On my study guide for synthesis reactions (specifically for the synthesis of disubstituted benzene derivatives), two of the problems look extremely similar.

The question on the guide asks us to synthesize each of these by using benzene as a starting material.

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They appear to be the same molecule, just with a different orientation of the ethyl group. And in the solutions section of the guide, both have the same exact synthesis reaction listed.

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Is there actually a way to determine the orientation of the substituent that isn't being explained by my study guide? Perhaps something done during the Friedel-Crafts acylation step? Or is this simply an effort to try and trip me up by asking the same question twice?

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    $\begingroup$ There is nothing to determine. They are the same molecule. $\endgroup$ – Ivan Neretin Oct 12 '17 at 4:33

Unless the temperature is really low, you will not be able to distinguish the two conformation isomers (conformers).

Rotation about single C-C bonds is rarely hindered. It can be, by using bulky groups or a tether, but neither is the case here. I expect the benzyl*-methyl bond and the phenyl*-ethyl bond to rotate freely at reasonable temperatures (say, above 250K).

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  • $\begingroup$ Arbitrary very high temperature you chose. Why $\pu{250K}$ and not $\pu{100K}$? $\endgroup$ – Jan Oct 12 '17 at 12:49
  • $\begingroup$ I have no data to back up the temperature claim. I was thinking of the typical conditions of the synthesis given in the question. $\endgroup$ – TAR86 Oct 12 '17 at 13:13

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