6
$\begingroup$

According to my textbook, 1-bromotriptycene is inert to any nucleophilic substitutions.

1-bromotriptycene

Due to steric hinderance we can say SN2 reactions cannot occur, but I think SN1 reactions should be favoured as formed carbocation will be resonance stabilised.

Can someone explain why it doesn’t undergo SN1 reactions?

Improve this question
$\endgroup$
2
  • 8
    $\begingroup$ Hard to get resonance stabilization. You can't align the vacant orbital on the carbocation with the $\pi$ electrons on the ring. $\endgroup$
    – Oscar Lanzi
    Commented Dec 9, 2017 at 10:55
  • $\begingroup$ Think of the consequences of an SN1 or SN2 reaction. $\endgroup$
    – user55119
    Commented Dec 9, 2017 at 21:43

2 Answers 2

Reset to default
3
$\begingroup$

SN1 reaction proceeds through carbocation intermediate which has planar structure. 1-Bromotriptycene posses a cage-like structure: the bridgehead carbon cannot assume planarity, hence the formation of a carbocation at the bridgehead position does not take place. Consequently, 1-bromotriptycene is inert to SN1 reactions.

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ It would be better if you make your answer more interesting by describing elaborately or adding some images. You could also mark some important sentences as bold, so that it becomes catchy. $\endgroup$
    – Soumik Das
    Commented May 16, 2018 at 12:34
0
$\begingroup$

Bredt's rule is a consequence of the fact that having a double bond on a bridgehead or carbocation is not stable for small rings (fewer than eight atoms) due to a combination of ring strain and angle strain (nonplanar alkene). Since SN1 reaction involves a carbocation intermediate, and such intermediate is not possible here, the SN1 mechanism is ruled out.

SN2 reaction involves a backside attack from the leaving group. Due to stearic hindrance to the backside attack, such reaction is also not possible. In general, nucleophilic substitution is not possible.

nucleophilic substitution

Improve this answer
$\endgroup$

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.