Difference Between Electrophile and Nucleophile: Definition, Reaction and Sample Questions

Ans. A nucleophile acts by donating a pair of electrons to another atom's nucleus. In general, a negatively charged compound is going to be a stronger nucleophile than a neutral compound. In addition, as one proceeds down a given column of the periodic table, the nucleophilicity increases because the electrons are not held as tightly to the nucleus (electronegativity decreases).

CH₃S⁻ is the best nucleophile, because it has a negative charge (more electron density), and its electrons are held less tightly than those of CH₃O⁻because sulfur is less electronegative than oxygen.

Ans. Electrophilic substitution reactions are chemical reactions in which an electrophile displaces a functional group in a compound, which is typically, but not always, a hydrogen atom. Electrophilic aromatic substitution reactions are characteristic of aromatic compounds and are important ways of introducing functional groups of benzene rings. The other main type of electrophilic substitution reaction is an electrophilic aliphatic substitution reaction.

In electrophilic substitution in aromatic compounds, an atom appended to the aromatic ring, usually, hydrogen is replaced by an electrophile. The most important reactions of this type that take place are aromatic nitration, aromatic halogenation, aromatic sulfonation, and acylation and alkylation Friedel-Crafts reactions. It further consists of alkylation and acylation.

In electrophilic substitution in aliphatic compounds, an electrophile displaces a functional group. This reaction is similar to nucleophilic aliphatic substitution where the reactant is a nucleophile rather than an electrophile.

Ans. There are two major types of nitrogen-containing moieties in this molecule.

First, there are the aromatic nitrogenated groups, such as purple, green, and gold. All three of these nitrogens, when reacted with an electrophile such as Boc anhydride, would produce positively charged species. 

This alone would be unfavorable, however, as these nitrogens each donate a lone pair to their aromatic systems, donating this lone pair to an electrophile would break the aromaticity of the system. Breaking aromaticity is always highly unfavorable, and hence, none of these three would readily react with Boc anhydride.

Second, there are the aliphatic nitrogenated groups, such as red and blue. Of these two, the red is tertiary and the blue is secondary. This means the red would produce a positively charged, tetrasubstituted product when reacting with Boc anhydride, whereas the blue would not form a charged product. The blue amine is also more sterically available, and is the correct answer, as it has the best ability to act as a nucleophile. 

Ans. The periodic trends of electronegativity and charge stability are useful tools for predicting nucleophilic strength. First, it is important to recognize that the two charged species, Cl− and F− are the two strongest nucleophiles. 

This is because the destabilizing negative charge present in these species may be neutralized by donating a lone pair to the formation of a chemical bond. As we know, opposite charges attract, so species bearing a full negative charge are drawn to electron-poor regions. Uncharged species such as water and ammonia carry a lone pair capable of bonding but are less energetically drawn towards positive charges. 

Ammonia is a stronger nucleophile than water because nitrogen is less electronegative than oxygen. What this means is that the nitrogen-bound lone pair of ammonia is more loosely contained than the oxygen-bound lone pairs of water. As a result, they are more easily donated to form a bond at an electron-poor carbon.

From this trend, one might expect that fluoride ions would be less nucleophilic than chloride ions since fluorine is more electronegative. However, moving down a group of the periodic table, the atomic radius increases. Anions are stabilized by spreading electron density across an electron cloud of greater volume, such as that of Cl− compared to the smaller F−. As such, the correct ordering of species is II, I, III, IV.

Ans. (A) The given reaction involves two steps. The first step is the nucleophilic addition and the second step is dehydration.

The hydroxide ion abstracts acidic proton to form carbonation. The carbanion attacks another carbonyl group (this is an intramolecular nucleophilic attack). This is followed by protonation of oxygen and dehydration to form the product.

(B) The given reaction is a nucleophilic substitution reaction in which chloride ion is substituted with a methyl group. The product is obtained by the rearrangement of the epoxide ring. The methyl group of the Grignard reagent attacks the terminal carbon atom of the epoxide ring to form an oxide anion. The oxide anion then attacks the carbon-bearing chlorine atom.

(C) The given reaction involves two steps. The first step is the nucleophilic addition and the second step is hydration. The aldehyde oxygen atom is protonated. This is followed by the nucleophilic attack of the amine nitrogen on carbonyl carbon. The −OH group is then protonated and H₃O+ is lost.

(D) The given reaction involves electrophilic aromatic substitution and involves dehydration (loss of a molecule of water). It also involves the formation of a carbocation intermediate.

The −OH group is protonated. This is followed by electrophilic aromatic substitution.