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In the above question I know that the three reactions involved are $\mathrm{S_N2}$ reactions and therefore there shall thrice inversion of the substitute. So, I got the solution for the MCQ as:

3‐methyl‐cyclopentanol

However, the solution to the question says that the first structure is only correct and the second one is wrong.

As far I have learnt I know that the orientation of the whole compound doesn't matter because we can rotate the compound in our mind. Is this wrong?

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  • $\begingroup$ Whilst you're not wrong in the fact that "we can rotate the compound in our mind", do you reach the exact same molecule when doing the transformations on the second one? Do you see the relationship between the two potential starting materials? $\endgroup$
    – sjb
    Apr 2, 2020 at 6:04
  • $\begingroup$ i'm a bit confused still, the end product we get on rotating the compound not the same? $\endgroup$ Apr 2, 2020 at 6:09
  • $\begingroup$ Yes you can rotate the compound in your mind, but if you do it accurately, you'll see they are different. $\endgroup$ Apr 2, 2020 at 6:30
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    $\begingroup$ The 5 member ring is planar. The solid arrows mean the substituent is above the plane, the dashed arrorws on the $\ce{N3}$ functional group mean that the group is below the plane. The reactant only has one configuration since you can flip the molecule over. So you could draw the methyl group down and the $\ce{N3}$ group up and it would be the same molecule. $\endgroup$
    – MaxW
    Apr 2, 2020 at 6:55
  • $\begingroup$ @porphyrin It would be 108°, not 72. So after all, the ring is relatively close to being planar. In other words, it is not because its non-planarity that the two structures differ. $\endgroup$ Apr 2, 2020 at 8:11

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