(a)
Interpretation:
To explain the following result obtain from theHammett equation.
Concept introduction:
Hammett reaction constant (
The negative value of the
(b)
Interpretation:
To explain the following result obtain from theHammett equation.
Concept introduction:
Hammett reaction constant (
The negative value of the
(c)
Interpretation:
To explain the following result obtain from theHammett equation.
Concept introduction: Hammett reaction constant (
The negative value of the
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Chapter 15 Solutions
Organic Chemistry (8th Edition)
- Alkynes do not react directly with aqueous acid as do alkenes, but will do so in the presence of mercury(II) sulfate as a Lewis acid catalyst. The reaction occurs with Markovnikov regiochemistry, so the OH group adds to the more highly substituted carbon and the H adds to the less highly substituted carbon. The initial product of the reaction is a vinyl alcohol, also called an enol. The enol immediately rearranges to a more stable ketone via tautomerization. Draw curved arrows to show the movement of electrons in this step of the mechanism. Arrow-pushing Instructions -X티 Hö: H-O -CH3 -CH3 H30*arrow_forwardSubstituents on an aromatic ring can have several effects on electrophilic aromatic substitution reactions. Substituents can activate or deactivate the ring to substitution, donate or withdraw electrons inductively, donate or withdraw electrons through resonance, and direct substitution either to the ortho/para or to the meta positions. From the lists of substituents, select the substituents that correspond to each indicated property. The substituents are written as -XY, where X is the atom directly bound to the aromatic ring.arrow_forwardAzulene, an isomer of naphthalene, has a remarkably large dipole moment for a hydrocarbon (µ = 1.0 D). Explain using resonance structures.arrow_forward
- Fill the blank space. Compounds containing a phenol group may work as ANTIOXIDANTS to prevent free radical damage. This is accomplished when a free radical (or UV light) encounters a phenol group, turning the phenol group into a radical. However, contrary to typical radical behavior, the structure of the phenol radical can neutralize (or quench) the unpaired electron. Specifically, the phenol structure neutralizes (or quenches) the unpaired radical electron by doing the following: taking the electron and ---------. The correct name (or abbreviation) of an example compound containing a phenol group with antioxidant properties is: ---------.arrow_forwardAlcohols are acidic in nature. Therefore, a strong base can abstract the acidic hydrogen atom of the alcohol in a process known as deprotonation. The alcohol forms an alkoxide ion by losing the proton attached to the oxygen atom of the hydroxyl ( -OH) group. The alkoxide formed can act as a base or a nucleophile depending on the substrate and reaction conditions. However, not all bases can abstract the acidic proton of alcohols and not all alcohols easily lose the proton. Deprotonation depends on the strength of the base and the acidity of the alcohol. Strong bases, such as NaNH2, can easily abstract a proton from almost all alcohols. Likewise, more acidic alcohols lose a proton more easily. Determine which of the following reactions would undergo deprotonation based on the strength of the base and the acidity of the alcohol. Check all that apply. ► View Available Hint(s) CH3CH,OH + NH3 →CH,CH,O-NH CH3 CH3 H3C-C-H+NH3 → H3 C-C-H OH O-NH CH3CH2OH + NaNH, → CH3CH,O-Na* + NH3 CHC12 Cl₂…arrow_forwardNonconjugated , -unsaturated ketones, such as 3-cyclohexenone, are in an acid-catalyzed equilibrium with their conjugated , -unsaturated isomers. Propose a mechanism for this isomerization.arrow_forward
- Ethers can be prepared by reaction of an alkoxide or phenoxide ion with a primary alkyl halide. Draw the structure of the expected organic product of the reaction of iodoethane with the following alkoxide ion: H3C CH3 + Na You do not have to consider stereochemistry. You do not have to explicitly draw H atoms. Do not include lone pairs in your answer. They will not be considered in the grading. • Do not include counter-ions, e.g., Na", I, in your answer. P opy aste [*arrow_forwardWhen 2-bromo-2-methylbutane is treated with a base, a mixture of 2-methyl-2-butene and 2-methyl-1- butene is produced. When potassium hydroxide is the base, 2-methyl-1-butene accounts for 45% of the product mixture. However, when potassium tert-butoxide is the base, 2-methyl-1-butene accounts for 70% of the product mixture. What percent of 2-methyl-1-butene would be in the mixture if potassium propoxide were the base? base Br A. Less than 45% B. C. 45% Between 45% and 70% D. More than 70%arrow_forwardCompounds containing a phenol group may work as ANTIOXIDANTS to prevent free radical damage. This is accomplished when a free radical (or UV light) encounters a phenol group, turning the phenol group into a radical. However, contrary to typical radical behavior, the structure of the phenol radical can neutralize (or quench) the unpaired electron. Specifically, the phenol structure neutralizes (or quenches) the unpaired radical electron by doing the following: taking the electron and The correct name (or abbreviation) of an example compound (discussed in the lecture videos) containing a phenol group with antioxidant properties is:arrow_forward
- Consider the following compounds that vary from nearly nonacidic to strongly acidic. Draw the conjugate bases of these compounds, and explain why the acidity increases so dramatically with substitution by nitro groups.refer to the photoarrow_forwardFuran undergoes electrophilic aromatic substitution more readily than benzene; mild reagents and conditions are sufficient.For example, furan reacts with bromine to give 2-bromofuran. Explain why furan undergoes bromination (and other electrophilic aromatic substitutions) primarily at the 2-position.arrow_forward*20. The two isomeric compounds with the formula C10H14 have NMR spectra shown below. Make no attempt to interpret the aromatic proton area between 7.1 and 7.3 ppm except to determine the number of protons attached to the aromatic ring. Draw the structures of the isomers. C1,H14 5.10 7.5 7.0 1.10 2.15 3.04 2.94 4.0 3.0 2.0 1.0 (а) C1,H14 5.10 7.5 7.0 2.14 2.06 2.06 2.98 4.0 3.5 3.0 2.5 2.0 1.5 1.0 (b)arrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning