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Chapter 7: Problem 19

Rank the nucleophiles in each group in order of increasing nucleophilicity. a. \({ }^{-} \mathrm{OH},{ }^{-} \mathrm{NH}_{2}, \mathrm{H}_{2} \mathrm{O}\) b. - \(\mathrm{OH}, \mathrm{Br}^{-}, \mathrm{F}^{-}\) (polar aprotic solvent) c. \(\mathrm{H}_{2} \mathrm{O},{ }^{-} \mathrm{OH}, \mathrm{CH}_{3} \mathrm{COO}^{-}\)

Short Answer

Expert verified
In increasing order of nucleophilicity: a) H2O < OH- < NH2-; b) Br- < F- < OH-; c) H2O < CH3COO- < OH-.

Step by step solution

01

Understand nucleophilicity

Nucleophilicity is the ability of an atom or molecule to donate a pair of electrons to an electrophile during a chemical reaction. In general, negatively charged species are more nucleophilic than their neutral counterparts, and nucleophilicity can also depend on the solvent.
02

Analyze Group (a)

In group (a), we have - ext{-} ext{OH} - ext{-} ext{NH}_{2} - ext{H}_{2} ext{O} All three species increase in nucleophilicity from left to right. Among these, ext{-} ext{NH}_{2} is the strongest nucleophile because nitrogen is less electronegative than oxygen, allowing it to hold a negative charge more effectively and donate electrons more easily. Next is ext{-} ext{OH}, and lastly, ext{H}_{2} ext{O} is the weakest since it is neutral.
03

Analyze Group (b)

In a polar aprotic solvent, nucleophilicity generally increases as the base strength increases, but we also have to consider the size of the halide ion. In group (b), - ext{-} ext{OH} - ext{Br}^{-} - ext{F}^{-} In a polar aprotic solvent, nucleophilicity decreases with electronegativity. Here, ext{-} ext{OH} is the most nucleophilic, followed by ext{F}^{-}, and finally, ext{Br}^{-} as the least.
04

Analyze Group (c)

In group (c), - ext{H}_{2} ext{O} - ext{-} ext{OH} - ext{CH}_{3} ext{COO}^{-} Nucleophilicity increases with negative charge and the ability to donate electrons. ext{-} ext{OH} is the strongest nucleophile due to its negative charge, ext{CH}_{3} ext{COO}^{-} is weaker due to resonance which delocalizes its negative charge, and ext{H}_{2} ext{O} is the weakest because it is neutral.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Nucleophiles
A nucleophile is a chemical species that donates an electron pair to another atom or ion, known as an electrophile, to form a chemical bond. Nucleophiles are typically rich in electrons, making them negatively charged, and they are always in search of positive centers to donate their electrons. This characteristic stems from the fact that nucleophiles have a pair of electrons they are willing to donate. In general:
  • Nucleophiles can be neutral or negatively charged species.
  • They are often associated with lone pair-bearing atoms such as oxygen, nitrogen, or halogens.
  • The strength of a nucleophile, known as nucleophilicity, can be influenced by the molecular structure and surrounding environment like solvents.
Understanding nucleophiles and their behavior in reactions is fundamental to predicting the outcome of many organic reactions.
Electronegativity
Electronegativity is the tendency of an atom to attract electrons towards itself when it is in a chemical bond. This concept plays a crucial role in determining the reactivity of nucleophiles. In order of electronegativity: fluorine is more electronegative than oxygen, which is more electronegative than nitrogen.
  • Lower electronegativity means an atom holds onto its electrons less tightly and can be a better nucleophile.
  • Nitrogen, less electronegative than oxygen, helps amide ions (\( ext{-} ext{NH}_2\)) to act as stronger nucleophiles compared to hydroxide ions (\( ext{-} ext{OH}\)).
  • Similarly, oxygen is less electronegative than fluorine, making hydroxide a better nucleophile than fluoride in some circumstances.
Electronegativity differences are essential in predicting how nucleophiles will behave in chemical reactions.
Polar Aprotic Solvents
Polar aprotic solvents are solvents that dissolve ionic compounds but do not donate hydrogen bonds themselves. Examples include acetone, dimethyl sulfoxide (DMSO), and acetonitrile. These solvents can significantly influence the strength and behavior of nucleophiles in reactions.
  • In polar aprotic solvents, nucleophilicity increases as the strength of the base increases.
  • Smaller ion size often leads to stronger nucleophiles because they are less strongly solvated.
  • In a polar aprotic environment, \( ext{-} ext{OH}\) acts as a stronger nucleophile than halides like \( ext{F}^{-}\) and \( ext{Br}^{-}\).
This is because polar aprotic solvents do not form hydrogen bonds with the nucleophile, meaning the nucleophile remains available and reactive in the solution.
Chemical Reactions
Chemical reactions involve the transformation of reactants into products, and nucleophilicity plays a key role in many of these transformations. Understanding the concept of nucleophilicity can help predict which species will be more reactive in a given reaction.
  • Nucleophiles participate actively in substitution and addition reactions, attacking positively polarized regions of molecules.
  • They are central in reactions such as SN2 (bimolecular nucleophilic substitution) and E2 (bimolecular elimination) reactions.
  • The rate and outcome of these reactions can depend on factors like the type of nucleophile, solvent, and the nature of leaving groups.
Having a grasp of when and how nucleophiles behave in chemical reactions provides insight into reaction mechanisms and helps in predicting reaction pathways.

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Most popular questions from this chapter

Which compound in each pair reacts faster in nucleophilic substitution? a. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{I}\) b. \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}\) or \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CI}\) c. \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}\) or \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}_{2}{ }^{+}\) d. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OCOCH}_{3}\)

Classify each solvent as protic or aprotic. a. \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHOH}\) b. \(\mathrm{CH}_{3} \mathrm{NO}_{2}\) c. \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) d. \(\mathrm{NH}_{3}\) e. \(\mathrm{N}\left(\mathrm{CH}_{3}\right)_{3}\) f. \(\mathrm{HCONH}_{2}\)

Draw the eight constitutional isomers having the molecular formula \(\mathrm{C}_{5} \mathrm{H}_{11} \mathrm{Cl}\). a. Give the IUPAC name for each compound (ignoring \(R\) and \(S\) designations). b. Label any stereogenic centers. c. For each constitutional isomer that contains a stereogenic center, draw all possible stereoisomers, and label each stereogenic center as \(R\) or \(S\).

Draw the substitution product that results when \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br}\) reacts with each nucleophile. a. \({ }^{-} \mathrm{OH}\) b. \({ }^{-} \mathrm{SH}\) c. \({ }^{-} \mathrm{CN}\) d. \(-\mathrm{OCH}\left(\mathrm{CH}_{3}\right)_{2}\) e. \({ }^{-} \mathrm{C} \equiv \mathrm{CH}\) f. \(\mathrm{H}_{2} \mathrm{O}\) g. \(\mathrm{NH}_{3}\) h. NaI i. \(\mathrm{NaN}_{3}\)

Classify each solvent as protic or aprotic. a. \(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) b. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OCH}_{2} \mathrm{CH}_{3}\) c. \(\mathrm{CH}_{3} \mathrm{COOCH}_{2} \mathrm{CH}_{3}\)
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