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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

Expert verified
Option 3 (Na, then CH鈧僆)

Step by step solution

01

Identify the target functional groups

Examine the initial and final compounds. The initial compound is PhC鈮H (a terminal alkyne) and the target compound is PhC=CHCH鈧 (an alkene). The goal is to convert a triple bond to a double bond with the addition of a CH鈧 group.
02

Analyze the given options

Consider each of the provided reagents and their typical reactions with an alkyne. Determine which set of reagents can achieve the conversion from a triple bond to a double bond while adding a CH鈧 group.
03

Evaluate Option 1

Option 1 suggests using Br鈧/CCl鈧 followed by KOH. Br鈧/CCl鈧 would add bromine to the alkyne (Br鈧 addition), forming a dibromoalkane, and KOH would be used for a dehydrohalogenation to form an alkyne or alkene. This path does not add a CH鈧 group, so it doesn't match our goal.
04

Evaluate Option 2

Option 2 involves using Na first, followed by CH鈧僀H鈧僆. Sodium can deprotonate the terminal alkyne forming a sodium acetylide, but CH鈧僀H鈧僆 does not introduce a new bonding site directly related to the desired change, pointing to the wrong path.
05

Evaluate Option 3

Option 3 uses Na followed by CH鈧僆. Sodium acetylide reacts with methyl iodide (CH鈧僆) to form a new C-C bond, resulting in PhC鈮CH鈧. A partial hydrogenation then may achieve the desired alkene (PhC=CHCH鈧) formation. This seems promising.
06

Evaluate Option 4

Option 4 involves CH鈧侼鈧, which is typically used for cyclopropanation reactions, not for converting triple bonds to an alkene with the addition of a methyl group.
07

Conclude the correct reagent

Based on the evaluations, Option 3 (Na, then CH鈧僆) is the correct reagent combination to achieve the conversion of PhC鈮H to PhC=CHCH鈧.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

reagents in organic chemistry
Reagents are substances or compounds added to a system to cause a chemical reaction or used to see if a reaction occurs. In organic chemistry, they are essential to performing transformations and synthesizing new compounds. For the conversion of an alkyne (PhC鈮H) to an alkene (PhC=CHCH鈧), specific reagents must be used. One important reagent is sodium (Na), which can deprotonate the terminal alkyne to form a highly reactive acetylide ion. This acetylide ion can then react with an alkyl halide like methyl iodide (CH鈧僆) to form a new carbon-carbon bond. Ultimately, choosing the right combination of reagents is crucial to achieving the desired chemical transformation.

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
Functional group transformations are central to organic synthesis. These transformations involve modifying the functional groups of molecules to achieve a specific target compound. For instance, converting an alkyne to an alkene involves:
  • 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
A reaction mechanism describes the step-by-step process by which reactants are converted into products. Understanding these mechanisms allows chemists to predict the outcome of reactions and to design pathways for synthesis.
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.

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Most popular questions from this chapter

Which of the following statements is correct? (1) alkynes are more reactive than alkenes towards halogen addition (2) alkynes are less reactive than alkenes towards halogen addition (3) both alkynes and alkenes are cqually reactive towards halogen addition (4) primary vinylic cation \((\mathrm{RCII}=\mathrm{CII})\) is more rcactive than sccondary vinylic cation \(\left(\mathrm{RC}=\mathrm{CII}_{2}\right)\)

Which of the following will convert \(\mathrm{HC}=\mathrm{CCH}_{2} \mathrm{CH}_{3}\) to \(\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{CH}_{3} ?\) (1) \(\mathrm{H}_{2} \mathrm{O} / \mathrm{H}\) (2) \(\mathrm{Hg}^{21} / \mathrm{H}_{2} \mathrm{SO}_{4}\) (3) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (conc) \(/ \mathrm{H}_{3} \mathrm{PO}_{4}\) (4) \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} / \mathrm{KMnO}_{4}\)

When \(\mathrm{C}_{2} \mathrm{II}_{2}, \mathrm{CH}_{4}\) and \(\mathrm{C}_{2} \mathrm{H}_{4}\) pass through a test tube which contains ammonical \(\mathrm{Cu}_{2} \mathrm{Cl}_{2}\), find out which gas comes out unaffected from the test tube. (1) \(\mathbf{C}_{2} \mathrm{H}_{2}\) and \(\mathrm{CH}_{4}\) (2) \(\mathrm{C}_{2} \mathrm{H}_{2}\) and \(\mathrm{C}_{2} \mathrm{H}_{4}\) (3) \(\mathbf{C}_{2} \mathrm{H}_{2}\) (4) \(\mathrm{C}_{2} \mathrm{H}_{4}\) and \(\mathrm{CH}_{4}\)

Using which of the following reagents one can perform a simple test that can be uscd to differentiate between \(\mathrm{C}_{6} \mathrm{II}_{5} \mathrm{C} \equiv \mathrm{CII}\) and \(\mathrm{C}_{6} \mathrm{I}_{5} \mathrm{CII}=\mathrm{CH}_{2} ?\) (1) \(\mathrm{NaOIL} / \mathrm{II}_{2} \mathrm{O}\) (2) \(\mathrm{Br}_{2} / \mathrm{CCl}_{4}\) (3) \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{OII}\) (4) \(\mathrm{CrO}_{3} / \mathrm{II}_{2} \mathrm{SO}_{4}\)

Butyne on reaction with hot alkaline \(\mathrm{KMnO}_{4}\) gives (1) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}\) (2) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\) (3) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}+\mathrm{CO}_{2}\) (4) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}+\mathrm{HCOOH}\)

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