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\(P\) ) is the only benzene dicarboxylic acid which on strong heating produces anhydride, then choose the incorrect option: $$ (\mathrm{P})+\mathrm{NH}_{3} \rightleftharpoons(\mathrm{Q}) \longrightarrow \Delta(\mathrm{R}) \stackrel{\Delta}{-\mathrm{NH}_{3}} \longrightarrow(\mathrm{S}) $$ (A) (S) has common name phthalamide (B) (Q) is known as ammonium phthalate (C) Formation of (Q) is acid-base interaction (D) (S) can be used in Hofmann bromamide degradation reaction

Short Answer

Expert verified
The incorrect option is (D) because phthalamide (S) cannot be used in Hofmann bromamide degradation reaction, as it is not a primary amide.

Step by step solution

01

Identify the benzene dicarboxylic acid (P)

The given dicarboxylic acid "P" is an aromatic benzene derivative and on strong heating, it forms an anhydride. The only benzene dicarboxylic acid capable of forming an anhydride upon heating is phthalic acid:$$ \mathrm{P} = \ce{C6H4(COOH)2}$$
02

Formation of Q

When phthalic acid (P) reacts with ammonia, it forms compound Q. This is an acid-base interaction involving NH3 as the base and phthalic acid (P) as the acid:$$ \mathrm{P} + 2\;\mathrm{NH}_{3} \rightleftharpoons \mathrm{Q} \qquad Q = \ce{C6H4(COONH4)2}$$Compound Q is known as ammonium phthalate.
03

Formation of R

Phthalic anhydride (R) is formed when compound Q (ammonium phthalate) is heated: $$ \mathrm{Q} \stackrel{\Delta}{\longrightarrow} \Delta \mathrm{R} \qquad R = \ce{C6H4(CO)2O}$$
04

Formation of S

Compound S is formed by heating compound R in the presence of ammonia (NH3). The reaction produces a new compound with an internal ring called phthalamide. This ring is formed through an imide formation reaction that replaces an amide group with an NH3 molecule:$$ \mathrm{R} + \mathrm{NH}_{3} \stackrel{\Delta}{\longrightarrow} \mathrm{S} - \mathrm{NH}_{3}\qquad S = \ce{C6H4(CO)NHCO}$$Phthalamide has the common name "phthalamide."
05

Analyzing the Options

Now that we have followed the reaction and identified each compound, let's analyze the given options: (A) (S) has the common name phthalamide - This statement is correct. (B) (Q) is known as ammonium phthalate - This statement is correct. (C) Formation of (Q) is acid-base interaction - This statement is correct. (D) (S) can be used in Hofmann bromamide degradation reaction - This statement is incorrect because Hofmann bromamide degradation reaction requires the presence of primary amide to create a primary amine, while phthalamide is not a primary amide. The correct answer is (D), as it is the incorrect option among the given options.

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Key Concepts

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

Phthalic Acid
Phthalic acid is an aromatic dicarboxylic acid with the formula \(C_6H_4(COOH)_2\). It is characterized by two carboxyl groups directly attached to a benzene ring. This compound is notable for its role in the production of many industrially important products, such as phthalate esters used as plasticizers in PVC.

In the context of the exercise, phthalic acid is the starting material \(P\) which readily forms phthalic anhydride \(R\) upon heating. This transformation is a key step in multiple synthetic pathways, including the production of dyes, fragrances, and polyesters.
Hofmann Bromamide Degradation Reaction
The Hofmann bromamide degradation reaction is a classical method used in organic chemistry to convert an amide to a primary amine with one fewer carbon atom. This reaction involves treating an amide with bromine in an aqueous or alkaline solution, followed by heating.

