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Chapter 13: Problem 264

Explain why \(\left(\mathrm{CH}_{3}\right)_{4} \mathrm{~N}^{+}\) is neither a nucleophile nor an electrophile.

Short Answer

Expert verified
The tetramethylammonium ion \((CH_3)_4N^+\) is neither a nucleophile nor an electrophile because it lacks a lone pair of electrons for donation as a nucleophile and cannot accept additional electrons as an electrophile. This is because the nitrogen atom is already participating in four covalent bonds, reaching its maximum bonding capacity within the molecule.

Step by step solution

01

Understand what nucleophiles and electrophiles are

Nucleophiles are chemical species that have a lone pair of electrons or a negatively charged atom and can donate electrons to an electrophilic center (typically positively charged or electron-deficient). Electrophiles, on the other hand, are chemical species that can accept electrons from a nucleophile and are often electron-deficient or sometimes positively charged.
02

Analyze the molecular structure of tetramethylammonium ion

The given molecule is the tetramethylammonium ion \((CH_3)_4N^+\). Let's examine the structure of this ion: The nitrogen atom is bonded to four methyl groups \((CH_3)\) and carries a positive charge. Since there are four bonds to the nitrogen atom, it doesn't have any lone pair of electrons available.
03

Determine if the molecule can act as a nucleophile

Since the nitrogen atom in the tetramethylammonium ion does not have a lone pair of electrons to donate and the molecule bears a positive charge, the tetramethylammonium ion cannot behave as a nucleophile. Nucleophiles are electron-rich species, but \((CH_3)_4N^+\) lacks accessible electrons for donation.
04

Determine if the molecule can act as an electrophile

Electrophiles are electron-deficient species that accept electrons from nucleophiles. While the tetramethylammonium ion carries a positive charge, which might suggest it is electrophilic, the nitrogen atom is already participating in four covalent bonds and reaches its maximum bonding capacity within the molecule. Therefore, this ion cannot act as an electrophile since it cannot accept any more electrons to form additional bonds with a nucleophile.
05

Conclusion

Based on our analysis, the tetramethylammonium ion \((CH_3)_4N^+\) is neither a nucleophile nor an electrophile because it doesn't have a lone pair of electrons to donate, and while carrying a positive charge, its nitrogen atom is unable to accept additional electrons due to its maximum bonding capacity already being reached.

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Most popular questions from this chapter

Under \(\mathrm{S}_{\mathrm{N}} 1\) conditions, 2 -bromo octane, of specific rotation \(-20.8^{\circ}\), was found to yield 2 -octanol of specific rotation \(+3.96^{\circ}\). If optically pure 2-bromooctane has a specific rotation of \(-34.6^{\circ}\) and optically pure 2-octanol has a specific rotation of \(-9.9^{\circ}\) calculate: (a) the optical purity of reactant and product; (b) the percentage of racemization and of inversion accompanying the reaction; (c) the percentage of front side and of back side attack on the carbonium ion.

Outline all steps in a possible laboratory synthesis of each of the following, using benzene, toluene, and any needed aliphatic or inorganic reagents. (a) p-bromobenzyl chloride (b) triphenylchloromethane (c) allyl iodide (d) benzal bromide (e) \(\mathrm{m}\) -nitrobenzotrichloride (f) 1,2 -dichloro-1- phenylethane (g) phenylacetylene (h) phenylcyclopropane

Each of the following cations is capable of rearranging to a more stable cation. Limiting yourself to a single 1,1 -shift, suggest a structure for the rearranged cation. (a) \(\mathrm{CH}_{3} \mathrm{CHCH}_{3} \mathrm{C}^{+} \mathrm{HCH}_{3}\) (b) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+} \mathrm{CHCH}_{3}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{C}^{+} \mathrm{HCH}\left(\mathrm{CH}_{3}\right) \mathrm{C}\left(\mathrm{CH}_{3}\right)_{3}\) (d) \(\mathrm{CH}_{2}=\mathrm{CHCH}_{2} \mathrm{C}^{+} \mathrm{HCH}_{2} \mathrm{CH}_{3}\) (e) \(\mathrm{CH}_{3} \mathrm{OCH}_{2} \mathrm{C}^{+} \mathrm{HC}\left(\mathrm{CH}_{3}\right)_{3}\)

Ethyl chloride \((0.1 \mathrm{M})\) reacts with potassium iodide \((0.1 \mathrm{M})\) in acetone solution at \(60^{\circ}\) to give ethyl iodide and potassium chloride at a rate of \(5.44 \times 10^{-7}\) mole/liter/sec (a) If the reaction proceeded by an \(\mathrm{S}_{\mathrm{N}} 2\) mechanism, what would the rate of the reaction be at \(0.01 \mathrm{M}\) concentrations of both reactants? Show your method of calculation. (b) Suppose the rate were proportional to the square of the potassium iodide concentration and the first power of the ethyl chloride \(\left(\mathrm{S}_{\mathrm{N}} 3\right)\). What would the rate be with \(0.01 \mathrm{M}\) reactants? (c) If one starts with solutions initially \(0.1 \mathrm{M}\) in both reactants, the rate of formation of ethyl iodide is initially \(5.44 \times 10^{-7}\) mole/liter/sec but falls as the reaction proceeds and the reactants are used up. Make plots of the rate of formation of ethyl iodide against the concentration of ethyl chloride as the reaction proceeds (remembering that one molecule of ethyl chloride consumes one molecule of potassium iodide) on the assumption that the rate of reaction is proportional to the first power of the ethyl chloride concentration; and to (1) the zeroth power, (2) the first power, and (3) the second power of the potassium iodide concentration. (d) What kind of experimental data would one need to tell whether the rate of the reaction of ethyl chloride with potassium iodide is first order in each reactant or first order in ethyl chloride and zero order in potassium iodide?

Benzyl bromide reacts with \(\mathrm{H}_{2} \mathrm{O}\) in formic acid solution to yield benzyl alcohol? the rate is independent of \(\left[\mathrm{H}_{2} \mathrm{O}\right]\). Under the same conditions \(\mathrm{p}\) -methylbenzyl bromide reacts 58 times as fast. Benzyl bromide reacts with ethoxide ion in dry alcohol to yield benzyl ethyl ether \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OC}_{2} \mathrm{H}_{5}\right) ;\) the rate depends upon both \([\mathrm{RBr}]\) and \(\left[\mathrm{OC}_{2} \mathrm{H}_{5}^{-}\right]\). Under the same conditions \(\mathrm{p}\) -methylbenzyl bromide reacts \(1.5\) times as fast. Interpret these results. What do they illustrate concerning the effect of: (a) polarity of solvent, (b) nucleophilic power of the reagent, and (c) electron release by substituents?
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