Question

(P) \(\frac{\text { (i) } \mathrm{Hg}(\mathrm{OAc})_{2} / \mathrm{H}_{2} \mathrm{O}}{\text { (ii) } \mathrm{NaBH}_{4} / \mathrm{OH}}(\mathrm{Q}) \stackrel{\mathrm{O}_{3}}{\mathrm{Zn}}(\mathrm{R})\) Which of the following is not possible as (P)? (A) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{3}\) (B) \(\mathrm{HC} \equiv \mathrm{CH}\) (C) \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CH}\) (D) \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3}\)

Short answer

The answer is (D) \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3}\).

Step-by-step solution

Reaction 1: Reagent (i) Hg(OAc)2 / H2O

In the first reaction, mercuric acetate (Hg(OAc)2) catalyst in the presence of water (H2O) is used. This reagent setup functions as an oxymercuration-demercuration reagent, adding an alcohol (-OH) group to a carbon-carbon double bond or triple bond with Markovnikov's Rule.

Reaction 2: Reagent (ii) NaBH4 / OH-

In the second reaction, sodium borohydride (NaBH4) in the presence of a hydroxide ion (OH-) is used. This reagent setup reduces carbonyl groups (C=O) to alcohols (C-OH), while it does not affect other functional groups, especially alkenes or alkynes.

Reaction 3: Reagent O3 + Zn

In the third reaction, ozone (O3) and zinc (Zn) are used for ozonolysis, a reaction that cleaves the carbon-carbon double bond or triple bond to form carbonyl compounds like aldehydes and/or ketones. Now, we should analyze each reactant given in (A), (B), (C), and (D) to find which one is not possible as a reactant (P).

Analyze reactant (A)

Reactant (A) is CH3-CH=CH-CH3, which has a double bond. It will undergo oxymercuration-demercuration to form an alcohol at the second carbon from the left. Then, it will not react with NaBH4 since there is no carbonyl group. After that, ozonolysis will cleave the double bond and form two carbonyl groups. Thus, reactant (A) can undergo these reactions and is not the answer.

Analyze reactant (B)

Reactant (B) is HC≡CH, which has a triple bond. It will undergo oxymercuration-demercuration to form an alcohol attached to the first carbon. Then, it will not react with NaBH4 since there is no carbonyl group. After that, ozonolysis will cleave the triple bond to form two carbonyl groups. Thus, reactant (B) can undergo these reactions and is not the answer.

Analyze reactant (C)

Reactant (C) is CH3-C≡CH, which has a triple bond. It will undergo oxymercuration-demercuration to form an alcohol attached to the second carbon. Then, it will not react with NaBH4 since there is no carbonyl group. After that, ozonolysis will cleave the triple bond and form two carbonyl groups. Thus, reactant (C) can undergo these reactions and is not the answer.

Analyze reactant (D)

Reactant (D) is CH3-C≡C-CH3, which has a triple bond. It will undergo oxymercuration-demercuration. However, the alcohol will form at a quaternary carbon (carbon bonded to four other carbons), which is not possible. Therefore, reactant (D) cannot undergo the set of reactions, and it is not possible as a reactant (P). The answer is (D) \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3}\).

Key concepts

Markovnikov's Rule

Markovnikov's Rule is a principle in organic chemistry that helps predict the outcome of certain addition reactions involving alkenes and alkynes. It states that when a hydrogen halide (HX) is added to an unsymmetrical alkene, the hydrogen atom ( H) will attach to the carbon with more hydrogen atoms, and the halide (X) will attach to the carbon with fewer hydrogen atoms. This rule generally helps determine how an alcohol group ( -OH> ) will be added when reacting alkenes with water in the presence of mercuric acetate, a process known as oxymercuration.

For example, in the compound CH₃-CH=CH₂, according to Markovnikov's Rule, the -OH group would attach to the second carbon in the chain, which has fewer hydrogen atoms than the outer carbons. This means it selects the carbon with the fewer hydrogens for addition, making the overall reaction more predictable.

• Helps predict products of addition reactions.
• Applies to adding H and -OH> during oxymercuration.
• Ensures the -OH> is added to the less substituted carbon.

Ozonolysis

Ozonolysis is a chemical reaction that cleaves carbon-carbon double bonds or triple bonds using ozone ( O₃ ) followed by reduction, often with zinc ( Zn ). This process is a powerful method for breaking down alkenes and alkynes into smaller, more manageable pieces, specifically carbonyl compounds such as aldehydes and ketones.

When an unsaturated molecule undergoes ozonolysis, the reaction breaks each C=C or C≡C bond and substitutes each end of the broken bond with oxygen atoms, creating carbonyl groups. For example, treating an alkene like CH₃-CH=CH-CH₃ with ozonolysis will split the double bond, resulting in two carbonyl groups: CH₃CHO (acetaldehyde) and CH₃CHO.

• Cleaves double and triple bonds.
• Produces aldehydes or ketones.
• Utilizes ozone and a reducing agent like zinc.
• Useful for structural analysis and transformations.

Reduction with Sodium Borohydride

Reduction with sodium borohydride ( NaBH₄ ) is a common method to convert carbonyl groups like aldehydes and ketones into alcohols. This process is relatively mild, which makes NaBH₄ selective for reducing only certain functional groups. Significantly, it does not affect other unsaturated bonds such as alkenes or alkynes, which is why it is used extensively in organic synthesis.

This reaction involves the transfer of hydride ions to the carbonyl carbon, transforming a carbon-oxygen double bond ( C=O ) into a single bond ( C-OH ). For instance, if we have a molecule like acetone (propan-2-one), NaBH₄ will reduce it to isopropanol, an alcohol. The mildness of NaBH4 makes it an ideal choice when alkenes or alkynes might be present alongside carbonyls.

• Reduces aldehydes and ketones to alcohols.
• Does not reduce alkenes or alkynes.
• Used in selective organic transformations.
• Known for its mildness and specificity.