Question

18 multiple choice 2 points which of the following molecular formulas corresponds to a monocyclic saturated compound? c₆h₆ c₃h₇br c₃h₈o c₃h₇n previous

Explanation:

Step1: Calculate degree of unsaturation

Degree of unsaturation formula: \(DU=\frac{2C + 2+N - H - X}{2}\) (where \(C\) = number of carbon atoms, \(N\) = number of nitrogen atoms, \(H\) = number of hydrogen atoms, \(X\) = number of halogen atoms). For a monocyclic saturated compound, \(DU = 1\) (due to the ring, and no double/triple bonds).

• For \(C_6H_6\): \(DU=\frac{2\times6 + 2-6}{2}=\frac{12 + 2-6}{2}=4\)
• For \(C_3H_7Br\): \(DU=\frac{2\times3+ 2-7 - 1}{2}=\frac{6 + 2-7 - 1}{2}=0\)
• For \(C_3H_8O\): \(O\) (oxygen) does not affect degree of unsaturation. \(DU=\frac{2\times3+ 2-8}{2}=\frac{6 + 2-8}{2}=0\) (but if we consider a monocyclic - saturated (in terms of no double/triple bonds, just a ring). Wait, re - calculate: For non - halogen, non - nitrogen, non - oxygen (except formula adjustment). The general formula for a saturated acyclic \(C_nH_{2n + 2}\). For a monocyclic saturated (no double/triple bonds) \(C_nH_{2n}\).
• For \(C_3H_7N\): \(DU=\frac{2\times3+ 2 + 1-7}{2}=\frac{6+2 + 1-7}{2}=1\)

Wait, correction. The formula for a monocyclic saturated (in terms of no \(C = C\) or \(C\equiv C\) bonds) compound (hydrocarbon) is \(C_nH_{2n}\). If there are heteroatoms:

• For \(C_3H_8O\): If we assume a cyclic ether (a monocyclic saturated compound). The formula for a saturated acyclic \(C_nH_{2n + 2}O\) (e.g., \(CH_3CH_2CH_2OH\)). For a cyclic ether (monocyclic, saturated in terms of \(C - C\) bonds), we lose 2 \(H\) (similar to forming a ring in hydrocarbons). \(C_3H_8O\) can be \(C_3H_6O\) (if we consider ring - formation in a way similar to hydrocarbons, but oxygen can also be part of the ring. For example, a three - membered ring ether (\(C - O - C - C\) with ring - closure). The degree of unsaturation for \(C_3H_8O\): Using the formula \(DU=\frac{2C+2 - H}{2}\) (neglecting \(O\) as \(O\) does not contribute to \(DU\) calculation in the basic formula). \(DU=\frac{2\times3 + 2-8}{2}=0\). But if we consider the ring (a monocyclic structure), in terms of bond - saturation (no \(C = C\) or \(C\equiv C\)), for a monocyclic compound made from a saturated acyclic precursor (\(C_3H_8O\) acyclic: like \(CH_3CH_2CH_2OH\) or \(CH_3OCH_2CH_3\)). When forming a ring (e.g., a cyclic ether), we have a structure that is \(C - C - C - O\) ring - closed. The formula \(C_3H_6O\) (if we follow hydrocarbon - like ring - formation \(C_nH_{2n}\) for rings). But \(C_3H_8O\) can be thought of as a saturated (in \(C - C\) bonds) monocyclic compound when oxygen is part of the ring (because the extra \(H\) can be due to the oxygen's bonding.
• For \(C_3H_7N\): \(DU=\frac{2\times3+2 + 1-7}{2}=1\) (this \(DU = 1\) could be due to a double bond or a ring. But if we consider a saturated (in \(C - C\) bonds) monocyclic, for a compound with \(N\), the formula adjustment.
• For \(C_6H_6\): \(DU = 4\) (three double bonds and a ring)
• For \(C_3H_7Br\): \(DU=0\) (saturated acyclic)
• \(C_3H_8O\) can represent a monocyclic saturated (in \(C - C\) bonds) compound (e.g., a cyclic ether). The general idea is that for a monocyclic compound (analogous to \(C_nH_{2n}\) for hydrocarbons, when we have oxygen in the ring, the formula can be adjusted. If we consider the ring - formation (losing 2 \(H\) from the acyclic saturated formula \(C_3H_8O\) (acyclic:[LLM SSE On Failure]

Answer:

C. \(C_3H_8O\)