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Which one of the following pairs is not correctly matched? (a) \(>\mathrm{C}=\mathrm{O} \longrightarrow \mathrm{CH}_{2}\) Clemmensen reduction (b) \(-\mathrm{COCl} \longrightarrow \mathrm{CHO}\) Rosenmund reduction (c) \(>\mathrm{C}=\mathrm{O} \longrightarrow>\mathrm{CHOH}\) Wolff - Kishner reduction (d) \(-\mathrm{C} \equiv \mathrm{N} \longrightarrow-\mathrm{CHO}\) Stephen reduction

Short Answer

Expert verified
Option (c) is not correctly matched.

Step by step solution

01

Understanding the Question

The question provides four options, each describing a reaction type matched with a corresponding reduction method. We need to identify which pair is incorrectly matched.
02

Reviewing Clemmensen Reduction

Clemmensen reduction involves the reduction of a carbonyl group (C=O) to a methylene group (CH2) using zinc amalgam and hydrochloric acid. This confirms option (a) is correctly matched.
03

Examining Rosenmund Reduction

Rosenmund reduction is used to convert an acyl chloride (-COCl) to an aldehyde (-CHO) using palladium on barium sulfate, typically in the presence of hydrogen gas. Therefore, option (b) is correctly matched.
04

Analyzing Wolff-Kishner Reduction

Wolff-Kishner reduction converts a carbonyl group (C=O) to an alkane, typically to a fully reduced form (>CH2) without forming alcohols, using hydrazine and a strong base. The description provided in option (c) leads to an alcohol (CHOH), thus, this match is incorrect.
05

Understanding Stephen Reduction

Stephen reduction describes the conversion of a nitrile group (CquivN) to an aldehyde (-CHO) using stannous chloride, followed by acidic hydrolysis. Therefore, option (d) is also correctly matched.

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Key Concepts

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

Clemmensen Reduction
The Clemmensen Reduction is a method commonly employed to convert a carbonyl group (\( >\mathrm{C}=\mathrm{O} \)) into a methylene group (\( >\mathrm{CH}_{2} \)). This reaction is particularly useful in removing the oxygen from ketones or aldehydes in acidic conditions. The reaction takes place using a combination of zinc amalgam and concentrated hydrochloric acid as the reducing agents. The process is highly effective for substrates that are stable in the presence of strong acids.
Student can find this reduction valuable in organic synthesis as it helps in simplifying molecules by reducing their complexity. However, it's important to note that Clemmensen Reduction is not suitable for compounds that are sensitive to acidic environments, as this may lead to unwanted reactions or decomposition.
Rosenmund Reduction
Rosenmund Reduction is a technique used to reduce acyl chlorides (\( -\mathrm{COCl} \)) to aldehydes (\( -\mathrm{CHO} \)). The reaction involves the use of palladium on barium sulfate as a catalyst along with hydrogen gas. This setup allows for selective reduction to aldehydes without further hydrogenation to alcohols, which is crucial in preserving the partial oxidation state of the carbon.
In this reaction, the catalyst is critical as it prevents full reduction, ensuring the transformation stops at the aldehyde stage. The Rosenmund Reduction is a preferred method for preparing aldehydes from acyl chlorides because of its specificity and efficiency.
Wolff-Kishner Reduction
The Wolff-Kishner Reduction process is distinct because it typically reduces carbonyl groups to the corresponding hydrocarbons rather than alcohols. This method uses hydrazine (\( \text{N}_2\text{H}_4 \)) and a strong base such as potassium or sodium hydroxide under high temperature. During the reaction, the carbonyl group is first transformed into a hydrazone, followed by base-induced elimination of nitrogen to form the reduced hydrocarbon.
This method is beneficial when working with base-stable compounds, and it's particularly effective when the goal is to achieve a complete reduction without introducing polar functionalities. Unlike the given description in option (c), Wolff-Kishner does not lead to alcohol formation, which makes option (c) in the original exercise a mismatch.
Stephen Reduction
Stephen Reduction is a valuable reaction to convert nitriles (\( -\mathrm{C} \equiv \mathrm{N} \)) into aldehydes (\( -\mathrm{CHO} \)). This conversion takes place using stannous chloride (SnCl₂) followed by hydrolysis. The process begins with the formation of an iminium salt, which, after hydrolysis, yields the desired aldehyde product.
This reduction method is specially used when the starting material is a nitrile that needs to be selectively reduced to an aldehyde, without progressing to an alcohol. The Stephen Reduction provides a convenient and controlled way to achieve this transformation, making it an important tool in organic synthetic strategies.

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Most popular questions from this chapter

If 3 -hexanone is reacted with \(\mathrm{NaBH}_{4}\) followed by hydrolysis with \(\mathrm{D}_{2} \mathrm{O}\), the product will be (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CD}(\mathrm{OD}) \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}(\mathrm{OD}) \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CD}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\)

Treatment of propionaldehyde with dil. \(\mathrm{NaOH}\) solution gives (a) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}_{2} \mathrm{CH}_{2} \mathrm{CHO}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHOHCH}_{2} \mathrm{CH}_{2} \mathrm{CHO}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHOHCH}\left(\mathrm{CH}_{3}\right) \mathrm{CHO}\) (d) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\)

Phenyl magnesium bromide reacts with methanol to give (a) a mixture of anisole and \(\mathrm{Mg}(\mathrm{OH}) \mathrm{Br}\) (b) a mixture of benzene and \(\mathrm{Mg}(\mathrm{OMe}) \mathrm{Br}\) (c) a mixture of toluene and \(\mathrm{Mg}(\mathrm{OH}) \mathrm{Br}\) (d) a mixture of phenol and \(\mathrm{Mg}(\mathrm{Me}) \mathrm{Br}\)

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) \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\)

Match list I (organic compounds oxidized by \(\left.\mathrm{HIO}_{4}\right)\) with list II (products of \(\mathrm{HIO}_{4}\) oxidation) and select the correct answer. List II (i) \(2 \mathrm{H}_{2} \mathrm{C}=\mathrm{O}+\mathrm{HCOOH}\) (ii) \(2 \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{O}\) (iii) \(2 \mathrm{H}_{2} \mathrm{C}=\mathrm{O}\) (iv) \(\mathrm{PhCH}_{2} \mathrm{CH}=\mathrm{O}+\mathrm{CH}_{3} \mathrm{CH}=\mathrm{O}\) The correct matching is: 1 2 3 4 (a) (ii) (iv) (i) (iii) (b) (iv) (ii) (iii) (i) (c) (ii) (iii) (i) (iv) (d) (ii) (iv) (iii) (i)

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