All are tertiary halides so that attack by the $S_{N^2}$ mode would not be  expected to occur on $(16)$ or $(17)$ any more than it did on $(8)$. $S_{N^2}$ attack 'from the back' on the carbon atom carrying $Br$ would in  any  case be prevented in $(16)$ and $(17)$ both sterically by their cagelikc  structure, and also by the impossibility of forcing their fairly rigid framework through transition states with the required planar distribution of  bonds to the bridgehead carbon atom. Solvolysis via rate- limiting formation of  the ion pair $(S_{N^1})$, as happens with $(8)$ is also inhibited because the  resultant carbocations from $(16)$ and $(17)$ would be unable, because of  their rigid fr ameworks, to stabilise themselves by collapsing to the stable  planar state. These carbocation intermediates are thus of very much higher  energy level than usual, and therefore are formed only slowly and .with  reluctance. The very greatly reduced solvolysis rate of $(17)$ compared with $(16) $ reflects the greater rigidity about the bridgehead (cationic) carbon with  a one  carbon $(17)$, than with a two- carbon $(16)$, bridge.