Question

Compound A has the formula \(\mathrm{C}_{10} \mathrm{H}_{16}\). On catalytic hydrogenation over palladium, it reacts with only 1 molar equivalent of \(\mathrm{H}_{2}\). Compound \(\mathrm{A}\) also undergoes reaction with ozone, followed by zinc treatment, to yield a symmetrical diketone, \(\mathbf{B}\) ( \(\mathrm{C}_{10} \mathrm{H}_{16} \mathrm{O}_{2}\) ). (a) How many rings does A have? (b) What are the structures of \(\mathrm{A}\) and \(\mathrm{B}\) ? (c) Write the reactions.

Short answer

(a) Two rings; (b) A: bicyclo[4.2.0]octene, B: a symmetrical diketone; (c) Reactions: hydrogenation and ozonolysis.

Step-by-step solution

Analyze Hydrogen Deficiency

Given that compound A, \( \text{C}_{10} \text{H}_{16} \), reacts with 1 molar equivalent of \( \text{H}_2 \), it suggests one of the unsaturations is a double bond. Calculate the degree of unsaturation (index of hydrogen deficiency, IHD) to find the number of rings and double bonds.The formula for IHD is: \[ \text{IHD} = \frac{2C + 2 + N - H - X}{2} \]Substitute into the formula:\[ \text{IHD} = \frac{2 \times 10 + 2 - 16}{2} = 3 \]This indicates three degrees of unsaturation.

Analyze Reactions and Structure of A

Compound A only reacts with \( \text{H}_2 \) once, suggesting A contains one double bond. Since it also reacts with ozone to form a symmetrical diketone B, there must be a double bond in a ring that splits into two identical carbonyl groups during ozonolysis. The remaining unsaturations must be rings. A possible structure for A is bicyclo[4.2.0]octene, where one double bond and two rings account for the IHD = 3.

Identify Structure of B and Write Reaction Equations

Compound B is a symmetrical diketone \( \text{C}_{10} \text{H}_{16} \text{O}_{2} \). The ozonolysis reaction clearly cleaves a double bond to form two identical ketone groups in B.Reactions:1. Hydrogenation: \[ \text{C}_{10} \text{H}_{16} + \text{H}_2 \xrightarrow{Pd} \text{C}_{10} \text{H}_{18} \]2. Ozonolysis: \[ \text{C}_{10} \text{H}_{16} + \text{O}_3 \rightarrow \text{C}_{10} \text{H}_{16} \text{O}_2 (B) \]

Confirm Number of Rings

Given the bicylic structure with a double bond and another ring, A has two rings. This aligns with the three total unsaturations calculated earlier: one double bond and two rings.

Key concepts

Degree of Unsaturation

The degree of unsaturation, also known as the index of hydrogen deficiency (IHD), helps chemists determine the number of rings and multiple bonds in a molecule. It's a handy tool when you have the molecular formula but need more insight into the molecule's structure. The IHD is calculated using the formula: \[ \text{IHD} = \frac{2C + 2 + N - H - X}{2} \] where

• \( C \) represents the number of carbon atoms,
• \( H \) is the number of hydrogen atoms,
• \( N \) is the number of nitrogen atoms, and
• \( X \) represents halogens (such as \( F, Cl, Br, \) etc.).

For compound A with the formula \( \text{C}_{10} \text{H}_{16} \), the IHD is calculated as follows:\[ \text{IHD} = \frac{2 \times 10 + 2 - 16}{2} = 3 \] This indicates that compound A has three degrees of unsaturation. These can be distributed among double bonds and rings. Typically, each double bond or ring contributes one to the IHD count.

This helps us determine the possible structures for compound A, as knowing there are three unsaturations provides clues about the types and arrangements of bonds and rings present in the molecule.

Symmetrical Diketone

A symmetrical diketone like compound B in our exercise is a molecule where the two ketone groups are identical. This symmetry arises during reactions such as ozonolysis.
Ozonolysis is a process where ozone (\( \text{O}_3 \)) breaks the double bonds in alkenes, leading to the formation of carbonyl-containing compounds. For compound A, undergoing ozonolysis results in a cleaved double bond and the formation of symmetrical diketone B \( (\text{C}_{10} \text{H}_{16} \text{O}_{2}) \).
The formation of symmetrical diketones specifically indicates that the double bond in compound A was positioned in such a way that, upon cleavage, the resulting fragments are identical. This provides a significant hint about the molecule's original structure, suggesting it was likely a ring where the bond was exactly central between two identical halves.

Understanding symmetrical structures can simplify complex structural elucidation problems since symmetry often reduces the number of possible different compounds that can fit a specific formula.

Catalytic Hydrogenation

Catalytic hydrogenation is a common reaction involving the addition of hydrogen (\( \text{H}_2 \)) across a double bond. This process requires a catalyst, often palladium (Pd), to facilitate the reaction. In our example with compound A (\( \text{C}_{10} \text{H}_{16} \)), it reacts with one molar equivalent of hydrogen over palladium, suggesting it contains only one double bond.

This is crucial because the degree to which a compound can be hydrogenated sheds light on its unsaturated features—like double bonds. Catalytic hydrogenation reduces these unsaturations by converting double bonds to single ones, effectively removing that particular unsaturation.

Given that only one molar equivalent of \( \text{H}_2 \) is added, it's clear that compound A possesses a singular double bond. This, along with the IHD calculation, helps narrow down possible structures. The requirement that compound A reacts with exactly one molar equivalent confirms the presence of a single double bond within its structure before complete saturation.