Question: In an advanced organic chemistry laboratory, a student synthesizes a novel compound " $L$ " with a molecular formula of $\mathrm{C}_{10} \mathrm{H}_{12} \mathrm{O}_{2}$. This compound, when reacted with lithium aluminum...

In an advanced organic chemistry laboratory, a student synthesizes a novel compound "L" with a molecular formula of $\mathrm{C}_{10} \mathrm{H}_{12} \mathrm{O}_{2}$. This compound, when reacted with lithium aluminum hydride $\left(\mathrm{LiAlH}_{4}\right)$, produces a new compound "M" $\left(\mathrm{C}_{10} \mathrm{H}_{16} \mathrm{O}\right)$. Further, compound "M", upon treatment with bromine water, forms compound " $N$ " $\left(\mathrm{C}_{10} \mathrm{H}_{16} \mathrm{OBr}\right.$
2). Analyze these reactions and deduce the probable structures of compounds $L, M$, and N. Explain the type of chemical reactions involved in each transformation.

Question

Question: In an advanced organic chemistry laboratory, a student synthesizes a novel compound " $L$ " with a molecular formula of $\mathrm{C}_{10} \mathrm{H}_{12} \mathrm{O}_{2}$. This compound, when reacted with lithium aluminum...

In an advanced organic chemistry laboratory, a student synthesizes a novel compound "L" with a molecular formula of $\mathrm{C}_{10} \mathrm{H}_{12} \mathrm{O}_{2}$. This compound, when reacted with lithium aluminum hydride $\left(\mathrm{LiAlH}_{4}\right)$, produces a new compound "M" $\left(\mathrm{C}_{10} \mathrm{H}_{16} \mathrm{O}\right)$. Further, compound "M", upon treatment with bromine water, forms compound " $N$ " $\left(\mathrm{C}_{10} \mathrm{H}_{16} \mathrm{OBr}\right.$
2). Analyze these reactions and deduce the probable structures of compounds $L, M$, and N. Explain the type of chemical reactions involved in each transformation.

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Identify the molecular formula and initial reaction**: The molecular formula of compound "L" is $C_{10}H_{12}O_{2}$. When "L" reacts with lithium aluminum hydride ($LiAlH_{4}$), it produces compound "M" with the molecular formula $C_{10}H_{16}O$. This suggests a reduction reaction where the carbonyl groups in "L" are reduced to alcohols in "M". ‖ ‖ 2 ⋮ **Analyze the reduction reaction**: Lithium aluminum hydride ($LiAlH_{4}$) is a strong reducing agent that typically reduces esters, carboxylic acids, and ketones to alcohols. Given the molecular formula of "L" and the product "M", it is likely that "L" contains two carbonyl groups (possibly as esters or ketones) that are reduced to alcohols in "M". ‖ ‖ 3 ⋮ **Identify the structure of "M"**: The molecular formula of "M" is $C_{10}H_{16}O$, indicating that "M" has one oxygen atom, likely in the form of a hydroxyl group. This suggests that "M" is a diol (a compound with two hydroxyl groups). ‖ ‖ 4 ⋮ **Analyze the bromine water reaction**: When "M" is treated with bromine water, it forms compound "N" with the molecular formula $C_{10}H_{16}OBr_{2}$. Bromine water typically reacts with alkenes to form dibromo compounds. This suggests that "M" contains a double bond that reacts with bromine to form "N". ‖ ‖ 5 ⋮ **Determine the structure of "N"**: The molecular formula of "N" is $C_{10}H_{16}OBr_{2}$, indicating that "N" is a dibromo compound with one oxygen atom. This suggests that "N" is a dibromo alcohol. ‖ ∻Answer∻ ⚹ The probable structures of the compounds are as follows: - Compound "L": A diketone or diester with the formula $C_{10}H_{12}O_{2}$. - Compound "M": A diol with the formula $C_{10}H_{16}O$. - Compound "N": A dibromo alcohol with the formula $C_{10}H_{16}OBr_{2}$. ⚹ ∻Key Concept∻ ⚹ Reduction and bromination reactions in organic chemistry ⚹ ∻Explanation∻ ⚹ The reduction of carbonyl compounds to alcohols using lithium aluminum hydride and the addition of bromine to alkenes are key reactions in determining the structures of the compounds. ⚹◊What type of reaction occurs when compound "L" reacts with lithium aluminum hydride to form compound "M"?⍭ Generate me a similar question◊

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