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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Organic Chemistry: Principles and Mechanisms (Second Edition)
- Help please. Draw the curved arrow notation and products for the each elementary step described by the sequence shown here.Note that the products of each step should be used as the reactants in the next step, and you may need to draw in additional reagents as directed. Remember to click on each box to see step-specific instructions. Draw H3CH2C–, and then add the curved arrow notation showing a nucleophilic addition.arrow_forwardDraw the major organic product for the reaction shown...arrow_forwarddraw the product for both reactions...arrow_forward
- Draw the structure of product, substrate, or condition in the following reactions (should clearly show the stereochemistry).arrow_forwardplease draw detailed mechanism. mention which one is the nucleophile, electrophile,base/ acid. Also talk about stereochemistry/ regioselectivity. In terms of stereochemistry why are we getting trans alkene. Why is the trans alkene Produced is it due choosing PPh2 as regeant? Also This reaction is the Horner wadsworth emmond wittig.arrow_forwardDraw the complete, detailed E1 mechanism for each of the following reactions, and show all resonance structures, where applicable.arrow_forward
- Draw the complete mechanism of each pair of reactants including any favorable rearrangements and all important resonance structures of all intermediates. a. Which reaction has a lower PE carbocation intermediate? b. Draw an energy diagram showing the reaction profiles of both reactions in the previous question. Use a dotted line for the first pair of reactants and a solid line for the second pair of reactants. (Assume the energy of the starting materials and products are the same for both pairs and the reactions are neither uphill nor downhill on net. c. Mark points on the energy diagram corresponding to each carbocation in your mechanisms.arrow_forwardPlace in order of increasing nucleophile, and explainarrow_forward
- Organic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage Learning