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Assess the possibility of O-alkylation in the reaction of \(\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{Br}\) with \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{2} \mathrm{NH}^{\ominus} \mathrm{Na}^{\oplus} .\) Give your reasoning.

Short Answer

Expert verified
O-Alkylation is unlikely; N-alkylation is favored due to stronger nucleophilicity of nitrogen.

Step by step solution

01

Identify the Reactants

We have two reactants: 1) Allyl bromide, which is \(\mathrm{CH}_2=\mathrm{CH}-\mathrm{CH}_2 \mathrm{Br}\), and 2) Sodium N-phenylsulfamide, \(\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{SO}_{2}\mathrm{NH}^{\ominus} \mathrm{Na}^{\oplus}\). The sodium phenylsulfonamide acts as the nucleophile here.
02

Consider Reaction Type

In this reaction, the sodium N-phenylsulfamide can act as a nucleophile. We need to decide whether it will perform O-alkylation (any reaction where an oxygen atom is alkylated) or N-alkylation.
03

Analyze the Nucleophile

The nucleophile in this reaction is \(\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{SO}_{2}\mathrm{NH}^{\ominus}\). The nitrogen atom carries a negative charge, making it a strong Lewis base capable of donating electrons.
04

Evaluate Reactivity of Allyl Bromide

The \(\mathrm{CH}_2=\mathrm{CH}-\mathrm{CH}_2\) group in allyl bromide can stabilize a positive charge through resonance, making it a relatively reactive electrophile. The bromine leaves easily in a substitution reaction.
05

Determine Likely Alkylation Site

In this scenario, N-alkylation (instead of O-alkylation) is more likely due to the nucleophilic nitrogen in \(\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{SO}_{2}\mathrm{NH}^{\ominus}\). The nitrogen is more nucleophilic than oxygen in this instance and will preferentially attack the electrophilic carbon in \(\mathrm{CH}_2=\mathrm{CH}-\mathrm{CH}_2\) to form an N-alkylated product.

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Key Concepts

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

O-alkylation
In organic chemistry, alkylation is a process that involves the transfer of an alkyl group (a part of a molecule made up of carbon and hydrogen) to another molecule. O-alkylation specifically involves the transfer of an alkyl group to an oxygen atom. This type of reaction can be seen in many synthetic procedures, particularly when forming ethers, where the alkyl group attaches to the oxygen atom.

O-alkylation is often considered when you have oxygen-containing compounds like alcohols or ethers that can act as nucleophiles. The oxygen atom in these compounds typically bears a partial negative charge, making it capable of attacking positively charged or electron-deficient species. However, the preference for O-alkylation over other types depends greatly on the structure of the reactants involved. In the exercise given, although O-alkylation is considered, the presence of a more nucleophilic nitrogen atom suggests that N-alkylation is more favorable.
Nucleophile
A nucleophile is a chemical species that donates an electron pair to form a chemical bond. Nucleophiles are "nucleus-loving" and seek out positively charged or electron-deficient centers in molecules, usually represented by an atom with a positive charge or a partial positive charge.

In the context of the exercise, sodium N-phenylsulfamide acts as the nucleophile. The structure \(\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{SO}_{2}\mathrm{NH}^{\ominus}\) indicates that the nitrogen atom is negatively charged, making it highly attractive to positively charged species. Its Lewis base characteristics enable it to readily donate its electrons in substitution reactions such as those involving allyl bromide. Understanding what makes a species a nucleophile—like possessing a lone pair of electrons or a negative charge—can help predict how it will react.
  • The strength of a nucleophile is influenced by charge, electronegativity, and the solvent.
  • Negatively charged species generally make stronger nucleophiles.
  • Nucleophiles are key in many organic reactions, including substitutions and additions.
Allyl Bromide
Allyl bromide, with the formula \(\mathrm{CH}_2=\mathrm{CH}-\mathrm{CH}_2 \mathrm{Br}\), is an organic compound and part of the class called allylic halides. The structure is characterized by a bromine atom attached to an allylic carbon, which is adjacent to a double bond. These molecules are known for their reactivity, which is largely due to the ability to stabilize the resulting positive charge after the bromine leaves in a chemical reaction.

When allyl bromide is used in reactions, its key characteristic is its electrophilic tendency. After bromine detaches as a leaving group—a step facilitated by the surrounding electron-rich double bond—the carbon previously bonded to bromine becomes an attractive site for nucleophiles. This reactivity makes allyl bromide a common reagent in organic synthesis to introduce allyl groups into various molecules.

