Question

Question: Draw the products formed when ${\mathbf{CH}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{2}}\mathbf{\equiv }{\mathbf{C}}^{\mathbf{-}}{\mathbf{Na}}^{\mathbf{+}}$ reacts with each compound.

$$\begin{gathered} \mathbf{a}\mathbf{.}\mathbf{}\mathbf{}{\mathbf{CH}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{2}}{\mathbf{CH}}_{\mathbf{2}}\mathbf{Br} \\ \mathbf{b}\mathbf{.}\mathbf{}\mathbf{}{\mathbf{\left(}{\mathbf{CH}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{2}}{\mathbf{CHCH}}_{\mathbf{2}}{\mathbf{CH}}_{\mathbf{2}}\mathbf{Cl} \\ \mathbf{c}\mathbf{.}\mathbf{}\mathbf{}{\mathbf{\left(}{\mathbf{CH}}_{\mathbf{3}}{\mathbf{CH}}_{\mathbf{2}}\mathbf{\right)}}_{\mathbf{3}}\mathbf{CCl} \end{gathered}$$

d.

e.

Short answer

Answer

a.

b.

c.

d.

e.

Step-by-step solution

Step-by-Step Solution Step 1: Reactions of acetylide ions

Acetylide ions are formed as a result of proton abstraction from an acetylene group.The anion hence formed can be used as a reagent for nucleophilic substitution.

The reactions are shown below:

Reaction of  CH3CH2≡C-Na+

a.

Reaction a.

b.

Reaction b.

c.

In this case, the electrophilic carbon center is a $3°$carbocation, and thus does not favor attack. Instead, elimination occurs.

Reaction c.

d.

The carbon centers are equivalent, thus the attack can take place from any of the two sides.

Reaction d

e.

Favoring localid="1648805732814" ${\mathrm{S}}_{\mathrm{N}}2$attack, the ion would tend to attack at the less substituted carbon center.

Reaction e