The final product A, formed in the following reaction sequence is:
- \( \text{Ph} - \text{CH}_2 - \text{CH}_2 - \text{CH}_3 \)
- \( \text{Ph} - \text{CH}_2 - \text{CH}_2 - \text{CH}_2 - \text{OH} \)
The Correct Option is D
Approach Solution - 1
To determine the final product A, we need to analyze each step of the given reaction sequence applied to the compound \( \text{Ph} - \text{CH} = \text{CH}_2 \).
- Hydroboration-Oxidation:
- The first step involves the reaction of the alkene with \( \text{BH}_3 \), which adds across the double bond to form a trialkylborane intermediate.
- The subsequent reaction with \( \text{H}_2\text{O}_2 \) and \( \text{OH}^- \) converts the borane intermediate into an alcohol. This is syn addition, meaning the hydrogen and hydroxyl (OH) groups add to the same side of the alkene.
- For an unsymmetrical alkene like \( \text{Ph} - \text{CH} = \text{CH}_2 \), the hydroxyl group attaches to the less substituted carbon (anti-Markovnikov addition), resulting in \( \text{Ph} - \text{CH}_2 - \text{CH}_2 - \text{OH} \).
- Hydrobromination:
- The product from the hydroboration-oxidation step does not undergo further change in this context. However, if it had been a typical halogenation, it would have added bromine across a double bond if present.
- Since no additional reactions or mechanisms are specifically indicated for further processing here, the product remains as \( \text{Ph} - \text{CH}_2 - \text{CH}_2 - \text{OH} \).
- Grignard Reaction and Formaldehyde:
- The final steps include the formation of a Grignard reagent and its subsequent reaction with formaldehyde followed by hydration.
- The reaction typically adds an additional carbon to the chain, but in this setup with only primary alcohols involved, the product does not get elongated further and stays as \( \text{Ph} - \text{CH}_2 - \text{CH}_2 - \text{OH} \).
Thus, the final product of the reaction is:
\( \text{Ph} - \text{CH}_2 - \text{CH}_2 - \text{CH}_2 - \text{OH} \)
This corresponds to the correct option: \( \text{Ph} - \text{CH}_2 - \text{CH}_2 - \text{CH}_2 - \text{OH} \), indicating successful completion of hydroboration-oxidation and Grignard reactions.
Approach Solution -2
Step (i) involves hydroboration-oxidation of the double bond in \( \text{Ph-CH=CH}_2 \), resulting in the anti-Markovnikov addition of water to form \( \text{Ph-CH}_2\text{-CH}_2\text{-OH} \).
Step (ii) converts the alcohol \( (\text{Ph-CH}_2\text{-CH}_2\text{-OH}) \) to the corresponding alkyl halide \( (\text{Ph-CH}_2\text{-CH}_2\text{-Br}) \) using \( \text{HBr} \).
Step (iii) involves the formation of a Grignard reagent with \( \text{Mg} \), producing \( \text{Ph-CH}_2\text{-CH}_2\text{-MgBr} \).
Step (iv) reacts the Grignard reagent with formaldehyde (\( \text{HCHO} \)) followed by hydrolysis to yield the final primary alcohol, \( \text{Ph-CH}_2\text{-CH}_2\text{-CH}_2\text{-OH} \).
Thus, the final product is:
\(\text{Ph-CH}_2\text{-CH}_2\text{-CH}_2\text{-OH}.\)
The Correct answer is : \(\text{Ph-CH}_2\text{-CH}_2\text{-CH}_2\text{-OH}.\)
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