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Chapter 5: Problem 163

The product (P) formed in the following reaction is \(\mathrm{CH}_{3}-\mathrm{C}=\mathrm{CH} \frac{\text { (i) NaNH }_{2}}{\text { (ii) HCHo }}(\mathrm{P})\) (a) \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{2} \mathrm{OH}\) (b) \(\mathrm{CH}_{3}-\mathrm{C}=\mathrm{C}-\mathrm{COONa}\) (c) \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CHO}\) (d) \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CHO}\)

Short Answer

Expert verified
Based on the given reaction sequence, the product formed is (a) CH3-C≡C-CH2OH.

Step by step solution

01

Reaction with Sodium Amide (NaNH\(_{2}\))

In the first step, the reactant is treated with sodium amide (NaNH\(_{2}\)). Sodium amide is a strong base and will extract the acidic hydrogen from the molecule, leading to the formation of a carbanion. In this case, the carbanion would have a negative charge on the carbon adjacent to the double bond. The reaction is: \(\mathrm{CH}_{3}-\mathrm{C}=\mathrm{CH} + \mathrm{NaNH}_{2} \rightarrow \mathrm{CH}_{3}-\mathrm{C}^-\mathrm{CH} \equiv \mathrm{Na}^+ + \mathrm{NH}_{3}\).
02

Reaction with Aldehyde (HCHO)

In the second step, the formed carbanion (which is now a nucleophile) reacts with the aldehyde (HCHO). The nucleophile attacks the electrophilic carbonyl carbon of the aldehyde, leading to the formation of a new C-C bond, and results in the conversion of carbonyl group (C=O) to hydroxyl group (C-OH). The reaction is: \(\mathrm{CH}_{3}-\mathrm{C}^-\mathrm{CH} \equiv \mathrm{Na}^+ + \mathrm{HCHO} \rightarrow \mathrm{CH}_{3}-\mathrm{C}=\mathrm{CH}-\mathrm{CH}_{2}\mathrm{OH} + \mathrm{Na}^+\).
03

Product Identification

By combining steps 1 and 2, we get the final product: \(\mathrm{CH}_{3}-\mathrm{C}=\mathrm{CH} + \mathrm{NaNH}_{2} + \mathrm{HCHO} \rightarrow \mathrm{CH}_{3}-\mathrm{C}=\mathrm{CH}-\mathrm{CH}_{2}\mathrm{OH}\) Comparing this product with the given options, we see that option (a) is the correct answer: (a) \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{2} \mathrm{OH}\)

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