Organic Chemistry
6th Edition
ISBN: 9781936221349
Author: Marc Loudon, Jim Parise
Publisher: W. H. Freeman
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Chapter 11, Problem 11.15P
Interpretation Introduction
Interpretation:
The reason as to why
Concept introduction:
Reaction of a halohydrin with a base yields an
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When trans-2-chloro-1-cyclohexanol is treated with a base, cyclohexene oxide is the product. However, when cis-2-chloro-1-cyclohexanol is treated with a base, the product is cyclohexanone.
Write the mechanism for each of the two reactions.
When cis-2-decalone is dissolved in ether containing a trace of HCI, an equilibrium is
established with trans-2-decalone. The latter ketone predominates in the equilibrium
mixture.
H
H
HCI
cis-2-Decalone
trans-2-Decalone
Propose a mechanism for this isomerization and account for the fact that the trans iso-
mer predominates at equilibrium.
When trans-2-chloro-1-cyclohexanol is treated with a base, cyclohexene oxide is the product. However, when cis-2-chloro-1-cyclohexanol is treated with a base, the product is cyclohexanone.
Why doesn’t the cis isomer yield the oxide?
Chapter 11 Solutions
Organic Chemistry
Ch. 11 - Prob. 11.1PCh. 11 - Prob. 11.2PCh. 11 - Prob. 11.3PCh. 11 - Prob. 11.4PCh. 11 - Prob. 11.5PCh. 11 - Prob. 11.6PCh. 11 - Prob. 11.7PCh. 11 - Prob. 11.8PCh. 11 - Prob. 11.9PCh. 11 - Prob. 11.10P
Ch. 11 - Prob. 11.11PCh. 11 - Prob. 11.12PCh. 11 - Prob. 11.13PCh. 11 - Prob. 11.14PCh. 11 - Prob. 11.15PCh. 11 - Prob. 11.16PCh. 11 - Prob. 11.17PCh. 11 - Prob. 11.18PCh. 11 - Prob. 11.19PCh. 11 - Prob. 11.20PCh. 11 - Prob. 11.21PCh. 11 - Prob. 11.22PCh. 11 - Prob. 11.23PCh. 11 - Prob. 11.24PCh. 11 - Prob. 11.25PCh. 11 - Prob. 11.26PCh. 11 - Prob. 11.27PCh. 11 - Prob. 11.28PCh. 11 - Prob. 11.29PCh. 11 - Prob. 11.30PCh. 11 - Prob. 11.31PCh. 11 - Prob. 11.32PCh. 11 - Prob. 11.33PCh. 11 - Prob. 11.34PCh. 11 - Prob. 11.35PCh. 11 - Prob. 11.36PCh. 11 - Prob. 11.37PCh. 11 - Prob. 11.38PCh. 11 - Prob. 11.39PCh. 11 - Prob. 11.40PCh. 11 - Prob. 11.41PCh. 11 - Prob. 11.42PCh. 11 - Prob. 11.43PCh. 11 - Prob. 11.44APCh. 11 - Prob. 11.45APCh. 11 - Prob. 11.46APCh. 11 - Prob. 11.47APCh. 11 - Prob. 11.48APCh. 11 - Prob. 11.49APCh. 11 - Prob. 11.50APCh. 11 - Prob. 11.51APCh. 11 - Prob. 11.52APCh. 11 - Prob. 11.53APCh. 11 - Prob. 11.54APCh. 11 - Prob. 11.55APCh. 11 - Prob. 11.56APCh. 11 - Prob. 11.57APCh. 11 - Prob. 11.58APCh. 11 - Prob. 11.59APCh. 11 - Prob. 11.60APCh. 11 - Prob. 11.61APCh. 11 - Prob. 11.62APCh. 11 - Prob. 11.63APCh. 11 - Prob. 11.64APCh. 11 - Prob. 11.65APCh. 11 - Prob. 11.66APCh. 11 - Prob. 11.67APCh. 11 - Prob. 11.68APCh. 11 - Prob. 11.69APCh. 11 - Prob. 11.70APCh. 11 - Prob. 11.71APCh. 11 - Prob. 11.72APCh. 11 - Prob. 11.73APCh. 11 - Prob. 11.74APCh. 11 - Prob. 11.75APCh. 11 - Prob. 11.76APCh. 11 - Prob. 11.77APCh. 11 - Prob. 11.78APCh. 11 - Prob. 11.79APCh. 11 - Prob. 11.80APCh. 11 - Prob. 11.81AP
