Question

The compound having molecular formula \(\mathrm{C}_{6} \mathrm{H}_{6}\). The correct statement is/are (A) It must be benzene (B) It must be cyclic (C) It must be hydrocarbon (D) It must have 4 double bond equivalent

Short answer

The correct statements for the compound with the molecular formula \(C_6H_6\) are: (B) It must be cyclic (C) It must be hydrocarbon

Step-by-step solution

Statement A: It must be benzene

Benzene has the molecular formula C6H6. While this matches the given formula, we cannot say that the given compound must be benzene, as there can be other isomers with the same formula. So, this statement is not necessarily true.

Statement B: It must be cyclic

For a molecule to have the formula C6H6 and follow the general formula for alkanes (CnH(2n+2)), it must be cyclic. A cyclic compound has the same number of carbon atoms as hydrogen atoms in its ring, making the hydrocarbon a closed loop. In this case, with C6H6, the compound would need to be cyclic to fulfill the molecular formula. So, this statement is true.

Statement C: It must be hydrocarbon

The given compound C6H6 contains only carbon (C) and hydrogen (H) atoms, which makes it a hydrocarbon. So, this statement is true.

Statement D: It must have 4 double bond equivalents

In a hydrocarbon, double bond equivalents (DBE) can be calculated using the formula: \[DBE = 1 + \frac{N_C - N_H + 2}{2}\] Where NC is the number of carbon atoms, and NH is the number of hydrogen atoms. In this case, NC = 6 and NH = 6, so the formula becomes: \[DBE = 1 + \frac{6 - 6 + 2}{2} = 1 + \frac{2}{2} = 1 + 1 = 2\] So the compound has 2 double bond equivalents, not 4. Therefore, this statement is false. In conclusion, the correct statements for the compound with the molecular formula C6H6 are: (B) It must be cyclic (C) It must be hydrocarbon

Key concepts

Molecular Formula Analysis

Molecular Formula Analysis is a fundamental starting point in organic chemistry. It gives us the ratio of atoms in a molecule. For the molecule with molecular formula \(\text{C}_{6}\text{H}_{6}\), this indicates there are six carbon atoms and six hydrogen atoms within the molecule. This formula is important because:

• It allows us to determine the basic elements and their ratio without considering the arrangement of those atoms.
• We can use it to calculate the molecular weight by summing the atomic weights of the atoms involved.
• The molecular formula helps us predict different possible structures, known as isomers.

It's key to note that molecular formulas do not provide information about the chemical structure. Therefore, a given molecular formula can correspond to multiple distinct structures, which leads us to the concept of isomerism.

Isomer Identification

In organic chemistry, isomers are molecules that have the same molecular formula, yet differ in their structural configuration, bonding or spatial orientation. The compound \(\text{C}_{6}\text{H}_{6}\) can represent multiple isomers including benzene and several other less common structures.

• Structural Isomers: These differ in the connectivity of atoms within the molecule. For \(\text{C}_{6}\text{H}_{6}\), most isomers are cyclic structures due to the specific ratio of carbons to hydrogens.
• Stereoisomers: Though rare for \(\text{C}_{6}\text{H}_{6}\), these involve the same connectivity but differ in spatial orientation.

Identifying isomers requires considering the possible structural arrangements that satisfy the molecular formula. For \(\text{C}_{6}\text{H}_{6}\), benzene is the most recognized isomer due to its aromatic stability.

Double Bond Equivalence

Double Bond Equivalence (DBE) is a concept used to infer the level of unsaturation in a molecule by assessing how many double bonds, triple bonds, or rings a compound may have. The formula for calculating DBE is:\[ \text{DBE} = 1 + \frac{N_C - N_H + 2}{2} \]Where \(N_C\) is the number of carbon atoms and \(N_H\) is the number of hydrogen atoms. For \(\text{C}_{6}\text{H}_{6}\), this becomes:\[ \text{DBE} = 1 + \frac{6 - 6 + 2}{2} = 2 \]This tells us that the molecule can be thought of as having 2 "units" of unsaturation, where each unit could be a double bond or a ring.

• A benzene ring, for example, has three double bonds contributing to these units of unsaturation.
• The DBE can sometimes mislead if not interpreted correctly, as it gives an overall count but not the exact location or type of unsaturation.

Understanding DBE is crucial for determining possible structures, especially in complex molecules where multiple isomeric forms exist.