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

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Organic Chemistry, 7e by L. G. Wade, Jr. Chapter 7 Structure and Synthesis of Alkenes Christine Hermann Radford University Radford, VA Copyright © 2010 Pearson Education, Inc.

7.1 Describe the geometry around the carbon–carbon double bond. a. Tetrahedral b. Trigonal pyramidal c. Trigonal planar d. Bent e. Linear

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

7.2 Give the formula for an alkene. a. CnH2n-4 b. CnH2n-2 c. CnH2n d. CnH2n+2 e. CnH2n+4

7.2 Answer a. CnH2n-4 b. CnH2n-2 c. CnH2n d. CnH2n+2 e. CnH2n+4

7.3 Name CH3CH=CHCH=CH2. a. 2,4-butadiene b. 1,3-butadiene c. 2,4-pentadiene d. 1,3-pentadiene e. 1,4-pentadiene

7.3 Answer a. 2,4-butadiene b. 1,3-butadiene c. 2,4-pentadiene d. 1,3-pentadiene e. 1,4-pentadiene

7.4 Calculate the unsaturation number for C6H10BrCl. d. 3

7.4 Answer a. 0 b. 1 c. 2 d. 3 U = 0.5 [2(6) + 2 – (12)] = 1

7.5 Name . a. Trans-2-pentene b. Cis-2-pentene c. Trans-3-methyl-2-pentene d. Cis-3-methyl-2-pentene

7.5 Answer a. Trans-2-pentene b. Cis-2-pentene c. Trans-3-methyl-2-pentene d. Cis-3-methyl-2-pentene The longest chain is five carbons with numbering is left to right.

7.6 Name . a. E-2-pentene b. Z-2-pentene c. E-3-methyl-2-pentene d. Z-3-methyl-2-pentene e. Z-2-methyl-2-pentene

7.6 Answer a. E-2-pentene b. Z-2-pentene c. E-3-methyl-2-pentene d. Z-3-methyl-2-pentene e. Z-2-methyl-2-pentene The longest chain is five carbons and the highest ranking groups are on opposite sides.

7.7 a. CH3COOH b. CH3CHO c. CH3CH2OH d. HOCH2CH2OH e. CH3CH(OH)2

7.7 Answer a. CH3COOH b. CH3CHO c. CH3CH2OH d. HOCH2CH2OH e. CH3CH(OH)2 Ethylene oxide is formed first, followed by a ring opening to form ethylene glycol.

7.8 a. ClCH2CH2Cl b. ClCH=CHCl c. CH2=CH2 d. CH2=CHCl

7.8 Answer a. ClCH2CH2Cl b. ClCH=CHCl c. CH2=CH2 d. CH2=CHCl Chlorine is added across the double bond, then HCl is lost.

7.9 a. (CH3)2CHOH b. CH3CH2CH2OH c. HOCH2CH2CH2OH d. CH3CH(OH)CH2OH

7.9 Answer a. (CH3)2CHOH b. CH3CH2CH2OH c. HOCH2CH2CH2OH d. CH3CH(OH)CH2OH Water adds by Markovnikov’s orientation across the double bond.

7.10 a. [CH2CH(CH3)]n b. [CH2CH2]n c. [CH2=CH(CH3)]n d. [CH2=CH2]n

7.10 Answer a. [CH2CH(CH3)]n b. [CH2CH2]n c. [CH2=CH(CH3)]n d. [CH2=CH2]n

7.11 Identify the product formed from the polymerization of tetrafluoroethylene. a. Polypropylene b. Poly(vinyl chloride), (PVC) c. Polyethylene d. Poly(tetrafluoroethylene), Teflon

7.11 Answer a. Polypropylene b. Poly(vinyl chloride), (PVC) c. Polyethylene d. Poly(tetrafluoroethylene), Teflon Teflon is formed from the polymerization of tetrafluoroethylene.

7.12 a. CH3CCCH3 b. CH2=CHCH=CH2 c. CH3CH=CHCH3 d. CH3CH2CH2CH3

7.12 Answer a. CH3CCCH3 b. CH2=CHCH=CH2 c. CH3CH=CHCH3 d. CH3CH2CH2CH3 Hydrogen adds across the double bond to form an alkane.

7.13 a. (CH3)2C(Br)CH3 b. (CH3)2C=CHBr c. (CH3)2C=CH2 d. (CH3)2CHCH3

7.13 Answer a. (CH3)2C(Br)CH3 b. (CH3)2C=CHBr c. (CH3)2C=CH2 d. (CH3)2CHCH3 Both bromides are lost to form an alkene.

7.14 Debromination of vicinal dibromides occurs by what mechanism? a. SN1 b. SN2 c. E1 d. E2

7.14 Answer a. SN1 b. SN2 c. E1 d. E2 The debromination of vicinal dibromides occurs by an E2 mechanism.

7.15 a. (CH3)2CHOSO3H b. CH3CH=CH2 c. (CH3)2C=O d. CH3CH2COOH

7.15 Answer a. (CH3)2CHOSO3H b. CH3CH=CH2 c. (CH3)2C=O d. CH3CH2COOH Acid dehydrates alcohols to form alkenes.

7.16 Dehydration of alcohols occurs by what mechanism? a. SN1 b. SN2 c. E1 d. E2

7.16 Answer a. SN1 b. SN2 c. E1 d. E2 The dehydration of alcohols occurs by an E1 mechanism.

7.17 Give the products from the catalytic cracking of alkanes. a. Alkanes b. Alkenes c. Alkynes d. Alkanes + alkenes e. Alkanes + alkynes

7.17 Answer a. Alkanes b. Alkenes c. Alkynes d. Alkanes + alkenes e. Alkanes + alkynes

7.18 Give the products from the dehydrogenation of alkanes. a. Alkanes b. Alkenes c. Alkynes d. Alkanes + alkenes e. Alkanes + alkynes

7.18 Answer a. Alkanes b. Alkenes c. Alkynes d. Alkanes + alkenes e. Alkanes + alkynes

7.19 a. (CH3)3CO-, (CH3)3COH b. CH3CH2O-, CH3CH2OH c. NaI, acetone d. H2, Pd

7.19 Answer a. (CH3)3CO-, (CH3)3COH b. CH3CH2O-, CH3CH2OH c. NaI, acetone d. H2, Pd The Hofmann product (least substituted) is favored with a bulky base.

7.20 a. Pt, 500o C b. H2, Pt c. H2SO4, 150o C d. NaI, acetone e. NaOH

7.20 Answer a. Pt, 500o C b. H2, Pt c. H2SO4, 150o C d. NaI, acetone e. NaOH Dehydrogenation occurs with a metal catalyst and heat.