Organic Chemistry : Help with Cis-Trans Isomers
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Example Questions
Example Question #1 : Isomers
Which of the following lists the product(s) of the presented reaction?
None of the other answers is correct.
Both trans and cis-1-4-dimethylcyclohexane
Methylcyclohexane
Only cis-1-4-dimethylcyclohexane
Only trans-1-4-dimethylcyclohexane
Both trans and cis-1-4-dimethylcyclohexane
Hydrogen and a catalyst like paladium reduce the double bond to a single bond. There is no equal steric hinderance on each side. The hydrogen can bond from either side. That means the methyl group can either be oriented into the page or out of the page. One form is cis, and one form is trans.
Example Question #2 : Isomers
I.
II.
III.
Which of the given molecules is(are) chiral?
I, II, and III
I and II
I only
II and III
I and III
I and III
For a molecule to be chiral, it must have a stereocenter and no axis of symmetry. An atom with a stereocenter has no identical bonds; it is a carbon atom with four unique substituents. There are two stereocenters in each of the three molecules. Notice that if you take the second molecule and draw a line connecting the top carbon and the point between the the two carbons with hydroxy groups, it has an axis of symmetry and therefore cannot be chiral. There is no way to draw that axis of symmetry for molecules one and three.
Example Question #1 : Isomers
How many stereocenters does the given molecule have?
Three
Four
One
Two
Five
Two
A stereocenter exists when the central atom is bound to four unique substituents. In the given molecule, the carbons are numbers from left to right. Carbons 1, 3, 5, 6, and 8 are all bound to at least two hydrogen atoms and cannot be stereocenters. Similarly, the carbons in the two methyl groups (bound to C4 and C7) do not qualify as stereocenters. Carbon 7 has two identical methyl substituents. This leaves only C2 and C4. The molecule has two stereocenters.
Example Question #1 : Isomers
What is the IUPAC name of the molecule shown?
E-5-carboxy-2-pentene
Z-pent-3-enoic acid
Z-5-carboxy-2-pentene
E-pent-3-enoic acid
Z-pent-3-enoic acid
Carboxylic acid is highest priority, so carbon chain labelled from right to left. Since highest priority groups are on the same side of the double bond, it's given the "Z" designation.
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