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Alkene Geometry Answers Alkene geometry-ANSWER KEY Chemistry 260 Organic Chemistry 1. The answer is 2. The two carbons of the double bond are sp2 hybridized (planar). Therefore, all carbons lie in the same plane. H3C CH3 C C H3C CH3 2. The answer is 4. CH2CH3 CH3CH C 1 2 3 CH2CH3 4 5 3-ethyl-2-pentene 3. Give the IUPAC name of 2-ethyl-1-butene 4 3 CH3CH2 H 2 1 C C CH3CH2 H 4. Give the structural formula of: (a) 2,3-dimethyl-2-butene (b) 2,3-dimethyl-2-butanol CH3 CH3 OH C C CH3C CHCH3 CH3 CH3 CH3 CH3 (c) 3-chloropropene (d) 3-methyl-2-pentanol H CH2Cl OH C C CH3CHCHCH2CH3 H H 1 CH3 5. (E)-2-pentene 1 CH3 H 2 3 C C H CH2CH3 4 5 6. The answer is 1. H H H CH3 H2C CHCH2CH3 H3C CH3 H3C H H3C H C H2C CH2 2 CH2 CHCH3 H2C CH2 H3C H2C 7. Give the IUPAC name of: (a) cis-3-hexene or (Z)-3-hexene 1 2 5 6 CH3CH2 CH2CH3 3 4 C C H H (b) trans-2,5-dimethyl-3-hexene or (E)-2,5-dimethyl-3-hexene CH3 CH3 CH H 1 2 C C 3 4 H 5CH CH3 CH3 6 2 8. The answer is 2. 9. The answer is 1. 10. The answer is 2. 2 2' 1 2' 2 2' H H H3CH2C H H3C H C C C C C C H3C CH2CH3 H CH2CH3 H3CH2C CH2CH3 1 1' 2 1' 1 1' (Z) (E) (Z) 11. The answer is 2. 12. The answer is 2. 13. The answer is 3. 1 1' 2 2' 1 2' Br Br H CO2CH3 C C C C Cl Cl C C H3C H Cl Cl Cl H 2 2' 1 1' (Z) (Z) 1 1' Neither – left carbon has two identical substituents 14. The answer is 3. H CH3 H H H H H C C H C C H3C C C C C H C C CH3 C C CH3 H3C H H3C H H H (E,E) (E,Z) (Z,Z) 15. (a) one isomer (d) two isomers H CH2CH3 C l CH3 H CH3 C C C C C C H H H H Cl H (Z) (E) 3 (b) two isomers (e) one isomer H3C CH3 H CH3 H3CH2C CH2CH3 C C C C C C H H H C H 3 H3C CH2CH3 (Z) (E) (c) one isomer H3C CH3 C C H3C H 16. Give the structural formula of cis-2-methyl-3-heptene CH3 CH3CH CH2CH2CH3 C C H H 17. cis- and trans-Isomers of 1,2-dichloroethene can be isolated because C=C (σ + π) bonding is quite rigid. For 1,2-dichloroethane, only a single C-C (σ) bond exists and this is not rigid. Free rotation about the σ bond in 1,2- dichloroethane removes any possibility of distinct cis- and trans-isomers: H Cl H H H H Cl C C C C H Cl H Cl Cl Cl H (E)-1,2-dichloroethene (Z)-1,2-dichloroethene 1,2-dichloroethane (rotation about σ-bond is easier than about π-bond) 18. The answer is 2. 1 2' H3C CH3 C C H F 2 1' (E) 4 19. E/Z Isomers are only possible when there is a C=C double bond present in the molecule. (a) NO - no double bond present (b) NO - only one orientation possible if triple bond present CH3 H3C C C CH3 H3C C CH3 Cl (c) NO - double bond present, but (d) YES - two geometric isomers: one C has two identical groups H3C H H H H CH3 C C C C C C H CH3 H (Z) (E) (e) YES - two geometric isomers: (f) YES - two geometric isomers: 2 1' 2 2' 1 1' 2 1' F Cl F H H3CH2C CH3 H CH3 C C C C C C C C Cl H Cl Cl H H H3CH2C H 1 2' 1 1' 2 2' 1 2' (E) (Z) (Z) (E) 20. Name the following compounds according to the IUPAC rules nomenclature. Indicate stereochemistry where appropriate. 5 A: (a) trans-4-isopropylcyclohexanol or trans-4-(1-methylethyl)cyclohexanol. (b) E-4-bromo-3-fluoro-5-methyl-3-heptene or E-4-bromo-3-fluoro-5- methylhept-3-ene. (c) (Z)-3-iodo-5-methyl-3-octene or (Z)-3-iodo-5- methyloct-3-ene. 21. Draw the structure of 2-chloro-1-hexene 22. Name the following compounds. Answers:: (a) (E)-5-bromo-4-isopropylhept-3-ene OR (E)-5-bromo-4-isopropyl-3- heptene (b) (E)-3-bromo-2-fluoro-4,6,6-trimethylhept-2-ene OR (E)-5-bromo-4-isopropyl- 3-heptene 6 .
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