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- Predict the products formed when cyclohexanone reacts with the following reagents. h) sodium acetylide, then mild H3O+arrow_forwardWhen trans-2-chloro-1-cyclohexanol is treated with a base, cyclohexene oxide is the product. However, when cis-2-chloro-1-cyclohexanol is treated with a base, the product is cyclohexanone –arrow_forwardThe Stork reaction is a condensation reaction between an enamine donor and an α,β-unsaturated carbonyl acceptor. The overall reaction consists of a three-step sequence of formation of an enamine from a ketone, Michael addition to an α,β-unsaturated carbonyl compound, and hydrolysis of the enamine in dilute acid to regenerate the ketone. Consider the Stork reaction between cyclohexanone and propenal Draw the structure of the product of the enamine formed between cyclohexanone and dimethylamine. - Michael addition to an α,β-unsaturated carbonyl compound, and - hydrolysis of the enamine in dilute acid to regenerate the ketone.arrow_forward
- 7. For compound A in acid, it is the hydroxyl that is protonated to a greater extent than the carbonyl oxygen. However, for compound B in acid, it is the carbonyl oxygen that is protonated to a greater extent than the hydroxyl oxygen. Explain this difference. H2SO4 ОН ОН2 ОН H2SO4 ОН ОНarrow_forwardDetermining the Regioselectivity of Opening an Epoxide Ring What product is formed when 2,2-dimethyloxirane is treated with each set of reagents: −OCH3 followed by H2O, or CH3OH and H2SO4?arrow_forwardPropose a mechanism for the reaction of(a) 1-methylcyclohexanol with HBr to form 1-bromo-1-methylcyclohexane.(b) 2-cyclohexylethanol with HBr to form 1-bromo-2-cyclohexylethane.arrow_forward
- Identify products A and B from the given 1H NMR data. Treatment of acetone [(CH3)2C=O] with dilute aqueous base forms B. Compound B exhibits four singlets in its 1H NMR spectrum at 1.3 (6 H), 2.2 (3 H), 2.5 (2 H), and 3.8 (1H) ppm. What is the structure of B?arrow_forwardWhen (R)-6-bromo-2,6-dimethylnonane is dissolved in CH3OH, nucleophilic substitution yields an optically inactive solution. When the isomeric halide (R)-2-bromo-2,5-dimethylnonane is dissolved in CH3OH under the same conditions, nucleophilic substitution forms an optically active solution. Draw the products formed in each reaction, and explain why the difference in optical activity is observed.arrow_forwardWhen treated with an equivalent of methanol, compound A, with molecular formula C4H6Cl2O, forms the compound whose 1H NMR spectrum is shownhere. Identify compound A.arrow_forward
- Treatment of cis-4-bromocyclohexanol with HO− affords compound Aand cyclohex-3-en-1-ol. Treatment of trans-4-bromocyclohexanol under the same conditions forms compound B and cyclohex-3-en-1-ol. A and Bcontain different functional groups and are not isomers of each other.Propose structures for A and B and offer an explanation for theirformation.arrow_forwardTreatment of cis-4-bromocyclohexanol with HO– affords compound A and cyclohex-3-en-1-ol. Treatment of trans-4- bromocyclohexanol under the same conditions forms compound B and cyclohex-3-en-1-ol. A and B contain different functional groups and are not isomers of each other. Propose structures for A and B and offer an explanation for their formation.arrow_forwardTreatment of (CH3)2CHCH(OH)CH2CH3 with TsOH affords two products (M and N) with molecular formula C6H12. The 1H NMR spectra of M and N are given below. Propose structures for M and N and draw a mechanism to explain their formation.arrow_forward
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