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The end product in the following sequence of reactions is \(\mathrm{HC} \equiv \mathrm{CH} \frac{1 \% \mathrm{HgSO}_{4}}{20 \% \mathrm{H}_{2} \mathrm{SO}_{4}}-\mathrm{A} \stackrel{\mathrm{CH}_{3} \mathrm{Mg} \mathrm{X}}{\mathrm{H}_{2} \mathrm{O}} \longrightarrow \mathrm{B}[\mathrm{O}]\) (a) ethanol (b) acetone (c) acetic acid (d) isopropyl alcohol

Short Answer

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(d) isopropyl alcohol

Step by step solution

01

Analyzing the given sequence of reactions

We are given a sequence of reactions starting with acetylene, \[ \text{HC} \equiv \text{CH} \xrightarrow{1\% \text{HgSO}_4/20\% \text{H}_2\text{SO}_4} \text{A} \] These conditions involve mercury(II) sulfate in sulfuric acid, which typically converts alkynes to ketones via a hydration reaction.
02

Hydration of acetylene to find compound A

The addition of water (hydration) in the presence of mercury(II) sulfate and sulfuric acid will cause the acetylene (\(\text{HC} \equiv \text{CH}\)) to form acetaldehyde. However, it generally creates an enol first, which tautomerizes to form acetaldehyde:\[(\text{CH}_3\text{CHO}) \] Here, compound A is then acetaldehyde (ethanal).
03

Reaction of compound A with Grignard reagent to find compound B

Next, the reaction is:\[ \text{CH}_3\text{CHO} \xrightarrow{\text{CH}_3\text{MgX}/\text{H}_2\text{O}} \text{B} \] A Grignard reagent like the methyl magnesium halide \(\text{CH}_3\text{MgX}\) adds to the carbonyl carbon of acetaldehyde forming a secondary alcohol upon hydrolysis. After the Grignard reaction, compound B is isopropyl alcohol \((\text{CH}_3)_2\text{CHOH}\).
04

Final identification of the product

The sequence of reactions forms isopropyl alcohol as the end product. The given reactions first hydrate acetylene to ethanol, which after reacting with a Grignard reagent, results in the final product.

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Key Concepts

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

Alkyne Hydration
Alkyne hydration is a chemical process where alkynes such as acetylene (\( \text{HC} \equiv \text{CH} \)) undergo hydration, meaning they add water across the triple bond. This reaction typically requires certain conditions to proceed, involving catalysts like mercury(II) sulfate (\( \text{HgSO}_4 \)) in the presence of strong acids such as sulfuric acid (\( \text{H}_2\text{SO}_4 \)).

When acetylene is hydrated, it initially forms an enol, a type of molecule that contains both a carbon-carbon double bond and a hydroxyl group (alcohol). However, enols are generally unstable and quickly rearrange to form more stable compounds through a process called tautomerization.
  • Initial product: Enol
  • Stable product: Acetaldehyde (\( \text{CH}_3\text{CHO} \))

This transformation is important in organic chemistry because it converts simple linear alkynes into aldehydes or ketones, which serve as key intermediates in various chemical syntheses.
Grignard Reaction
The Grignard reaction is an essential tool in organic chemistry, primarily used to form carbon-carbon bonds. This reaction involves Grignard reagents, which are organomagnesium compounds typically represented as \( \text{R-MgX} \), where \( \text{R} \) is an alkyl or aryl group and \( \text{X} \) is a halide.

These reagents are highly reactive, especially towards carbonyl compounds like aldehydes and ketones. When acetaldehyde (\( \text{CH}_3\text{CHO} \)) reacts with a Grignard reagent (\( \text{CH}_3\text{MgX} \)), the Grignard reagent adds to the carbonyl group forming an intermediate, which upon hydrolysis, results in a secondary alcohol:
  • Intermediate: Alkoxide
  • Final product: Isopropyl alcohol (\( (\text{CH}_3)_2\text{CHOH} \))

The Grignard reaction is favored for its ability to effectively build complex organic molecules by forming new carbon-carbon bonds, essential in synthesizing alcohols like isopropyl alcohol.
Tautomerization
Tautomerization is a chemical phenomenon where compounds called tautomers exist in dynamic equilibrium due to the shift of a hydrogen atom and the rearrangement of a double bond. This process is vital in the transformation of enols to more stable compounds like aldehydes or ketones.

In the context of alkyne hydration, after hydrating acetylene, the product is an enol. This enol form is not as stable and rapidly converts through tautomerization to acetaldehyde:
  • Enol: Less stable, transient
  • Aldehyde: More stable, long-lived

Tautomerization is crucial as it dictates the final structure of the hydrated product. In many organic pathways, this concept helps stabilize molecules, ultimately affecting the direction and efficiency of reactions. Understanding this process includes realizing its impact on molecular structure and reaction mechanisms in chemistry.

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Most popular questions from this chapter

Consider the following statements; Acetophenone can be prepared by I. oxidation of 1 -phenylethanol II. reaction of benzaldehyde with methyl magnesium bromide III. Friedal Crafts reaction of benzene with acetyl chloride IV. Distillation of calcium benzoate Which of the above statements are correct? (a) I and III (b) III and IV (c) II and III (d) I and IV

From which of the following tertiary butyl alcohol is obtained by the action of methyl magnesium iodide? (a) \(\mathrm{CO}_{2}\) (b) \(\mathrm{HCHO}\) (c) \(\mathrm{CH}_{3} \mathrm{CHO}\) (d) \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\)

\(\mathrm{Ph}-\mathrm{CHO}+\mathrm{NH}_{2} \mathrm{OH} \rightarrow(\mathrm{A})\) Which of the following is/are true about compound (A)? (a) The syn- (A) on treatment with \(\mathrm{PCl}_{5}\) in ether gives N- phenyl formamide. (b) The product (A) exists in 2 isomeric forms. (c) The suitable \(\mathrm{pH}\) for the reaction is between 4 to 6 (d) The anti- (A) on treatment with \(\mathrm{PCl}_{5}\) in ether followed by hydrolysis gives aniline and formic acid.

Which one of the following statements incorrect? Benzaldehyde can be produced by the (a) reduction of benzoyl chloride (b) oxidation of toluene (c) hydrolysis of benzal chloride (d) reduction of benzyl chloride

In cannizaro reaction given below \(\mathrm{Ph} \mathrm{CHO} \stackrel{\odot}{\stackrel{\circ} \mathrm{OH}}{\longrightarrow} \mathrm{Ph} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{Ph} \ddot{\mathrm{C}} \mathrm{O}_{2}\), the slowest step is: (a) the transfer of hydride to the carbonyl group (b) the abstraction of proton from the carboxylic group (c) the deprotonation of \(\mathrm{Ph} \mathrm{CH}_{2} \mathrm{OH}\) (d) the attack of : \(\mathrm{OH}\) at the carboxyl group

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