Question

\(p K_{\mathrm{a}}\) value of ethane is (A) \(4.5\) (B) \(6.4\) (C) \(15.7\) (D) 50

Short answer

The \(p K_{\mathrm{a}}\) value of ethane is relatively high due to its low acidity as an alkane. Among the given options, option (C) with a \(p K_{\mathrm{a}}\) value of \(15.7\) aligns with our understanding of alkanes' low acidity and is the correct answer.

Step-by-step solution

Identifying the ethane molecule

Ethane is an alkane with the molecular formula \(C_2H_6\). It is an organic compound composed of two carbon atoms each bonded to three hydrogen atoms.

Understanding acidity and \(p K_{\mathrm{a}}\)

In chemistry, the acidity of a substance is generally characterized by its ability to donate a proton (hydrogen ion) to a solution. A high acidity represents a greater ability to donate a proton, whereas a low acidity signifies a lower ability. The \(p K_{\mathrm{a}}\) value is used to quantitatively express this acidity. Lower \(p K_{\mathrm{a}}\) values correspond to higher acidity, while higher values indicate a lower acidity.

Analyzing ethane's acidity

In the case of ethane, it's important to know that alkanes, in general, are weak acids. The acidity of alkanes is primarily due to the sp3 hybridization of carbon atoms, which means the electron density on each carbon-hydrogen bond is relatively high. This high electron density severely reduces ethane's ability to lose a proton, leading to low acidity levels.

Evaluating the given options

Since ethane's acidity is relatively low, we expect its \(p K_{\mathrm{a}}\) value to be relatively high. Among the given options, the highest \(p K_{\mathrm{a}}\) value is 50 (option D), and the second-highest value is 15.7 (option C). This means we can eliminate options A and B, as their \(p K_{\mathrm{a}}\) values are too low to represent the acidity of ethane.

Selecting the correct option

Here, we consider that option D with a \(p K_{\mathrm{a}}\) value of 50 seems too high, even for the low acidity of ethane. Option C, with a \(p K_{\mathrm{a}}\) value of 15.7, is a more reasonable value that aligns with our understanding of alkanes' low acidity. Therefore, we can conclude that the correct answer is: (C) \(15.7\)

Key concepts

Alkane Acidity

Alkanes, commonly perceived as paraffins, are organic compounds consisting solely of hydrogen and carbon atoms arranged in a tree structure in which all the carbon–carbon bonds are single bonds. Understanding their acidity requires a look into their molecular structure. Generally, alkanes are regarded as very weak acids, which is intriguing considering they contain hydrogen atoms—elements commonly associated with acids.

However, the hydrogen atoms in alkanes are connected to carbon atoms by a bond that is not easily broken to release a proton (H+). This resistance arises from the stability of alkanes and the strength of the carbon-hydrogen bond. In a sense, the bond is too 'happy' where it is, making the prospect of losing its hydrogen not very favorable. As a result, alkanes like ethane have very high pKa values, signifying their low acidity. When evaluating problems related to the acidity of alkanes, one should remember that a higher pKa value indicates less acidity - typical for hydrocarbons with a simple, single-bonded carbon arrangement.

Proton Donation and Acidity

In the grand scheme of chemistry, acidity is a concept that revolves around an atom or molecule's tendency to donate a proton to its environment—a reaction that's fundamentally an exchange with a willing partner, typically a base. Acids are thus proton donors, and their strength is measured by how readily they give up this proton.

In organic chemistry, this 'willingness' is quantified by the pKa value, standing for the negative base-10 logarithm of the acid dissociation constant, Ka. Lower pKa values suggest a substance is a strong acid, eager to part with its proton, while higher values hover around the less willing, weak acids. This concept is crucial in understanding reactivity, stability, and mechanism pathways in organic chemistry. It's also beneficial in predicting the outcome of chemical reactions, particularly those involving acid-base equilibria.

Hybridization and Acidity

In the world of atoms and bonds, hybridization explains how atomic orbitals combine to form new orbitals that can create the optimal bonds for a molecule's geometry. It has a profound effect on a molecule's acidity. For example, in alkanes, the carbon atoms are sp3 hybridized, meaning one s-orbital and three p-orbitals merge to form four equivalent sp3 orbitals, each forming a single bond with a hydrogen atom or another carbon.

This hybridization results in a bond that firmly holds the cloud of electrons close to the carbon atom, making it difficult for the carbon to let go of its attached hydrogen as a proton. Since acid strength is linked to how easy it is to lose that proton, a firm grasp equates to weaker acidity. Moving along the spectrum, sp2 and sp hybridizations—seen in alkenes and alkynes, respectively—allow for a more willing proton release due to the increased s-character, which pulls the electron cloud closer to the atom and makes the hydrogen atoms more acidic. Understanding hybridization is essential not only to grasp acidity differences but also electronegativity and bond strengths in organic compounds.