JEE Carboxylic Acids PYQs With Video Solutions, Download PDF

JEE Carboxylic Acids PYQ

Carboxylic Acids is an important chapter in JEE Organic Chemistry. In this chapter, you study compounds that contain the carboxyl group, written as –COOH.

The chapter covers the preparation, properties, acidity, and reactions of carboxylic acids. It is also connected with alcohols, aldehydes, ketones, amines, and hydrocarbons.

Most questions from this chapter are based on acidity order, named reactions, preparation methods, conversions, and reaction products. Some questions may look easy, but a small change in the reagent or reaction condition can change the final answer.

Solving previous-year questions is one of the best ways to prepare this chapter. It helps you understand which topics are asked regularly and which reactions need more practice.

On this page, you can revise the main concepts, download the PYQ PDF, learn important formulas and reactions, avoid common mistakes, and practise chapter-wise questions.

With regular practice, Carboxylic Acids can become a simple and scoring topic in JEE Chemistry.

JEE Carboxylic Acids Previous-Year Questions

Previous-year questions help you understand the most important parts of the chapter. Questions are often asked from acidity, esterification, reduction, decarboxylation, the Hell–Volhard–Zelinsky reaction, and preparation methods.

While solving a JEE Main Previous Paper, first identify the functional group, reagent, and reaction condition.

Do not select the product only by looking at the starting compound. The same compound can give different products with different reagents.

For acidity questions, compare the stability of the conjugate base. A carboxylic acid is stronger when the carboxylate ion formed after losing H⁺ is more stable.

Electron-withdrawing groups increase acidity because they help stabilise the negative charge on the carboxylate ion.

Electron-donating groups usually decrease acidity. Also remember that the inductive effect becomes weaker as the group moves farther away from the –COOH group.

For reaction-based questions, check whether the acid is undergoing reduction, esterification, decarboxylation, or conversion into another functional group.

Try to solve every question within a fixed time. After completing the set, check your incorrect answers and find the exact reason behind each mistake.

Common reasons may include:

• You used the wrong acidity order.
• You missed the position of a substituent.
• You confused reduction with decarboxylation.
• You forgot an important reaction condition.
• You selected an intermediate instead of the final product.

Making a simple reaction chart can help you connect carboxylic acids with esters, alcohols, amides, acid chlorides, and alkanes.

JEE Carboxylic Acids PYQs PDF

You can download the JEE Carboxylic Acids PYQs PDF from the section below.

First, solve the PDF without checking the answers. Try to complete it in one sitting so that you can understand your current preparation level.

After solving the questions, divide them into three groups:

• Questions you solved easily
• Questions that took more time
• Questions you could not solve

Revise the weak topics and try the difficult questions again after a few days.

While practising these JEE Questions, do not only check the correct option. Read the full solution and understand why the other options are incorrect.

Pay close attention to the reagent and final work-up conditions.

For example, LiAlH₄ reduces a carboxylic acid to a primary alcohol. Soda lime, on the other hand, removes the carboxyl group and gives an alkane with one carbon atom less.

You can keep this PDF with your revision notes and use it before school tests, coaching tests, or the final exam.

When you solve the PDF again, focus mainly on the questions that were wrong or took too much time.

It is also helpful to practise conversions in both directions because JEE may ask the same reaction using a different starting compound.

Important Formulas for JEE Carboxylic Acids PYQs

Carboxylic Acids does not have many numerical formulas. However, it includes several important equations, reaction conditions, and acidity rules.

You can add the following points to your JEE Mains Formula notes for quick revision.

Do not memorise these reactions without understanding what is happening.

For example, soda-lime decarboxylation removes one carbon atom from the molecule. Esterification does not reduce the carbon chain because both the acid and alcohol become part of the ester.

In the HVZ reaction, the carboxylic acid must have at least one alpha-hydrogen.

Before applying the reaction, always check the carbon atom next to the –COOH group.

Common Mistakes to Avoid in JEE Carboxylic Acids PYQs

One common mistake is comparing acidity only by looking at the size of the compound.

The correct method is to compare the stability of the carboxylate ion.

Electron-withdrawing groups increase acidity because they stabilise the negative charge. Electron-donating groups usually decrease acidity.

Students also forget that the inductive effect decreases with distance.

For example, a chlorine atom near the –COOH group has a stronger effect than a chlorine atom placed farther away.

Another common mistake is using the HVZ reaction for every carboxylic acid. This reaction only works when an alpha-hydrogen is present.

