Which of these molecules can have cis and trans isomers? For...

Which of these molecules can have cis and trans isomers? For those that do, write the structural formulas of the two isomers and label each cis or trans. For those that cannot have these isomers, explain why. (a) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{BrC}=\mathrm{CBrCH}_{3}$ (b) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{C}\left(\mathrm{CH}_{3}\right)_{2}$ (c) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{lC}=\mathrm{ClCH}_{2} \mathrm{CH}_{3}$ (d) $\mathrm{CH}_{3} \mathrm{ClC}=\mathrm{CHCH}_{3}$ (e) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CHCH}_{3}$

Which of these molecules can have cis and trans isomers? For those that do, write the structural formulas of the two isomers and label each cis or trans. For those that cannot have these isomers, explain why.
(a) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{BrC}=\mathrm{CBrCH}_{3}$
(b) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{C}\left(\mathrm{CH}_{3}\right)_{2}$
(c) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{lC}=\mathrm{ClCH}_{2} \mathrm{CH}_{3}$
(d) $\mathrm{CH}_{3} \mathrm{ClC}=\mathrm{CHCH}_{3}$
(e) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CHCH}_{3}$

Key Concepts

E/Z Nomenclature and Priority Rules

In cases where the simple cis-trans descriptors are inadequate or ambiguous, the E/Z system is used. This method applies the Cahn–Ingold–Prelog priority rules to assign priorities to the substituents on each carbon of the double bond, thereby determining the stereochemical configuration of the molecule.

Substituent Requirements on the Double Bond

For cis-trans isomerism to occur, each carbon atom involved in the double bond must have two different substituents. If a carbon has identical substituents, then no distinct geometric isomers can form because the relative positioning of substituents cannot vary.

Cis?Trans (Geometric) Isomerism

This is a form of stereoisomerism that occurs in compounds with double bonds, where the restricted rotation around the double bond leads to different spatial arrangements of substituents. It is characterized by 'cis' and 'trans' designations depending on whether similar or higher priority groups are on the same side or opposite sides of the double bond.

Double Bond Rigidity

A carbon–carbon double bond consists of a sigma and a pi bond. The pi bond creates a region of electron density above and below the plane of the molecule, preventing free rotation about the double bond. This rigidity is fundamental to the existence of geometric (cis-trans) isomers.

Transcript

00:02 All right, our part a is ch3, ch2, brc, double bond c, br, ch3.

00:15 And the way it's typed out really isn't super clear about who's attached to who.

00:21 So the way that i would interpret this would be that you've got a cc double bond.

00:27 And then on the left hand side, you've got two groups.

00:31 One of them is that ch3, ch2, and then the other one is a bromine.

00:36 All right.

00:37 So you've got bromine and then down below it's ch3, ch2.

00:43 I don't think it's ch3, then ch2, then br all separate because the carbon would have too many bonds.

00:51 That would put that carbon in the double bond having five and that's not possible.

00:54 And then i do the same thing on the other side.

00:57 Okay, separate it out.

00:59 I think that's what was intended here.

01:01 All right.

01:01 Now, in order to be a cis and trans isomer, the biggest thing is that on either side of the double bond, so these two groups and then these two groups, the bromine and then the ch3 -ch2, those cannot be identical.

01:18 And then the same thing on the left hand side.

01:21 Those two and those two cannot be the same thing.

01:24 You have to be able to tell them apart.

01:26 So here you could definitely do that.

01:29 You could, to get the other form, to get the other isomer, you would just flip one side and then this side, you would just flip it.

01:44 It doesn't actually matter which side you flip.

01:47 What i mean is you swap either the two that are on the right or the two that are on the left.

01:53 Okay.

01:54 And that creates the other form.

01:56 In this case where both the bromines are sticking up, that would be your cis isomer.

02:02 Come on, little pen.

02:03 There we go.

02:04 And then the one where they're opposites, that is trans.

02:11 All right.

02:12 Next one.

02:13 We have got ch3 in parentheses two, c, c, c, and then ch3, parentheses, 2.

02:23 What this usually interprets as is the carbon that's in the double bond has two different ch3 groups attached to it.

02:33 So we are looking at something like this.

02:43 And putting together what we had just talked about, these two groups would have to be different because you'd have to be able to tell them apart.

02:51 And then the same thing for the two right -hand groups.

02:53 They could not be the same thing.

02:55 So this one cannot be.

02:59 Isomers because you can't tell the sides apart.

03:05 C was a little, i think there was a typo in c.

03:09 I think i figured out what it was trying to say, though.

03:13 I think what the formula should have been, should have looked something like this...