Concept explainers
(a)
Interpretation: The
Concept introduction: Cleavage of
Answer to Problem 15.30P
The
Explanation of Solution
The energy required to break the
Thus, the increasing order of bond strength is,
The
(b)
Interpretation: The radicals resulting from the cleavage of each
Concept introduction: Primary
Answer to Problem 15.30P
The radicals resulting from the cleavage of each
Figure 1
Explanation of Solution
Primary
The homolytic cleavage of the
The radicals resulting from the cleavage of each
Figure 1
The radicals resulting from the cleavage of each
(c)
Interpretation: The radicals resulting from the cleavage of each
Concept introduction:
Answer to Problem 15.30P
The radicals in increasing order of stability are
Explanation of Solution
The radicals resulting from the cleavage of each
Figure 1
The stability of radical depends upon the number of alkyl groups attached to the radical carbon. Therefore, stability of tertiary radical is more than secondary and primary radical. The radicals in increasing order of stability are
The radicals in increasing order of stability are
(d)
Interpretation: The
Concept introduction:
Answer to Problem 15.30P
The
Explanation of Solution
The radicals resulting from the cleavage of each
Figure 1
The stability of radical depends upon the number of alkyl groups attached to the radical carbon. Therefore, stability of tertiary radical is more than secondary and primary radical. The radicals in increasing order of stability are
Hydrogen atoms are less polarizable than alkyl groups. Therefore, alkyl group can easily donate electron density to the electron deficient carbon radical. Therefore, the increasing ease of
The
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Chapter 15 Solutions
PKG ORGANIC CHEMISTRY
- The following is a concerted, bimolecular reaction: CH3Br + NaCN → CH3CN + NaBr. a.What is the rate equation for this reaction? b.What happens to the rate of the reaction if [CH3Br] is doubled? c.What happens to the rate of the reaction if [NaCN] is halved? d.What happens to the rate of the reaction if [CH3Br] and [NaCN] are both increased by a factor of five?arrow_forward1. Free radicals are generated by subjecting a molecule such as chlorine or bromine gas to A. high temperature conditions. B. photolysis. C. ionizing conditions. D. choices A and B 2. Which of the following radical intermediates is the most stable? q •CF3 A. B. C. D. 3. In the reaction shown below the intermediate that is formed at the fastest rate is Cl₂ A. uv CH, CH₂CHCH₂CH₂CH₂ B. CH, CH.CCH₂CH₂CH, CH₂ C. CH.CHCHCH.CH, D. CH₂ CH₂CHCH₂CHCHarrow_forward1. Rank the indicated hydrogen atoms in order of increasing ease of abstraction in a radical halogenation reaction (most difficult to easiest). Ha H₂ H H H H Hc Hdarrow_forward
- In the alkene below, why is HB more easily abstracted by a halogen atom than HẠ ? НА HB A. Abstraction of Hg forms a carbocation, while abstraction of HA forms an carbanion. B. Abstraction of HB produces a more stable free radical. C. HB is abstracted as a stable H* ion, but HA is not. D. Abstraction of HB involves less steric hindrance.arrow_forward11) Which one of the following is the missing reagent in the Kolbe-Schmitt reaction? OH & ONa A. HCO₂Et B. (EtO)₂C-O C. CO₂ D. HCO₂Na + ? S heat CO₂Naarrow_forwardAs we will learn in Section 15.12, many antioxidants-compounds that prevent unwanted radical oxidation reactions from occurring-are phenols, compounds that contain an OH group bonded directly to a benzene ring. a. Explain why homolysis of the O-H bond in phenol requires considerably less energy than homolysis of the O-H bond in ethanol (362 kJ/mol vs. 438 kJ/mol). b. Why is the C-O bond in phenol shorter than the C-O bond in ethanol? -O-H CH,CH2-0-H phenol ethanolarrow_forward
- 4. Which of the following reactions is NOT a method of preparation of alkylhalide? A. addition of halogens to alkenes B. replacement of the -OH in alcohol by a halide C. dehydrohalogenation of hydrocarbon D. addition of hydrogen halide to alkene 5. In SN₂ reaction of alkylhalide(substrate) with NaOH(nucleophile), the rate of the reaction is tripled if the concentration of: A. Substrate and nucleophile are both doubled B. Nucleophile is decreased by 1/3 and substrate doubled C. Substrate is tripled and nucleophile doubled D. Nucleophile is tripled 6. Arrange the following alkylhalide in order of decreasing reactivity in an SN₁ reaction: I. CH3CH(Br)CH₂CH3 A. I>II>III>IV II. CH3CH(C1)CH₂CH3 III. CICH₂CH₂CH₂CH3 IV. (CH3)2C(Br)CHCH3 D. IV>II>I>III B. IV>III>II>I C. IV>I>II>IIIarrow_forwardWhy does acrylonitrile (A) react faster with 1,3-butadiene in a pericyclic reaction than methyl vinylether (B)? Check all that apply. CEN A O 1. The bond lenght between carbon and nitrogen is greater than the bond lenght between carbon and oxygen which brings the bulky cyano group further away from the carbon-carbon double bond. That reduces sterical O 2. The LUMO of acrylonitrile (A) is has a lower energy than the LUMO of methyl vinylether (B) which makes the reaction go faster O 3. The high electronegativity of the oxygen atom in ether B polarizes the carbon-carbon double bond which reduces the reaction rate. O 4. The boiling point of nitrile A is higher than the boiling point of ether B, therefore the reaction can be carried out at a higher temperature which accelerates the reaction. hindrance and accelerates the reaction.arrow_forwardAddition of HCl to alkene X forms two alkyl halides Y and Z.a.Label Y and Z as a 1,2-addition product or a 1,4-addition product. b. Label Y and Z as the kinetic or thermodynamic product and explain why. c.Explain why addition of HCl occurs at the indicated C=C (called an exocyclic double bond), rather than the other C=C (called an endocyclic double bond).arrow_forward
- 6. Select the incorrect statement conceming the reaction you conducted during the isomerization experiment. Choose one. H,CO. Brz, light H,CO. rOCH rOCH methylene chioride dimethyl maleate dimethyl fumarate A. The major intermediate formed during the reaction contains both a bromine atom and a radical. B. Bromine is added catalytically instead of stoichiometrically because the bromine radical is regenerated when the product is formed. C. After twenty minutes, the reaction mixture contained both reactant and product. D. Both reactant and product are soluble in methylene chloride. E. If the reaction mixture turned colorless, that was an indication that the bromine radicals were consumed. F. When cold hexanes was added, only dimethyl fumarate precipitated out of solution because dimethyl maleate is soluble in cold hexanes.arrow_forwardAs we will learn in Section 13.12, many antioxidants–compounds that prevent unwanted radical oxidation reactions from occurring–are phenols, compounds that contain an OH group bonded directly to a benzene ring. a. Explain why homolysis of the O–H bond in phenol requires considerably less energy than homolysis of the O–H bond in ethanol (362 kJ/mol vs. 438 kJ/mol). b.Why is the C–O bond in phenol shorter than the C–O bond in ethanol?arrow_forwardRank the nucleophiles in each group in order of increasing nucleophilicity. a.−OH, −NH2, H2O b.−OH, Br−, F− (polar aprotic solvent) c.H2O, −OH, CH3CO2−arrow_forward
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