A major component of Borsch reagent is obtained by reacting hydrazine hydrate with which of the following ?
1
2
3
4
Official Solution
Correct Option: (3)
The major component of Borsch reagent is 2,4- dinitrophenyl hydrazine which can be obtained by reaction of 2,4-dinitrochloro benzene and hydrazine
02
PYQ 2013
medium
chemistryID: jee-main
A solution of phenylethane in toluene racemises slowly in the presence of a small amount of , due to the formation of
1
carbanion
2
carbene
3
carbocation
4
free radical
Official Solution
Correct Option: (3)
(c)
03
PYQ 2013
medium
chemistryID: jee-main
The Wurtz-Fittig reaction involves condensation of:
1
two molecules of aryl halides
2
one molecule of each of aryl-halide and alkyl-halide
3
one molecule of each of aryl-halide and phenol
4
two molecules of aralkyl-halides
Official Solution
Correct Option: (2)
The Wurtz-Fittig reaction is a coupling reaction that involves the condensation of different types of halides. Let's go through the options and determine which one correctly describes this reaction:
Two molecules of aryl halides: The Wurtz reaction typically involves coupling two alkyl halides, not aryl halides. Therefore, this option is incorrect for the Wurtz-Fittig reaction, which specifically includes both an aryl and an alkyl component.
One molecule of each of aryl-halide and alkyl-halide: This option describes the Wurtz-Fittig reaction correctly. In this reaction, an aryl halide and an alkyl halide are condensed in the presence of sodium to form an aryl-alkyl compound.
One molecule of each of aryl-halide and phenol: Phenols are not involved in the Wurtz-Fittig reaction, as it only involves halides. Hence, this option is incorrect.
Two molecules of aralkyl-halides: This is not the mechanism of the Wurtz-Fittig reaction. The Wurtz reaction, when involving aralkyl halides, differs and does not specifically refer to the Wurtz-Fittig reaction.
In conclusion, the correct answer is one molecule of each of aryl-halide and alkyl-halide. The Wurtz-Fittig reaction takes advantage of the reactivity of both types of halides to form a new carbon-carbon bond in the resulting compound. This reaction is particularly useful in organic synthesis where mixed alkyl-aryl products are needed.
04
PYQ 2014
medium
chemistryID: jee-main
In S 2 reactions, the correct order of reactivity for the following compounds and (CH is
1
2
3
4
Official Solution
Correct Option: (2)
Rate of reaction depends on steric crowding of alkyl halide. So order is
05
PYQ 2014
medium
chemistryID: jee-main
For the compounds and the correct order of increasing -halogen bond length is :
1
2
3
4
Official Solution
Correct Option: (3)
As the radius of halogen increases ?halogen bond length increased.
06
PYQ 2017
medium
chemistryID: jee-main
-Methylpent- -ene on reaction with in presence of peroxide forms an addition product. The number of possible stereoisomers for the product is :
1
Two
2
Four
3
Six
4
Zero
Official Solution
Correct Option: (2)
Since product (X) contains two chiral centres and it is unsymmetrical.
So, its total stereoisomers
07
PYQ 2021
medium
chemistryID: jee-main
Among the following compounds I-IV, which one forms a yellow precipitate on reacting sequentially with (i) NaOH (ii) dil. (iii) ?
1
I
2
II
3
III
4
IV
Official Solution
Correct Option: (4)
Step 1: Understanding the Concept: This sequence of reagents is a test for halide ions in organic compounds. NaOH induces nucleophilic substitution or elimination to release halide ions ( ). Addition of dil. neutralizes the base, and reacts with to form a silver halide precipitate. Step 2: Detailed Explanation: 1. Analysis of Compounds: Compound I and II are aryl halides (chlorobenzenes). Chlorines attached directly to the benzene ring do not undergo nucleophilic substitution easily due to partial double bond character. Compound III is an aryl bromide. Similarly, the Br is inert to NaOH at room temperature. Compound IV contains an iodo-methyl group (benzyl-type iodide) and a chloro group on the ring. The group is highly reactive towards nucleophilic substitution. 2. Reaction of IV: Reacting IV with NaOH releases iodide ions ( ).
3. Precipitation: After neutralizing with , the addition of results in: (Yellow precipitate) Silver chloride ( ) is white, and silver bromide ( ) is pale yellow, but silver iodide ( ) is distinctively yellow. Step 3: Final Answer: Compound IV forms the yellow precipitate.
