11
$\begingroup$

Why is a bridgehead atom bearing a negative or a positive charge unstable? Is it something with the structure of bridge bonds? I am unable to understand why.

enter image description here

$\endgroup$
4
$\begingroup$

Since the answer by Aditya Dev has already dealt with the carbocation part, I'll write about the carbanion one. The OP states that:

Why is a bridgehead atom bearing a negative or a positive charge unstable?

This is partially incorrect. Carbanions in bridgehead position are infact stable. From March Advanced Organic Chemistry, 6th ed, Chapter 5, Carbanions:

Carbons at bridgeheads, although extremely reluctant to undergo reactions in which they must be converted to carbocations, undergo with ease reactions in which they must be carbanions and stable bridgehead carbanions are known.

(my emphasis)

It also says that:

For other evidence that carbanions are pyramidal, see Streitwieser, Jr., A.; Young, W.R. J. Am. Chem. Soc. 1969, 91, 529; Peoples, P.R.; Grutzner, J.B. J. Am. Chem. Soc. 1980, 102, 4709.

This makes the picture completely clear. Bridgehead carbanions are not planar, but in fact, pyramidal. So, they are definitely not sp2 and hence, the Bredt rule isn't violated by their being stable.

$\endgroup$
6
$\begingroup$

The hybridisation of a bridgehead carbon when its not charged is sp3. Now, if you remove a hydrogen atom, its hybridisation becomes sp2.

Normally, in sp2 hybridisation, the 3 hybridised orbitals lie in a single plane for minimised repulsions. But the bridgehead carbon with positive charge would not have hybridised orbitals in a single plane. This increases repulsion and makes the compound unstable.

Here is a link - Bredt's Rule - that explains the same thing, but instead of a positively charged bridgehead, they discuss a double bond on the bridgehead atom.

$\endgroup$

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.