1. Organic Chemistry, 7e by L. G. Wade, Jr.
Chapter 9
Alkynes
Christine Hermann
Radford University
Radford, VA
Copyright © 2010 Pearson Education, Inc.

2. 9.1 Name . a. 6-hepten-3-yne b. 5-penten-3-yne c. 1-hepten-4-yne
d. 1-hepten-5-yne

3. 9.1 Answer a. 6-hepten-3-yne b. 5-penten-3-yne c. 1-hepten-4-yne
d. 1-hepten-5-yne
The double bond has higher priority than the triple bond.

4. 9.2 Name . a. Pent-2-yn-4-ol b. Pent-2-en-4-ol c. Pent-3-en-2-ol
d. Pent-3-yn-2-ol

5. 9.2 Answer a. Pent-2-yn-4-ol b. Pent-2-en-4-ol c. Pent-3-en-2-ol
d. Pent-3-yn-2-ol
The OH has priority over the carbon–carbon triple bond.

6. 9.3 Give the hybridization of a carbon in an alkyne.
a. sp
b. sp2
c. sp3
d. sp4

7. 9.3 Answer
a. sp
b. sp2
c. sp3
d. sp4

8. 9.4 Give the shape of a carbon in an alkyne.
a. Trigonal planar
b. Trigonal pyramidal
c. Tetrahedral
d. Linear
e. Bent

9. 9.4 Answer a. Trigonal planar b. Trigonal pyramidal c. Tetrahedral
d. Linear
e. Bent

10. 9.5
a. CH3CH=CH2
b. CH3CCCl
c. CH3CCCH3
d. CH3(Cl)=CHCH3

11. 9.5 Answer a. CH3CH=CH2 b. CH3CCCl c. CH3CCCH3 d. CH3(Cl)=CHCH3
The methyl replaces the sodium.

12. 9.6 Give the mechanism for the alkylation of an acetylide ion.
a. E1
b. E2
c. SN1
d. SN2

13. 9.6 Answer
a. E1
b. E2
c. SN1
d. SN2

14. 9.7
a. CH3CCCH2OH
b. CH3CH=CHCH2OH
c. CH3C(OH)=CH2
d. HOCH2C(CH3)=CH2

15. 9.7 Answer a. CH3CCCH2OH b. CH3CH=CHCH2OH c. CH3C(OH)=CH2
d. HOCH2C(CH3)=CH2
The aldehyde adds to the acetylide ion.

16. 9.8 a. CH3C(Cl)=CHCH3 b. CH3CH(OH)CH(OH)CH3 c. CH3CH(OH)CH(Cl)CH3
d. CH3CCCH3

17. 9.8 Answer a. CH3C(Cl)=CHCH3 b. CH3CH(OH)CH(OH)CH3
c. CH3CH(OH)CH(Cl)CH3
d. CH3CCCH3
Double dehydrohalogenation of a vicinal dihalide occurs under extreme basic conditions to form an alkyne.

18. 9.9 a. CH3CH=CHCH3 (cis) b. CH3CH=CHCH3 (trans) c. CH3CH2CH2CH3
d. CH3CCCH3

19. 9.9 Answer a. CH3CH=CHCH3 (cis) b. CH3CH=CHCH3 (trans) c. CH3CH2CH2CH3
d. CH3CCCH3
Catalytic hydrogenation reduces an alkyne to an alkane.

20. 9.10 a. CH3CH=CHCH3 (cis) b. CH3CH=CHCH3 (trans) c. CH3CH2CH2CH3
d. CH3CCCH3

21. 9.10 Answer a. CH3CH=CHCH3 (cis) b. CH3CH=CHCH3 (trans)
c. CH3CH2CH2CH3
d. CH3CCCH3
In the presence of the Lindlar catalyst, hydrogen adds syn across the double bond.