08
PYQ 2021
medium
chemistryID: jee-main
The major product of the following reaction is :
1
A
2
B
3
C
4
D
Official Solution
Correct Option: (2)
Step 1: Understanding the Concept: The reaction involves a secondary alkyl halide (1-chloro-2-methylcyclopentane) reacting with a strong base (sodium hydroxide, NaOH) in a protic solvent (ethanol, ). Under these conditions, the (bimolecular elimination) pathway is favored over substitution. Step 2: Key Formula or Approach: 1. Reaction Type: Elimination. 2. Regioselectivity Rule: Zaitsev's Rule (the most substituted alkene is the major product). Step 3: Detailed Explanation: In the given cyclic structure, the chlorine atom is at the position and a methyl group is at the position. During the mechanism, the base (hydroxide ion) abstracts a proton from a -carbon relative to the chlorine. There are two types of -hydrogens available: 1. At : Abstraction of the tertiary hydrogen at leads to the formation of a double bond between and . This results in 1-methylcyclopentene, which is a trisubstituted alkene. 2. At : Abstraction of a secondary hydrogen at leads to the formation of a double bond between and . This results in 3-methylcyclopentene (or 2-methylcyclopentene depending on numbering), which is a disubstituted alkene. According to Zaitsev's rule, the trisubstituted alkene is thermodynamically more stable and therefore forms as the major product. Step 4: Final Answer: The major product is 1-methylcyclopentene, which corresponds to option (B).
09
PYQ 2021
medium
chemistryID: jee-main
In the following sequence of reactions the P is:
1
A
2
B
3
C
4
D
Official Solution
Correct Option: (2)
Step 1: Understanding the Question:
We are given a two-step reaction sequence starting with cyclopentyl chloride and need to identify the major organic product P. Step 2: Key Formula or Approach:
The reaction sequence involves: Formation of a Grignard reagent: An alkyl halide reacts with magnesium metal in dry ether. R-Cl + Mg R-MgCl. Reaction of Grignard reagent with a protic solvent: Grignard reagents are very strong bases and will react with any compound that has an acidic proton (like water, alcohols, carboxylic acids) in an acid-base reaction.
Step 3: Detailed Explanation: Step 1: Formation of A
Cyclopentyl chloride reacts with magnesium (Mg) in dry ether to form the Grignard reagent, cyclopentylmagnesium chloride.
The structure [A] contains a highly basic cyclopentyl carbanion. Step 2: Formation of P
The Grignard reagent [A] is then reacted with ethanol (CH CH OH). Ethanol has an acidic proton on the oxygen atom. The Grignard reagent acts as a strong base and abstracts this proton.
The cyclopentyl anion (C H ) takes the proton (H ) from ethanol to form the alkane, cyclopentane (C H ). Step 4: Final Answer:
The major organic product P is cyclopentane. Option (B) shows the structure of cyclopentane.
10
PYQ 2021
medium
chemistryID: jee-main
The correct order of reactivity of the given chlorides with acetate in acetic acid is :
1
A
2
B
3
C
4
D
Official Solution
Correct Option: (3)
Step 1: Understanding the Concept: The reaction of chlorides with acetate in acetic acid is a solvolysis-type reaction, which typically follows an mechanism. The rate-determining step is the formation of a carbocation. Therefore, reactivity depends on carbocation stability. Step 2: Detailed Explanation: 1. Structure (2): Removal of gives a tertiary allylic carbocation stabilized by resonance and the inductive effect of the methyl group. This is the most stable intermediate. 2. Structure (3): Removal of gives a secondary allylic carbocation. It is resonance-stabilized but less so than the tertiary allylic one. 3. Structure (1): This gives an allylic carbocation, but the positioning of the methyl group relative to the double bond in the starting material affects the stability compared to (3). 4. Structure (4): This involves a primary-like environment or less stabilized allylic system compared to others. Reactivity order in : Tertiary allylic Secondary allylic Primary allylic. Step 3: Final Answer: Following the stability of carbocations, the correct order is (2) (3) (1) (4).
11
PYQ 2021
medium
chemistryID: jee-main
Identify A and B in the chemical reaction. p-Nitrotoluene A B
1
A
2
B
3
C
4
D
Official Solution
Correct Option: (1)
Step 1: Reaction of p-nitrotoluene with (side-chain chlorination) results in the substitution of a hydrogen on the methyl group. This forms p-nitrobenzyl chloride (A).
