Chapter 19: Problem 2
By using which of the following can the conversion of \(\mathrm{PhC} \equiv \mathrm{CH} \longrightarrow \mathrm{PhC}=\mathrm{CHCH}_{3}\) be achieved? (1) \(\mathrm{Br}_{2} \mathrm{CCl}_{4}\), then \(\mathrm{KOH}\) (2) \(\mathrm{Na}\), then \(\mathrm{CH}_{3} \mathrm{CH}_{3} \mathrm{I}\) (3) \(\mathrm{Na}\), then \(\mathrm{CH}_{3} \mathrm{I}\) (4) \(\mathrm{CH}_{2} \mathrm{~N}_{2}\)
Short Answer
Step by step solution
Identify the target functional groups
Analyze the given options
Evaluate Option 1
Evaluate Option 2
Evaluate Option 3
Evaluate Option 4
Conclude the correct reagent
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
reagents in organic chemistry
In the example provided:
- Na forms a sodium acetylide.
- CH鈧僆 reacts with the sodium acetylide to add a CH鈧 group, forming PhC鈮CH鈧.
functional group transformations
- Understanding the functional groups involved (e.g., an alkyne with a triple bond and an alkene with a double bond).
- Knowing the steps to convert these groups (often through reagents and specific reaction conditions).
In our exercise, the transformation of PhC鈮H (an alkyne) to PhC=CHCH鈧 (an alkene) requires the addition of a methyl group (CH鈧) and the reduction of the triple bond to a double bond. This step-by-step progression is essential in achieving the desired functional group change.
For the conversion:
- Add a CH鈧 group to form PhC鈮CH鈧 using Na and CH鈧僆.
- Hydrogenate the triple bond partially to convert it into an alkene, leading to PhC=CHCH鈧.
reaction mechanisms
For the conversion from an alkyne to an alkene, the mechanism can be outlined as:
- Deprotonation: Sodium (Na) removes a hydrogen atom from the terminal alkyne (PhC鈮H), creating a sodium acetylide (PhC鈮鈦籒a鈦).
- Nucleophilic substitution: The acetylide ion formed reacts with methyl iodide (CH鈧僆) through a nucleophilic attack, resulting in the formation of a new C-C bond, giving PhC鈮CH鈧.
- Partial hydrogenation: The triple bond in PhC鈮CH鈧 is partially hydrogenated, reducing it to a double bond, and resulting in the target compound PhC=CHCH鈧.
By understanding each step in the mechanism, we can see how the reactions are carried out and what intermediates are formed, which is crucial for mastering organic synthesis.