Organic Chemistry, Binder Ready Version
2nd Edition
ISBN: 9781118454312
Author: David R. Klein
Publisher: WILEY
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Question
Chapter 17, Problem 62IP
Interpretation Introduction
Interpretation:
The product of the given reaction should be identified, if 1,2-dibromocycloheptane reacts with excess of potassium tert-butoxide in the given reaction.
Concept introduction:
- The conjugated dienes absorb UV light.
- In an elimination reaction,
alkenes are formed whenalkyl halides are treated with bases via eliminating one proton and one halo group of the alkyl halide. - E2 reaction is a bimolecular elimination reaction in which only one step is involved and it is favored by the strong bases.
- The product of the elimination reaction is depends upon the two β-positions of alkyl halide. If the β-positions are identical and the products formed are also identical.
- Potassium tert-butoxide is a strong bulky base.
To identify: the product of the given reaction of 1,2-dibromocycloheptane and excess of potassium tert-butoxide and it absorbs UV light.
#assumes E2 reactions are involved in the given reaction (because the product should have double bonds to absorb UV light).
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Can the molecule on the right-hand side of this organic reaction be made in good yield from no more than two reactants, in one
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?
Δ
If your answer is yes, then draw the reactant or reactants in the drawing area below. You can draw the reactants in any
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• If your answer is no, check the box under the drawing area instead.
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Diels Alder Cycloaddition: Focus on regiochemistry (problems E-F) –> match + of thedienophile and - of the diene while also considering stereochemistry (endo).
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Chapter 17 Solutions
Organic Chemistry, Binder Ready Version
Ch. 17.1 - Prob. 1CCCh. 17.2 - Prob. 2CCCh. 17.2 - Prob. 3CCCh. 17.2 - Prob. 4CCCh. 17.2 - Prob. 5CCCh. 17.3 - Prob. 6CCCh. 17.4 - Prob. 1LTSCh. 17.4 - Prob. 7PTSCh. 17.4 - Prob. 8ATSCh. 17.4 - Prob. 9ATS
Ch. 17.5 - Prob. 2LTSCh. 17.5 - Prob. 10PTSCh. 17.5 - Prob. 11ATSCh. 17.5 - Prob. 12ATSCh. 17.5 - Prob. 13CCCh. 17.7 - Prob. 3LTSCh. 17.7 - Prob. 14PTSCh. 17.7 - Prob. 15ATSCh. 17.7 - Prob. 16ATSCh. 17.7 - Prob. 17CCCh. 17.7 - Prob. 18CCCh. 17.7 - Prob. 19CCCh. 17.8 - Prob. 20CCCh. 17.9 - Prob. 21CCCh. 17.9 - Prob. 4LTSCh. 17.9 - Prob. 22PTSCh. 17.9 - Prob. 23ATSCh. 17.9 - Prob. 24ATSCh. 17.10 - Prob. 25CCCh. 17.10 - Prob. 26CCCh. 17.10 - Prob. 27CCCh. 17.10 - Prob. 28CCCh. 17.11 - Prob. 5LTSCh. 17.11 - Prob. 29PTSCh. 17.11 - Prob. 30ATSCh. 17.12 - Prob. 31CCCh. 17 - Prob. 32PPCh. 17 - Prob. 33PPCh. 17 - Prob. 34PPCh. 17 - Prob. 35PPCh. 17 - Prob. 36PPCh. 17 - Prob. 37PPCh. 17 - Prob. 38PPCh. 17 - Prob. 39PPCh. 17 - Prob. 40PPCh. 17 - Prob. 41PPCh. 17 - Prob. 42PPCh. 17 - Prob. 43PPCh. 17 - Prob. 44PPCh. 17 - Prob. 45PPCh. 17 - Prob. 46PPCh. 17 - Prob. 47PPCh. 17 - Prob. 48PPCh. 17 - Prob. 49PPCh. 17 - Prob. 50PPCh. 17 - Prob. 51PPCh. 17 - Prob. 52PPCh. 17 - Prob. 53PPCh. 17 - Prob. 54PPCh. 17 - Prob. 55PPCh. 17 - Prob. 56PPCh. 17 - Prob. 57PPCh. 17 - Prob. 58PPCh. 17 - Prob. 59PPCh. 17 - Prob. 60IPCh. 17 - Prob. 61IPCh. 17 - Prob. 62IPCh. 17 - Prob. 63IPCh. 17 - Prob. 64IPCh. 17 - Prob. 65IPCh. 17 - Prob. 66IPCh. 17 - Prob. 67IPCh. 17 - Prob. 68IPCh. 17 - Prob. 69IP
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