Step 2: Reaction of A with in dry acetone is the Finkelstein reaction. Step 3: The chloride is substituted by iodide to form p-nitrobenzyl iodide (B).
12
PYQ 2021
medium
chemistryID: jee-main
Given below are two statements : Statement I : C₂H₅OH and AgCN both can generate nucleophile. Statement II : KCN and AgCN both will generate nitrile nucleophile with all reaction conditions. Choose the most appropriate option :
1
Both statement I and statement II are true.
2
Both statement I and statement II are false.
3
Statement I is true but statement II is false.
4
Statement I is false but statement II is true.
Official Solution
Correct Option: (3)
Step 1: Evaluate Statement I. acts as a weak nucleophile (via oxygen lone pairs), and provides the cyanide/isocyanide group. Both are sources of nucleophilic species. (Statement I is True).
Step 2: Evaluate Statement II. is ionic and typically produces nitriles ( ). However, is primarily covalent; the lone pair on Nitrogen is more available for attack, typically producing isocyanides ( ). Therefore, they do not both generate nitrile nucleophiles under all conditions. (Statement II is False).
13
PYQ 2021
medium
chemistryID: jee-main
The product formed in the first step of the reaction of with excess is :
1
A
2
B
3
C
4
D
Official Solution
Correct Option: (1)
Step 1: Magnesium reacts with alkyl halides in dry ether to form Grignard reagents ( ).
Step 2: Since there are two bromine atoms and "excess" Magnesium is used, both bromine sites will react.
Step 3: The resulting product will have two groups attached to the carbon chain.
14
PYQ 2021
medium
chemistryID: jee-main
The correct order of the following compounds showing increasing tendency towards nucleophilic substitution reaction is : (Note: The compounds are chloro benzene with different nitro positions: (i) Chlorobenzene, (ii) 4-nitrochlorobenzene, (iii) 2,4-dinitrochlorobenzene, (iv) 2,4,6-trinitrochlorobenzene)
1
2
3
4
Official Solution
Correct Option: (3)
Step 1: Nucleophilic aromatic substitution ( ) is difficult in chlorobenzene because of resonance.
Step 2: Electron-withdrawing groups (EWG) like at ortho and para positions stabilize the carbanion intermediate (Meisenheimer complex).
Step 3: More groups = greater stabilization = faster reaction. Step 4: Thus, Trinitrochlorobenzene (Picryl chloride) is the most reactive, and chlorobenzene is the least.
15
PYQ 2021
medium
chemistryID: jee-main
The major product (A) formed in the reaction given below is :
\includegraphics[width=0.6\linewidth]{image43.png}
Step 1: Understanding the Question:
We are asked to predict the major product of the reaction between 1-bromo-2-phenylbutane and sodium methoxide in methanol. This involves a competition between substitution (S 2) and elimination (E2) reactions. Step 2: Detailed Explanation: 1. Analysis of Reactants:
- Substrate: 1-bromo-2-phenylbutane is a primary (1 ) alkyl halide. Primary halides generally favor S 2 reactions.
- Reagent: Sodium methoxide (CH O Na ) provides the methoxide ion (CH O ), which is a strong nucleophile and also a strong, unhindered base.
- Solvent: Methanol (CH OH) is a polar protic solvent, which can favor both S 1 and E1 for tertiary halides, but for primary halides, it doesn't prevent S 2 or E2. 2. S 2 vs. E2 Competition:
For a primary halide with a strong base/nucleophile, both S 2 and E2 are possible.
- S 2 Path: The methoxide ion attacks the primary carbon bearing the bromine, displacing Br to form an ether. Product: 1-methoxy-2-phenylbutane.
- E2 Path: The methoxide ion acts as a base and removes a proton from the -carbon (the carbon adjacent to the one with the bromine). In this substrate, the -carbon is the one bonded to the phenyl group. 3. Favoring the E2 Pathway:
Although the substrate is primary, several factors strongly favor the E2 elimination in this specific case:
- Acidity of -Proton: The proton on the -carbon is benzylic, meaning it is adjacent to the phenyl ring. Benzylic protons are more acidic than typical alkyl protons, making them easier to remove by a base.
- Stability of the Product: The alkene formed by the E2 elimination is 2-phenylbut-1-ene. In this molecule, the newly formed double bond is conjugated with the aromatic -system of the phenyl ring. This conjugation provides significant stabilization to the product.
