Concept explainers
Videos
Draw the products of homolysis or heterolysis of each indicated bond. Use electronegativity
differences to decide on the location of charges in the heterolysis reaction. Classify each
carbon reactive intermediate as a radical, carbocation, or carbanion.
a.
b. 
(a)
Interpretation: The products of homolysis or heterolysis of the indicated bond is to be drawn by using the electronegativity differences. Each carbon reactive intermediate is to be classified as a radical, carbocation, or carbanion.
Concept introduction: In organic chemistry, the formation of carbocation or carbanion occurs due to the heterolysis or homolysis process. Carbocations possess six electrons around them, whereas carbanions possess the lone pair of electrons. Carbocation behaves as electrophile due to lack of electrons and incomplete octet.
Answer to Problem 26P
In the given indicated bond, homolysis takes place that results in the formation of the radicals.
The first product is,
The second product is,
Explanation of Solution
The heterolysis does not take place in the given compound due to the less electronegativity difference between atoms.
The product of homolysis is shown below.
Figure 1
In the above reaction, cyclohexane forms cyclohexyl radical and hydrogen radical by homolysis. Homolysis is opposite to the heterolysis. It forms radical with unpaired electron because the electrons are not attracted toward one element in the homolysis.
In the given indicated bond, homolysis takes place that results in the formation of the radical. The product of homolysis is shown in Figure 1.
(b)
Interpretation: The products of homolysis or heterolysis of the indicated bond is to be drawn by using the electronegativity differences. Each carbon reactive intermediate is to be classified as a radical, carbocation, or carbanion.
Concept introduction: In organic chemistry, the formation of carbocation or carbanion occurs due to the heterolysis or homolysis process. Carbocations possess six electrons around them, whereas carbanions possess the lone pair of electrons. Carbocation behaves as electrophile due to lack of electrons and incomplete octet.
Carbanion behaves as a nucleophile in the chemical reaction due to the presence of excess electrons. Heterolysis is the process in which unequal sharing of electrons results in the breaking of the bond.
Answer to Problem 26P
In the given indicated bond, heterolysis takes place that results in the formation of the carbocation.
The first product is,
The second product is,
Explanation of Solution
Heterolysis in the compound takes place due to the more electronegativity difference. The product of heterolysis is shown below.
Figure 2
In the above reaction, ethanol forms ethyl carbocation and hydroxide ion by heterolysis. The heterolysis in the chemical reaction leads to the formation of ionic species because electrons are attracted toward more electronegative atom. Oxygen is more electronegative than carbon. Therefore, heterolysis and the formation of carbocation take place in the reaction.
In the given indicated bond, heterolysis takes place that results in the formation of the carbocation. The product of heterolysis is shown in Figure 2.
Want to see more full solutions like this?
Chapter 6 Solutions
Study Guide/solutions Manual For Organic Chemistry
- 1. Which among these can make a molecule nucleophilic? a.double bondsb.positive chargec. incomplete octet 2. Which among these can make a molecule electrophilic? a.Triple bondsb.positive chargec. radicalsarrow_forwardConsider 3-methylpent-2-ene, if it undergoes hydroboration, which of the following final product is formed? a. 3-methylpent-2-en-2-ol b. 3-methylpentan-3-ol c. 3-methylpentan-2-ol d. 2-methylpentan-3-olarrow_forwardHow would melting point determination help us narrow down the identity of a product or products of an organic chemistry reaction? a. The higher the melting point, the more of the desired product has formed. b. The lower the melting point, the more of the desired product has formed. c. A narrow melting point range would suggest that only one product has formed, a wide melting point range would suggest a mixture of products. d. A wide melting point range would suggest that only one product has formed, a narrow melting point range would suggest a mixture of products.arrow_forward
- In the reaction of cyclopentene with HCl, hydrogen acts as the a. nucleophile. b. intermediate. c. electrophile. d. carbonium ion.arrow_forwardHow many moles of Bra are required to completely halogenate the alkene?A. One moleB. Two molesC. Three molesD. Four moles What is the expected arrangement of the bromine atoms relative to each other amongthe carbon involved in pi bonding?A. anti-conformationB. syn-conformationC. trans-configurationD. cis-configuration What happens to bromine when it is adjacent to an alkene during a chemical reaction?A. Bromine becomes stable. (? kasi before brown siya/acidic tas naging colorless? Jk ewan)B. Bromine becomes polarized.C. Bromine becomes hybridized.D. Bromine becomes acidic. The relative arrangement of bromine atoms in the product is primarily due to:A. ElectronegativityB. RepulsionC. Hydrogen bondingD. Atomic weightWhat is your observation after the reaction?A. A yellow flame is produced.B. Bromine water decolorizes.C. The alkene becomes denser.D. A brown precipitate forms.arrow_forwardIn the addition reactions of alkynes, the reagents undergo transformation to form first. a. a nucleophile b. an electrophile c. a carbocation d. a radicalarrow_forward
- 4-methyl-1 -pentene reacts with hydrogen bromide. a. Draw a structure for the carbocation intermediate that leads to the major Markovnikov product in the reaction of hydrogen bromide with 4-methyl-1-pentene. b. Using Markovnikov's rule, draw a structure of the major organic product formed in the reaction of hydrogen bromide with 4-methyl-1-pentene.arrow_forward7. Enols react as a. Electrophiles.....alkenes b. Electrophiles....alkanes c. Nucleophiles...alkenes d. Nucleophiles...alkanes in the same way that 8. Acylation of enolate ions proceeds by a a. E1 b. E2 C. SN1 d. SN2 c. Anhydride d. Beta-ketoacid | 9. The following molecule would be classified as a a. Alpha-ketoamide b. Beta-aldobase do. mechanism. HOarrow_forwardHydrocarbons like benzene are metabolized in the body to arene oxides, which rearrange to form phenols. This is an example of a general process in the body, in which an unwanted compound (benzene) is converted to a more water-soluble derivative called a metabolite, so that it can be excreted more readily from the body. a. Classify each of these reactions as oxidation, reduction, or neither. b. Explain why phenol is more water soluble than benzene. This means that phenol dissolves in urine, which is largely water, to a greater extent than benzene.arrow_forward
- Hydrocarbons like benzene are metabolized in the body to arene oxides, which rearrange to form phenols. This is an example of a general process in the body, in which an unwanted compound (benzene) is converted to a more water-soluble derivative called a metabolite, so that it can be excreted more readily from the body.a.Classify each of these reactions as oxidation, reduction, or neither. b. Explain why phenol is more water soluble than benzene. This means that phenol dissolves in urine, which is largely water, to a greaterextent than benzene.arrow_forwardHomolysis of the indicated C–H bond in propene forms a resonancestabilized radical. a.Draw the two possible resonance structures for this radical. b.Use half-headed curved arrows to illustrate how one resonance structure can be converted to the other. c. Draw a structure for the resonance hybrid.arrow_forwardClassify each carbocation as primary, secondary, or tertiary. a. b. 人 C. d.arrow_forward
Organic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage Learning