Which of the following compounds can exist as pairs of...

Which of the following compounds can exist as pairs of cis-trans isomers? Draw each cis-trans pair, and indicate the geometry of each isomer. (a) $\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}_{2}$ (b) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CHCH}_{3}$ (c) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CHCH}_{3}$ (d) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{C}\left(\mathrm{CH}_{3}\right) \mathrm{CH}_{2} \mathrm{CH}_{3}$ (e) $\mathrm{ClCH}=\mathrm{CHCl}$ (f) $\mathrm{BrCH}=\mathrm{CHCl}$

Which of the following compounds can exist as pairs of cis-trans isomers? Draw each cis-trans pair, and indicate the geometry of each isomer.
(a) $\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}_{2}$
(b) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CHCH}_{3}$
(c) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CHCH}_{3}$
(d) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{C}\left(\mathrm{CH}_{3}\right) \mathrm{CH}_{2} \mathrm{CH}_{3}$
(e) $\mathrm{ClCH}=\mathrm{CHCl}$
(f) $\mathrm{BrCH}=\mathrm{CHCl}$

Key Concepts

Geometric (cis-trans/E-Z) Isomerism

This concept involves the existence of different spatial arrangements around a double bond when rotation is restricted. For an alkene to display cis-trans (or E-Z) isomerism, each carbon atom of the double bond must have two different substituents. The relative positions of these substituents—being on the same side (cis/Z) or opposite sides (trans/E)—result in distinct isomers with different physical and sometimes chemical properties.

Substituent Requirements for Stereogenic Double Bonds

A double bond becomes stereogenic and capable of geometric isomerism only when both of its carbons are bonded to two different groups. If either of the carbons carries two identical substituents, the double bond cannot adopt two distinct spatial arrangements, and as a result, cis-trans isomerism is not possible. This criterion is fundamental in predicting which alkenes can or cannot display geometric isomerism.

Cahn-Ingold-Prelog Priority Rules

These rules determine the order of priority among substituents attached to the double bond carbons, which is crucial when assigning the configurations (E or Z) to the isomers. According to these rules, the substituent with the higher atomic number is given higher priority. When the two higher priority substituents lie on the same side of the double bond, the configuration is designated as Z (from the German 'zusammen'); if they are on opposite sides, it is designated as E (from 'entgegen').

Transcript

00:01 We have to distinguish here the given alkenes into cis and trans isomers if possible.

00:07 So first of all, we have ch3, ch double bond ch2.

00:16 Okay, so here you can see that this terminal carbon, it has two identical hydrogen atom.

00:22 So since terminal carbon has two identical hydrogen.

00:37 Hydrogen atoms cis trans isomarism is not possible then we have b part ch3 old twice c double bond chchc3 so now it can be represented as we have two carbon atoms containing double bond now one side i have hydrogen atom and one side i have methyl growth.

01:28 For this carbon also one side i have methyl group and another side also i have methyl group.

01:36 So now you can see that because of identical ch3 group, no cis trans isomer is.

02:09 Now see part ch3 ch2 ch2 double bond chch ch ch ch ch 3 now this can be represented as i have the carbon atoms containing double bond and now to one carbon i have this hydrogen atomatives and then i have this ch2 ch3 group attached ethyl group and then to this carbon i have hydrogen atom attached and to another side i have this methyl group attached so here you can see that on one side i have same kind of hydrogen atoms and on another side i have these methyl and ethyl group so when the same kind of group present on the same side okay so that configuration is called cis configuration okay when same same kind of atoms or group are present adjacent to each other.

03:15 So that configuration is sis.

03:18 And now it can also be represented as i have the double bonded carbon atoms.

03:25 Now the hydrogen atoms are present opposite to each other and this ethyl group and methyl group so that is also present.

03:38 So now this will be trans isomer because the hydrogen atoms are present opposite to each other.

03:49 D part i have ch3, whole twice, c, double bond c, c, h3, ch3, ch2, ch3.

04:02 So this can be represented as i have the double bonded carbon atoms.

04:07 Now on one side i have the methyl groups attached and on another side i have this methyl group test and on another side i have this ethyl group a test okay so this will be the complete structure so because of identical ch3 group no cis trans isomerism e part we have cl ch double bond ch c h c s so now we have the double bonded carbon atoms and we have hydrogen atoms on one side, cl atoms on one side...

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