The formation of a highly stable, conjugated product is a powerful thermodynamic driving force that makes the E2 pathway the major one, even for a primary halide. 4. Identifying the Product:
The E2 product is 2-phenylbut-1-ene (structure shown in option B). The S 2 product is 1-methoxy-2-phenylbutane (structure shown in option A). Due to the formation of a conjugated system, the E2 product is the major product. Step 3: Final Answer:
The major product (A) is 2-phenylbut-1-ene.
16
PYQ 2022
easy
chemistryID: jee-main
The IUPAC name of ethylidene chloride is:
1
1-Chloroethene
2
1-Chloroethyne
3
1, 2-Dichloroethane
4
1, 1-Dichloroethane
Official Solution
Correct Option: (4)
Ethylidene chloride is , its IUPAC name is 1,1-Dichloromethane.
So, the correct option is (D).
17
PYQ 2022
medium
chemistryID: jee-main
The major product of the following reaction contains ______ bromine atom(s).
Fig
Official Solution
Correct Option: (1)
The correct answer is 1
Fig.
18
PYQ 2022
hard
chemistryID: jee-main
The interhalogen compound formed from the reaction of bromine with excess of fluorine is a :
1
hypohalite
2
halate
3
perhalate
4
halite
Official Solution
Correct Option: (2)
So, the correct option is (B) halate.
19
PYQ 2022
medium
chemistryID: jee-main
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Fluorine forms one oxoacid. Reason R: Fluorine has smallest size amongst all halogens and is highly electronegative. In the light of the above statements, choose the most appropriate answer from the option given below.
1
Both A and R are correct and R is the correct explanation of A.
2
Both A and R are correct but R is NOT the correct explanation of A.
3
A is correct but R is not correct.
4
A is not correct but R is correct.
Official Solution
Correct Option: (1)
Due to smaller size, fluorine forms only one oxoacid. Both the Assertion and Reason are correct and Reason is the correct explanation.
20
PYQ 2022
medium
chemistryID: jee-main
Which one of the following compounds is inactive towards SN1 reaction?
1
2
3
4
Official Solution
Correct Option: (3)
The carbocation fromed is very unstable. Therefore, it is inactive towards SN1 as halogen is attached to bridge head carbon atom, where formation of carbocation is not possible.
So, the correct answer is (C).
21
PYQ 2022
easy
chemistryID: jee-main
‘A’ and ‘B’ respectively are:
1
2
3
4
Official Solution
Correct Option: (4)
Solution :
22
PYQ 2023
easy
chemistryID: jee-main
The correct order of melting points of dichlorobenzenes is
1
2
3
4
Official Solution
Correct Option: (4)
So, the correct answer is (D)
23
PYQ 2023
medium
chemistryID: jee-main
The total number of monobromo derivatives formed by the alkanes with molecular formula is (excluding stereo isomers) _______
Official Solution
Correct Option: (1)
The correct answer is 8. The Alkanes and their mono bromo dervative are
24
PYQ 2023
easy
chemistryID: jee-main
Which of the following statements is correct about freons?
1
All radicals are known as freons.
2
Freons cause skin cancer.
3
Freons are chlorofluoro carbon.
4
Freons are used in sunscreen lotion.
Official Solution
Correct Option: (3)
Freons are chlorofluorocarbon (CFC) compounds, primarily used as refrigerants and aerosol propellants. These compounds contain chlorine, fluorine, and carbon.
Option 3: True. Freons are CFC compounds.
Option 2: False. While freons contribute to ozone layer depletion, which indirectly increases UV radiation exposure (a cause of skin cancer), freons themselves are not chemicals directly causing skin cancer.
Option 4: False. Freons are not radicals but stable compounds.
Option 1: False. Not all radicals are freons. Conclusion: The only correct option is (3).
25
PYQ 2023
easy
chemistryID: jee-main
CH3CH2-Br CH3-CH2-I+NaBr Which of the following statement is correct ?
1
Acetic acid solvent can take in above reaction.
2
NaI is soluble in acetone but NaBr is precipitate in acetone
3
NaI is precipitated in acetone but NaBr is soluble in acetone
4
When acetone is taken in solvent transition state is highly polar
Official Solution
Correct Option: (2)
CH3CH2-Br + NaI (in acetone) → CH3-CH2-I + NaBr
Correct Statement:
NaI is soluble in acetone, but NaBr precipitates in acetone.
Explanation:
The reaction occurs via the Finkelstein reaction, which is used to prepare alkyl iodides from alkyl bromides or chlorides.
