Using wedge-and-dash bonds, draw both the cis and trans stereoisomers for each of the following compounds:
(b) 1,3-diethylcyclobutane

Question

Using wedge-and-dash bonds, draw both the cis and trans stereoisomers for each of the following compounds:

(b) 1,3-diethylcyclobutane

Accepted Answer

Welcome back everyone. If the is and trans stereo isomers of one fl three isom bey Cyclo Beane using watch and dash bonds, what we want to do in this problem is first, we'll draw cycle be saying it is a four member ring. And as we can see, position number one contains an ethyl group. So what we want to do is just draw a two membered LK group and position number three contains an iso beetle group. So what we want to do is draw a three member chain and add a metal group to carbon. Number two, we have the structure of one ethyl three isometry cyclo Beane. But now we want to draw its cysts and trans stereo isomers. Let's recall that this means that the substituents bonded to the ring are on the same side relative to the plane of the ring. So we can either draw two wedges or two dashes. And the reason why we can draw either of them is simply due to the plane of symmetry. We can see that there's a diagonal line passing through the four membered ring, right. So essentially our structure has an internal plane of symmetry let's suppose that we draw two wedges. And this means that our substituents are pointing up relative to the viewer, right. So relative to our view, these substitutions are pointing towards us if we are rotated around the diagonal, well, essentially we will get two dashed bonds and that be the same structure. So we only have one possible structure for a cis arrangement. Now, for the trans arrangement, we essentially want to add a one, a wedged bond and one dashed bond because trans means that the two substations are on opposite sides relative to the plane of the ring. So we're choosing one substituent to be on a wedge and the other one on a dashed bond. And now if we rotate it, we will get a dash and a wedge respectively, which will not be a unique structure, right? Generally, if we don't have an internal plane of symmetry, we can draw 22 C and two transforms. But in this case, due to the plane of symmetry, we essentially can rotate our structure. And, and we can see that there's only one cell form and one transform. Thank you for watching.

Using wedge-and-dash bonds, draw both the cis and trans stereoisomers for each of the following compounds:
(b) 1,3-diethylcyclobutane

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Key Concepts

Stereoisomers

Cis and Trans Isomerism

Wedge-and-Dash Notation