8 Question: In a challenging laboratory experiment, an organic chemist synthesizes a novel compound "Alpha" with a molecular formula of $\mathrm{C}_{9} \mathrm{H}_{10} \mathrm{O}_{2}$. When "Alpha" is treated with sodium...
In a challenging laboratory experiment, an organic chemist synthesizes a novel compound "Alpha" with a molecular formula of $\mathrm{C}_{9} \mathrm{H}_{10} \mathrm{O}_{2}$. When "Alpha" is treated with sodium borohydride $\left(\mathrm{NaBH}_{4}\right)$, it yields a compound "Beta" with a molecular formula $\mathrm{C}_{9}$ $\mathrm{H}_{12} \mathrm{O}_{2}$. Further, "Beta" reacts with acidified potassium permanganate $\left(\mathrm{KMnO}_{4} / \mathrm{H}^{+}\right)$to form a compound "Gamma" $\left(\mathrm{C}_{9} \mathrm{H}_{12} \mathrm{O}_{3}\right)$. Determine the structures of compounds Alpha, Beta, and Gamma, and explain the type of chemical reactions involved in each transformation.

Question

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Identify the reaction**: The molecular formula of "Alpha" is $C_9H_{10}O_2$. When treated with sodium borohydride ($NaBH_4$), it yields "Beta" with a molecular formula $C_9H_{12}O_2$. This indicates a reduction reaction where $NaBH_4$ reduces a carbonyl group (likely an aldehyde or ketone) to an alcohol. ‖ ‖ 2 ⋮ **Determine the structure of "Alpha"**: Given the molecular formula $C_9H_{10}O_2$ and the reduction reaction, "Alpha" likely contains a carbonyl group. A plausible structure is a benzene ring with a formyl group (aldehyde) and a methoxy group, such as 4-methoxybenzaldehyde. ‖ ‖ 3 ⋮ **Identify the reaction for "Beta"**: The molecular formula of "Beta" is $C_9H_{12}O_2$. This suggests that the aldehyde group in "Alpha" has been reduced to a primary alcohol, resulting in 4-methoxybenzyl alcohol. ‖ ‖ 4 ⋮ **Determine the structure of "Beta"**: Based on the reduction, "Beta" is likely 4-methoxybenzyl alcohol ($C_9H_{12}O_2$). ‖ ‖ 5 ⋮ **Identify the reaction for "Gamma"**: "Beta" reacts with acidified potassium permanganate ($KMnO_4 / H^+$) to form "Gamma" with a molecular formula $C_9H_{12}O_3$. This indicates an oxidation reaction where the primary alcohol is oxidized to a carboxylic acid. ‖ ‖ 6 ⋮ **Determine the structure of "Gamma"**: Based on the oxidation, "Gamma" is likely 4-methoxybenzoic acid ($C_9H_{12}O_3$). ‖ ∻Answer∻ ⚹ The structures of the compounds are as follows: - "Alpha": 4-methoxybenzaldehyde ($C_9H_{10}O_2$) - "Beta": 4-methoxybenzyl alcohol ($C_9H_{12}O_2$) - "Gamma": 4-methoxybenzoic acid ($C_9H_{12}O_3$) ⚹ ∻Key Concept∻ ⚹ Reduction and oxidation reactions in organic chemistry ⚹ ∻Explanation∻ ⚹ Sodium borohydride ($NaBH_4$) is a reducing agent that converts aldehydes to primary alcohols, while acidified potassium permanganate ($KMnO_4 / H^+$) is an oxidizing agent that converts primary alcohols to carboxylic acids. ⚹◊What is the structural difference between compound "Beta" and compound "Alpha" based on their molecular formulas?⍭ Generate me a similar question◊

Sia