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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.
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7.1 Describe the geometry around the carbon–carbon double bond.
a. Tetrahedral b. Trigonal pyramidal c. Trigonal planar d. Bent e. Linear
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7.1 Answer a. Tetrahedral b. Trigonal pyramidal c. Trigonal planar
d. Bent e. Linear
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7.2 Give the formula for an alkene.
a. CnH2n-4 b. CnH2n-2 c. CnH2n d. CnH2n+2 e. CnH2n+4
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7.2 Answer a. CnH2n-4 b. CnH2n-2 c. CnH2n d. CnH2n+2 e. CnH2n+4
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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
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7.3 Answer a. 2,4-butadiene b. 1,3-butadiene c. 2,4-pentadiene
d. 1,3-pentadiene e. 1,4-pentadiene
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7.4 Calculate the unsaturation number for C6H10BrCl.
d. 3
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7.4 Answer a. 0 b. 1 c. 2 d. 3 U = 0.5 [2(6) + 2 – (12)] = 1
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7.5 Name . a. Trans-2-pentene b. Cis-2-pentene
c. Trans-3-methyl-2-pentene d. Cis-3-methyl-2-pentene
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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.
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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
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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.
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7.7 a. CH3COOH b. CH3CHO c. CH3CH2OH d. HOCH2CH2OH e. CH3CH(OH)2
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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.
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7.8 a. ClCH2CH2Cl b. ClCH=CHCl c. CH2=CH2 d. CH2=CHCl
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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.
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7.9 a. (CH3)2CHOH b. CH3CH2CH2OH c. HOCH2CH2CH2OH d. CH3CH(OH)CH2OH
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7.9 Answer a. (CH3)2CHOH b. CH3CH2CH2OH c. HOCH2CH2CH2OH
d. CH3CH(OH)CH2OH Water adds by Markovnikov’s orientation across the double bond.
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7.10 a. [CH2CH(CH3)]n b. [CH2CH2]n c. [CH2=CH(CH3)]n d. [CH2=CH2]n
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7.10 Answer a. [CH2CH(CH3)]n b. [CH2CH2]n c. [CH2=CH(CH3)]n
d. [CH2=CH2]n
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7.11 Identify the product formed from the polymerization of tetrafluoroethylene.
a. Polypropylene b. Poly(vinyl chloride), (PVC) c. Polyethylene d. Poly(tetrafluoroethylene), Teflon
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7.11 Answer a. Polypropylene b. Poly(vinyl chloride), (PVC)
c. Polyethylene d. Poly(tetrafluoroethylene), Teflon Teflon is formed from the polymerization of tetrafluoroethylene.
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7.12 a. CH3CCCH3 b. CH2=CHCH=CH2 c. CH3CH=CHCH3 d. CH3CH2CH2CH3
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7.12 Answer a. CH3CCCH3 b. CH2=CHCH=CH2 c. CH3CH=CHCH3
d. CH3CH2CH2CH3 Hydrogen adds across the double bond to form an alkane.
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7.13 a. (CH3)2C(Br)CH3 b. (CH3)2C=CHBr c. (CH3)2C=CH2 d. (CH3)2CHCH3
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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.
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7.14 Debromination of vicinal dibromides occurs by what mechanism?
a. SN1 b. SN2 c. E1 d. E2
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7.14 Answer a. SN1 b. SN2 c. E1 d. E2 The debromination of vicinal dibromides occurs by an E2 mechanism.
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7.15 a. (CH3)2CHOSO3H b. CH3CH=CH2 c. (CH3)2C=O d. CH3CH2COOH
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7.15 Answer a. (CH3)2CHOSO3H b. CH3CH=CH2 c. (CH3)2C=O d. CH3CH2COOH
Acid dehydrates alcohols to form alkenes.
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7.16 Dehydration of alcohols occurs by what mechanism?
a. SN1 b. SN2 c. E1 d. E2
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7.16 Answer a. SN1 b. SN2 c. E1 d. E2 The dehydration of alcohols occurs by an E1 mechanism.
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7.17 Give the products from the catalytic cracking of alkanes.
a. Alkanes b. Alkenes c. Alkynes d. Alkanes + alkenes e. Alkanes + alkynes
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7.17 Answer a. Alkanes b. Alkenes c. Alkynes d. Alkanes + alkenes
e. Alkanes + alkynes
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7.18 Give the products from the dehydrogenation of alkanes.
a. Alkanes b. Alkenes c. Alkynes d. Alkanes + alkenes e. Alkanes + alkynes
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7.18 Answer a. Alkanes b. Alkenes c. Alkynes d. Alkanes + alkenes
e. Alkanes + alkynes
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7.19 a. (CH3)3CO-, (CH3)3COH b. CH3CH2O-, CH3CH2OH c. NaI, acetone
d. H2, Pd
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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.
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7.20 a. Pt, 500o C b. H2, Pt c. H2SO4, 150o C d. NaI, acetone e. NaOH
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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.
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