup$ What mechanism does the slow reaction undergo? $\endgroup$ – RobChem Jan 21 '15 at 8:23
  • $\begingroup$ One could run additional experiments to find out, but without additional data I'm not sure, either E1 or E1CB. $\endgroup$ – ron Jan 21 '15 at 16:25
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Yes, the fast reaction is a syn-periplanar E2 reaction. If we are ranking the rates of E2 based on the orientation of the C-H and C-LG bonds, anti-periplanar would be the fastest, syn-periplanar would be slower, and perpendicular would be the slowest. The anti-periplanar arrangement is the fastest because it provides the best overlap of the (filled) bonding C-H orbital with the (empty) anti-bonding C-LG orbital. The syn-periplanar arrangement is slower because there is much less productive overlap between these orbitals. However, the small lobe of the bonding C-H molecular orbital does provide some overlap with the anti-bonding C-LG orbital. A perpendicular arrangement results in no productive overlap.

enter image description here

The slow reaction must occur by some other mechanism, such as an E1 or E2-conjugate base mechanism.

enter image description here

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  • $\begingroup$ Why would E1 be slower? Is E2 absolutely impossible, they're not completely perpendicular are they? Could the structure be somewhat distorted? Also E1cb can't work because the lone pair isn't in the same plane is it? $\endgroup$ – RobChem Jan 20 '15 at 22:32
  • $\begingroup$ E1 would be slower because carbocation formation is inherently slow. I think a concerted E2 is unlikely. You're right that the C-H and C-LG bonds aren't completely perpendicular, but they are close enough that it's unlikely that E2 can operate. I think E2cb is unlikely because the base isn't strong enough. However, if that deprotonation did occur, the elimination of chloride would be fast. $\endgroup$ – jerepierre Jan 20 '15 at 22:48
  • $\begingroup$ But the lone pair orbital would not be in the same plane as the C-LG bond in E2cb (same problem as E2) surely? If deprotonation were to occur that is. $\endgroup$ – RobChem Jan 20 '15 at 22:56
  • $\begingroup$ This is a guess, but the anion would probably be high enough in energy to overcome the non-ideal orbital alignment. $\endgroup$ – jerepierre Jan 20 '15 at 23:14
  • $\begingroup$ Wouldn't the empty p orbital of the carbocation in E1 also be in the incorrect or orientation? Also, surely the base is too strong for E1 $\endgroup$ – RobChem Jan 20 '15 at 23:17

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