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Which substance in each of the following pairs is more reactive as a nucleophile? Explain. (a) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~N}^{-}\) or \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}\) (b) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~B}\) or \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}\) (c) \(\mathrm{H}_{2} \mathrm{O}\) or \(\mathrm{H}_{2} \mathrm{~S}\)

Short Answer

Expert verified
(a) \\(\mathrm{(CH}_{3})_{2} \mathrm{~N}^{-}\\); (b) \\(\mathrm{(CH}_{3})_{3} \mathrm{~N}\\); (c) \\(\mathrm{H}_{2} \mathrm{~S}\\) are more nucleophilic.

Step by step solution

01

Understanding the Nucleophilicity

Nucleophilicity refers to the ability of a chemical species to donate an electron pair to an electrophile. Generally, stronger bases are better nucleophiles. Factors affecting nucleophilicity include charge, electronegativity, steric hindrance, and solvent effects.
02

Step 2(a): Analyze \\((\mathrm{CH}_{3})_{2} \mathrm{~N}^{-}\\) vs. \\((\mathrm{CH}_{3})_{2} \mathrm{NH}\\)

The dimethylamide ion \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~N}^{-}\) carries a negative charge, making it a stronger base and thus a more effective nucleophile compared to the neutral dimethylamine \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}\). Charged nucleophiles are typically more reactive than their neutral counterparts.
03

Step 3(b): Analyze \\((\mathrm{CH}_{3})_{3} \mathrm{~B}\\) vs. \\((\mathrm{CH}_{3})_{3} \mathrm{~N}\\)

Trimethylborane \(\mathrm{(CH}_{3})_{3} \mathrm{~B}\) acts as an electron acceptor due to its empty p-orbital, whereas trimethylamine \(\mathrm{(CH}_{3})_{3} \mathrm{~N}\) has a lone pair of electrons available for donation, making it a better nucleophile. Therefore, \(\mathrm{(CH}_{3})_{3} \mathrm{~N}\) is more nucleophilic than \(\mathrm{(CH}_{3})_{3}~B\).
04

Step 4(c): Analyze \\((\mathrm{H}_{2} \mathrm{O})\\) vs. \\((\mathrm{H}_{2} \mathrm{~S})\\)

Hydrogen sulfide \(\mathrm{H}_{2} \mathrm{~S}\) is more nucleophilic than water \(\mathrm{H}_{2} \mathrm{O}\) due to sulfur being less electronegative than oxygen, which makes HS less likely to retain its electron pair and more likely to donate it. Additionally, sulfur's larger atomic radius allows for better overlap with substrates.

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Key Concepts

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

Reactivity
In organic chemistry, reactivity is a crucial concept that refers to how readily a substance can engage in a chemical reaction. The reactivity of a nucleophile is influenced by several factors:
  • **Charge:** Anions are usually more reactive nucleophiles compared to their neutral versions. For instance, in part (a) of the exercise, \((\mathrm{CH}_{3})_{2} \mathrm{~N}^{-}\) is a negatively charged ion, making it more reactive than its neutral counterpart, \((\mathrm{CH}_{3})_{2} \mathrm{NH}\).
  • **Electronegativity:** Less electronegative atoms are more likely to donate electrons, enhancing their nucleophilicity. This is demonstrated in part (c), where sulfur in \(\mathrm{H}_{2} \mathrm{S}\) is less electronegative than oxygen in \(\mathrm{H}_{2} \mathrm{O}\), resulting in higher nucleophilicity for hydrogen sulfide.
  • **Steric Hindrance:** Larger substituents can block access to the nucleophile, decreasing its reactivity. While not a direct factor in the provided exercise, it's essential to note in general reactions.
These factors determine how a nucleophile will react with an electrophile, emphasizing the diverse nature of chemical interactivity.
Organic Chemistry
Organic chemistry focuses on carbon-containing compounds and their reactions. It's a vast field dealing with molecular structure, properties, and chemical transformations. Within this domain, nucleophiles play a pivotal role by participating in nucleophilic reactions.

In our exercise, we observed the reactivity of different nucleophiles. Part (b) shows how \(\mathrm{(CH}_{3})_{3} \mathrm{~N}\) acts as a better nucleophile compared to \(\mathrm{(CH}_{3})_{3} \mathrm{~B}\), highlighting the need for available electron pairs in nucleophilic attacks.

Understanding nucleophilicity helps in predicting outcomes of organic reactions like the formation of bonds in substitutions or additions. Recognizing the electrons' movement through shared pairs, unshared pairs, and potential changes in bonding is key to mastering many organic mechanisms.

Thus, grasping nucleophilicity in organic chemistry aids in understanding the dynamics between molecular species and how they transform during reactions.
Nucleophile-Electrophile Interactions
The interactions between nucleophiles and electrophiles are foundational in organic reactions. Nucleophiles, typically electron-rich species, seek electron-deficient partners, the electrophiles, to form bonds.

In the discussed solutions, nucleophilicity is highlighted by the availability of electron pairs. \(\mathrm{(CH}_{3})_{3} \mathrm{~N}\), a potent nucleophile, has a lone pair ready to attack an electrophile with a suitable vacant orbital. Such interactions drive many chemical processes, including substitution reactions where bonds are broken and formed as nucleophiles replace other groups on the electrophile.

A good way to visualize this interaction is to imagine a dance where the nucleophile and electrophile come together to form a new pair. The nucleophile's electrons "reach out" to the electrophile's positively charged areas. This process is governed by principles like bond strength, energy changes, and the specific chemical environment, which can influence reaction rates and outcomes.

Understanding these interactions helps chemists design efficient pathways for synthesizing new compounds, crucial in fields ranging from pharmaceuticals to materials science.

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Most popular questions from this chapter

Propose structures for compounds that fit the following descriptions: (a) An alkyl halide that gives a mixture of three alkenes on E2 reaction (b) An organic halide that will not undergo nucleophilic substitution (c) An alkyl halide that gives the non-Zaitsev product on E2 reaction (d) An alcohol that reacts rapidly with \(\mathrm{HCl}\) at \(0{ }^{\circ} \mathrm{C}\)

What product would you expect to obtain from \(\mathrm{S}_{\mathrm{N}} 2\) reaction of \(\mathrm{OH}^{-}\) with (R)-2-bromobutane? Show the stereochemistry of both reactant and product.

Why do you suppose it's not possible to prepare a Grignard reagent from a bromo alcohol such as 4 -bromopentan-1-ol?

How might you prepare each of the following molecules using a nucleophilic substitution reaction at some step? (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CN}\) (b) \(\mathrm{H}_{3} \mathrm{C}-\mathrm{O}-\mathrm{C}\left(\mathrm{CH}_{3}\right)_{3}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\)

Which reaction in each of the following pairs would you expect to be faster? (a) The \(\mathrm{S}_{\mathrm{N}} 2\) displacement by \(\mathrm{I}^{-}\) on \(\mathrm{CH}_{3} \mathrm{Cl}\) or on \(\mathrm{CH}_{3}\) OTos (b) The \(\mathrm{S}_{\mathrm{N}} 2\) displacement by \(\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}\) on bromoethane or on bromocyclohexane (c) The \(\mathrm{S}_{\mathrm{N}}^{2}\) displacement on 2-bromopropane by \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}\) or by \(\mathrm{CN}^{-}\) (d) The \(S_{N} 2\) displacement by \(H S^{-}\) on bromomethane in toluene or in acetonitrile

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