While solving JEE Carboxylic Acids Questions, always count the carbon atoms before and after decarboxylation.

In soda-lime decarboxylation, the alkane formed has one carbon atom less than the original acid salt.

Do not confuse esterification with neutralisation.

Esterification gives an ester and water. Neutralisation gives a salt and water.

Also remember that carboxylic acids react with sodium bicarbonate and release carbon dioxide gas. This test can help distinguish carboxylic acids from phenols.

Read the complete reagent and reaction condition before choosing your answer. A missed condition can easily lead to the wrong product.

List of JEE Carboxylic Acids PYQs

Use the questions given below as a short chapter test.

Try to solve them like a JEE Main Mock Test. Set a time limit, avoid using notes, and check the answers only after completing the full set.

Question 1

The spin-only magnetic moment value of the ion among $$Ti^{2+}, V^{2+}, Co^{3+}$$ and $$Cr^{2+}$$, that acts as strong oxidising agent in aqueous solution is ______ BM (Near integer). (Given atomic numbers : $$Ti : 22, V : 23, Cr : 24, Co : 27$$)

Show Answer Explanation Video Solution

correct answer:- 5

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Question 2

The sodium salt of an organic acid X produces effervescence with concentrated H$$_{2}$$SO$$_{4}$$. X reacts with the acidified aqueous CaCl$$_{2}$$ solution to give a white precipitate which decolourises acidic solution of KMnO$$_{4}$$. X is

A

HCOONa

B

CH$$_{3}$$COONa

C

Na$$_{2}$$C$$_{2}$$O$$_{4}$$

D

C$$_{6}$$H$$_{5}$$COONa

Show Answer Explanation

correct answer:- 3

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Question 3

The products A and B formed in the following reaction scheme are respectively:

A

B

C

D

Show Answer Explanation Video Solution

correct answer:- 3

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Question 4

The correct order of acid character of the following compounds is:

A

IV > III > II > I

B

II > III > IV > I

C

III > II > I > IV

D

I > II > III > IV

Show Answer Explanation Video Solution

correct answer:- 2

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Question 5

Maleic anhydride
Maleic anhydride can be prepared by:

A

Heating trans-but-2-enedioic acid

B

Heating cis-but-2-enedioic acid

C

Treating cis-but-2-enedioic acid with alcohol and acid

D

Treating trans-but-2-enedioic acid with alcohol and acid

Show Answer Explanation Video Solution

correct answer:- 2

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Question 6

The number of elements from the following that do not belong to lanthanoids is $$Eu, Cm, Er, Tb, Yb$$ and $$Lu$$

A

3

B

4

C

1

D

5

Show Answer Explanation Video Solution

correct answer:- 3

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Question 7

Consider the following molecules and statements related to them:

(a) (B) is more likely to be crystalline than (A)
(b) (B) has higher boiling point than (A)
(c) (B) dissolves more readily than (A) in water
Identify the correct option from below:

A

(a) and (b) are true

B

(a) and (c) are true

C

only (a) is true

D

(b) and (c) are true

Show Answer Explanation Video Solution

correct answer:- 4

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Question 8

A cyclic anhydride reacts with H$$_2$$O to give 'A' (major product).

'A' formed in the above reaction is

A

B

C

D

Show Answer Explanation Video Solution

correct answer:- 4

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Question 9

An organic compound [A], molecular formula $$C_{10}H_{20}O_2$$ was hydrolyzed with dilute sulphuric acid to give a carboxylic acid [B] and an alcohol [C]. Oxidation of [C] with $$CrO_3 - H_2SO_4$$ produced [B]. Which of the following structures are not possible for [A]?

P)

Q)

R) $$CH_3CH_2CH_2COOCH_2CH_2CH_2CH_3$$

S) $$(CH_3)_3 - C - COOCH_2C(CH_3)_3$$

A

P and Q

B

Q and R

C

P and R

D

Q and S

Show Answer Explanation Video Solution

correct answer:- 3

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Question 10

The number of compound/s given below which contain/s $$-COOH$$ group is ______. (Integer answer)
(A) Sulphanilic acid (B) Picric acid (C) Aspirin (D) Ascorbic acid

Show Answer Explanation

correct answer:- 1

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Question 11

Arrange the following elements in the increasing order of number of unpaired electrons in it. (A) Sc (B) Cr (C) V (D) Ti (E) Mn. Choose the correct answer from the options given below :

