Question

The boiling points of hex-1-ene (64 °C) and hex-1-yne (71 °C) are sufficiently close that it is difficult to achieve a clean separation by distillation. Show how you might use the acidity of hex-1-yne to remove the last trace of it from a sample of hex-1-ene.

Short answer

The last trace of hex-1-yne is removed from hex-1-ene by reacting with a strong base like sodium amide.

Step-by-step solution

Step-by-Step SolutionStep 1: Acidity

Acidity is defined as the chemical compound's property to release its $\left[{\mathbf{H}}^{+}\right]$ (hydrogen ion) to the other chemical compound.

Two concepts explain acidy:

• Bronsted Lowrey concept
• Lewis acid-base concept

Acidity of Hex-1-yne

As the acidity of the C-H bond increases with the increasing percent of s character of the orbitals$\underset{\underset{\text{Acidity}}{\to }}{\left({\text{sp}}^{\text{3}}{\text{<sp}}^{\text{2}}\text{<sp}\right)}$ , hex-1-yne has terminal hydrogen, which is acidic due to sp hybrid$\equiv \text{C-H}$ bonds.

 Step 3: Removing the last trace of hex-1-yne from hex-1-ene

Hex-1-yne has more acidic proton than hex-1-ene due to more % s character. A strong base like sodium amide ${\text{NaNH}}_{\text{2}}$, when reacted with both, will form a sodium salt of hex-1-yne and leave the hex-1-ene without reacting.

The reaction of hex-1-yne and hex-1-ene with sodium amide is represented as:

$\begin{array}{c}\underset{\text{Hex-1-yne}}{{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{C}\equiv \text{H}}\text{+}\underset{\text{Sodium\hspace{0.17em}amide}}{{\text{NaNH}}_{\text{2}}}\to \underset{\text{Sodium\hspace{0.17em}salt\hspace{0.17em}ofhex-1-yne}}{{\text{Na}}^{\text{+}}{\text{\hspace{0.17em}CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{C}\equiv \stackrel{\text{-}}{\text{C}}}\\ \\ \underset{\text{Hex-1-ene}}{{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{C=H}}\text{+}\underset{\text{Sodium\hspace{0.17em}amide}}{{\text{NaNH}}_{\text{2}}}\to \underset{\text{Hex-1-ene}}{{\text{CH}}_{\text{3}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\text{C=H}}\end{array}$