Acetone is a polar aprotic solvent that helps dissolve NaI and facilitates the reaction.
When NaI reacts with CH3CH2-Br:
NaI is soluble in acetone.
NaBr, the byproduct, is not soluble in acetone and precipitates out.
The precipitation of NaBr drives the equilibrium of the reaction toward the formation of the product, CH3-CH2-I.
Conclusion:
The correct statement is: NaI is soluble in acetone, but NaBr precipitates in acetone.
26
PYQ 2023
easy
chemistryID: jee-main
Match List I with List II
List I
List II
A
I
Fittig reaction
B
II
Wurtz Fittig reaction
C
III
Finkelstein reaction
D
IV
Sandmeyer reaction
Choose the correct answer from the options given below:
1
A - III, B - II, C - IV, D - I
2
A - II, B - I, C - IV, D - III
3
A - IV, B - II, C - III, D - I
4
A - II, B - I, C - III, D - IV
Official Solution
Correct Option: (2)
kk
So, the correct answer is (B): A - II, B - I, C - IV, D - III
27
PYQ 2023
medium
chemistryID: jee-main
Number of compounds giving (i) red colouration with ceric ammonium nitrate and also (ii) positive iodoform test from the following is_____
Official Solution
Correct Option: (1)
Correct answer is (c) Three compounds meet the criteria for both tests, indicating the presence of specific functional groups reactive to these tests. These compounds contain functional groups that react with ceric ammonium nitrate and the iodoform test:
Ceric ammonium nitrate is used to detect alcohols and phenols, where the reaction forms a color change.
The iodoform test is specific for methyl ketones, where the formation of a yellow precipitate (iodoform) indicates the presence of the functional group.
Thus, the three compounds must contain alcohols, phenols, or methyl ketones to show a positive result for both tests.
28
PYQ 2023
hard
chemistryID: jee-main
Compound undergoes following sequence of reactions to give the Lactone
1
2
3
4
Official Solution
Correct Option: (3)
The correct answer is (C) :
29
PYQ 2023
easy
chemistryID: jee-main
A compound A on reaction with X and Y produces the same major product but different by product a and b Oxidation of a gives a substance produced by ants X and Y respectively are :
Official Solution
Correct Option: (1)
30
PYQ 2023
medium
chemistryID: jee-main
Compound from the following that will not produce precipitate on reaction with Compound from the following that will not produce precipitate on reaction with AgNO3 is:
1
2
3
4
Official Solution
Correct Option: (4)
The reaction with is used to test for halide ions. A precipitate of (pale yellow) will form if a bromide ion is available to react with .
Analysis of the Options:
Option 1: Cyclopropyl bromide
The bond in cyclopropyl bromide is not easily broken.
The carbocation formed upon the removal of bromide is stabilized by aromatic nature.
Hence, it does not readily form a precipitate with .
Option 2: Bromobenzene
The bond in aryl halides is strong due to resonance and the hybridization of carbon.
This bond does not readily break to release bromide ions.
The carbocation is not formed as it is highly unstable.
Thus, bromobenzene does not react with to form a precipitate.
Option 3: Benzylic bromide
In benzylic bromide, the benzylic carbocation formed after the removal of bromide ion is stabilized by resonance with the benzene ring.
This makes the bromide ion readily available to react with , forming a precipitate.
Option 4: Allylic bromide
In allylic bromide, the allylic carbocation formed is stabilized by resonance.
This allows the bromide ion to react with , forming a precipitate.
Conclusion:
Final Answer: Bromobenzene will not produce a precipitate with .
31
PYQ 2023
easy
chemistryID: jee-main
Product [X] formed in the above reaction is:
1
2
H3C-CH2-CH=CH2
3
H3C-CH=CH-CH3
4
Official Solution
Correct Option: (4)
Step 1: Reaction with and
The alcohol reacts with and to form the corresponding iodoalkane via a nucleophilic substitution reaction. In this process:
acts as an acid catalyst.
from acts as the nucleophile, replacing the hydroxyl group with an iodine atom.
The reaction is as follows:
Step 2: Formation of the Grignard Reagent
The iodoalkane reacts with magnesium in dry ether to form a Grignard reagent:
Here:
Dry ether is used as a solvent because it stabilizes the Grignard reagent and prevents it from reacting with water.
