# Why does 1,3-dimethylcyclopent-1-ene have geometric isomerism while 1-bromo-2-chlorobenzene doesn't?

The bottom compound has bromine and an R group connected to one of its SP2 double bond carbons, and chlorine and the same R group connected to the other. This fact, it would seem, indicates cis-trans isomerism.

The top compound has a methyl group and an R group connected to one of its SP2 double bond carbons, and a hydrogen atom and a different R' group connected to the other. Conversely, it appears that this precludes cis-trans isomerism.

First, Why does the top compound has geometric isomerism while the bottom compound doesn't given these facts?

Secondly, given that benzene is planar and its substituents and hydrogen atoms are level with the plane of the ring, are the methyl group and the hydrogen atom around the double bond of the top compound also flatly level as in benzene, or are they equatorial like in cyclohexane, slating above or under the plane of the ring?

• Benzene ring is planar all right, but cyclopentene isn't. – Ivan Neretin Oct 13 '17 at 19:13
• So, is the cis-trans isomerism a result of the Methyl groups' at positions 1 and 3 possibly being at different sides of the ring's plane? Also, can the Methyl group at position 1 be axial as well as equatorial? – Frank Oct 13 '17 at 20:02
• There is no cis/trans isomerism here. – Ivan Neretin Oct 13 '17 at 20:27
• I'm confused. In the assignment it's noted that 1,3-Dimethylcyclopent-1-ene has cis-trans isomerism. From what I could dig up, cycloalkenes can have only a trans configuration. If so, is the methyl group at the double bond perfectly planar or does it slant in such a manner as to form a trans isomer? – Frank Oct 13 '17 at 21:54

IUPAC defines geometric isomerism as an obsolete synonym for cis-trans isomerism. This makes it a subset of diastereoisomerism, meaning that cis and trans isomers exhibit different properties such as melting point or density. The descriptors cis and trans themselves are also obsolete for double bonds and proper nomenclature should instead use (E) and (Z) as stereodescriptors since these are also applicable to tri or tetrasubstituted double bonds.

While linear or branched alkene chains can — as soon as the double bond is substituted asymmetrically — typically exist in both the (E) (or trans) and (Z) (or cis) isomers, this is not true for any cycloalkenes with less than eight carbon atoms. (E)-Cyclooctene is the smallest and most strained (E) configured cyclic alkene. You can use a molecular modelling kit to find out why this is the case.

Therefore, neither of your two compounds can exhibit cis-trans isomerism. Both ring sizes — the six-membered benzene ring and the five-membered cyclopentene ring — are smaller than eight and can therefore not accomodate an (E) configured double bond.

In an additional exercise, we may investigate whether other types of stereoisomerism can apply to the compounds in question — basically this boils down to whether enantiomers can exist. The flat benzene ring is $C_\mathrm s$ symmetric and thus no enantiomers are possible. However, the 3-methyl group on 1,3-dimethylcyclopentene can point both above (R) or below (S) the paper plane.

Neither has geometric isomers. 1,3-Dimethylcyclopent-1-ene is necessarily in its (Z)-configuration, because a five membered ring can't contain a trans double bond. The smallest stable trans cycloalkene is cyclooctene.

• Think about what the (E)-isomer of 1,3-dimethylcyclopent-1-ene would look like, if it existed! – WMe6 Oct 13 '17 at 20:35
• In the assignment it's noted that 1,3-Dimethylcyclopent-1-ene has cis-trans isomerism. I was thinking it could be due to the methyls' at 1 and 3 being on either different sides or on the same side of the ring's plane. – Frank Oct 13 '17 at 21:27
• This is not the typical understanding of the term geometric isomerism, which is synonymous (according to the IUPAC) with cis-trans isomerism. You can't have a trans double bond in a ring that small. – WMe6 Oct 13 '17 at 21:40
• So, the geometric isomerism only concerns with the double bond? Also, will the methyl group at the double bond slant or be planar? – Frank Oct 13 '17 at 22:07

Because the bottom one is a benzene, which is flat and aromatic. All its substituents would be in the same plane.

• But, even so, the mentioned double bond of the Benzene resembles an Alkene's double bond, which is two-dimensionally planar as well. Doesn't this qualify it as having geometric isomerism? – Frank Oct 13 '17 at 20:07
• Frank it is easy to see that is the same compound. You can virtually put two of these molecules literally on top of each other. Answer by WM6 and Jan are correct. There is an error in your assignment. – Alchimista Oct 14 '17 at 10:16