The following compound is a carbocation.

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Clearly, the given compound does not have any element of symmetry. The compound is non planar due to the presence of ethyl group. Thus it must be chiral. But my book says the compound is achiral. How is this possible? Please explain.

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    $\begingroup$ It is planar.$\,$ $\endgroup$ – Ivan Neretin Aug 13 '19 at 6:19
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    $\begingroup$ The entire compound is not planar, but the bond geometry around the positive carbon is (trigonal) planar. All orientations of the ethyl group are equivalent. $\endgroup$ – electronpusher Aug 13 '19 at 6:24
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    $\begingroup$ Yes, that's right. $\endgroup$ – Ivan Neretin Aug 13 '19 at 6:27
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    $\begingroup$ @GuruVishnu Essentially correct. You may want to look into the difference between Rotational Isomers and Optical Isomers. $\endgroup$ – electronpusher Aug 13 '19 at 6:35
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    $\begingroup$ @electronpusher, Ivan Neretin, Thank you for clearing my doubt. $\endgroup$ – Guru Vishnu Aug 13 '19 at 6:36

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