Step 3: Reaction with
The Grignard reagent reacts with deuterated water ( ) to form the deuterated alkane. The bond is replaced by a bond:
Final Answer:
The product formed is .
32
PYQ 2023
medium
chemistryID: jee-main
Which of the following compounds is an example of Freon?
1
C2Cl2F2
2
C2HF3
3
C2F4
4
C2H2F2
Official Solution
Correct Option: (1)
Freons are chlorofluorocarbons (CFCs) or hydrochlorofluorocarbons (HCFCs), which are hydrocarbons where some or all of the hydrogen atoms have been replaced by chlorine and fluorine atoms.
Key Characteristics of Freons:
They contain carbon (C), chlorine (Cl), and fluorine (F).
Freons are widely used as refrigerants and propellants due to their stability and low toxicity.
Analysis of the Given Options:
Among the given options, only (dichlorodifluoroethylene) contains both chlorine and fluorine along with carbon, making it a Freon.
Final Answer:
is an example of a Freon.
33
PYQ 2023
medium
chemistryID: jee-main
Compound that will give positive Lassaigne’s test for both nitrogen and halogen is
1
CH3NH2.HCl
2
NH2OH .HCl
3
NH4Cl
4
N2H4 . HCl
Official Solution
Correct Option: (1)
1. Lassaigne’s Test: This test detects nitrogen, sulfur, and halogens in organic compounds. - The compound is fused with sodium, converting nitrogen into NaCN and halogen into NaX ( ). 2. Compound Analysis: : Contains nitrogen but no halogen. : Contains both nitrogen (from the group) and halogen ( from ). : Contains nitrogen and halogen but is not an organic compound. : Contains nitrogen and halogen but does not form the required products in Lassaigne’s test. 3. Conclusion: Only gives a positive Lassaigne’s test for both nitrogen and halogen. Final Answer: (2) .
34
PYQ 2024
hard
chemistryID: jee-main
What will be the reactivity order of following compounds towards electrophilic substitution reaction ?
1
1>3>2>4
2
4>1>2>3
3
3>2>1>4
4
4>3>1>2
Official Solution
Correct Option: (1)
The correct option is (A): 1>3>2>4
35
PYQ 2024
easy
chemistryID: jee-main
Which of the following sets of elements can be detected by Lassaigne's Test ?
1
N and S only
2
N, P and S only
3
P and halogens only
4
N, P, S and halogens
Official Solution
Correct Option: (4)
The correct option is (D): and halogens
36
PYQ 2026
medium
chemistryID: jee-main
First and second ionization enthalpies of lithium are and respectively. Energy required to convert lithium (g) into is ________ . (nearest integer) [Molar mass of ]
Official Solution
Correct Option: (1)
This problem involves calculating the total energy required to perform successive ionizations on a specific mass of gaseous lithium. Ionization enthalpy is the energy required to remove an electron from a gaseous atom or ion. For lithium to become , it must lose two electrons sequentially.
First, let's determine the total energy required to convert 1 mole of to . This is the sum of the first ionization enthalpy ( ) and the second ionization enthalpy ( ): $ 3.5\text{ mg} 7\text{ g mol}^{-1} 4\text{ kJ}$. Note: While the question blank ends with 'kJ mol⁻¹', it is asking for the energy for the specified mass, which is typically expressed in kJ. Based on standard numerical patterns, the answer is 4.
37
PYQ 2026
medium
chemistryID: jee-main
Match the LIST-I with LIST-II.Choose the correct answer from the options given below:
1
A-IV, B-II, C-I, D-III
2
A-IV, B-I, C-II, D-III
3
A-III, B-II, C-I, D-IV
4
A-IV, B-I, C-III, D-II
Official Solution
Correct Option: (2)
To solve this matching problem, we need to identify the specific functional groups present in the side chains of the given amino acids and match them with the chemical tests that identify those functional groups.
1. Glutamine (A): The side chain of glutamine contains an amide group ( ). The Hoffman bromamide degradation is a specific reaction for primary amides, where they react with and to form a primary amine with one fewer carbon atom. Thus, A matches with IV.
2. Lysine (B): Lysine has a side chain containing a primary aliphatic amino group ( at the epsilon position). Hinsberg's test (using benzene sulfonyl chloride) is used to distinguish between primary, secondary, and tertiary amines. A primary amine reacts to form a sulfonamide that is soluble in alkali. Thus, B matches with I.