The overall reaction can be simplified as \(\ce{RCONH2 + Br2 + 4NaOH -> RNH2 + Na2CO3 + 2NaBr + 2H2O}\). This method is particularly useful for identifying the structure of an unknown amide and for reducing the carbon chain length of a compound in synthesis pathways. However, it requires a primary amide; thus, compounds like phthalamide \(S\) with an imide functional group do not undergo this reaction.
Acid-Base Interaction
An acid-base interaction is a chemical reaction between an acid and a base which can result in the formation of a salt and usually water. In the context of the given problem, phthalic acid (a weak acid) reacts with ammonia (a weak base) to form ammonium phthalate \(Q\), the ammonium salt of phthalic acid.

Acid-base interactions are vital for understanding a variety of chemical reactions and are ubiquitous in both organic and inorganic chemistry. In this case, the acid (phthalic acid) donates a proton (\(H^+\)) to the base (ammonia), which accepts the proton, leading to the production of the salt \(Q\). This interaction is fundamental to the formation of many compounds and is greatly utilized in industrial chemistry.
Imide Formation
Imide formation is a reaction where an anhydride reacts with ammonia or a primary amine to form an imide. This is a specific type of condensation reaction that involves the combination of two carbonyl groups with an amine group to form a five-membered ring compound containing two carbonyl functionalities adjacent to the nitrogen atom.

The imide functional group is present in many biologically active molecules and synthetic materials, such as polyimides used in high-performance plastics. Imides are typically less reactive than amides due to resonance stabilization. In the exercise, phthalic anhydride \(R\) reacts with ammonia to form phthalamide \(S\), showcasing imide formation where the loss of water molecule is integral to the reaction mechanism.

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Most popular questions from this chapter

\(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} / \mathrm{H}^{+}, \Delta\) is a strong oxidising agent which can oxidise \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OH}\) into \(\mathrm{CH}_{3}-\mathrm{COOH}\) but still \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{O}\) is obtained by oxidation of \(\mathrm{CH}_{3}-\mathrm{CH}_{3}-\mathrm{OH}\) by using distillation flask at temperature \(55^{\circ} \mathrm{C}_{\text {with }}\) \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}, \mathrm{H}^{+}, \Delta\). It is possible because (A) Acetaldehyde distills out at this temperature and escape further oxidation (B) At this temperature, acetaldehyde is very stable and escape further oxidation (C) Oxidation of \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OH}\) is slower than oxidation of \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{O}\) (D) None

Correct option(s) regarding reaction, reactant and product respectively given is : (A) Schmidt reaction Acid Primary amine (B) Hoffmann bromamide Primary acid amide Primary amine (C) Schmidt reaction Ketone Secondary acid amide (D) Beckmann rearrangement Ketoxime Secondary acid amide

Which of the following tetracarboxylic acid gives two isomeric monoanhydrides on addition of an equivalent of \(\left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{O} ?\)

\((\mathrm{P}) \stackrel{\mathrm{Red} \mathrm{P} / \mathrm{X}_{2}}{\longrightarrow}(\mathrm{Q})\) (P) has M.F. \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{2}\) which gives fruity smell with alcohol and ' \(\mathrm{X}^{\prime}\) is halogen. Choose the incore: option: (A) Product (Q) is \(\beta\)-halocarboxylic acid (B) Reaction is possible when \(\alpha\)-hydrogen is present (C) Reaction is known as HVZ reaction (D) Red \(\mathrm{P}\) is taken in small amount and ' \(\mathrm{X}\) ' can be chlorine or bromine.

Choose the correct option(s): $$ \mathrm{NBS} \stackrel{\mathrm{NaOBr}}{\longrightarrow} \mathrm{P} \stackrel{\mathrm{NaNO}_{2} / \mathrm{HCl}}{\longrightarrow} \mathrm{Q} \stackrel{\Delta}{\longrightarrow} \mathrm{R} $$ (A) (P) on strong heating will produce (R) (B) Formation of (Q) involves \(\mathrm{N}_{2}\) as leaving group (C) Formation of (P) is Hofmann bromamide degradation (D) Rearrangment step is rate determining step to form (P)

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