CHAPTER 11: Unit 10. Shapes & Isomerism: Alkenes and Alkynes

Cis-trans isomerism: Alkenes and for certain distributed cycloalkanes, cis-trans isomers exist. Cis-trans isomers are compounds that have the same molecular and structural formulas but different arrangement of atoms in space due to restricted rotation about bonds. Cis means same side and tans means opposite sides across the double bond.

Cis trans isomerism is not possible when one of the double bonded carbon contains two identical group. That is why there is no cis-trans isomers possible for 1-butene and 3-methylpropane.To determine whether an alkene has cis or trans isomers, draw the alkene structure in a manner that emphasizes the four attachments to the double-bonded carbon atoms.If each of the two carbons of the double bond has two different groups attached to it, cis and trans isomers exist.The simplest alkene for which cis and trans isomers exist is 2-butene cis and trans(figure below)                                                                                       2-butene

Cis trans isomerism is not possible when one of the double bonded carbon contains two identical group. That is why there is no cis-trans isomers possible for 1-butene and 3-methylpropane.                                                                                   2-methyl propene1-buteneAnother example of cis-trans isomerism is dinitrogen difluoride N2F2 ( blue balls= Nitrogen, Green balls= Fluorine)

Conversion of cis isomer into trans isomers

HOMEWORK EXERCISE: NOMENCLARURE OF ALKENE

1. Draw the structure for each compound.a. 2-methyl-2-penteneb. 2,3-dimethyl-1-butenec. cyclohexene
2. Draw the structure for each compound.a. 5-methyl-1-hexeneb. 3-ethyl-2-pentenec. 4-methyl-2-hexene
3. Name each compound according to the IUPAC system.a.b.c. 
4. Name each compound according to the IUPAC system.a.b.c.

Answers

1.

a.

b.

c.

3.

a. 2-methyl-2-pentene

b. 3-methyl-2-heptene

c. 2,5-dimethyl-2-hexene

HOMEWORK EXERCISE: CIS-TRANS ISOMERISM

1. Draw the structures of the cis-trans isomers for each compound. Label them cis and trans. If no cis-trans isomers exist, write none.a. 2-bromo-2-penteneb. 3-heptenec. 4-methyl-2-pentened. 1,1-dibromo-1-butenee. 2-butenoic acid (CH₃CH=CHCOOH)
2. Draw the structures of the cis-trans isomers for each compound. Label them cis and trans. If no cis-trans isomers exist, write none.a. 2,3-dimethyl-2-penteneb. 1,1-dimethyl-2-ethylcyclopropanec. 1,2-dimethylcyclohexaned. 5-methyl-2-hexenee. 1,2,3-trimethylcyclopropane

Answer

1. a.b.c.       noned.       nonee.