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All acid derivatives on hydrolysis certainly give (A) Carboxylic acid (B) Alcohol (C) Aldehyde (D) Carbon dioxide

Short Answer

Expert verified
The correct option is (A) Carboxylic acid. The hydrolysis of all acid derivatives, such as amides, esters, anhydrides, and acyl halides, results in the formation of a carboxylic acid as one of the products.

Step by step solution

01

Understand acid derivatives

Acid derivatives are organic compounds that can be derived from carboxylic acids by replacing the hydroxyl (-OH) group with other functional groups. Common acid derivatives include amides, esters, anhydrides, and acyl halides.
02

Determine the general hydrolysis reaction of acid derivatives

In general, the hydrolysis of acid derivatives involves breaking the bond connecting the functional group and the carbonyl carbon, resulting in two products - one of which is a carboxylic acid: - For amides: R-C(O)-NR'2 + H2O ⟶ R-C(O)OH + HNR'2 - For esters: R-C(O)-OR' + H2O ⟶ R-C(O)OH + R'OH - For anhydrides: R-C(O)-O-C(O)R' + H2O ⟶ R-C(O)OH + R'-C(O)OH - For acyl halides: R-C(O)-X + H2O ⟶ R-C(O)OH + HX
03

Analyze each given option

(A) Carboxylic acid: As seen in the hydrolysis reactions above, a carboxylic acid is formed in each case. (B) Alcohol: Alcohol is produced only in the hydrolysis of esters (R'OH), which means it is not a general product for all acid derivatives. (C) Aldehyde: No hydrolysis reaction produces an aldehyde among the products. (D) Carbon dioxide: Hydrolysis of acid derivatives does not produce carbon dioxide as a product.
04

Choose the correct option

From the analysis in step 3, we can conclude that the correct option is (A) Carboxylic acid, as it is produced in the hydrolysis reaction of all acid derivatives.

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Key Concepts

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

Carboxylic Acid Formation
Carboxylic acid formation is a fundamental concept in organic chemistry, particularly important for students preparing for examinations like JEE Main and Advanced. Through hydrolysis, various acid derivatives are converted into carboxylic acids, involving the addition of water to break chemical bonds.

When it comes to the structure of carboxylic acids, all share the same functional group -COOH, which is the carboxyl group. During hydrolysis, the bond between the carbonyl carbon and the substituent group in acid derivatives is cleaved, ultimately yielding a carboxylic acid and another compound.

The hydrolysis process of acid derivatives is catalyzed by either acidic or basic conditions, which assist in the cleavage of the bond. Acid derivatives include substances like amides, esters, anhydrides, and acyl halides, each reacting differently yet conclusively producing carboxylic acids. Understanding this transformation is crucial for students as it forms the basis for more complex reactions and mechanisms in organic chemistry.
Acid Derivatives in Organic Chemistry
In the context of organic chemistry, acid derivatives are classes of compounds that are structurally related to carboxylic acids. They retain the acyl group (R-CO-) and differ by the substituent attached to the carbonyl carbon.

  • Amides (R-CO-NR'2) have an amino group attached to the carbonyl carbon.
  • Esters (R-CO-OR') feature an alkoxy group.
  • Anhydrides (R-CO-O-CO-R') contain two acyl groups bonded to an oxygen atom.
  • Acyl halides (R-CO-X) have a halogen atom attached.
Each type of derivative undergoes hydrolysis with water, but the conditions (acidic or basic) and products formed can differ. For instance, esters will produce alcohols alongside carboxylic acids upon hydrolysis, while amides yield amines. This variance in products is a critical consideration during synthesis and mechanism determination in advanced organic chemistry courses.
JEE Main and Advanced Organic Chemistry
For students preparing for JEE Main and Advanced, proficiency in organic chemistry is a must. This topic not only requires understanding of fundamental concepts such as carboxylic acid formation and the properties of acid derivatives but also demands the ability to apply these concepts to solve complex problems.

Including hydrolysis reactions in the JEE syllabus reflects the examination's emphasis on mechanism-based understanding. Hydrolysis questions often appear in the JEE Main and Advanced papers, where students need to deduce the products formed from different organic compounds when reacting with water. Mastering such topics involves practicing reaction mechanisms, recognizing functional groups, and predicting the outcome of hydrolysis reactions under various conditions.

To excel in these exams, it's essential to grasp the intricacies of acid derivatives and their reactivity. By understanding the step-by-step reactions and being able to deduce products correctly, as in the given exercise, aspirants can significantly improve their organic chemistry proficiency for the JEE examinations.

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Most popular questions from this chapter

The correct set of reagent to carry out following conversion is $$\mathrm{CH}_{3}-\mathrm{COOH} \longrightarrow \mathrm{HCOOH}$$ (A) (i) \(\mathrm{N}_{3} \mathrm{H}, \mathrm{H}_{2} \mathrm{SO}_{4}\) (ii) \(\mathrm{CHCl}_{3}, \mathrm{NaOH}\) (iii) \(\mathrm{H}_{3} \mathrm{O}^{+}\) (B) (i) \(\mathrm{SOCl}_{2}\) (ii) \(\mathrm{CH}_{2} \mathrm{~N}_{2}\) (iii) \(\mathrm{Ag}_{2} \mathrm{O}\) (iv) \(\mathrm{H}^{+}\) (C) (i) \(\mathrm{NH}_{3}, \Delta\) (ii) \(\mathrm{NaOBr}\) (iii) \(\mathrm{NaNO}_{2} / \mathrm{HCl}\) (D) None of these

\((\mathrm{P}) \stackrel{\mathrm{Red} \mathrm{P} / \mathrm{X}_{2}}{\longrightarrow}(\mathrm{Q})\) (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.

The correct statement(s) regarding final major product ' \(\mathrm{R}\) ' of the following reaction sequence is/are: (A) Can give yellow ppt. with \(\mathrm{NaOI}\) (B) Cannot reduce \(\mathrm{H}_{5} \mathrm{IO}_{6}\) (C) Exist in enol form (D) Michael reaction is used for its formation

\(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} / \mathrm{H}^{+}, \Delta\) 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

Sodium salt of which carboxylic acid on electrolysis gives organic product which can show steroisomerism?

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