Question

Consider the following reaction sequence. \( \text{CH}_3\text{CHO} \xrightarrow[\text{(ii) } \text{H}_2\text{O} / \text{H}^+]{\text{(i) } \text{CH}_3\text{MgBr}} \text{(A)} \xrightarrow{\text{H}_2\text{SO}_4, \Delta} \text{(B)} \xrightarrow[\text{(ii) } \text{H}_2\text{O}_2 / \text{OH}^-]{\text{(i) } \text{B}_2\text{H}_6} \text{(C)} \) (A) and (C) are

• A. Functional isomers
• B. Metamers
• C. Optical isomers
• D. Position isomers

Hint:

- Grignard + Formaldehyde \(\to\) 1\(^\circ\) Alcohol. - Grignard + Other Aldehydes \(\to\) 2\(^\circ\) Alcohol. - Grignard + Ketones \(\to\) 3\(^\circ\) Alcohol. - Hydroboration-Oxidation yields Anti-Markovnikov Alcohol (1\(^\circ\) from terminal alkene).

Answer 1

The Correct Option is D

Step 1: Reaction Analysis:

1. Step 1: Acetaldehyde (\( \text{CH}_3\text{CHO} \)) + Methyl Magnesium Bromide (\( \text{CH}_3\text{MgBr} \)) followed by hydrolysis. Nucleophilic addition of Grignard reagent to aldehyde gives a secondary alcohol. \[ \text{CH}_3\text{CHO} + \text{CH}_3\text{MgBr} \to \text{CH}_3\text{CH(OMgBr)CH}_3 \xrightarrow{\text{H}_3\text{O}^+} \text{CH}_3\text{CH(OH)CH}_3 \] (A) is Propan-2-ol (Isopropyl alcohol). 2. Step 2: Dehydration with \( \text{H}_2\text{SO}_4, \Delta \). Alcohol (A) loses water to form an alkene. \[ \text{CH}_3\text{CH(OH)CH}_3 \xrightarrow{\text{H}_2\text{SO}_4, \Delta} \text{CH}_3\text{-CH=CH}_2 \] (B) is Propene. 3. Step 3: Hydroboration-Oxidation (\( \text{B}_2\text{H}_6, \text{H}_2\text{O}_2/\text{OH}^- \)). This reaction adds water across the double bond according to Anti-Markovnikov's rule. Propene (\( \text{CH}_3\text{-CH=CH}_2 \)) \(\to\) Primary Alcohol. \[ \text{CH}_3\text{-CH=CH}_2 \to \text{CH}_3\text{-CH}_2\text{-CH}_2\text{-OH} \] (C) is Propan-1-ol.
Step 2: Comparison of A and C:

- (A): Propan-2-ol (\( \text{CH}_3\text{CH(OH)CH}_3 \)) - OH at position 2. - (C): Propan-1-ol (\( \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \)) - OH at position 1. They have the same molecular formula (\( \text{C}_3\text{H}_8\text{O} \)) and functional group (Alcohol), but different positions of the functional group. Thus, they are Position Isomers.
Step 4: Final Answer:

Position isomers.