A

$$(A) < (D) < (C) < (B) < (E)$$

B

$$(C) < (E) < (B) < (A) < (D)$$

C

$$(A) < (D) < (C) < (E) < (B)$$

D

$$(B) < (C) < (D) < (E) < (A)$$

Show Answer Explanation Video Solution

correct answer:- 3

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Question 12

A

B

C

D

Show Answer Explanation

Solution

correct answer:- 1

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Question 13

Compound(s) which will liberate carbon dioxide with sodium bicarbonate solution is/are:

A

A and B only

B

B and C only

C

C only

D

B only

Show Answer Explanation Video Solution

correct answer:- 2

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Question 14

The major product of the following reaction is:
CH$$_3$$CH = CHCO$$_2$$CH$$_3$$ $$\xrightarrow{\text{LiAlH}_4}$$

A

CH$$_3$$CH$$_2$$CH$$_2$$CO$$_2$$CH$$_3$$

B

CH$$_3$$CH = CHCH$$_2$$OH

C

CH$$_3$$CH$$_2$$CH$$_2$$CH$$_2$$OH

D

CH$$_3$$CH$$_2$$CH$$_2$$CHO

Show Answer Explanation Video Solution

correct answer:- 2

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Question 15

The sum of total number of carbonyl groups $$(>C=O)$$ present in the major products $$\mathbf{X}$$ and $$\mathbf{Y}$$ in the following reactions is ___.

$$\mathbf{X}$$ is the major product from heating a 2-methyl-substituted malonic-type substrate having two $$-CO_2H$$ groups on the central carbon which also bears two $$H_3C-C(=O)-$$ (acetyl) ketone arms (i.e. $$3$$-methyl-$$3$$-(carboxy/diacid) pentane-$$2,4$$-dione type) on heating.
$$\mathbf{Y}$$ is the major product from heating a cyclopentane ring bearing a ketone $$(O=)$$ and two adjacent $$-CO_2H$$ groups on heating.

Show Answer

correct answer:- 4.00

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Question 16

The sum of total number of carbonyl groups $$(>C=O)$$ present in the major products $$\mathbf{X}$$ and $$\mathbf{Y}$$ in the following reactions is ___.

$$\mathbf{X}$$ is the major product from heating a 2-methyl-substituted malonic-type substrate having two $$-CO_2H$$ groups on the central carbon which also bears two $$H_3C-C(=O)-$$ (acetyl) ketone arms (i.e. $$3$$-methyl-$$3$$-(carboxy/diacid) pentane-$$2,4$$-dione type) on heating.
$$\mathbf{Y}$$ is the major product from heating a cyclopentane ring bearing a ketone $$(O=)$$ and two adjacent $$-CO_2H$$ groups on heating.

Show Answer

correct answer:- 4.00

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Question 17

The sum of total number of carbonyl groups $$(>C=O)$$ present in the major products $$\mathbf{X}$$ and $$\mathbf{Y}$$ in the following reactions is ___.

Show Answer Explanation

correct answer:- 4.00

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Question 18

Consider the following sequence of reactions:

1. Benzoic acid reacts with $$Br_2/FeBr_3$$ to give product $$[A]$$.
2. Product $$[A]$$ is then treated with soda lime $$(NaOH/CaO,\ \Delta)$$ to produce product $$[B]$$.
3. Product $$[B]$$ is subjected to nitration using $$conc.\ HNO_3/conc.\ H_2SO_4$$ to give the final product $$[C]$$.

Identify the final product $$[C]$$.

A

$$m$$-Bromonitrobenzene

B

$$o$$-Bromonitrobenzene

C

$$p$$-Bromonitrobenzene

D

$$2,4$$-Dibromonitrobenzene

Show Answer Explanation

Solution

Benzoic acid contains the $$-COOH$$ group, which is a deactivating and meta-directing group.

Therefore, bromination of benzoic acid occurs predominantly at the meta position:

$$C_6H_5COOH \xrightarrow{Br_2/FeBr_3} m\text{-}BrC_6H_4COOH$$

Hence,

$$[A]=m\text{-bromobenzoic acid}$$

On treatment with soda lime, decarboxylation occurs. The $$-COOH$$ group is removed and replaced by a hydrogen atom:

$$m\text{-}BrC_6H_4COOH \xrightarrow{NaOH/CaO,\ \Delta} C_6H_5Br+CO_2$$

Thus,

$$[B]=\text{Bromobenzene}$$

Bromine is a deactivating but ortho/para-directing group due to resonance donation.