3. Tyrosine (C): Tyrosine contains a phenolic hydroxyl group ( ) in its side chain. Phenols react with neutral ferric chloride ( ) solution to produce characteristic colored complexes (usually violet, blue, or green). Thus, C matches with II.
4. Serine (D): Serine contains an aliphatic alcoholic hydroxyl group ( ) in its side chain. The Ceric ammonium nitrate (CAN) test is a standard test for the detection of the alcohol functional group, resulting in a color change from yellow to red. Thus, D matches with III.
Correct matching: A-IV, B-I, C-II, D-III.
38
PYQ 2026
hard
chemistryID: jee-main
Read the following statements: (A) Q has more on chlorine than P. (B) Q has more dipole moment than P. (C) In Q, C–Cl bond has double bond character. (D) In Q, Cl is attached to sp hybridised carbon but in P, Cl is attached to sp . (E) In Q, C–Cl bond length is more due to repulsion between lone pair on chlorine and electron in aromatic ring.
1
A, B, D, E
2
C, D
3
B, C, D
4
B, C, D, E
Official Solution
Correct Option: (2)
Step 1: Analyzing Statement (A). In Q, the chlorine is attached to a sp hybridised carbon, which is more electronegative than sp hybridised carbon in P. As a result, chlorine in Q has more electron density ( ) than chlorine in P. Therefore, Statement (A) is correct.
Step 2: Analyzing Statement (B). Since Q has more electron density on chlorine due to the sp hybridisation of the carbon, Q will have a higher dipole moment compared to P. This makes Statement (B) correct.
Step 3: Analyzing Statement (C). In Q, the C–Cl bond exhibits partial double bond character due to resonance with the aromatic ring. This leads to an increase in bond strength and decreases the bond length. Hence, Statement (C) is correct.
Step 4: Analyzing Statement (D). In Q, chlorine is attached to a sp hybridised carbon (from the benzene ring), which is more electronegative and forms a stronger bond with chlorine. In P, chlorine is attached to sp hybridised carbon, making it less electronegative. Thus, Statement (D) is correct.
Step 5: Analyzing Statement (E). In Q, due to the resonance and the proximity of chlorine's lone pair and the electron density of the aromatic ring, there is repulsion that increases the C–Cl bond length. This makes Statement (E) incorrect.
Step 6: Conclusion. The correct statements are C and D. Thus, the correct answer is (2).
39
PYQ 2026
medium
chemistryID: jee-main
CH –Br _3 CH OH Correct order of rate of this reaction for given nucleophile:
1
F >PhO >C H O >I
2
C H O >PhO >I >F
3
I >C H O >PhO >F
4
PhO >C H O >F >I
Official Solution
Correct Option: (3)
Step 1: Understanding nucleophilicity. Nucleophilicity refers to the ability of a nucleophile to donate electrons to form a new bond with an electrophile (in this case, CH –Br). The stronger the nucleophile, the faster the nucleophilic substitution reaction will occur.
Step 2: Comparing the nucleophilicity of the given nucleophiles. - I is the most nucleophilic due to its larger size and lower electronegativity, which makes it more willing to donate electrons. - C H O (ethoxide) is a good nucleophile, but not as strong as I , because oxygen is more electronegative, making it less willing to donate electrons. - PhO (phenoxide) is also a strong nucleophile, but the resonance stabilization of the phenoxide ion reduces its nucleophilicity compared to C H O and I . - F is the least nucleophilic because fluorine is highly electronegative, making it reluctant to donate electrons.
Step 3: Conclusion. The correct order of nucleophilicity is I >C H O >PhO >F , which corresponds to option (3).
40
PYQ 2026
medium
chemistryID: jee-main
Compare rate of S
1
P>Q>R>S
2
R>S>P>Q
3
R>S>Q>P
4
S>R>Q>P
Official Solution
Correct Option: (2)
Step 1: Understanding S reactions. In an S reaction, the rate depends on the stability of the carbocation formed during the reaction. A more stable carbocation will lead to a faster reaction.
Step 2: Comparing the structures. - P: The carbocation formed here will be a benzylic carbocation (due to the phenyl group) which is highly stable. - Q: The carbocation formed here will also be a benzylic carbocation, but its stability will be less than P due to the difference in substituents. - R: This structure will form a tertiary carbocation, which is more stable than a benzylic carbocation, making the reaction rate the highest. - S: This structure will form a secondary carbocation, which is less stable than the tertiary carbocation, but more stable than a primary carbocation.
