Question

The organic compounds P, Q, and R are: 
\(\underline{\hspace{3cm}}\).

• A. P = CH\(_3\)-CH(OH)-CH\(_2\)-CH\(_3\), Q = CH\(_3\)-C(=O)-CH\(_3\), R = CH\(_3\)-C(OH)(CH\(_3\))-CH\(_2\)-CH\(_3\)
• B. P = CH\(_3\)-CH\(_2\)-CH\(_2\)-OH, Q = CH\(_3\)-CH\(_2\)-CHO, R = CH\(_3\)-CH\(_2\)-CH(OH)-CH\(_3\)
• C. P = CH\(_3\)-CH\(_2\)-CH\(_2\)-OH, Q = CH\(_3\)-CH\(_2\)-COOH, R = CH\(_3\)-CH\(_2\)-C(=O)-OCH\(_3\)
• D. P = CH\(_3\)-CH(OH)-CH\(_3\), Q = CH\(_3\)-C(=O)-CH\(_3\), R = CH\(_3\)-CH(OCH\(_3\))-CH\(_3\)

Hint:

In multi-step reaction questions, look for inconsistencies. If you find one, try to identify the most likely typo by checking which parts of the options are chemically sound. Here, the conversion from a corrected Q (butan-2-one) to R (2-methylbutan-2-ol) is a standard Grignard reaction, making the error likely in the provided structure of Q.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The reaction sequence involves hydroboration-oxidation, controlled oxidation of an alcohol, and Grignard addition to a carbonyl compound.
Step 2: Key Formula or Approach:
Map the reagents to their specific organic transformations: hydroboration-oxidation yields anti-Markovnikov alcohols, $CrO_3$ in anhydrous media stops at the aldehyde stage, and Grignard addition creates a higher-order alcohol. (Note: The starting material is indicated as $C_4H_{10}$ in OCR, but hydroboration applies to alkenes; we proceed based on the products derived from an implied terminal alkene like propene or butene. Based on the options, the starting alkene is propene $C_3H_6$ since the products have 3 carbon chains before Grignard.)
Step 3: Detailed Explanation:
Let's analyze the options. The options (B) and (C) suggest the chain "P" has 3 carbons ($CH_3-CH_2-CH_2-OH$). This implies the starting alkene was Propene ($CH_3-CH=CH_2$), not $C_4H_{10}$.
1. Reaction 1 (Hydroboration-Oxidation): $CH_{3}-CH=CH_{2}$ reacts with $(BH_{3})_{2}$ followed by $H_{2}O_{2}/NaOH$. This is an anti-Markovnikov addition of water.
Product P: $CH_{3}-CH_{2}-CH_{2}-OH$ (Propan-1-ol).

2. Reaction 2 (Mild Oxidation): P (Propan-1-ol) reacts with $CrO_{3}$ in anhydrous medium. Primary alcohols are oxidized to aldehydes.
Product Q: $CH_{3}-CH_{2}-CHO$ (Propanal).

3. Reaction 3 (Grignard Reaction): Q (Propanal) reacts with $CH_{3}MgBr$ followed by hydrolysis. The nucleophilic $CH_{3}^{-}$ attacks the carbonyl carbon of the aldehyde.
Product R: $CH_{3}-CH_{2}-CH(OH)-CH_{3}$ (Butan-2-ol).
Step 4: Final Answer:
Matching the derived structures with the options leads to choice (B).