Key reactions involving allyl bromide include:
  • Substitution reactions, where nucleophiles replace the bromine.
  • Allylation processes, where it forms bonds with nucleophilic sites.
Sodium N-phenylsulfamide
Sodium N-phenylsulfamide is an organosulfur compound with the formula \(\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{SO}_{2}\mathrm{NH}^{\ominus} \mathrm{Na}^{\oplus}\). It plays an important role in the field of organic chemistry as a useful nucleophile. The N-phenylsulfamide ion is negatively charged at the nitrogen atom, making it versatile for various organic reactions.

The presence of the phenyl group \(\mathrm{C}_{6}\mathrm{H}_{5}\) attached to the sulfonamide group \(\mathrm{SO}_{2}\mathrm{NH}^{\ominus}\) contributes to its ability to stabilize charge, yet the nitrogen still acts as a stronger site for nucleophilic attack compared to the oxygen (as might be considered for O-alkylation).

Reasons sodium N-phenylsulfamide is utilized include:
  • The stability and reactivity balance it offers in nucleophilic substitution reactions.
  • Its ability to form stable N-alkylation products with alkylating agents like allyl bromide.
Understanding the properties of this compound aids in predicting the sites and outcomes of nucleophilic reactions, such as favoring N-alkylation over O-alkylation.

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Most popular questions from this chapter

Explain why the nitration of benzenamine to give 2- and 4-nitrobenzenamines is unsatisfactory with nitric acid-sulfuric acid mixtures. Show how this synthesis could be achieved by suitably modifying the amine function.

Show how the following transformations may be achieved. List reagents and approximate reaction conditions. a. 3-bromopropene to 3-butenamine b. cyclohexanone to cyclohexamine c. benzenecarboxylic acid to phenylmethanamine (not N-phenylmethanamine) d. benzenecarbaldehyde to N-methylphenylmethanamine \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{NHCH}_{3}\right)\)

Decide which member in each of the following pairs of compounds is the stronger base. Give your reasoning. a. \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}\) or \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{~N}\) b. \(\left(\mathrm{CH}_{2}\right)_{5} \mathrm{NH}\) or \(\left(\mathrm{CH}_{2}\right)_{2} \mathrm{NH}\) c. \(\mathrm{CF}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) d. \(\left(\mathrm{CH}_{3}\right)_{3} \stackrel{\oplus}{\mathrm{N}} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) or \(\stackrel{\ominus}{\mathrm{O}}_{2} \mathrm{CCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\)

Write structural formulas for substances (one for each part) that fit the following descriptions: a. an aromatic amine that is a stronger base than benzenamine b. a substituted phenol that would not be expected to couple with benzenediazonium chloride in acid, alkaline, or neutral solution c. a substituted benzenediazonium chloride that would be a more active coupling agent than benzenediazonium chloride itself d. methyl \(Z\) -benzenediazoate e. the important resonance structures of the ammonium salt of N-nitroso-N- phenylhydroxylamine (Cupferron)

Benzenediazonium chloride solvolyzes in water to give a mixture of benzenol and chlorobenzene. Some of the facts known about this and related reactions are 1\. The ratio \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Cl} / \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\) increases markedly with \(\mathrm{Cl}^{\ominus}\) concentration but the rate hardly changes at all. 2\. There is no rearrangement observed with 4-substituted benzenediazonium ions, and when the solvolysis is carried out in \(\mathrm{D}_{2} \mathrm{O}\), instead of \(\mathrm{H}_{2} \mathrm{O}\), no \(\mathrm{C}-\mathrm{D}\) bonds are formed to the benzene ring. 3\. 4-Methoxybenzenediazonium chloride solvolyzes about 30 times faster than 4-nitrobenzenediazonium chloride. 4\. Benzenediazonium salts solvolyze in \(98 \% \mathrm{H}_{2} \mathrm{SO}_{4}\) at almost the same rate as in \(80 \% \mathrm{H}_{2} \mathrm{SO}_{4}\) and, in these solutions, the effective \(\mathrm{H}_{2} \mathrm{O}\) concentration differs by a factor of 1000 . Show how these observations support an \(S_{\mathrm{N}} 1\) reaction of benzenediazonium chloride, and can be used to argue against a benzyne-type elimination-addition reaction with water acting as the \(E 2\) base (Section \(14-6 \mathrm{C}\) ) or an \(S_{\mathrm{N}} 2\) reaction with water as the nucleophile (Section 8-4, Mechanism B, and Section 14-6).

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