[Fla27]

Hi

My text book asked whether propene and 2-methylbut-2-ene can exist as cis-trans isomers ? I said they can't exist as cis-trans isomers because if you rotate the groups you get a mirror image of the of the compound you just had before rotation of the bonds. Is it right ?

Also when they are mirror images of each other can't they be classified as Enantiomers ?

Aren't Enantiomers a type of geometric isomerism ?

Can someone please confirm with me ?

Thank You

[Hunter2]

You need a molecule ABC=CAB, then A and B can be cis/tans-isomers. In your case you have AAC=CAB and there can be no cis/transisomers.

[mjc123]

You are basically right, but your expression is a little unclear. You cannot physically rotate groups about a double bond (which would interchange cis and trans), but you can rotate the molecule about an axis including the double bond. For a molecule with two identical groups on one side of the double bond (AAC=CAB), the result of the two processes would be the same, so there are not two distinct isomers.

Your molecules have a plane of symmetry (the plane containing all carbon atoms, assuming free rotation of the methyl groups), so the mirror image is the same as the original molecule. The plane perpendicular to this and containing the C=C bond is not a plane of symmetry; reflection in it would exchange the substituents on the same C atoms, but the result would be the same as physically rotating the molecule, so it is not an enantiomer. Enantiomers are non-superimposable mirror images, such as you get with a C atom with 4 different substituents. They are not classified as geometrical isomers.

[Fla27]

You are basically right, but your expression is a little unclear. You cannot physically rotate groups about a double bond (which would interchange cis and trans), but you can rotate the molecule about an axis including the double bond. For a molecule with two identical groups on one side of the double bond (AAC=CAB), the result of the two processes would be the same, so there are not two distinct isomers.

Your molecules have a plane of symmetry (the plane containing all carbon atoms, assuming free rotation of the methyl groups), so the mirror image is the same as the original molecule. The plane perpendicular to this and containing the C=C bond is not a plane of symmetry; reflection in it would exchange the substituents on the same C atoms, but the result would be the same as physically rotating the molecule, so it is not an enantiomer. Enantiomers are non-superimposable mirror images, such as you get with a C atom with 4 different substituents. They are not classified as geometrical isomers.

Thank You very much. Your explanation is very thorough. I appreciate it. 

[Fla27]

You need a molecule ABC=CAB, then A and B can be cis/tans-isomers. In your case you have AAC=CAB and there can be no cis/transisomers.

Thank You.