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) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{C} \equiv \mathrm{CH}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCH}_{2} \mathrm{CH}=\mathrm{CH}_{2}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{CCH}\left(\mathrm{CH}_{3}\right)_{2}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCH}=\mathrm{CHCH}_{3}\)

Short answer

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

Step-by-step solution

Understanding the Reaction

The molecular formula of the hydrocarbon is \( \mathrm{C}_7 \mathrm{H}_{12} \). On catalytic hydrogenation, it becomes \( \mathrm{C}_7 \mathrm{H}_{16} \). This indicates the presence of four hydrogen atoms being added, suggesting the presence of two double bonds or one triple bond.

Analyzing Reaction with Bromine

The fact that the compound adds bromine indicates it has double or triple bonds. All options must be checked for unsaturations, which allow bromine addition.

Reaction with Silver Ammoniacal Solution

The compound reacts with \( \left[ \mathrm{Ag} \left( \mathrm{NH}_3 \right)_2 \right] \mathrm{OH} \) to form a precipitate. This reaction indicates the presence of a terminal alkyne which can form a silver salt precipitate.

Evaluating Options

Let's evaluate each provided option:- Option (a): \( \left( \mathrm{CH}_3 \right)_3 \mathrm{CCH}_2 \mathrm{C} \equiv \mathrm{CH} \). This is a terminal alkyne and can react with the silver solution to give a precipitate. It is unsaturated, aligns with the hydrogenation reaction, and the addition of bromine. This matches all the criteria.- Option (b): \( \mathrm{CH}_3 \mathrm{CH} = \mathrm{CHCH}_2 \mathrm{CH} = \mathrm{CH}_2 \). This is a diene, not a terminal alkyne, thus unlikely to follow all observed reactions.- Option (c): \( \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{C} \equiv \mathrm{CCH} \left( \mathrm{CH}_3 \right)_2 \). This is also a terminal alkyne but doesn't match the hydrogenation product due to the molecular structure.- Option (d): \( \mathrm{CH}_3 \mathrm{CH} = \mathrm{CHCH} = \mathrm{CHCH}_3 \). This is a hexadiene, and wouldn't react with the silver solution as specified.

Conclusion

The best matching compound according to all the reactions and transformations described is option (a): \( \left( \mathrm{CH}_3 \right)_3 \mathrm{CCH}_2 \mathrm{C} \equiv \mathrm{CH} \).

Key concepts

Catalytic Hydrogenation

Catalytic hydrogenation is a chemical reaction where hydrogen is added to a hydrocarbon in the presence of a catalyst, such as platinum. The primary role of the catalyst is to speed up the reaction without being consumed in the process.
When we talk about hydrocarbons, they may contain unsaturated bonds such as double or triple bonds. During hydrogenation, these bonds are saturated by adding hydrogen atoms. For instance, in the exercise provided, the initial hydrocarbon with the formula \( \mathrm{C}_7 \mathrm{H}_{12} \) transforms to \( \mathrm{C}_7 \mathrm{H}_{16} \) during hydrogenation. This means four hydrogen atoms are added, showing that the original compound had a combination of bonds equivalent to a triple bond (since one triple bond or two double bonds can take up this amount of hydrogen).
This kind of reaction is crucial in the production of saturated fats and oils from plant oils, among other industrial applications.

Bromine Addition Reaction

The bromine addition reaction is a key test for the presence of unsaturation in organic compounds, particularly alkenes and alkynes. Bromine, which is initially a reddish-brown liquid, reacts with these unsaturated compounds leading to a colorless solution. This color change serves as an indicator of the reaction.
In the context of the exercise, the parent hydrocarbon can add bromine, suggesting the presence of a double or triple bond. When bromine is added to the compound, it breaks the bond and attaches to the carbon atoms, leading to a dibromo addition product. Therefore, for the compound to engage in a bromine addition reaction, it must be unsaturated, like alkenes or alkynes, that have these types of bonds.

• Reddish-brown bromine becomes colorless.
• Primarily involves the breaking of double or triple bonds.
• Confirms unsaturation in organic molecules.

Terminal Alkyne

A terminal alkyne refers to an alkyne that has a triple bond at the end of the carbon chain. It is characterized by the presence of a hydrogen atom bonded directly to the sp-hybridized carbon involved in the triple bond.
One significant property of terminal alkynes is their reactivity with certain reagents. When terminal alkynes encounter \( \left[ \mathrm{Ag} \left( \mathrm{NH}_3 \right)_2 \right] \mathrm{OH} \), they form a silver precipitate. This reaction is useful for both detection and analysis of terminal alkynes.
The compound in option (a) from the exercise is an excellent example of a terminal alkyne. It not only fits the reaction criteria due to its structure but also matches the other reactions described in the exercise. Terminal alkynes are essential in various synthetic applications, from pharmaceuticals to organic materials.

Silver Ammoniacal Test

The silver ammoniacal test is a laboratory process used to identify terminal alkynes. This test involves the reaction of a terminal alkyne with Tollens' reagent, which is silver ammonium hydroxide \( \left[ \mathrm{Ag} \left( \mathrm{NH}_3 \right)_2 \right] \mathrm{OH} \).
When a terminal alkyne reacts with this reagent, a silver acetylide precipitate forms. This appears as a solid compound, visually confirming the presence of a terminal alkyne in the tested sample. This is crucial in the exercise when considering the possible identities of the parent hydrocarbon.
This test not only helps in confirming the structure of compounds but also aids in comprehensively understanding reaction mechanisms. Its utility is vast in organic chemistry, particularly in distinguishing between internal and terminal alkynes.

• Specific to terminal alkynes.
• Forms a silver precipitate typically after the reaction.
• A key method for identifying specific bond types in hydrocarbons.