Problem 178 - A Complete Resource Book in Chemistry for JEE Main

Question

Match the following \begin{tabular}{ll} Column-I & Column-II \\ \hline (a) \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{2}-\) & \((\mathrm{p})\left(\mathrm{CH}_{2}\right)_{2} \mathrm{C}\) \end{tabular} \(\mathrm{CH}_{3} \frac{\mathrm{Hg}^{2+}, \mathrm{H}_{2} \mathrm{SO}_{4}}{330 \mathrm{~K}}\) (b) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}=\) (q) \(\mathrm{CH}_{2} \frac{(1) \mathrm{Hg}(\mathrm{OAc})_{2}}{(2) \mathrm{NaBH}_{4}, \mathrm{OH}}\) (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}=\) (r) \(\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) \(\mathrm{CH}_{2} \frac{\mathrm{B}_{2} \mathrm{H}_{6}}{\mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{OH}}\) (d) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}=\) (s) \(\mathrm{CH}_{2} \frac{\mathrm{H}_{2} \mathrm{SO}_{4}}{\mathrm{H}_{2} \mathrm{O}} \longrightarrow\) (t) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}_{2} \mathrm{CH}_{3}\)

Step-by-Step Solution

Answer

(a) - (t), (b) - (q), (c) - (p), (d) - (s) possible by deduction.

Step 1: Analyze Reaction (a)

For reaction (a), we have 2-butyne \[ \text{\(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{2}\-\)} \] undergoing hydrolysis with \(\mathrm{Hg}^{2+}, \mathrm{H}_{2} \mathrm{SO}_{4}\) as the reagent. The mercury(II)-catalyzed hydration of an alkyne results in the formation of a ketone. The product will have an oxygen double-bonded to the more substituted carbon. Hence, the product is \[ \text{ketone: \(\mathrm{CH}_{3}\mathrm{COCH}_{2}\mathrm{CH}_{3}\)} \] This matches with \(t\) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}_{2} \mathrm{CH}_{3}\).

Step 2: Analyze Reaction (b)

For reaction (b), we have \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}=\mathrm{CH}_{2}\) undergoing oxymercuration with \(\mathrm{Hg}(\mathrm{OAc})_{2}/\mathrm{NaBH}_{4}, \mathrm{OH}\). The oxymercuration-demercuration reaction proceeds with Markovnikov's rule, adding an OH group to the more substituted carbon. Resulting in \[\text{product: \((\mathrm{CH}_{3})_{3}\mathrm{C}-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}_{3}\)}\]As this is not listed in Column II, let's verify and match in subsequent analysis.

Step 3: Analyze Reaction (c)

For reaction (c), \[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}=\mathrm{CH}_{2} \] undergoes hydroboration with \(\mathrm{B}_{2} \mathrm{H}_{6}/\mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{OH}\). The reaction follows Anti-Markovnikov's rule, where the OH group is added to the less substituted carbon. Thus, the OH is added to the primary carbon: \[\mathrm{product: \left(\mathrm{CH}_{3}\right)_{3}\mathrm{C}-\mathrm{CH}_{2}-\mathrm{CH}_{2}\left(\mathrm{OH}\right)}\]The plausible match is \(p\) \((\mathrm{CH}_{2})_{2} \mathrm{C}\).

Step 4: Analyze Reaction (d)

For reaction (d), \[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{CH}=\mathrm{CH}_{2}\] undergoes acid-catalyzed hydration with \(\mathrm{H}_{2} \mathrm{SO}_{4}/\mathrm{H}_{2} \mathrm{O}\). This will result in a Markovnikov addition of water across the double bond, the OH group attaches to the more substituted carbon, creating an alcohol: \[(\mathrm{CH}_{3})_{3}\mathrm{C}-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}_{3}\]This matches best with no exact option in Column II implying logical elimination indicates a review of consistency or phenomena such as consecutive examination revisits.

Key Concepts

Markovnikov's Rule

Markovnikov's Rule is a guiding principle in organic chemistry, often applied to predict the outcome of addition reactions that involve unsaturated hydrocarbons like alkenes and alkynes. This rule states that when a protic acid (an acid that donates a proton) adds to an unsymmetrical alkene, the hydrogen atom from the acid attaches to the less substituted carbon atom, while the other substituent from the acid (such as a halide, hydroxyl group, or other nucleophile) attaches to the more substituted carbon atom. This tends to occur due to the stability of the carbocation intermediate formed during the reaction process. For example, in the hydration of propene (CH₃-CH=CH₂), hydrogen will bond to the CH₂ end of the double bond, and the hydroxyl group (OH) will attach to the more substituted CH site, resulting in the formation of 2-propanol.

• When applying Markovnikov's Rule, always identify the more and less substituted carbon in the alkene or alkyne.
• Consider additional factors like rearrangements that might stabilize the carbocation further.

Hydration Reactions

Hydration reactions are a subset of addition reactions where water is added across the multiple bonds of a molecule. This is a common transformation, particularly in the conversion of alkenes and alkynes into alcohols. During these reactions, the components of water (H-OH) add separately across the double or triple bond. A typical hydration process involves the formation of a carbocation intermediate when proceeding through Markovnikov addition. The structured reaction often involves catalysts such as acids, and can include reagents like mercury(II) sulfate to aid specific reactions, such as the hydration of alkynes.

• The hydration process is essential in converting simple hydrocarbons into more functionalized alcohols.
• This reaction helps in the synthesis of a wide range of chemical compounds used in various industries.

Oxymercuration-Demercuration

Oxymercuration-Demercuration is an important method in organic chemistry for the conversion of alkenes into alcohols. This two-step reaction circumvents the formation of a carbocation, thereby avoiding the rearrangements seen with other reactions like simple alkene hydration. 1. **Oxymercuration**: The alkene reacts with mercuric acetate in the presence of water, leading to the formation of a mercurinium ion intermediate. This intermediate is attacked by water, placing an OH group on the more substituted carbon in line with Markovnikov's Rule. 2. **Demercuration**: Subsequently, the mercury is removed using a reducing agent such as sodium borohydride (NaBH₄), replacing mercury with a hydrogen atom to finally create an alcohol.

• This reaction is preferred for situations where controlling side products and avoiding rearrangements is critical.
• Offers a more predictable outcome compared to direct acid-catalyzed hydration.

Hydroboration-Oxidation

Hydroboration-Oxidation is a two-step reaction mechanism used to achieve the anti-Markovnikov hydration of alkenes and alkynes. This method adds a hydrogen atom to the more substituted carbon and an alcohol group to the less substituted carbon, thus yielding the opposite regioselectivity compared to typical Markovnikov reactions. 1. **Hydroboration**: The alkene undergoes a syn addition with diborane (or a related borane reagent) where both the boron and hydrogen attach to the same side of the alkene, leading to trialkylborane. 2. **Oxidation**: The intermediate trialkylborane is then oxidized using hydrogen peroxide in a basic medium, ultimately replacing the boron with an OH group, and yielding an alcohol.

• This method is particularly important when synthesizing alcohols at sites distant from other substituents.
• It provides high regioselectivity and avoids isomerization.
• The reaction is typically carried out under mild conditions, making it efficient and favorable.