EBK ORGANIC CHEMISTRY-PRINT COMPANION (
4th Edition
ISBN: 9781119776741
Author: Klein
Publisher: WILEY CONS
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Question
Chapter 7, Problem 73PP
(a)
Interpretation Introduction
Interpretation:
Newman projections must be utilized to explain why trans-stilbene is the major product when (S)-1-bromo-1,2-diphenylethane is used.
Concept Introduction :
During the elimination reaction, anti-periplanar groups are removed to form a double bond. Newman projection is the representation of a molecule when it is viewed through a particular bond between two atoms.
(b)
Interpretation Introduction
Interpretation:
Newman projections must be utilized to explain why trans-stilbene is the major product when (R)-1-bromo-1,2-diphenylethane is employed.
Concept Introduction :
During the elimination reaction, anti-periplanar groups are removed to form a double bond. Newman projection is the representation of a molecule when it is viewed through a particular bond between two atoms.
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2. For the following nucleophilic substitution reaction:
NaOC(CH3)3
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'CH3
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TS₁
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TS2
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CH3
Br
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10. A compound X of molecular formula C8H12 with no triple bonds reacts with one
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B)
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Compound X has 1 ring and 2 pi bonds.
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Chapter 7 Solutions
EBK ORGANIC CHEMISTRY-PRINT COMPANION (
Ch. 7.2 - Prob. 1CCCh. 7.3 - Prob. 2CCCh. 7.3 - Prob. 1LTSCh. 7.3 - Prob. 3PTSCh. 7.3 - Prob. 4ATSCh. 7.3 - Prob. 2LTSCh. 7.3 - Prob. 5PTSCh. 7.3 - Prob. 6ATSCh. 7.3 - Prob. 7CCCh. 7.4 - Prob. 8CC
Ch. 7.5 - Prob. 9CCCh. 7.6 - Prob. 10CCCh. 7.6 - Prob. 11CCCh. 7.7 - Prob. 12PTSCh. 7.7 - Prob. 13PTSCh. 7.7 - Prob. 14ATSCh. 7.7 - Prob. 4LTSCh. 7.7 - Prob. 16ATSCh. 7.7 - Prob. 17CCCh. 7.7 - Prob. 18CCCh. 7.7 - Prob. 5LTSCh. 7.7 - Prob. 19PTSCh. 7.7 - Prob. 20ATSCh. 7.8 - Prob. 21PTSCh. 7.8 - Prob. 22ATSCh. 7.8 - Prob. 23CCCh. 7.8 - Prob. 24CCCh. 7.8 - Prob. 25CCCh. 7.8 - Prob. 26CCCh. 7.8 - Prob. 27CCCh. 7.9 - Prob. 7LTSCh. 7.9 - Prob. 29ATSCh. 7.9 - Prob. 30ATSCh. 7.9 - Prob. 31ATSCh. 7.10 - Prob. 32CCCh. 7.10 - Prob. 33CCCh. 7.10 - Prob. 34CCCh. 7.11 - Prob. 8LTSCh. 7.11 - Prob. 35PTSCh. 7.11 - Prob. 36PTSCh. 7.11 - Prob. 37ATSCh. 7.11 - Prob. 9LTSCh. 7.11 - Prob. 40PTSCh. 7.11 - Prob. 41ATSCh. 7.12 - Prob. 42CCCh. 7.12 - Prob. 43CCCh. 7.12 - Prob. 44CCCh. 7.12 - Prob. 45CCCh. 7.12 - Prob. 46CCCh. 7 - Prob. 47PPCh. 7 - Prob. 48PPCh. 7 - Prob. 49PPCh. 7 - Prob. 50PPCh. 7 - Prob. 51PPCh. 7 - Prob. 52PPCh. 7 - Prob. 53PPCh. 7 - Prob. 54PPCh. 7 - Prob. 55PPCh. 7 - Prob. 56PPCh. 7 - Prob. 57PPCh. 7 - Prob. 58PPCh. 7 - Prob. 59PPCh. 7 - Prob. 60PPCh. 7 - Prob. 61PPCh. 7 - Prob. 64PPCh. 7 - Indicate whether you would use NaOEt or tBuOK to...Ch. 7 - Prob. 68PPCh. 7 - Draw a plausible mechanism for each of the...Ch. 7 - Prob. 70PPCh. 7 - Prob. 71PPCh. 7 - Prob. 72PPCh. 7 - Prob. 73PPCh. 7 - Prob. 74PPCh. 7 - Prob. 77PPCh. 7 - Prob. 78PPCh. 7 - Prob. 81ASPCh. 7 - Prob. 87ASPCh. 7 - Prob. 90ASPCh. 7 - Prob. 91IPCh. 7 - Prob. 92IPCh. 7 - Prob. 93IPCh. 7 - Prob. 94IPCh. 7 - Prob. 95IPCh. 7 - Prob. 96IPCh. 7 - Prob. 97IPCh. 7 - Prob. 98IPCh. 7 - Prob. 99IPCh. 7 - Prob. 100IPCh. 7 - Prob. 101IPCh. 7 - Prob. 102IPCh. 7 - Prob. 103IPCh. 7 - Prob. 105IPCh. 7 - Prob. 106IPCh. 7 - Prob. 107IPCh. 7 - Prob. 109IPCh. 7 - Prob. 110CPCh. 7 - Prob. 112CPCh. 7 - Prob. 114CP
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- Draw the products of the following reactions. If the products can exist as stereoisomers show what stereoisomers are formed. a. cis-2-pentene + Br2/CH2Cl2 b. trans-2-pentene + Br2/CH2Cl2 c. 1- butene + HCl d. methylcyclohexene + HBr e. trans-3-hexene + Br2/CH2Cl2 f. cis-3-hexene + Br2/CH2Cl2 g. 3,3-dimethyl-1-pentene + HBr h. cis-2-butene + HBr i. (Z)-2,3-dichloro-2-butene + H2, Pd/C j. (E)-2,3-dichloro-2-butene + H2, Pd/C k. (Z)-3,4-dimethyl-3-hexene + H2, Pd/C l. (E)-3,4-dimethyl-3-hexene + H2, Pd/Carrow_forwarda. Identify two alkenes that react with HBr to form 1-bromo-1-methylcyclohexane without undergoing a carbocation rearrangement. b. Would both alkenes form the same alkyl halide if DBr were used instead of HBr? (D is an isotope of H, so D+ reacts like H+.)arrow_forward10. Which of these describe a set of enantiomers? a. (2R,3R)-2-bromo-3-chloropentane and (2S,3S)-3-bromo-2-chloropentane b. (2R,3R)-2-bromo-3-chlorpentane and (2R,3S)-2-bromo-3-chloropentane c. (2S,3S)-2-bromo-3-chloropentane and (2S,3R)-2-bromo-3-chloropentane d. (2S,3S)-2-bromo-3-chloropentane and (2R,3R)-2-bromo-3-chloropentane e. None of these.arrow_forward
- 3. Draw the structures of the products, showing stereochemistry where relevant. a) 1-methylcyclohexene HBr b) Br2 1-methylcyclohexene Br2 1-methylcyclohexene → H2O 4. When bromine adds to the double bond in cyclohexene, the resulting dibromide is trans. This stereochemistry is due to: a) The intermediate bromonium ion d) Markovnikov's rule b) the intermediate carbocation d) IUPAC rulesarrow_forwardDraw the structure of the major organic product of the reaction below. CH3 CH₂CHCH₂CH₂CH=CH₂ + CHC13 KOH • Use the wedge/hash bond tools to indicate stereochemistry where it exists. Show stereochemistry in a meso compound. If the reaction produces a racemic mixture, just draw one stereoisomer.arrow_forward10. Draw ALL possible stereoisomers possible for 8i. Give the configurations of each stereocenter in 8i. (NOTE: Ph is the abbreviation for an aromatic ring and is assigned a priority of 4 in the structure for 8i, below) H₂C U CH3 MAIII OH Product 8iarrow_forward
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