Therefore, nitration of bromobenzene gives mainly ortho- and para-bromonitrobenzene, with the para product being the major product because of lower steric hindrance:

$$C_6H_5Br \xrightarrow{conc.\ HNO_3/conc.\ H_2SO_4} p\text{-bromonitrobenzene\ (major)}$$

Hence, the final major product $$[C]$$ is:

$$\boxed{p\text{-Bromonitrobenzene}}$$

Therefore, the correct answer is:

$$\boxed{\text{(c) }p\text{-Bromonitrobenzene}}$$

correct answer:- 3

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Instruction for set :

Question 19

Identify the major organic product Compound Y?

A

$$\text{C}_6\text{H}_5-\text{CH}_2-\text{CH}_2-\text{CH}_2\text{OH}$$

B

$$\text{C}_6\text{H}_5-\text{CH}=\text{CH}-\text{CHO}$$

C

$$\text{C}_6\text{H}_5-\text{CH}_2-\text{CH}_2-\text{CHO}$$

D

$$\text{C}_6\text{H}_5-\text{CH}_2-\text{CH}_2-\text{COOH}$$

Show Answer Explanation

Solution

Let us analyze the unique chemical behaviors and exceptions associated with the reagents used in this multi-step sequence:

Step 1: Reduction of Cinnamic Acid with $$\text{LiAlH}_4$$ (The Conjugation Exception)
The starting material is cinnamic acid ($$\text{C}_6\text{H}_5-\text{CH}=\text{CH}-\text{COOH}$$). Generally, Lithium Aluminium Hydride ($$\text{LiAlH}_4$$) is chemoselective and does not reduce regular, isolated carbon-carbon double bonds ($$\text{C}=\text{C}$$). However, there is a classic exception to this rule:

• When a $$\text{C}=\text{C}$$ double bond is in direct conjugation with both a phenyl ring on one side and a strongly polarizing carbonyl group on the other side, the olefinic bond becomes electronically activated.
• The extended resonance conjugation allows the hydride ion ($$\text{H}^-$$) to attack via a conjugate 1,4-addition mechanism, reducing the alkene along with the carboxylic acid group.

Consequently, both the double bond and the carboxylic acid functional group are completely reduced to form a saturated primary alcohol, 3-phenylpropan-1-ol:

$$\text{Compound X} = \text{C}_6\text{H}_5-\text{CH}_2-\text{CH}_2-\text{CH}_2\text{OH}$$

Step 2: Controlled Oxidation with Pyridinium Chlorochromate ($$\text{PCC}$$)
Compound X is treated with pyridinium chlorochromate ($$\text{PCC}$$) dissolved in anhydrous dichloromethane ($$\text{CH}_2\text{Cl}_2$$):

• $$\text{PCC}$$ is a mild and selective oxidizing agent.
• It oxidizes primary alcohols cleanly into aldehydes, stopping the oxidation precisely at that stage and preventing further oxidation into a carboxylic acid because there is no water present to form an aldehyde hydrate intermediate.

$$\text{C}_6\text{H}_5-\text{CH}_2-\text{CH}_2-\text{CH}_2\text{OH} \xrightarrow{\text{PCC}} \text{C}_6\text{H}_5-\text{CH}_2-\text{CH}_2-\text{CHO}$$

The structural layout of the product obtained is 3-phenylpropanal.

Conclusion:
Due to the specific reduction of the activated double bond by $$\text{LiAlH}_4$$, followed by selective oxidation with $$\text{PCC}$$, the final product is 3-phenylpropanal. This matches Option C perfectly.

Answer:
The correct choice is Option C .

correct answer:- 3

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Question 20

The molecular formula of second homologue in the homologous series of mono carboxylic acids is :

A

$$C_3H_6O_2$$

B

$$C_2H_4O_2$$

C

$$CH_2O$$

D

$$C_2H_2O_2$$

Show Answer Explanation Video Solution

correct answer:- 2

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Question 21

Number of colourless lanthanoid ions among the following is ________
$$Eu^{3+}, Lu^{3+}, Nd^{3+}, La^{3+}, Sm^{3+}$$

Show Answer Explanation Video Solution

correct answer:- 2

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Question 22

Match the dicarboxylic acids in List I with their respective major products formed upon strong heating $$(\Delta)$$ in List II.

