Question

The correct nucleophilicity order among the following is (a) \(\mathrm{NH}_{2}^{-}>\mathrm{OH}^{-}>\mathrm{F}^{-}>\mathrm{CH}_{3}^{-}\) (b) \(\mathrm{CH}_{3}^{-}>\mathrm{NH}_{2}^{-}>\mathrm{OH}^{-}>\mathrm{F}^{-}\) (c) \(\mathrm{OH}^{-}>\mathrm{NH}_{2}^{-}>\mathrm{CH}_{3}^{-}>\mathrm{F}^{-}\) (d) \(\mathrm{CH}_{3}^{-}>\mathrm{OH}^{-}>\mathrm{F}^{-}>\mathrm{NH}_{2}^{-}\)

Short answer

Question: Arrange the given species in the order of increasing nucleophilicity: F-, NH2-, OH-, CH3-. Choose the correct option: (a) F- < OH- < NH2- < CH3- (b) CH3- > NH2- > OH- > F- (c) NH2- > CH3- > OH- > F- (d) OH- > NH2- > CH3- > F- Answer: (b) CH3- > NH2- > OH- > F

Step-by-step solution

Factors affecting nucleophilicity

It is important to understand the factors affecting nucleophilicity to solve this exercise: 1. Charge: Higher charge (more negative) makes a species more nucleophilic 2. Size and polarizability: Larger and more polarizable species are more nucleophilic 3. Electronegativity: Less electronegative species are more nucleophilic.

Evaluate each species

Let's examine each species to determine its nucleophilicity: 1. \(\mathrm{NH}_{2}^{-}\): Has a -1 charge and is relatively small. 2. \(\mathrm{OH}^{-}\): Has a -1 charge and is relatively small. 3. \(\mathrm{F}^{-}\): Has a -1 charge and is relatively small. 4. \(\mathrm{CH}_{3}^{-}\): Has a -1 charge and is larger and more polarizable than the others.

Compare nucleophilicity of different species

Now, we can compare the species and determine the correct nucleophilicity order: 1. Charges are all equal (-1), so this factor does not help us rank the species. 2. \(\mathrm{CH}_{3}^{-}\) is the largest and most polarizable species, making it the most nucleophilic. 3. Considering electronegativity, nitrogen (in \(\mathrm{NH}_{2}^{-}\)) is less electronegative than oxygen (in \(\mathrm{OH}^{-}\)) which is less electronegative than fluorine (in \(\mathrm{F}^{-}\)). 4. Based on the factors affecting nucleophilicity, the order of nucleophilicity should be \(\mathrm{CH}_{3}^{-}>\mathrm{NH}_{2}^{-}>\mathrm{OH}^{-}>\mathrm{F}^{-}\).

Choose the correct option

The correct nucleophilicity order (as determined in step 3) is given by: (b) \(\mathrm{CH}_{3}^{-}>\mathrm{NH}_{2}^{-}>\mathrm{OH}^{-}>\mathrm{F}^{-}\)

Key concepts

Factors Affecting Nucleophilicity

Understanding nucleophilicity is crucial in organic chemistry as it determines how a nucleophile will interact with an electrophile. A nucleophile is a chemical species that donates an electron pair to an electrophile to form a chemical bond in a reaction. Several factors influence the strength or nucleophilicity of a molecule:

• Charge: Generally, a negatively charged species (anions) are more nucleophilic than their neutral counterparts. They have extra electrons that they can donate readily to electrophiles.
• Electronegativity: The tendency of an atom to attract electrons to itself can affect nucleophilicity. Atoms with lower electronegativity are less tight on their valence electrons, making these electrons available for bonding.
• Steric hindrance: Bulkier molecules with large groups attached can be less nucleophilic because they encounter physical obstacles in approaching the electrophile.
• Solvent effects: The environment in which a reaction takes place can have a significant impact. For example, polar protic solvents can hinder nucleophilicity by surrounding and solvating the nucleophile, making it less available for the reaction.

In the exercise above, we can apply these principles to rank the nucleophilicity of the given anions. All species involved carry the same charge (-1), so we must look at other factors, such as atomic size, polarizability, and electronegativity, to establish the correct order of nucleophilicity.

Polarizability and Nucleophilicity

Polarizability is a measure of how easily an electron cloud around an atom or molecule can be distorted. This is an essential factor in nucleophilicity because a more polarizable ion has a more easily distortable electron cloud, which can facilitate the formation of a bond with an electrophile.

In the progression of an organic reaction, a polarizable nucleophile, such as a large, less electronegative atom, can form a bond more quickly with an electrophile. In our exercise, the methyl anion (\r\(\mathrm{CH}_{3}^{-}\)\r) is the most polarizable owing to its larger size and relatively low electronegativity, compared to the other anions like \r\(\mathrm{NH}_{2}^{-}\)\r, \r\(\mathrm{OH}^{-}\)\r, and \r\(\mathrm{F}^{-}\)\r. As a result, the methyl anion would be the most nucleophilic, followed by the amide, hydroxide, and fluoride anions respectively.

Electronegativity and Nucleophilicity

Electronegativity is the ability of an atom to attract electrons toward itself in a chemical bond. It plays a pivotal role in determining nucleophilicity. Atoms with lower electronegativity are less greedy in holding onto their electrons and therefore can share them more readily as a nucleophile in chemical reactions.

When comparing nucleophiles with the same charge, the less electronegative one will typically be more reactive towards electrophiles. This is simply because it's more willing to let go of its valence electrons to form a new bond. In the context of our exercise, nitrogen in the amide ion (\r\(\mathrm{NH}_{2}^{-}\)\r) is less electronegative than oxygen in the hydroxide (\r\(\mathrm{OH}^{-}\)\r), which in turn is less electronegative than fluorine in the fluoride ion (\r\(\mathrm{F}^{-}\)\r). Consequently, the nucleophilicity order reflects decreased electronegativity: \r\(\mathrm{CH}_{3}^{-}>\r\mathrm{NH}_{2}^{-}>\r\mathrm{OH}^{-}>\r\mathrm{F}^{-}\). The methyl anion is the most nucleophilic, despite carbon's electronegativity, because its larger size and polarizability compensate in this case.