Problem 35 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Compound ' \(X\) ' (molecular formula, \(\mathrm{C}_{5} \mathrm{H}_{8} \mathrm{O}\) ) does not react appreciably with Lucas reagent at room temperature but gives a precipitate with ammonical silver nitrate. With excess of \(\mathrm{MeMgBr}, 0.42 \mathrm{~g}\) of \(^{‘} X\) gives \(224 \mathrm{~mL}\) of \(\mathrm{CH}_{4}\) at STP. Treatment of ' \(X\) ' with \(\mathrm{H}_{2}\) in presence of \(\mathrm{Pt}\) catalyst followed by boiling with excess HI, gives \(n\)-pentane. Suggest structure for ' \(X\) and write the equation involved.

Step-by-Step Solution

Answer

Compound 'X' is propargyl alcohol (\(\text{CH}_3\text{-C}\equiv\text{C-CH}_2\text{OH}\)).

Step 1: Analyzing The Given Information

Compound 'X' has the molecular formula \(\text{C}_5\text{H}_8\text{O}\). It does not react with Lucas reagent at room temperature, which indicates it is likely not a tertiary alcohol. It reacts with ammonical silver nitrate, suggesting the presence of a terminal alkyne. Upon hydrogenation followed by treatment with HI, it converts to \(n\)-pentane.

Step 2: Reaction With Excess MeMgBr

Treating 0.42 g of 'X' with excess \(\text{MeMgBr}\) gives 224 mL of \(\text{CH}_4\) at STP. Using the molar volume at STP (22.4 L/mol), we can find the moles of \(\text{CH}_4\) produced: \(224 \text{ mL} = 0.224 \text{ L}\). Thus, moles of \(\text{CH}_4 = 0.224 / 22.4 = 0.01\) mol.

Step 3: Determine the Molar Mass of 'X'

The moles of \(\text{CH}_4\) tell us the same number of moles of 'X' was used in the reaction. Given mass of 'X' is 0.42 g, molar mass of 'X' is \(0.42 \text{ g}/0.01 \text{ mol} = 42 \text{ g/mol}\). From the molecular formula \(\text{C}_5\text{H}_8\text{O}\), calculate the actual molar mass, \(5(12) + 8(1) + 16 = 84 \text{ g/mol}\). The given data suggests complete conversion to a product half the molecular weight of 'X', indicating doubly bonded or cyclic structure.

Step 4: Identify Structure of 'X'

Considering all information, the structure that fits is an enyne or alkyne-alcohol combination. Since the compound does not react with Lucas reagent, but reacts with \(\text{Ag(NH}_3\text{)}_2\text{NO}_3\), it indicates a terminal alkyne. Structural formula can be: \(\text{CH}_3\text{-C}\equiv\text{C-CH}_2\text{OH}\).

Step 5: Write Equations For Reactions

- With \(\text{MeMgBr}\): \[\text{CH}_3\text{-C}\equiv\text{C-CH}_2\text{OH} + \text{MeMgBr} \rightarrow \text{CH}_4 + \text{other products} \]. - Hydrogenation: \[\text{CH}_3\text{-C}\equiv\text{C-CH}_2\text{OH} + 2\text{H}_2 \xrightarrow[\text{Pt}]{\text{catalyst}} \text{pentanol} \]. - Boiling with HI: \[\text{pentanol} + \text{HI} \rightarrow \text{n-pentane} + \text{I}_2\].

Key Concepts

Lucas Reagent

Lucas reagent is a classic tool in organic chemistry, especially when dealing with alcohol classification. It consists of concentrated hydrochloric acid (HCl) and zinc chloride (ZnCl₂). This reagent helps determine whether an alcohol is primary, secondary, or tertiary based on the reaction rate and solubility changes in the alcohol's solution. At room temperature, tertiary alcohols react rapidly, while secondary alcohols take longer, and primary alcohols show no reaction.
This makes Lucas reagent an excellent qualitative test for alcohols. It's important to note, as observed in the original exercise, that compound 'X' does not react with Lucas reagent at room temperature, indicating it is likely not a tertiary alcohol. Thus, knowing the correct classification helps students identify possible structural clues of an unknown compound.

Ammonical Silver Nitrate

Ammonical silver nitrate, often denoted as Tollens' reagent, is another powerful reagent used in organic chemistry. It is typically used to test for aldehydes and terminal alkynes. Composed of silver nitrate dissolved in aqueous ammonia, this reagent reacts with terminal alkynes to form a silver acetylide precipitate.
This specific reaction is crucial, as seen in the original exercise. Compound 'X' forms a precipitate with ammonical silver nitrate, signaling the presence of a terminal alkyne group. This reaction helps pinpoint the position of the triple bond in a compound’s structure, which is highly informative for structural elucidation problems in chemistry.

Grignard Reaction

The Grignard reaction is one of the most versatile and widely used reactions in organic synthesis. It involves organomagnesium halides, known as Grignard reagents, which are used to form carbon-carbon bonds. Grignard reagents can react with carbonyl groups to form alcohols after protonation, making this reaction crucial for alcohol synthesis.
In the original exercise, compound 'X' reacts with MeMgBr, a Grignard reagent, to produce methane (\(\text{CH}_4\)). This reaction shows how Grignard reagents interact with different functional groups, providing insights into the functional structure of 'X'. Understanding Grignard reactions is essential for identifying how compounds can be transformed into different products during synthesis.

Hydrogenation

Hydrogenation is a crucial method in organic chemistry, referring to the addition of hydrogen (\(\text{H}_2\)) to other compounds, typically in the presence of a metal catalyst like palladium, platinum, or nickel. This process is widely used to convert alkenes, alkynes, or other unsaturated compounds into saturated alkanes.
In the context of the given exercise, hydrogenation is used to convert the alkyne section of 'X' to a saturated alkyl group, ultimately leading to the formation of n-pentane after further reactions. Understanding hydrogenation helps with learning about the reduction of unsaturated compounds and their catalytic conversion to more stable structures in organic chemistry.

Terminal Alkyne

A terminal alkyne is a type of alkyne in which the carbon-carbon triple bond is located at the terminal (end) part of the carbon chain. These alkynes are characterized by their acidic hydrogen atoms bonded to the terminal carbon, making them reactive with specific reagents, like ammonical silver nitrate.
This distinct property of terminal alkynes allows them to form precipitates with silver salts, as seen with compound 'X', providing clear evidence of their presence in a compound. This fundamental concept is vital for recognizing and understanding the structural nature of organic compounds, particularly when solving problems regarding unknown molecular structures.