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 ORGANIC CHEMISTRY: PRINCIPLES AND M
- We learned four factors (ARIO) for comparing the relative acidity of compounds. When two of these factors are in competition, the order of priority is the order in which these factors were covered ("atom" being the most important factor and "orbital" being the least important). However, we also mentioned that there are exceptions to this order of priority. Compare the two compounds and identify the exception. OH PK-4.75 SH PK-10.6 5. "Resonance" is more important than "atom" because the conjugate base of first compound is more stable than the second. "Atom" is more important than "resonance" because the conjugate base of first compound is more stable than the second. "Resonance" is more important than "atom" because the conjugate base of second compound is more stable than the first. "Atom" is more important than "resonance" because the conjugate base of second compound is more stable than the first.arrow_forwardThe relative fitnesses of three genotypes are WA/A= 1.0, WA/a = 0.7, and Wa/a = 0.3. If the population starts at the allele frequency p = 0.5, what is the value of p in the next generation? (3 pts) 12pt v Paragraph V BIU ALarrow_forwardIdentify the most acidic proton in the compound: a d b Оа Ob Ос ○ darrow_forward
- A Standard Reference Material is certified to contain 94.6 ppm of an organic contaminant in soil. Your analysis gives values of 98.6, 98.4, 97.2, 94.6, and 96.2. Do your results differ from the expected results at the 95% confidence interval?arrow_forwardThe percentage of an additive in gasoline was measured six times with the following results: 0.13, 0.12, 0.16, 0.17, 0.20, and 0.11%. Find the 95% confidence interval for the percentage of additive.arrow_forwardExplain why this data led Rayleigh to look for and to discover Ar.arrow_forward
- 5) Confidence interval. Berglund and Wichardt investigated the quantitative determination of Cr in high-alloy steels using a potentiometric titration of Cr(VI). Before the titration, samples of the steel were dissolved in acid and the chromium oxidized to Cr(VI) using peroxydisulfate. Shown here are the results (as %w/w Cr) for the analysis of a reference steel. 16.968, 16.922, 16.840, 16.883, 16.887, 16.977, 16.857, 16.728 Calculate the mean, the standard deviation, and the 95% confidence interval about the mean. What does this confidence interval mean?arrow_forwardIn the Nitrous Acid Test for Amines, what is the observable result for primary amines? Group of answer choices nitrogen gas bubbles form a soluble nitrite salt yellow oily layer of nitrosoaminearrow_forward3. a. Use the MS to propose at least two possible molecular formulas. For an unknown compound: 101. 27.0 29.0 41.0 50.0 52.0 55.0 57.0 100 57.5 58.0 58.5 62.0 63.0 64.0 65.0 74.0 40 75.0 76.0 20 20 40 60 80 100 120 140 160 180 200 220 m/z 99.5 68564810898409581251883040 115.0 116.0 77404799 17417M 117.0 12.9 118.0 33.5 119.0 36 133 0 1.2 157.0 2.1 159.0 16 169.0 219 170.0 17 171.0 21.6 172.0 17 181.0 1.3 183.0 197.0 100.0 198.0 200. 784 Relative Intensity 2 2 8 ō (ppm) 6 2arrow_forward
- Organic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage Learning
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