**List I (Dicarboxylic Acid)**

| Column I | Compound |
| -------- | ------------- |
| P | Malonic acid |
| Q | Succinic acid |
| R | Adipic acid |
| S | Phthalic acid |

**List II (Major Product upon Heating)**

| Column II | Product |
| --------- | ------------------------------------ |
| 1 | Cyclopentanone + $$CO_2$$ + $$H_2O$$ |
| 2 | Acetic acid + $$CO_2$$ |
| 3 | Phthalic anhydride + $$H_2O$$ |
| 4 | Succinic anhydride + $$H_2O$$ |

Choose the correct answer from the options given below.

A

P-2, Q-4, R-1, S-3

B

P-2, Q-1, R-4, S-3

C

P-4, Q-2, R-1, S-3

D

P-1, Q-4, R-2, S-3

Show Answer Explanation

Solution

On heating, dicarboxylic acids undergo different reactions depending on the number of carbon atoms separating the two carboxyl groups.

**(P) Malonic acid**

Malonic acid has the structure:

$$HOOC-CH_2-COOH$$

It readily undergoes decarboxylation on heating to give acetic acid.

$$HOOC-CH_2-COOH \xrightarrow{\Delta} CH_3COOH + CO_2$$

Hence,

$$P \rightarrow 2$$

**(Q) Succinic acid**

Succinic acid has the structure:

$$HOOC-(CH_2)_2-COOH$$

On heating, it undergoes dehydration to form a stable five-membered cyclic anhydride.

$$\text{Succinic acid} \xrightarrow{\Delta} \text{Succinic anhydride} + H_2O$$

Hence,

$$Q \rightarrow 4$$

**(R) Adipic acid**

Adipic acid has the structure:

$$HOOC-(CH_2)_4-COOH$$

On strong heating, it undergoes decarboxylation and dehydration to form cyclopentanone.

$$\text{Adipic acid} \xrightarrow{\Delta} \text{Cyclopentanone} + CO_2 + H_2O$$

Hence,

$$R \rightarrow 1$$

**(S) Phthalic acid**

Phthalic acid is benzene-1,2-dicarboxylic acid. On heating, it readily forms phthalic anhydride.

$$\text{Phthalic acid} \xrightarrow{\Delta} \text{Phthalic anhydride} + H_2O$$

Hence,

$$S \rightarrow 3$$

Therefore, the correct matching is:

$$P-2,\ Q-4,\ R-1,\ S-3$$

Hence, the correct option is (a).

correct answer:- 1

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Question 23

Aspirin is known as:

A

Acetyl salicylic acid

B

Phenyl salicylate

C

Acetyl salicylate

D

Methyl salicylic acid

Show Answer Explanation

correct answer:- 1

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Question 24

The correct order of their reactivity towards hydrolysis at room temperature is:

A

(A) > (B) > (C) > (D)

B

(D) > (A) > (B) > (C)

C

(D) > (B) > (A) > (C)

D

(A) > (C) > (B) > (D)

Show Answer Explanation

correct answer:- 1

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Question 25

On mixing ethyl acetate with aqueous sodium chloride, the composition of the resultant solution is

A

$$CH_3 COOC_2 H_5 + NaCl$$

B

$$CH_3 Cl + C_2 H_5 COONa$$

C

$$CH_3 COCl + C_2 H_5 OH + NaOH$$

D

$$CH_3 COONa + C_2 H_5 OH$$

Show Answer Explanation

correct answer:- 1

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Question 26

The product "A" in the above reaction is:

A

B

C

D

Show Answer Explanation

correct answer:- 2

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Question 27

Identify A in the following reaction.

A

B

C

D

Show Answer Explanation Video Solution

correct answer:- 4

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Question 28

Acid D formed in above reaction is:

A

Gluconic acid

B

Succinic acid

C

Oxalic acid

D

Malonic acid

Show Answer Explanation Video Solution

correct answer:- 2

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Question 29

Number of compounds from the following which will not dissolve in cold NaHCO$$_3$$ and NaOH solutions but will dissolve in hot NaOH solution is ______.

Show Answer Explanation

correct answer:- 3

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Question 30

A liquid was mixed with ethanol and a drop of concentrated $$H_2SO_4$$ was added. A compound with a fruity smell was formed. The liquid was:

A

$$CH_3OH$$

B

$$HCHO$$

C

$$CH_3COCH_3$$

D

$$CH_3COOH$$

Show Answer Explanation Video Solution

correct answer:- 4