Rates of hydrolysis in bicyclic and monocyclic compounds


> 
> Which of the following compounds, 1-bromo-1-methylcyclohexane **1** or 1-bromobicyclo[2.2.2]octane **2**, hydrolyzes faster (under comparable conditions)? It may be acid catalyzed or base catalyzed hydrolysis, but the conditions must remain the same for each pair.
> 
> 
> [![enter image description here](https://i.stack.imgur.com/ffajC.png)](https://i.stack.imgur.com/ffajC.png)
> 
> 
> 


In my view because of the steric hindrance in first compound, hydrolysis should be easier for second, but I am not sure about it.


See part one of my question [here](https://chemistry.stackexchange.com/q/64574/32360).




I’m going to shamelessly steal [electronpusher’s drawing](https://chemistry.stackexchange.com/a/64822) for clarification:


[![conformations of both molecules in question](https://i.stack.imgur.com/3rExc.png)](https://i.stack.imgur.com/3rExc.png)


Two general hydrolysis mechanisms are possible: $\mathrm{S\_N2}$ and $\mathrm{S\_N1}$. First, let us consider $\mathrm{S\_N2}$ since the analysis is more simple.


Both of the two species are tertiary bromides meaning that $\mathrm{S\_N2}$ reactions are already severely inhibited. However, the second compound is even worse: if the nucleophile tries to attack from behind to interact with the $\sigma^\*\_{\ce{C-Br}}$ orbital, it would need to come from *inside* the bicyclo[2.2.2]octane system. This is completely impossible, not only due to sterics. Thus, while compound **1** is *very* slow in $\mathrm{S\_N2}$ reactions, it is not possible for **2** to react in an $\mathrm{S\_N2}$ manner.


Conclusion: **1** should react faster by $\mathrm{S\_N2}$ — albeit still unmeasurably slow.




---


To undergo an $\mathrm{S\_N1}$ reaction, first the halide must dissociate to give a carbocation. Both species are, as noted above, tertiary, meaning that carbocations are well stabilised by triple hyperconjugation. Initially, we expect both $\mathrm{S\_N1}$ mechanisms to be fast.


A carbocation when formed must practically always adopt a planar, $\mathrm{sp^2}$ hybridised configuration. Thereby, the unoccupied orbital has the highest energy and all bonding electrons are as low as possible in energy. Any distortion away from the perfect $\mathrm{sp^2}$ geometry requires the empty orbital to be stabilised at the cost of populated orbitals — very unfavourable.


For bridgehead carbons as in **2**, especially those that bridge small rings (which is the case here), the general rule is that they *do not* undergo reactions that require planar transition states for exactly that reason: they are too strained to adopt an $\mathrm{sp^2}$ configuration. (This configuration would not only require the bridgehead to ‘move inwards’, but would also strain the bonds of the three neighbouring carbons, whose angles would have to be decreased.)


Conclusion: while **1** reacts rapidly *via* $\mathrm{S\_N1}$, compound **2** does not react *via* $\mathrm{S\_N1}$.




---


Complete conclusion: **2** is unreactive towards hydrolysis. **1** therefore reacts much faster.


 5

Both hydrolysis reactions will almost certainly occur by $\mathrm{S\_N1}$, forming carbocation intermediates. I'm not sure if sterics will play the biggest role here. One thing you should consider is the actual conformations of these molecules. The bicyclic is slightly more rigid and may not adjust to the trigonal planar geometry of a carbocation as easily as the cyclohexane, and this will likely be the deciding factor. Here are the conformations, as I hope you know. 


[![enter image description here](https://i.stack.imgur.com/3rExc.png)](https://i.stack.imgur.com/3rExc.png)


 3

Hydrolysis of the first compound is easier than that of the second compound. Because the second compound is the bridged compound, it is constrained to be non-planar. 


The carbocations are $\ce{sp^2}$ hybridized, so the normal geometry should be trigonal planar. But because of multiple rings in different planes, it exhibits pyramidal type geometry (i.e., non-planar.)


Because of this non-planar geometry, adjacent $\ce{p}$ orbitals of two carbons are very staggered which further results in poor overlap of those $\ce{p}$ orbitals.


 1