Question

e. Of course, cyclohexanol never exists in a flat hexagonal form. It is just drawn this way for convenience. It exists in one of two chair conformations that are rapidly interconverting at room temperature, the conformation with the hydroxyl in the equatorial position being favored. Гон — Ян Make a model of cyclohexanol with the OH in the equatorial position. Now do a chair flip, so that the OH is in the axial position. Can you find a plane of symmetry for each configuration? Is cyclohexanol chiral? 1. Now consider a disubstituted cyclohexane derivative, cis-1,2-dibromocyclohexane: Br Looking at the flat, hexagonal representation of cis-1,2-dibromocyclohexane, does this molecule contain a plane of symmetry? Yes No Once again, we know that cyclohexane rings never exist in a flat hexagonal form. They exist as rapidly interconverting chair conformations. Only at absolute zero can you isolate a single chair conformation of any cyclohexane derivative. If you convert cis-1,2-dibromocyclohexane to a chair conformation, however, both chair conformations are chiral: H Br Br : Br Both chair conformations are non-superimposable mirror images of eachother and there is no plane of symmetry, BUT they are interconvertible by a ring flip!

Answer 1

f. The molecule does possess a plane of symmetry for the two confirmations. The molecule is not chiral since the carbon having OH group has two -CH2- groups attached to it. Thus the four substituents are nonidentical.

Explanation: The two molecule possess plane of symmetry which can symmetrically divide the molecule into two halves. This is shown in the following figure.

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f. The planar molecule does have a plane of symmetry.

This is shown in the following figure.

Bor C IT