(a)
To draw: The resonance forms of three possible allylic intermediates that are formed during the reaction of NBS with
Interpretation: The resonance forms of three possible allylic intermediates that are formed during the reaction of NBS with
Concept introduction: The reaction that includes the addition of one or more bromine atoms in a compound is known as bromination reaction. Bromination reaction is a type of halogenation reaction. In this reaction, NBS (source of
A molecule that is uncharged and contains an unpaired electron is termed as free radical. Free radicals are highly reactive materials, but they have a short life span.
(b)
To determine: The ranking of three intermediates formed during the reaction of NBS with
Interpretation: The ranking of three intermediates formed during the reaction of NBS with
Concept introduction: The reaction that includes the addition of one or more bromine atoms in a compound is known as bromination reaction. Bromination reaction is a type of halogenation reaction. In this reaction, NBS (source of
A molecule that is uncharged and contains an unpaired electron is termed as free radical. Free radicals are highly reactive materials, but they have a short life span.
(c)
To draw: The products that are obtained from each of the free radical intermediate.
Interpretation: The products that are obtained from each of the free radical intermediate are to be drawn.
Concept introduction: The reaction that includes the addition of one or more bromine atoms in a compound is known as bromination reaction. Bromination reaction is a type of halogenation reaction. In this reaction NBS (source of
A molecule that is uncharged and contains an unpaired electron is termed as free radical. Free radicals are highly reactive materials, but they have short life span.
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Chapter 6 Solutions
ORGANIC CHEMISTRY MASTERINGCHEM ACCESS
- As 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_forwardi need help with this please.arrow_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
- 19. Which of the following radical intermediates is the least stable? A. methane radical B. primary radical C. secondary radical D. tertiary radical 20. Compound J undergoes a carbonium ion rearrangement to yield compounds K and L. Based on the potential energy diagram below which of the following statements is TRUE. Potential energy A. K is formed faster and is more stable than L. B. K is formed faster and is less stable than L C. L is formed faster and is less stable than K L Reaction coordinate 21. Which of the following is the expected major product for the reaction below. 000 000 00-50 D. L is formed faster and is more stable than K D.arrow_forward1. Alkene Addition Reaction Mechanisms: Draw the product for the addition reactions below in the boxes. Then, draw the FULL electron-pushing mechanism for the reactions, including all intermediates (with formal charges and lone pairs of electrons) and all electron pushing arrows. For each step, label the electrophile and nucleophile and name what type of elementary step it is. a. Addition of H-Br to an alkene: of b. Addition of H₂O using an acid catalyst: H-CI: H-Ö-SO₂H (cat.) H₂Oarrow_forwardDraw the major organic product of each reaction. Indicate the stereochemistry a. Draw product 1. If the C−OC−O bond breaks, assume inversion of configuration. b. Draw product 2.arrow_forward
- Which is a better nucleophile? a. Br− or Cl− in H2O b. Br− or Cl− in DMSO c. CH3O− or CH3OH in H2O d. CH3O− or CH3OH in DMSO e. HO− or −NH2 in H2O f. HO− or −NH2 in DMSO g. I− or Br− in H2O h. I− or Br− in DMSOarrow_forward1. In a Birch reaction, an aromatic compound can be treated with Na/CH3OH, it can also be treated with Li/NH3 and have the same product form. One of the intermediates in the Birch reaction is a radical anion. In the example below, anisole could possibly form a radical anion A or B, then will go into C and D. Li/NH3 A B C D A. With the knowledge you have gained from what we have discussed, which anion A or B is more stable? What is your reasoning? B. Is a Birch reaction and oxidation or reduction? Why?arrow_forwardThe 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_forward
- A reaction flask contains 2-bromopentane in an ethanolic solution of sodium ethoxide at room temperature and result in the formation of two olefnic products. What is responsible for the formation of major and minor products. A. Different activated complex involved in the mechanism. B. Bimolecular nucleophilic substitution reaction. C. Bimolecular elimination reaction D. The presence of sodium ethoxide. E. The hybridization nature of secondary carbocationarrow_forward6. 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_forwardDraw an energy diagram for the SN2 reaction on the previous page. Your reaction coordinate diagram should: Include structures of starting materials, products, and transition states at correct relative energies. Indicate the activation energy of the rate determining step and whether the reaction is endothermic or exothermic. What makes this reaction endothermic or exothermic?arrow_forward
Organic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage Learning