22. 9.11 Describe how hydrogen is added in the reaction of an alkene with Lindlar’s catalyst.
a. Anti
b. Syn
c. Both
d. Neither

23. 9.11 Answer
a. Anti
b. Syn
c. Both
d. Neither

24. 9.12 a. CH3CH=CHCH3 (cis) b. CH3CH=CHCH3 (trans) c. CH3CH2CH2CH3
d. CH3CCCH3

25. 9.12 Answer a. CH3CH=CHCH3 (cis) b. CH3CH=CHCH3 (trans)
c. CH3CH2CH2CH3
d. CH3CCCH3
Sodium metal in liquid ammonia reduces alkynes, resulting in anti addition of the hydrogens.

26. 9.13 a. CH3CH2C(Br)2C(Br)2 b. CH3C(Br)2C(Br)2CH3
c. CH(Br)2CH2CH2CH(Br)2
d. CH3CH2CH(Br)2CH3
e. CH3CH(Br)CH(Br)CH3

27. 9.13 Answer a. CH3CH2C(Br)2C(Br)2 b. CH3C(Br)2C(Br)2CH3
c. CH(Br)2CH2CH2CH(Br)2
d. CH3CH2CH(Br)2CH3
e. CH3CH(Br)CH(Br)CH3

28. 9.14 a. CH3C(Cl)2CH3 b. CH3CH2CHCl2 c. CH3C(Cl)=CH2
d. CH3CH=CHCl (cis)
e. CH3CH=CHCl (trans)

29. 9.14 Answer a. CH3C(Cl)2CH3 b. CH3CH2CHCl2 c. CH3C(Cl)=CH2
d. CH3CH=CHCl (cis)
e. CH3CH=CHCl (trans)
HCl adds across the triple bond twice in a Markovnikov orientation.

30. 9.15
a. CH3C(Br)2CH3
b. CH3CH2CHBr2
c. CH3C(Br)=CH2
d. CH3CH=CHBr

31. 9.15 Answer a. CH3C(Br)2CH3 b. CH3CH2CHBr2 c. CH3C(Br)=CH2
d. CH3CH=CHBr

32. 9.16 a. CH3C(OH)=CH2 b. (CH3)2C=O c. CH3CH2CHO d. CH3CH=CHOH (cis)
e. CH3CH=CHOH (trans)

33. 9.16 Answer a. CH3C(OH)=CH2 b. (CH3)2C=O c. CH3CH2CHO
d. CH3CH=CHOH (cis)
e. CH3CH=CHOH (trans)
Water adds across the triple bond in a Markovnikov orientation, followed by tautomerism to form a ketone.

34. 9.17 a. CH3C(OH)=CH2 b. (CH3)2C=O c. CH3CH2CHO d. CH3CH=CHOH (cis)
e. CH3CH=CHOH (trans)

35. 9.17 Answer a. CH3C(OH)=CH2 b. (CH3)2C=O c. CH3CH2CHO
d. CH3CH=CHOH (cis)
e. CH3CH=CHOH (trans)
Water adds across the triple bond in a syn anti-Markovnikov orientation, followed by tautomerism to form an aldehyde.

36. 9.18 Describe the hydroboration–oxidation of an alkyne.
a. Markovnikov; anti
b. Markovnikov; syn
c. Anti-Markovnikov; anti
d. Anti-Markovnikov; syn

37. 9.18 Answer a. Markovnikov; anti b. Markovnikov; syn
c. Anti-Markovnikov; anti
d. Anti-Markovnikov; syn

38. 9.19 Identify the product(s) from the permanganate oxidation of 1-butyne.
a. Keto-acid
b. Diketone
c. Dialdehyde
d. Diacid
e. Aldo-acid

39. 9.19 Answer a. Keto-acid b. Diketone c. Dialdehyde d. Diacid
e. Aldo-acid

40. 9.20 Identify the product(s) from the ozonolysis of an alkyne.
a. Two aldehydes
b. Two ketones
c. Ketone + aldehyde
d. Two carboxylic acids
e. Ketone + carboxylic acid

41. 9.20 Answer a. Two aldehydes b. Two ketones c. Ketone + aldehyde
d. Two carboxylic acids
e. Ketone + carboxylic acid