Question

The compound of molecular formula \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}\) may have functional group (A) Alcohol (B) Ether (C) Carboxylic acid (D) Aldehyde

Short answer

The compound with the molecular formula \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}\) may have functional groups (A) Alcohol and (B) Ether since they both match the given formula.

Step-by-step solution

List the general molecular formulas of given functional groups

First, let's list the general molecular formulas of the given functional groups: (A) Alcohol: \(\mathrm{C}_{n} \mathrm{H}_{2n+1} \mathrm{OH}\) (B) Ether: \(\mathrm{R}\mathrm{-O-}\mathrm{R'}\) (C) Carboxylic acid: \(\mathrm{C}_{n} \mathrm{H}_{2n} \mathrm{COOH}\) (D) Aldehyde: \(\mathrm{C}_{n} \mathrm{H}_{2n+1} \mathrm{CHO}\)

Compare the given molecular formula with the general molecular formulas for each functional group

Now, let's compare our molecular formula \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}\) with the general molecular formulas we listed in step 1: (A) Alcohol: If n = 3, it gives the formula \(\mathrm{C}_{3+1} \mathrm{H}_{2(3)+1+1} \mathrm{O} = \mathrm{C}_{4}\mathrm{H}_{10}\mathrm{O}\), which matches the given formula. (B) Ether: Since ethers have a general molecular formula of \(\mathrm{R}\mathrm{-O-}\mathrm{R'}\), there will always be two carbon chains or groups. In this case, one possibility could be \(\mathrm{C}_{2}\mathrm{H}_{5} - \mathrm{O} - \mathrm{C}_{2}\mathrm{H}_{5}\) which gives the formula \(\mathrm{C}_{4}\mathrm{H}_{10}\mathrm{O}\), which matches the given formula. (C) Carboxylic acid: If n = 4, it gives the formula \(\mathrm{C}_{4} \mathrm{H}_{2(4)} \mathrm{COOH} = \mathrm{C}_{5}\mathrm{H}_{8}\mathrm{O}_{2}\), which does not match the given formula. (D) Aldehyde: If n = 3, it gives the formula \(\mathrm{C}_{3+1} \mathrm{H}_{2(3)+1+2} \mathrm{O} = \mathrm{C}_{4}\mathrm{H}_{12}\mathrm{O}\). Based on our comparison, we can see that the molecular formula \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}\) matches with the general molecular formulas of Alcohol (A) and Ether (B).

Determine the answer

Since the given molecular formula \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}\) matches with the general molecular formulas of both Alcohol (A) and Ether (B), the correct answer is that the compound may have functional groups: (A) Alcohol and (B) Ether

Key concepts

Alcohol

In organic chemistry, alcohols are characterized by the presence of a hydroxyl group (-OH) attached to a carbon atom. This carbon is usually part of a larger carbon chain or ring. The basic structure of an alcohol can be represented by the general formula

• \(\mathrm{C}_{n} \mathrm{H}_{2n+1}\mathrm{OH}\)

,where \( n \) is the number of carbon atoms in the molecule. This formula indicates a saturated hydrocarbon chain with an -OH group attached.

Alcohols play a vital role in both industrial chemistry and biochemistry. They range in complexity from simple methanol (one carbon atom) to complex multi-carbon alcohols found in nature, like cholesterol. Alcohols are polar molecules, which affects their solubility and boiling points, typically making them higher than their hydrocarbon counterparts due to hydrogen bonding.

In the compound with molecular formula \( \mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O} \), the presence of an alcohol group could be suggested by matching it to the general formula above, leading to structures like butanol, which could exist in different isomeric forms: n-butanol, isobutanol, etc.

Ether

Ethers are another important class of organic compounds characterized by an oxygen atom connected to two alkyl or aryl groups, represented by the general structure

• \( \mathrm{R}\mathrm{-O-}\mathrm{R'} \)

,where \( \mathrm{R} \) and \( \mathrm{R'} \) can be identical or different carbon-containing groups.

Ethers are known for their relatively low reactivity compared to alcohols. They are useful as solvents due to their ability to dissolve both polar and nonpolar substances. Ethers possess relatively low boiling points because they lack the hydrogen bonding seen in alcohols. However, the C-O-C linkage is more than adequate for versatile chemical interactions.

In the context of the \( \mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O} \) formula, one common ether structure is diethyl ether (\( \mathrm{C}_{2} \mathrm{H}_{5}\mathrm{-O-} \mathrm{C}_{2} \mathrm{H}_{5} \)), which fits the molecular formula perfectly. This substance was historically used as a common anesthetic, highlighting its significance in both chemical and medical fields.

Molecular formula

A molecular formula represents the exact number of each type of atom in a molecule without any indication of how these atoms are bonded or arranged. In the formula \( \mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O} \), we see four carbon atoms, ten hydrogen atoms, and one oxygen atom.

Molecular formulas are essential for identifying and distinguishing substances in chemistry. They convey crucial information regarding a molecule’s composition, aiding in determining its possible structure and properties. However, molecular formulas can represent different compounds with the same atoms but different bonds or arrangements, known as isomers.

When analyzing \( \mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O} \), the formula can map to multiple structures such as alcohols like butanol or ethers like diethyl ether. Each unique configuration significantly affects a compound's physical and chemical properties. Therefore, molecular formulas serve as a foundational step for deeper structural analysis.