Question

Identify the electrophile and the nucleophile in each of the following reaction steps. Then draw curved arrows to illustrate the bond-making and bond-breaking processes.

Short answer

An atom or molecule that seeks out another atom or molecule that has an electron pair available for bonding in a chemical process.

Lewis’s acids (compounds that receive electron pairs) are electrophilic chemicals, and many of them are Bronsted acids (compounds that donate protons).

Definition ofnucleophile-

The term "nucleophile" refers to compounds that transfer electron pairs to electrophiles in order to establish chemical bonds with them.

Any ion or molecule with a free electron pair or a pi bond carrying two electrons has the ability to act as a nucleophile.

Step-by-step solution

Step 1:Definition of electrophile- 

An atom or molecule that seeks out another atom or molecule that has an electron pair available for bonding in a chemical process.

Lewis’s acids (compounds that receive electron pairs) are electrophilic chemicals, and many of them are Bronsted acids (compounds that donate protons).

Definition ofnucleophile-

The term "nucleophile" refers to compounds that transfer electron pairs to electrophiles in order to establish chemical bonds with them.

Any ion or molecule with a free electron pair or a pi bond carrying two electrons has the ability to act as a nucleophile.

Addition of a Halogen to an Alkyne-

A second addition reaction happens in the presence of excess halogen.

• A pi-complex is formed when an alkyne (a nucleophile) interacts with an electrophile.

• The halo-substituted alkene is formed when the chloride ion joins the pi-complex.

Mechanism for the conversion of alkyne to a trans alkene- 

• The lone electron from sodium's s orbital is transported to the alkyne's sp carbon.
• This results in the formation of a radical anion, which has a negative charge and an unpaired electron.
• An arrowhead with a single barb is used to depict the movement of a single electron. (Recall from Section 1.3 that sodium has a significant tendency to lose the lone electron in its outer-shell s orbital.)
• Because the radical anion is such a powerful base, it can take a proton from ammonia. The unpaired electron is on a vinylic carbon, resulting in the production of a vinylic radical.
• A vinylic anion is formed by another single-electron transfer from sodium to the vinylic radical.
• The vinylic anion is likewise a strong base, as it eliminates a proton from the equation.

Addition of borane to alkyne-

• In the same way that it contributes to alkenes, BH₃or R₂BH (in THF) adds to alkynes. The electrophile is boron, and the nucleophile is H^-. When the addition reaction is completed, the aqueous phase is formed.
• To the reaction mixture, sodium hydroxide and hydrogen peroxide are added. As a result, the end result, as in the case of alkenes, the boron is replaced by an OH group. The enol that results is instantly usable.
• To a ketone it tautomerizes.