Animated Solution for Chemistry - Organic Chemistry: The correct order for acid strength of compounds $\text{CH}\equiv \text{CH}$, ${\text{CH}}_3-\text{C}\equiv \text{CH}$ and ${\text{CH}}_2={\text{CH}}_2$ is as follows :

The acidic strength of hydrocarbons is a fascinating topic that beautifully ties together the concepts of hybridization, electronegativity, and inductive effects. In this problem, we are asked to compare the acidic strength of three simple yet conceptually rich molecules: ethyne ($\text{HC}\equiv \text{CH}$), propyne (${\text{CH}}_3-\text{C}\equiv \text{CH}$), and ethene (${\text{CH}}_2={\text{CH}}_2$).

To determine the acidic strength of a compound, we must look at the stability of its conjugate base. When a hydrocarbon loses a proton (${\text{H}}^{+}$), it forms a carbanion. The more stable this carbanion is, the stronger the original acid. The stability of a carbanion is directly related to the electronegativity of the carbon atom bearing the negative charge.

But how do we determine the electronegativity of a carbon atom? The secret lies in its hybridization.

Hybridization and s-Character

The electronegativity of a hybridized carbon atom is directly proportional to its percentage s-character. The s-orbital is closer to the nucleus than the p-orbital. Therefore, a hybrid orbital with a higher s-character will hold its electrons closer to the positively charged nucleus, making the atom more electronegative.

Let's break down our three molecules: 1. Ethyne ($\text{HC}\equiv \text{CH}$): The carbon atoms are triple-bonded, meaning they are $sp$ hybridized. An $sp$ hybrid orbital has 50% s-character. 2. Ethene (${\text{CH}}_2={\text{CH}}_2$): The carbon atoms are double-bonded, meaning they are $s{p}^2$ hybridized. An $s{p}^2$ hybrid orbital has 33.3% s-character.

Since 50% is significantly greater than 33.3%, the $sp$ hybridized carbon in ethyne is much more electronegative than the $s{p}^2$ hybridized carbon in ethene. This makes it much easier for ethyne to release a proton. Thus, ethyne is more acidic than ethene.

Inductive Effect in Propyne

Now, let's introduce propyne (${\text{CH}}_3-\text{C}\equiv \text{CH}$) into the mix. Like ethyne, the terminal carbon in propyne is $sp$ hybridized. You might initially think they have the same acidic strength. However, there is a crucial difference: the attached methyl group ($-{\text{CH}}_3$).

Alkyl groups, like the methyl group, are electron-donating. They exert a $+I$ (positive inductive) effect. This means the methyl group pushes electron density towards the $sp$ hybridized carbon. This increased electron density makes the carbon less willing to accept the extra electrons left behind when a proton is released. In other words, the $+I$ effect destabilizes the resulting carbanion.

Because of this electron-donating effect, propyne is less acidic than ethyne, even though both have an $sp$ hybridized terminal carbon.

Combining all our observations, we can establish the final order of acidic strength: - Ethyne is the most acidic because it has a highly electronegative $sp$ carbon and no destabilizing $+I$ groups. - Propyne is next; it has an $sp$ carbon, but its acidity is slightly reduced by the $+I$ effect of the methyl group. - Ethene is the least acidic because its carbon is only $s{p}^2$ hybridized, making it the least electronegative of the three.

Therefore, the correct order is: $\text{HC}\equiv \text{CH}>{\text{CH}}_3-\text{C}\equiv \text{CH}>{\text{CH}}_2={\text{CH}}_2$

This perfectly matches option (c). By understanding the interplay between hybridization and inductive effects, you can easily conquer any question on acidic strength!