Concept explainers
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
Organic Chemistry (6th Edition)
- For each reaction below, decide if the first stable organic product that forms in solution will create a new CC bond, and check the appropriate box. Next, for each reaction to which you answered "Yes" to in the table, draw this product in the drawing area below. Note for advanced students: for this problem, don't worry if you think this product will continue to react under the current conditions - just focus on the first stable product you expect to form in solution. དྲ。 ✗MgBr ? O CI Will the first product that forms in this reaction create a new C-C bond? Yes No • ? Will the first product that forms in this reaction create a new CC bond? Yes No × : ☐ Xarrow_forwardPredict the major products of this organic reaction: OH NaBH4 H ? CH3OH Note: be sure you use dash and wedge bonds when necessary, for example to distinguish between major products with different stereochemistry. Click and drag to start drawing a structure. ☐ : Sarrow_forwardPredict the major products of this organic reaction: 1. LIAIHA 2. H₂O ? Note: be sure you use dash and wedge bonds when necessary, for example to distinguish between major products with different stereochemistry. Click and drag to start drawing a structure. X : ☐arrow_forward
- For each reaction below, decide if the first stable organic product that forms in solution will create a new C - C bond, and check the appropriate box. Next, for each reaction to which you answered "Yes" to in the table, draw this product in the drawing area below. Note for advanced students: for this problem, don't worry if you think this product will continue to react under the current conditions - just focus on the first stable product you expect to form in solution. NH2 tu ? ? OH Will the first product that forms in this reaction create a new CC bond? Yes No Will the first product that forms in this reaction create a new CC bond? Yes No C $ ©arrow_forwardAs the lead product manager at OrganometALEKS Industries, you are trying to decide if the following reaction will make a molecule with a new C-C bond as its major product: 1. MgCl ? 2. H₂O* If this reaction will work, draw the major organic product or products you would expect in the drawing area below. If there's more than one major product, you can draw them in any arrangement you like. Be sure you use wedge and dash bonds if necessary, for example to distinguish between major products with different stereochemistry. If the major products of this reaction won't have a new CC bond, just check the box under the drawing area and leave it blank. Click and drag to start drawing a structure. This reaction will not make a product with a new CC bond. G marrow_forwardIncluding activity coefficients, find [Hg22+] in saturated Hg2Br2 in 0.00100 M NH4 Ksp Hg2Br2 = 5.6×10-23.arrow_forward
- give example for the following(by equation) a. Converting a water insoluble compound to a soluble one. b. Diazotization reaction form diazonium salt c. coupling reaction of a diazonium salt d. indacator properties of MO e. Diazotization ( diazonium salt of bromobenzene)arrow_forward2-Propanone and ethyllithium are mixed and subsequently acid hydrolyzed. Draw and name the structures of the products.arrow_forward(Methanesulfinyl)methane is reacted with NaH, and then with acetophenone. Draw and name the structures of the products.arrow_forward
- 3-Oxo-butanenitrile and (E)-2-butenal are mixed with sodium ethoxide in ethanol. Draw and name the structures of the products.arrow_forwardWhat is the reason of the following(use equations if possible) a.) In MO preperation through diazotization: Addition of sodium nitrite in acidfied solution in order to form diazonium salt b.) in MO experiment: addition of sodium hydroxide solution in the last step to isolate the product MO. What is the color of MO at low pH c.) In MO experiment: addition of sodium hydroxide solution in the last step to isolate the product MO. What is the color of MO at pH 4.5 d.) Avoiding not cooling down the reaction mixture when preparing the diazonium salt e.) Cbvcarrow_forwardA 0.552-g sample of an unknown acid was dissolved in water to a total volume of 20.0 mL. This sample was titrated with 0.1103 M KOH. The equivalence point occurred at 29.42 mL base added. The pH of the solution at 10.0 mL base added was 3.72. Determine the molar mass of the acid. Determine the Ka of the acid.arrow_forward
- Organic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage Learning
