Strictly on theoretical basis, not based on experimental results.

I can very well be wrong, but I think that methanol should be optically inactive because it has a plane of symmetry passing through it. My teacher however insists that it is optically active as you can create two non-superimposable mirror images out of it.

Who is correct and why?

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    $\begingroup$ You can create two STATIC images of methanol that are non-superimposable mirror images but that does not make methanol optically active. Every conformation has a mirror image, i.e., a racemate, or a plane of symmetry. Both conditions lead to a lack of optical activity. Consider changing classes. $\endgroup$ – user55119 Jul 1 '19 at 13:20
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    $\begingroup$ The -OH group is pretty much free to rotate around. If not for this fact, your teacher would be wrong just once. $\endgroup$ – Ivan Neretin Jul 1 '19 at 15:39
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    $\begingroup$ Is your teacher an engineer? $\endgroup$ – electronpusher Jul 1 '19 at 15:47
  • $\begingroup$ Thanks @user55119 for your answer. I however for some reason can't tick your answer because I found it helpful. $\endgroup$ – Vivek Yadav Jul 3 '19 at 15:08

Just to sum up what has already been pointed out in the comments: Methanol can have infinite conformations in free space due to free rotation of the C-H and C-OH sigma bonds. Hence, for any particular spatial conformation you can think of for the molecule, it can always develop a perfect mirror image for the same almost instantly(in case you are wondering about some time lag between them) as conformers are in chemical equilibrium.

So what this essentially means is that, suppose some conformation A of methanol has polarized light in some direction (assuming the right-handed coordinate system, let's say the positive X direction) in some fashion(let's say circularly polarized, clockwise). Then, this particular interference pattern of light will be internally compensated by a conformation B of the molecule which will circularly polarize the light in the anticlockwise direction in the negative X direction. So, the emergent light ray will appear to be unpolarized for the observer


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