Question

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

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

Step-by-step solution

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: \(\mathrm{CH}_{3}-\mathrm{C}=\mathrm{CH} + \mathrm{NaNH}_{2} \rightarrow \mathrm{CH}_{3}-\mathrm{C}^-\mathrm{CH} \equiv \mathrm{Na}^+ + \mathrm{NH}_{3}\).

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: \(\mathrm{CH}_{3}-\mathrm{C}^-\mathrm{CH} \equiv \mathrm{Na}^+ + \mathrm{HCHO} \rightarrow \mathrm{CH}_{3}-\mathrm{C}=\mathrm{CH}-\mathrm{CH}_{2}\mathrm{OH} + \mathrm{Na}^+\).

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) \(\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{2} \mathrm{OH}\)

Key concepts

Sodium Amide's Role in Reactions

Sodium amide (NaNH\(_2\)) is a very strong base commonly used in organic chemistry reactions. Its primary purpose is to abstract acidic hydrogen atoms from organic molecules. This abstraction is critical because it helps to generate a more reactive intermediate known as a carbanion. Carbanions possess a negative charge localized on a carbon atom, which makes them extremely useful as a nucleophile in subsequent reactions. In the given exercise, the role of sodium amide is to act on a hydrocarbon chain and remove an acidic hydrogen atom. As detailed in the exercise solution, sodium amide effectively extracts the hydrogen, forming a carbanion with a negative charge on the carbon atom adjacent to a double bond:

• The negative charge on the carbon atom indicates increased reactivity.
• This carbanion formation prepares the molecule for further reactions with electrophiles like aldehydes.

Sodium amide's efficiency as a strong base allows such reactions to occur smoothly and facilitates the formation of key intermediates in synthetic pathways.

Understanding Aldehyde Reactions

Aldehydes are a class of highly reactive organic compounds characterized by a carbonyl group (C=O) on a terminal carbon. This carbonyl group features an electrophilic carbon atom which is prone to attack by nucleophiles. The reactivity of this group makes aldehydes prime candidates for various chemical reactions, including those involving carbanions. In the reaction sequence presented, the carbanion generated by sodium amide acts upon the electrophilic carbon of formaldehyde (HCHO), the simplest aldehyde. This nucleophilic attack facilitates the formation of a new carbon-carbon bond.

• The carbon atom in the aldehyde accepts electrons from the carbanion.
• This results in the reduction of the aldehyde's carbonyl group to a hydroxyl group, ultimately leading to the formation of an alcohol.

This transformation showcases the utility of aldehyde reactions in building more complex organic structures from simpler precursors, leveraging their inherent electrophilicity.

Carbanion Formation and Reactivity

Carbanions are negatively charged carbon species generated through the removal of a proton, typically facilitated by a strong base such as sodium amide. This negative charge is of crucial importance in organic chemistry due to its implications for reactivity and reaction mechanism. The stabilization of the charge generally depends on the surrounding molecular environment. However, in many instances like the presented exercise, the carbanion serves as a potent nucleophile ready to engage in further chemical transformations.

• Carbanions are inherently unstable, which accounts for their high levels of reactivity.
• They tend to seek out and bond with electrophilic centers, such as the carbon atom of an aldehyde.

In essence, carbanions act as intermediates that drive the formation of new compounds. In the specific case of the exercise, the carbanion generated from the initial sodium amide reaction undergoes a nucleophilic attack on formaldehyde, facilitating a valuable step in organic synthesis that constructs a compound with increased complexity and functionality.