Problem 113 - A Complete Resource Book in Chemistry for JEE Main

Question

A hydrocarbon of molecular formula \(\mathrm{C}_{7} \mathrm{H}_{12}\) on catalytic hydrogenation over platinum gives \(\mathrm{C}_{7} \mathrm{H}_{16^{-}}\)The parent hydrocarbon adds bromine and also reacts with \(\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right] \mathrm{OH}\) to give a precipitate. The parent hydrocarbon is (a) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{C}=\mathrm{CH}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCH}_{2} \mathrm{CH}=\mathrm{CH}_{2}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C}=\mathrm{CHCH}\left(\mathrm{CH}_{3}\right)_{2}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCH}=\mathrm{CHCH}_{3}\)

Step-by-Step Solution

Answer

The correct answer is (a) \(\left(\mathrm{CH}_3\right)_3\mathrm{CCH}_2\mathrm{C}\equiv\mathrm{CH}\).

Step 1: Identify the degree of unsaturation

The molecular formula of the hydrocarbon is \(\mathrm{C}_7\mathrm{H}_{12}\). To find the degree of unsaturation (DoU), use the formula: \[\text{DoU} = \frac{2\text{C} + 2 - \text{H}}{2}\]. Plug in the values: \(\text{DoU} = \frac{2(7) + 2 - 12}{2} = 3\). This means the compound can have three rings or double bonds (or any combination thereof).

Step 2: Analyze the catalytic hydrogenation info

Catalytic hydrogenation over platinum yielding \(\mathrm{C}_7\mathrm{H}_{16}\) indicates that the compound becomes fully saturated. Since \(\mathrm{C}_7\mathrm{H}_{16}\) is a saturated alkane, the DoU for the original hydrocarbon was 3, which we have previously calculated.

Step 3: Analyze bromine addition

Addition of bromine indicates the presence of a double or triple bond since Br2 adds across unsaturated sites. This is further evidence of the unsaturation already identified.

Step 4: Analyze reaction with \(\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]\mathrm{OH}\)

This reaction gives a precipitate, indicating the presence of a terminal alkyne. \(\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]\mathrm{OH}\) is known to give precipitates with terminal alkynes, so one of the units of unsaturation is likely a terminal triple bond.

Step 5: Evaluate options based on properties

Evaluate each option: (a) \(\left(\mathrm{CH}_3\right)_3\mathrm{CCH}_2\mathrm{C}\equiv\mathrm{CH}\) has one triple bond and matches reactions described.(b) \(\mathrm{CH}_3\mathrm{CH}\equiv\mathrm{CHCH}_2\mathrm{CH}\equiv\mathrm{CH}_2\) has two double bonds, does not match the reactions.(c) \(\mathrm{CH}_3\mathrm{CH}_2\mathrm{C}\equiv\mathrm{CHCH}\left(\mathrm{CH}_3\right)_2\) resembles (a), has a terminal triple bond.(d) Does not match due to different structure of double bonds.Option (a) is correct because it fits all given reactions including the triple bond needed for the reaction with \(\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]\mathrm{OH}\).

Key Concepts

Degree of Unsaturation

The degree of unsaturation (DoU), also known as the index of hydrogen deficiency, is a useful concept in organic chemistry for determining the level of unsaturation within a molecule. It tells us about the number of rings or multiple bonds present in a compound. To calculate the DoU, we use the formula: \[\text{DoU} = \frac{2\text{C} + 2 - \text{H}}{2}\]where \( \text{C} \) is the number of carbon atoms and \( \text{H} \) is the number of hydrogen atoms. For a hydrocarbon with the formula \(\mathrm{C}_7\mathrm{H}_{12}\), the degree of unsaturation calculates as:\[\text{DoU} = \frac{2(7) + 2 - 12}{2} = 3\]This result indicates there are three degrees of unsaturation, meaning the molecule can contain any combination of three rings or double bonds, including triple bonds.

Catalytic Hydrogenation

Catalytic hydrogenation is a chemical reaction where hydrogen molecules \(\mathrm{H}_2\) are added to unsaturated hydrocarbons in the presence of a catalyst, commonly platinum, palladium, or nickel. This process converts unsaturated hydrocarbons to saturated hydrocarbons by breaking double or triple bonds and adding hydrogen atoms.For instance, given the initial compound \(\mathrm{C}_{7}\mathrm{H}_{12}\), catalytic hydrogenation over platinum gives \(\mathrm{C}_{7}\mathrm{H}_{16}\). This transformation signals that the compound has become fully saturated, confirming that the original molecule had multiple double or triple bonds equal to the degree of unsaturation calculated previously, which was 3.Through this technique, chemists can determine the presence and number of unsaturated bonds, as all unsaturations are eradicated effectively by this type of reaction.

Bromine Addition

Bromine addition is a classical reaction used to identify unsaturation in organic molecules. When bromine \(\mathrm{Br}_2\) reacts with a compound containing double or triple bonds, the bromine atoms add across these unsaturated sites. This decolorizes the reddish-brown color of bromine, indicating a positive test for unsaturation.In our exercise, the hydrocarbon \( \mathrm{C}_7\mathrm{H}_{12} \) is found to add bromine. This suggests the presence of at least one multiple bond. Since bromine addition is indicative of unsaturation, this aligns with our previous discovery of three degrees of unsaturation—highlighting the existence of double or triple bonds in the molecule.

Terminal Alkyne Reaction

The terminal alkyne reaction involves the reaction of terminal alkynes with an agent such as \([\mathrm{Ag}(\mathrm{NH}_3)_2]\mathrm{OH}\), which results in the formation of a precipitate. Terminal alkynes are characterized by a triple bond at the end of the carbon chain, making them uniquely reactive with these agents.In the context of the given exercise, the hydrocarbon \(\mathrm{C}_{7}\mathrm{H}_{12}\) undergoes reaction with this reagent to form a precipitate, confirming the existence of a terminal alkyne. Among other unsaturations, having a terminal triple bond also explains why the compound could reduce to \(\mathrm{C}_7\mathrm{H}_{16}\) through catalytic hydrogenation, as all unsaturated positions are hydrogenated to saturated configurations.This reaction also reduces plausible structural options, enabling us to narrow down the correct hydrocarbon to one that aligns specifically with the presence of a terminal triple bond.