Step 3: Conclusion. Thus, the order of reactivity in the S reaction is R>S>P>Q, corresponding to option (2).
41
PYQ 2026
medium
chemistryID: jee-main
Match the LIST-I with LIST-II: Choose the correct answer from the options given below:
1
A-III, B-IV, C-I, D-II
2
A-III, B-IV, C-II, D-I
3
A-I, B-II, C-IV, D-III
4
A-IV, B-I, C-III, D-II
Official Solution
Correct Option: (2)
Step 1: Identifying Vinyl Chloride. Vinyl chloride has chlorine directly attached to a carbon involved in a double bond. Its structure is: Hence, A III. Step 2: Identifying Benzyl Chloride. In benzyl chloride, chlorine is attached to a carbon next to the benzene ring. The given aromatic structure corresponds to this case. Hence, B IV. Step 3: Identifying Alkyl Chloride. Alkyl chlorides have chlorine attached to a saturated carbon atom. The structure: is an alkyl chloride. Hence, C II. Step 4: Identifying Allyl Chloride. Allyl chloride has chlorine attached to a carbon adjacent to a carbon–carbon double bond. Its structure is: Hence, D I. Step 5: Final conclusion. The correct matching is A-III, B-IV, C-II, D-I, which corresponds to option (2).
42
PYQ 2026
hard
chemistryID: jee-main
Cycloalkene (X) reacts with bromine. During the reaction, 1 mole of cycloalkane consumes 1 mole of Br to form a product (Y). The product (Y) has C:Br ratio of 3:1. The percentage of bromine in product (Y) is:
1
66.11%
2
65.11%
3
76.11%
4
67.11%
Official Solution
Correct Option: (1)
Step 1: Analyzing the reaction. Cycloalkene (X) reacts with bromine (Br ), and 1 mole of cycloalkene consumes 1 mole of Br . The product (Y) has a carbon (C) to bromine (Br) ratio of 3:1. This means for every 3 carbon atoms, 1 bromine atom is present in the product.
Step 2: Determining the molecular composition. Let the molar mass of cycloalkene (X) be , and that of bromine (Br ) be . The molar mass of the product (Y) will be the sum of the molar masses of cycloalkene and bromine, adjusted for the given ratio. The molecular formula of the product (Y) can be expressed as , based on the given C:Br ratio of 3:1.
Step 3: Calculating the percentage of bromine. The molar mass of is: The mass of bromine in the product is 80 g (since 1 mole of Br is 80 g). Therefore, the percentage of bromine in the product (Y) is: Step 4: Conclusion. Thus, the percentage of bromine in product (Y) is 66.11%, which corresponds to option (1).
43
PYQ 2026
medium
chemistryID: jee-main
Correct statement is : (A) when reacted with gives (B) is best reducing agent (C) Methanoic acid gives iodoform test (D) Isopropyl alcohol gives iodoform test (E) gives iodoform test
1
A, B, C only
2
B, D only
3
D, E only
4
B, C, D only
Official Solution
Correct Option: (3)
The question asks which statements related to the iodoform test and halogen chemistry are correct. --- Statement (A): is a strong oxidizing agent and oxidizes iodide ( ). However, the direct product is not under normal conditions. (hypoiodite) is unstable and formed only in situ during iodoform reaction. Hence, Statement (A) is incorrect. --- Statement (B): is best reducing agent contains iodine in +1 oxidation state. It acts as an oxidizing agent, not a reducing agent. Hence, Statement (B) is incorrect. --- Statement (C): Methanoic acid gives iodoform test Iodoform test requires:
Methanoic acid ( ) has neither structure. So it does not give iodoform test. Statement (C) is incorrect. --- Statement (D): Isopropyl alcohol gives iodoform test Isopropyl alcohol:
It contains the required methyl carbinol group . On oxidation, it forms acetone, which gives yellow precipitate of . Hence, Statement (D) is correct. --- Statement (E): gives iodoform test This compound contains the methyl ketone group ( ). Presence of double bond does not affect the iodoform reaction. Hence, Statement (E) is correct. --- Final Conclusion: Correct statements are (D) and (E). Correct option is (3).
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![Product [X] formed in the above reaction is:](https://images.collegedunia.com/public/qa/images/content/2024_03_05/2_17e3c26d1709624662980.jpg)
![Product [X] formed in the above reaction is:](https://images.collegedunia.com/public/qa/images/content/2024_03_05/3_cd890db61709624825638.jpg)
