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Two sets of nucleophiles (i) and (ii) are given below. Arrange the nucleophiles in each set in the increasing order of reactivity towards an alkyl halide. (i) \(\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{~S}\) and \(\mathrm{NaSH}\) (ii) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}, \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{SNa}\)

Short Answer

Expert verified
Question: Arrange the given nucleophiles in increasing order of reactivity towards an alkyl halide. Set (i): \(H_2O, H_2S, NaSH\) Set (ii): \(CH_3CH_2NH_2, CH_3CH_2OH, CH_3CH_2SNa\) Answer: Set (i): \(H_2O < H_2S < NaSH\) Set (ii): \(CH_3CH_2NH_2 < CH_3CH_2OH < CH_3CH_2SNa\)

Step by step solution

01

Determine the nucleophile strength for set (i)

For set (i), we have \(\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{~S}\), and \(\mathrm{NaSH}\). The strength of a nucleophile is determined by its electron density and ability to donate a lone pair of electrons to the electrophilic center (alkyl halide). In this case, the nucleophilicity increases with the size of the atom. Since sulfur (\(\mathrm{H}_{2}\mathrm{S}\) and \(\mathrm{NaSH}\)) is larger than oxygen (\(\mathrm{H}_{2}\mathrm{O}\)), it will be a stronger nucleophile.
02

Determine the basicity for set (i)

Basicity refers to the ability of a nucleophile to donate a proton when reacting with an acid. In this set, \(\mathrm{H}_{2} \mathrm{O}\) is a stronger base than \(\mathrm{H}_{2} \mathrm{S}\). This is because oxygen is more electronegative than sulfur, leading to a stronger bond with hydrogen.
03

Arrange the nucleophiles in set (i)

Considering both nucleophile strength and basicity, we can arrange the nucleophiles in increasing order of reactivity towards an alkyl halide in set (i) as follows: \(\mathrm{H}_{2} \mathrm{O} < \mathrm{H}_{2} \mathrm{S} < \mathrm{NaSH}\).
04

Determine the nucleophile strength and basicity for set (ii)

For set (ii), we have \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}, \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) and \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{SNa}\). Here, we need to compare the nucleophilicity and basicity of nitrogen (\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2}\)), oxygen (\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\)), and sulfur (\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{SNa}\)). As we move down the periodic table, the nucleophilicity increases. Thus, sulfur will be the most nucleophilic, followed by oxygen and then nitrogen. In terms of basicity, nitrogen will be the most basic, followed by oxygen and then sulfur.
05

Arrange the nucleophiles in set (ii)

Taking into account the nucleophile strength and basicity, we can arrange the nucleophiles in increasing order of reactivity towards an alkyl halide in set (ii) as follows: \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2} < \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} < \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{SNa}\). To sum up, the increasing order of reactivity for the nucleophiles in each set are: (i) \(\mathrm{H}_{2} \mathrm{O} < \mathrm{H}_{2} \mathrm{S} < \mathrm{NaSH}\) (ii) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NH}_{2} < \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} < \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{SNa}\)

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Key Concepts

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

Nucleophile Strength
Nucleophiles are species that donate an electron pair to form a chemical bond in reaction mechanisms. The strength of a nucleophile is a crucial concept in organic chemistry; it determines the reactivity and rate of nucleophilic substitution reactions. Strong nucleophiles are crucial in displacing leaving groups, such as halides in alkyl halide reactions, leading to the formation of new bonds.

ucleophile strength is influenced by several factors, including the charge, electronegativity, steric hindrance, and the solvent. Negatively charged nucleophiles are generally stronger than their neutral counterparts due to the increased electron density available for bonding. In terms of atomic size, larger atoms tend to be better nucleophiles in polar aprotic solvents because their outer electrons are less tightly held, allowing for easier donation.

In aprotic solvents, nucleophilicity tends to increase down the periodic table within a group, which is why, in the given exercise, sulfur-containing nucleophiles are stronger nucleophiles than an oxygen-containing nucleophile. Understanding these nuanced characteristics of nucleophiles can lead to better predictions of the outcomes of organic reactions.
Basicity in Organic Chemistry
Basicity is another key property in organic chemistry that measures a molecule's ability to accept protons. It is closely related to nucleophilicity, but while nucleophilicity involves the donation of an electron pair to an electrophile, basicity is specifically concerned with proton acceptance. Factors affecting basicity include the atom's electronegativity, charge, and the stability of the resulting base.

