Concept explainers
(a)
Interpretation:
It is to be explained why the heterolysis step on the left does not occur readily than the one on the right.
Concept introduction:
Heterolysis is an elementary step in which a single bond is broken, and both the electrons from that bond end up on one of the atoms initially involved in the bond. In this step, when the bond breaks, the bonding pair of electrons gets distributed unequally. This results in the formation of a positively charged species and a negatively charged species. The alkyl groups attached to positively charged species stabilizes the positive charge. The increasing order for the stability of carbocations is:
Driving force is responsible for the elementary step to go to completion. The driving force for a reaction is the extent to which the reaction favors products over reactants, and that tendency increases with increasing stability of the products relative to the reactants. Charge stability and total bond energy are two major factors that contribute to a reaction’s driving force. The amount of energy required to break a bond is termed as bond energy.
(b)
Interpretation:
It is to be explained why the heterolysis step on the left does not occur readily than the one on the right.
Concept introduction:
Heterolysis is an elementary step in which a single bond is broken, and both the electrons from that bond end up on one of the atoms initially involved in the bond. In this step, when the bond breaks, the bonding pair of electrons gets distributed unequally. This results in the formation of a positively charged species and a negatively charged species. The alkyl groups attached to positively charged species stabilizes the positive charge. The increasing order for the stability of carbocations is:
Driving force is responsible for an elementary step to go to completion. The driving force for a reaction is the extent to which the reaction favors products over reactants, and that tendency increases with increasing stability of the products relative to the reactants. Charge stability and total bond energy are two major factors that contribute to a reaction’s driving force. The amount of energy required to break a bond is termed as bond energy.
(c)
Interpretation:
It is to be explained why the heterolysis step on the left does not occur readily than the one on the right.
Concept introduction:
Heterolysis is an elementary step in which a single bond is broken and both the electrons from that bond end up on one of the atoms initially involved in the bond. In this step, when the bond breaks, the bonding pair of electrons gets distributed unequally. This results in the formation of a positively charged species and a negatively charged species.
Polar protic solvents tend to solvate both cations and anions very strongly, whereas,
Driving force is responsible for an elementary step to go to completion. The driving force for a reaction is the extent to which the reaction favors products over reactants, and that tendency increases with increasing stability of the products relative to the reactants. Charge stability and total bond energy are two major factors that contribute to a reaction’s driving force. The amount of energy required to break a bond is termed as bond energy.
(d)
Interpretation:
It is to be explained why the heterolysis step on the left does not occur readily than the one on the right.
Concept introduction:
Heterolysis is an elementary step in which a single bond is broken, and both the electrons from that bond end up on one of the atoms initially involved in the bond. In this step, when the bond breaks, the bonding pair of electrons gets distributed unequally. This results in the formation of a positively charged species and a negatively charged species.
During nucleophilic substitution reactions, a nucleophile forms a bond to the substrate, and at the same time, the bond to the leaving group is broken. Leaving group comes off in the form of a negatively charged species. Larger atoms accommodate the negative charge better as compared to smaller atoms. Leaving groups are typically conjugate bases of strong acids. Driving force is responsible for an elementary step to go to completion. The driving force for a reaction is the extent to which the reaction favors products over reactants, and that tendency increases with increasing stability of the products relative to the reactants. Charge stability and total bond energy are two major factors that contribute to a reaction’s driving force. The amount of energy required to break a bond is termed as bond energy.

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Chapter 7 Solutions
EBK GET READY FOR ORGANIC CHEMISTRY
- Determine if the following salt is neutral, acidic or basic. If acidic or basic, write the appropriate equilibrium equation for the acid or base that exists when the salt is dissolved in aqueous solution. If neutral, simply write only NR. Be sure to include the proper phases for all species within the reaction LiNO3arrow_forwardAn unknown weak acid with a concentration of 0.410 M has a pH of 5.600. What is the Ka of the weak acid?arrow_forward(racemic) 19.84 Using your reaction roadmaps as a guide, show how to convert 2-oxepanone and ethanol into 1-cyclopentenecarbaldehyde. You must use 2-oxepanone as the source of all carbon atoms in the target molecule. Show all reagents and all molecules synthesized along the way. & + EtOH H 2-Oxepanone 1-Cyclopentenecarbaldehydearrow_forward
- R₂ R₁ R₁ a R Rg Nu R₂ Rg R₁ R R₁₂ R3 R R Nu enolate forming R₁ R B-Alkylated carbonyl species or amines Cyclic B-Ketoester R₁₁ HOB R R₁B R R₁₂ B-Hydroxy carbonyl R diester R2 R3 R₁ RB OR R₂ 0 aB-Unsaturated carbonyl NaOR Aldol HOR reaction 1) LDA 2) R-X 3) H₂O/H₂O ketone, aldehyde 1) 2°-amine 2) acid chloride 3) H₂O'/H₂O 0 O R₁ R₁ R R₁ R₁₂ Alkylated a-carbon R₁ H.C R₁ H.C Alkylated methyl ketone acetoacetic ester B-Ketoester ester R₁ HO R₂ R B-Dicarbonyl HO Alkylated carboxylic acid malonic ester Write the reagents required to bring about each reaction next to the arrows shown. Next, record any regiochemistry or stereochemistry considerations relevant to the reaction. You should also record any key aspects of the mechanism, such as forma- tion of an important intermediate, as a helpful reminder. You may want to keep track of all reactions that make carbon-carbon bonds, because these help you build large molecules from smaller fragments. This especially applies to the reactions in…arrow_forwardProvide the reasonable steps to achieve the following synthesis.arrow_forwardIdentify which compound is more acidic. Justify your choice.arrow_forward
- Provide the reasonable steps to achieve the following synthesis.arrow_forwardWhen anisole is treated with excess bromine, the reaction gives a product which shows two singlets in 1H NMR. Draw the product.arrow_forward(ii) Draw a reasonable mechanism for the following reaction: CI NaOH heat OH (hint: SNAr Reaction) :arrow_forward
- Organic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage Learning
