Problem 38 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

A hydrocarbon A (molecular formula \(\mathrm{C}_{5} \mathrm{H}_{10}\) ) yields 2-methylbutane on catalytic hydrogenation. A adds \(\mathrm{HBr}\) (in accordance with Markownikoff's rule) to form a compound B which on reaction with silver hydroxide forms an alcohol \(\mathrm{C}, \mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O} .\) Alcohol \(\mathrm{C}\) on oxidation gives a ketone D. Deduce the structures of \(A, B, C\) and \(D\) and show the reactions involved.

Step-by-Step Solution

Answer

A is 2-methyl-2-butene, B is 2-bromo-2-methylbutane, C is 2-methyl-2-butanol, and D is 2-methyl-2-butanone.

Step 1: Analyze Hydrocarbon A

Hydrocarbon A has the formula \( \text{C}_5\text{H}_{10} \), which suggests it is an alkene due to the unsaturation index. Upon catalytic hydrogenation, it forms 2-methylbutane, indicating A is 2-methylbutene, as hydrogenation of alkenes results in saturated alkanes.

Step 2: Determine Hydrocarbon A's Structure

Since hydrogenation yields 2-methylbutane, hydrocarbon A must specifically be 2-methyl-2-butene. This structure can explain the formation of 2-methylbutane upon hydrogenation.

Step 3: Addition of HBr to form Compound B

According to Markownikoff's rule, the hydrogen from HBr will add to the carbon with more hydrogens in A. Thus, the addition will occur at the less substituted carbon of the double bond in 2-methyl-2-butene, forming compound B, which is 2-bromo-2-methylbutane.

Step 4: Reaction of Compound B with Silver Hydroxide

When 2-bromo-2-methylbutane (compound B) is reacted with silver hydroxide (AgOH), the bromine is substituted with an -OH group, forming alcohol C, which is 2-methyl-2-butanol with the formula \( \text{C}_5\text{H}_{12}\text{O} \).

Step 5: Oxidation of Alcohol C to form Ketone D

The oxidation of the tertiary alcohol, 2-methyl-2-butanol (compound C), will result in the formation of a ketone. Therefore, structure D is 2-methyl-2-butanone.

Key Concepts

Hydrocarbons

Hydrocarbons are an essential area of organic chemistry consisting entirely of carbon and hydrogen atoms. They form the backbone of various organic compounds. The two main types are alkanes, which are saturated hydrocarbons with single bonds, and alkenes, which are unsaturated due to at least one double bond.
In the given exercise, hydrocarbon A is identified as an alkene with the molecular formula \( \text{C}_5\text{H}_{10} \). This formula hints at the presence of a double bond, necessary for the unsaturation indicated by the formula. When catalytically hydrogenated, the alkene's double bond is broken, and it gains additional hydrogen atoms, transforming into a saturated alkane. This process leads to the formation of 2-methylbutane, suggesting that A is originally 2-methylbutene.
This fundamental understanding of hydrocarbons and their reactions underpins much of organic chemistry, offering versatile routes in synthesis and modification of organic compounds.

Markownikoff's Rule

Markownikoff's Rule is a guideline in organic chemistry that predicts the outcome of addition reactions involving alkenes and unsymmetrical reagents, like HBr. The rule states that when a protic acid is added to an unsymmetrical alkene, the acid's hydrogen (H) atom will attach itself to the carbon with the most hydrogen atoms already bonded, while the other component of the acid (like Br in HBr) attaches to the other carbon.
In the context of the exercise, when HBr is added to the double bond of 2-methyl-2-butene, hydrogen attaches to the less substituted carbon. This adherence to Markownikoff's Rule explains the formation of 2-bromo-2-methylbutane (compared to any other possible structures from a non-Markownikoff addition).
This rule is crucial in predicting and explaining the product distribution in reactions involving alkenes.

Hydrogenation

Hydrogenation is a chemical reaction that adds hydrogen (H₂) to an unsaturated compound, like an alkene, typically in the presence of a catalyst such as platinum, palladium, or nickel. This process saturates the compound, converting double bonds into single bonds by "breaking" them and "filling" with hydrogen atoms.
In the solved problem, hydrogenation is employed to convert the alkene 2-methyl-2-butene into the alkane, 2-methylbutane. Here, the catalyst facilitates the breaking of the double bond, and the additional hydrogen atoms create a saturated hydrocarbon.
This transformation is not just significant in organic synthesis, but it has practical applications in the industrial production of margarine, turning liquid oils into solid fats by hydrogenating carbon-carbon double bonds.

Oxidation Reactions

Oxidation reactions play a pivotal role in organic chemistry, referring to a process where a molecule loses electrons, often involving an increase in the oxidation state, especially with the addition of oxygen or removal of hydrogen.
In this particular exercise, alcohol C, identified as 2-methyl-2-butanol, undergoes such an oxidation. On reacting with suitable oxidizing agents, the alcohol loses hydrogen atoms, transforming into a ketone (2-methyl-2-butanone).
This conversion is an example of oxidizing a secondary alcohol to a ketone, a frequent and crucial transformation in organic chemistry, vital for synthesizing various functional groups and compounds with differing properties. This exemplifies how oxidation is a tool for chemists to modify and build complex molecular structures efficiently.