As a general rule, the more electronegative the atom, the less basic it will be, as it holds onto its electrons more tightly and is less inclined to accept a proton. Additionally, the more stable the conjugate base, the stronger the original base will be. In aqueous solutions, basicity often decreases down a group on the periodic table, contrasting with nucleophilicity in polar aprotic solvents. In the exercise example, the oxygen in H2O is more basic than the sulfur in H2S. This difference in basicity can influence reaction mechanisms, especially acid-base reactions, a fundamental concept in understanding organic reaction pathways.
Reactivity Towards Alkyl Halides
Alkyl halides are compounds that contain a halogen atom attached to an alkyl group. Their reactivity toward nucleophiles is of significant importance in substitution reactions such as SN1 and SN2 mechanisms. The reactivity of alkyl halides with nucleophiles depends on the alkyl group's structure (primary, secondary, or tertiary) and the nature of the leaving group (the halide).

For an SN2 reaction, a good nucleophile is one that can effectively attack the electron-deficient carbon and displace the leaving group. Primary alkyl halides are typically most reactive in SN2 reactions due to less steric hindrance. Among halides, iodide is considered the best leaving group because of its larger size and weaker bond with carbon, making alkyl iodides highly reactive.

In the exercise provided, the arrangement of nucleophiles in each set toward an alkyl halide depends on their nucleophilic strength, which is a culmination of the factors discussed earlier. Understanding these reactivity trends provides students with the insight needed to predict the outcomes of a wide range of chemical syntheses involving alkyl halides.

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Most popular questions from this chapter

Reactions of Grignard reagents Organic compounds which contain atleast one carbon-metal bond are called organometallic compounds. Grignard reagents are very reactive organometallic compounds which are of great synthetic utility. They are used for the synthesis of alcohols, aldehydes, ketones, carboxylic acids, etc. They are however decomposed by compounds which have active hydrogen.

When isobutyl bromide loses the bromine to form a cation, it undergoes a 1,2 shift. Which of the following is true? (a) Hydride shift is preferred over methyl shift. (b) Methyl shift is preferred over hydride shift. (c) Methyl shift results in a tertiary carbocation (d) Hydride shift results in a secondary carbocation.

The correct decreasing order of reactivity of the halides for \(\mathrm{S}_{\mathrm{N}} 1\) reaction is (A) \(\mathrm{PhCH}_{2} \mathrm{Br}>\mathrm{CH}_{2}=\mathrm{CHBr}>\mathrm{CH}_{3}-\mathrm{CH}_{2} \mathrm{Br}>\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{Br}\) (B) \(\mathrm{CH}_{2}=\mathrm{CHBr}>\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{Br}>\mathrm{PhCH}_{2} \mathrm{Br}>\mathrm{CH}_{3}-\mathrm{CH}_{2} \mathrm{Br}\) (c) \(\mathrm{PhCH}_{2} \mathrm{Br}>\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{Br}>\mathrm{CH}_{3}-\mathrm{CH}_{2} \mathrm{Br}>\mathrm{CH}_{2}=\mathrm{CHBr}\) (d) \(\mathrm{PhCH}_{2} \mathrm{Br}>\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{Br}>\mathrm{CH}_{2}=\mathrm{CHBr}>\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}\)

2-Bromopentane \(\frac{\left(\mathrm{CH}_{3}\right)_{3} \operatorname{cok} /\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}}{\longrightarrow} \mathrm{P}\) The major product (P) formed in the above reaction is (a) pent-1-ene (b) cis pent-2-ene (c) trans pent-2-ene

The products formed in the above reaction are (a) \(\mathrm{CH}_{3}-\mathrm{CHO}, \mathrm{HCHO}, \mathrm{HCOOH}\) (b) \(\mathrm{CH}_{3}-\mathrm{CHO}, \mathrm{HCOOH}\) (c) \(\mathrm{CH}_{3}-\mathrm{CHO}, \mathrm{OHC}-\mathrm{CHO}\) (d) \(\mathrm{CH}_{3}-\mathrm{COOH}, \mathrm{HCOOH}\)

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