Question

Each molecule or ion can function as a base. Write a structural formula of the conjugate acid formed by reaction of each with \(\mathrm{HCl}\). (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) (b) \(\mathrm{HCH}\) (c) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}\) (d) \(\mathrm{HCO}_{3}^{-}\)

Short answer

(a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) (b) \(\mathrm{HCH}\) (c) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}\) (d) \(\mathrm{HCO}_{3}^{-}\) Answer: (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}_{2}^{+}\) (b) No conjugate acid for methane. (c) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}_{2}^{+}\) (d) \(\mathrm{H}_{2}\mathrm{CO}_{3}\)

Step-by-step solution

Identify the site for protonation

For each molecule or ion, we need to identify the atom that will accept the proton. This will usually be a lone pair electron on more electronegative elements, such as oxygen or nitrogen. (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\): The oxygen atom in the hydroxyl group has a lone pair and will accept the proton. (b) \(\mathrm{HCH}\): This is methane (\(\mathrm{CH}_4\)), and it cannot act as a base as there are no electronegative elements with lone pairs. (c) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}\): The nitrogen atom has a lone pair and will accept the proton. (d) \(\mathrm{HCO}_{3}^{-}\): The oxygen atoms in the bicarbonate ion have lone pairs and can accept a proton. In this case, it's the oxygen linked to the hydrogen that accepts the proton.

Add the proton and draw the conjugate acids

Now that we have identified where the proton will bond, we can add it to the formula and draw the conjugate acids. (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) + \(\mathrm{H}^{+}\) → \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}_{2}^{+}\) (b) \(\mathrm{HCH}\) does not have a site for protonation and cannot act as a base. (c) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}\) + \(\mathrm{H}^{+}\) → \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}_{2}^{+}\) (d) \(\mathrm{HCO}_{3}^{-}\) + \(\mathrm{H}^{+}\) → \(\mathrm{H}_{2}\mathrm{CO}_{3}\) So the conjugate acids are: (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}_{2}^{+}\) (b) No conjugate acid for methane. (c) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}_{2}^{+}\) (d) \(\mathrm{H}_{2}\mathrm{CO}_{3}\)

Key concepts

Protonation in Organic Chemistry

Protonation is a fundamental process in organic chemistry, where a proton (a hydrogen ion, \( \mathrm{H}^{+} \) ) is added to a molecule. This often occurs when a base reacts with an acid.

For instance, when organic bases are exposed to hydrochloric acid (\( \mathrm{HCl} \) ), the bases accept protons from the \( \mathrm{HCl} \) and transform into their respective conjugate acids. Protonation typically occurs on an atom with a lone pair of electrons, such as nitrogen or oxygen, making these sites more likely to accept protons due to their higher electronegativity.

Understanding where protonation occurs helps predict the behavior and reactivity of molecules in different chemical environments, a crucial skill in organic synthesis and analytical chemistry.

Bases and Conjugate Acids

In the context of acid-base chemistry, bases are substances that can accept protons, whereas acids are proton donors. When a base accepts a proton, it becomes its conjugate acid.

The strength of a base depends on its ability to accept a proton. Strong bases readily form their conjugate acids. For example, \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} \) (ethyl alcohol) when protonated forms \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}_{2}^{+} \) which is its conjugate acid.

Conjugate acids are notable in that they have one more \( \mathrm{H}^{+} \) and one less negative charge than their base forms. The concept of conjugation pairs is invaluable, as it helps chemists comprehend the various outcomes of acid-base reactions and their equilibrium states.

Drawing Structural Formulas

Structural formulas are representations of molecules that show how atoms are arranged and bonded in a molecule. They are crucial for understanding the geometry and reactivity of organic compounds.

While drawing structural formulas, it's essential to consider valences of the atoms and how they bond with other atoms. Single lines represent single bonds, double lines represent double bonds, and lone pairs are typically denoted by dots around the atom. Detailed drawings will include the 3D structure of the molecule, indicating the spatial distribution of atoms crucial for understanding stereochemistry.

When representing a molecule's transformation into its conjugate acid, adding a proton requires showing the new bond to the hydrogen ion. As we saw in the exercise, it is important to first identify the reactive site. For \( \left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH} \) dimethylamine, the lone pair on the nitrogen can form a new bond by protonation, yielding \( \left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}_{2}^{+} \) as its corresponding conjugate acid. Visualizing these changes helps in understanding overall molecular changes during reactions.