Organic Chemistry, 7e by L. G. Wade, Jr.

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Presentation transcript:

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

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

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.

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

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.

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

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

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

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

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

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

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

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

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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

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

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.

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

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.

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

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

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

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

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

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