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Basically I want to know what does the question mean when it says "taking into consideration conformational interconversions."

enter image description here

I assume it means, when I draw the mirror image of the given compound, the mirror image can be same as the compound itself (the case with option (a) and option (c) ) or it (the mirror image) can be same as confomer of the compound (the case for option (a) and option (b) ), for the compound to be achiral.

Note: option (a) shows both the features because option (a) is same as it's confomer and it is same as it's mirror image as well.

Is my assumption correct?

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Your analysis is correct. The stipulation "taking into consideration conformational interconversions" means that conformational interconversion is fast. That's important because compounds that we consider achiral can have chiral conformations. I used the example of butane in a previous question.

In this problem, three compounds are given that have multiple stereocenters, compounds b, c, and d. A common way to decide whether a compound with multiple stereocenters is chiral is to look for planes or points of symmetry within the compound. Compound c has a plane of symmetry, so it is clearly achiral. Neither of compounds b or d has a plane of symmetry in the drawn conformations, which means that those conformations are chiral. However, as you suggest, chair flip of b gives the mirror image of the original drawing, which means the original and mirror image are equivalent, assuming that the two conformations can interconvert.

There are many cases where conformations cannot interconvert, and that leads to very interesting compounds that don't even require a stereocenter to be chiral (again see earlier question).

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    $\begingroup$ I can add figures if that would be helpful. $\endgroup$ – jerepierre Jul 28 '15 at 15:26
  • $\begingroup$ Thanks a lot for your input @jerepierre. I believe I have it figured out but you can add figures if you like as other stackexchange users might find it helpful. Thanks again. :) $\endgroup$ – TheProgrammer Jul 28 '15 at 20:30
  • $\begingroup$ For this question, if I draw the planar ring form for each of the cyclohexane rings, only d) doesn't have a plane of symmetry. Would that be a correct way to approach the question? $\endgroup$ – kane9530 Apr 2 '16 at 15:21
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    $\begingroup$ @kane9530 yes, because the planar form is essentially the 'average' of the conformations. $\endgroup$ – jerepierre Apr 2 '16 at 15:48

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