Consider the following sequence of reaction: \(\mathrm{Ph}-\mathrm{COOH} \stackrel{\mathrm{PCl}_{5}}{\longrightarrow} \mathrm{A} \stackrel{\mathrm{NH}_{3}}{\longrightarrow} \mathrm{B} \stackrel{\mathrm{POCl}_{3} / \Delta}{\longrightarrow}\) \(\mathrm{C} \stackrel{\mathrm{H}_{2} / \mathrm{Ni}}{\longrightarrow} \mathrm{D}\) The final product \(D\) is : (1) Benzonitrile (2) Benzylamine (3) Aniline (4) Benzamide

Short Answer

Expert verified
The final product \(D\) is Benzylamine.

Step by step solution

01

Identify the First Reaction

The sequence begins with \(\text{Ph-COOH}\) reacting with \(PCl_5\). This converts the carboxylic acid into an acid chloride, so compound \(A\) is \(\text{Ph-COCl}\)
02

Identify the Second Reaction

Compound \(A = \text{Ph-COCl}\) reacts with \(NH_3\) (ammonia) to form an amide. Hence, compound \(B\) is \( \text{Ph-CONH}_2\), also known as benzamide.
03

Identify the Third Reaction

Compound \(B = \text{Ph-CONH}_2\) undergoes cyclization with \(POCl_3 / \Delta\) to form a nitrile. Thus, compound \(C\) is \( \text{Ph-CN}\) (benzonitrile).
04

Identify the Fourth Reaction

Compound \(C = \text{Ph-CN}\) undergoes hydrogenation in the presence of \(H_2 / Ni\) catalyst, converting the nitrile group to an amine. Therefore, the final product \(D\) is \( \text{Ph-CH}_2\text{NH}_2\) (benzylamine).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

acid chloride formation
The synthesis of acid chloride is a vital step often seen in organic chemistry. To form an acid chloride, one typically starts with a carboxylic acid. In our sequence, the carboxylic acid used is benzoic acid (\text{Ph-COOH}).
When benzoic acid reacts with phosphorus pentachloride (PCl\(_5\)), it converts into benzoyl chloride (\text{Ph-COCl}).
The reaction can be summarized as follows: You can remember that PCl\(_5\) is a potent chlorinating agent often used to replace the \text{OH} group in a carboxylic acid with a \text{Cl} group, forming an acid chloride.
This reaction is useful because acid chlorides are more reactive and can be further transformed into various other functional groups.
amide formation
The next important step involves converting the acid chloride into an amide. In our sequence, benzoyl chloride (\text{Ph-COCl}) reacts with ammonia (\text{NH\(_3\)}).
When ammonia reacts with benzoyl chloride, it replaces the chlorine atom, forming an amide called benzamide (\text{Ph-CONH\(_2\)}).
The reaction can be represented as: Recognizing this transformation is crucial because amides are a recurring functional group in both biological molecules and synthetic compounds.
The use of ammonia results in the formation of primary amides, which are simpler and often the target structure in basic organic syntheses.
nitrile formation
Following the formation of benzamide, the next step in our sequence is its conversion into a nitrile.
This can be achieved through dehydration, usually by reacting the amide with phosphorus oxychloride (POCl\(_3\)) under heat (Δ).
For benzamide (\text{Ph-CONH\(_2\)}), this reaction produces benzonitrile (\text{Ph-CN}).
The reaction can be generalized as follows: This process is important because nitriles are key intermediates in many organic reactions.
They can be converted into amines, carboxylic acids, and other valuable compounds. The presence of the triple bond in nitriles can also offer unique reactivity patterns.
hydrogenation
The final step in our sequence involves the reduction of the nitrile group to an amine.
Benzonitrile (\text{Ph-CN}) undergoes hydrogenation in the presence of a nickel (Ni) catalyst and hydrogen gas (H\(_2\)).
This reaction converts the nitrile into a primary amine, specifically benzylamine (\text{Ph-CH\(_2\)NH\(_2\)}).
The general reaction is as follows: Hydrogenation is a valuable method in organic synthesis allowing for the selective reduction of double and triple bonds. Catalysts like nickel play a crucial role in facilitating these reactions efficiently.
Hydrogenation can be applied to various functional groups, transforming them into more useful or targeted structures for further chemical manipulation.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The integral curve of equation \(x d y-y d x=y d y\), intersects the \(y\)-axis at \((0,1)\) and the line \(y=e\) at \((a, b)\). The value of \(a+b=\) (1) 0 (2) 1 (3) \(2 \mathrm{e}\) (4) \(e\)

A drop of liquid of surface tension \(\sigma\) is in between the two smooth parallel glass plates held at a distance d apart from each other in zero gravity. The liquid wets the plate so that the drop is a cylinder of diameter D with its curved surface at right angles to both the plates. Determine the force acting on each of the plates from drops under the following considerations. (1) \(\frac{\sigma \pi D}{2}\) (2) \(\frac{\sigma^{2} \pi D}{2}\) (3) zero (4) None of these

A particle moves along a constant curvature path between \(A\) and \(B\) (length of curve wire \(A B\) is \(100 \mathrm{~m}\) ) with a constant speed of \(72 \mathrm{~km} / \mathrm{hr}\). The acceleration of particle at mid point ' \(\mathrm{C}\) ' of curve \(\mathrm{AB}\) is : (1) Zero (2) \(3.14 \mathrm{~m} / \mathrm{s}^{2}\) (3) \(50.84 \mathrm{~m} / \mathrm{s}^{2}\) (4) \(50.84 \mathrm{~km} / \mathrm{hr}^{2}\)

A man running round a race course notes that the sum of the distance of two flag - posts from him is always 10 meters and distance between the flag - posts is 8 meters. The area of the path he encloses in square meters is (1) \(15 \pi\) (2) \(12 \pi\) (3) \(18 \pi\) (4) \(8 \pi\)

Water, having positive ions of charge '+e'm dissolved in it with concentration c (Number of ions/volume), falls from a tube of cross sectional area 'a' with a speed 'v' in a sieve such that water is not retained in the sieve. The value of \(\int \vec{B} \cdot \vec{d} \ell\) integrated over the upper circular part of the sieve will be : [ \(\overrightarrow{\mathrm{B}}\) indicates the magnetic field produced by \(\mathrm{Na}^{+}\)ions and \(\overrightarrow{\mathrm{d} \ell}\) is along the tangent on the periphery of circular part of the sieve]. (1) \(\mu_{0}\) avce (2) \(2 \mu_{0}\) avce (3) \(\mu_{0} \mathrm{ace} / \mathrm{v}\) (4) \(\mu_{0 /} /\) avce

See all solutions

Recommended explanations on English Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free