JEE Main 2026 April 06, Shift 1 Question Paper with Solutions

All 75 questions from the JEE Main 2026 (April 06, Shift 1) shift — Physics (25), Chemistry (25) and Mathematics (25) — with the correct answer and a step-by-step solution for every question.

Physics PYQs 2026 Chemistry PYQs 2026 Mathematics PYQs 2026

Chemistry25 questions

Q51Single correctSome Basic Concepts in Chemistry

An oxide of iron contains

69.9

% iron, its empirical formula, is: (Given : Molar mass of Fe and O are 56 and

16\,\text{g mol}^{-1}

respectively.)

(A)

\displaystyle \text{FeO}

(B)

\displaystyle \text{Fe}_2\text{O}_3

(C)

\displaystyle \text{Fe}_3\text{O}_4

(D)

\displaystyle \text{FeO}_3

Q52Single correctAtomic Structure

If shortest wavelength of hydrogen atom in Lyman series is x, then longest wavelength in Balmer series of

\text{He}^{+}

is:

(A)

\displaystyle \dfrac{9x}{5}

(B)

\displaystyle \dfrac{36x}{5}

(C)

\displaystyle \dfrac{x}{4}

(D)

\displaystyle \dfrac{5x}{9}

Q53Single correctAtomic Structure

Match the LIST-I with LIST-II

List-I (Orbital)	List-II (Radial nodes and nodal plane)
A. 2s	I. 1 Radial node + two nodal planes
B. 3s	II. 1 Radial node + one nodal plane
C. 3p	III. 2 Radial nodes + No nodal plane
D. 4d	IV. 1 Radial node + No nodal plane

(A)

A-IV, B-I, C-III, D-II

(B)

A-IV, B-II, C-III, D-I

(C)

A-III, B-I, C-IV, D-II

(D)

A-IV, B-III, C-II, D-I

Q54Single correctChemical Bonding and Molecular Structure

The pairs among

A = [\text{SO}_3^{2-},\,\text{CO}_3^{2-}],\ B = [\text{O}_2^{2-},\,\text{F}_2],\ C = [\text{CN}^{-},\,\text{CO}],\ D = [\text{NH}_3,\,\text{H}_3\text{O}^{+}]\ \text{and}\ E = [\text{MnO}_4^{2-},\,\text{CrO}_4^{2-}]

that do not have similar Lewis dot structure are

(A)

A, B and E

(B)

A and E

(C)

B, C and D

(D)

C and D

SolutionAnswer: Option 2A and E

Approach:

For each pair, count total valence electrons and decide whether the two species draw to the same Lewis dot pattern (same skeleton, same lone-pair distribution, same bond orders). The pairs that fail this isoelectronic / isostructural test are the answer.

Step 1:Pair A:

\text{SO}_3^{2-}

has

V = 6 + 3(6) + 2 = 26

valence

e^{-}

sp^{3}

S with one lone pair, pyramidal.

\text{CO}_3^{2-}

has

V = 4 + 3(6) + 2 = 24

valence

e^{-}

sp^{2}

C with no lone pair, trigonal planar. Different electron counts and different lone-pair counts on central atom.

\text{SO}_3^{2-}\ (26\,e^{-},\,sp^{3},\,1\,\text{LP}) \ne \text{CO}_3^{2-}\ (24\,e^{-},\,sp^{2},\,0\,\text{LP})

Step 2:Pair B:

\text{O}_2^{2-}

has

V = 2(6) + 2 = 14

valence

e^{-}

; one O

-

O single bond, three lone pairs on each O.

\text{F}_2

has

V = 2(7) = 14

valence

e^{-}

; one F

-

F single bond, three lone pairs on each F. Same Lewis pattern.

\text{O}_2^{2-}\ \equiv\ \text{F}_2\ (\text{14 e}^{-},\,\text{single bond},\,3\,\text{LP each})

Step 3:Pair C:

\text{CN}^{-}

has

V = 4 + 5 + 1 = 10

valence

e^{-}

; C

\equiv

N triple bond with one lone pair on each terminal. CO has

V = 4 + 6 = 10

valence

e^{-}

; C

\equiv

O triple bond with one lone pair on each terminal.

\text{CN}^{-}\ \equiv\ \text{CO}\ (\text{10 e}^{-},\,\text{triple bond},\,1\,\text{LP each})

Step 4:Pair D:

\text{NH}_3

has

V = 5 + 3 = 8

valence

e^{-}

; three N

-

H bonds with one lone pair on N (pyramidal).

\text{H}_3\text{O}^{+}

has

V = 6 + 3 - 1 = 8

valence

e^{-}

; three O

-

H bonds with one lone pair on O (pyramidal).

\text{NH}_3\ \equiv\ \text{H}_3\text{O}^{+}\ (\text{8 e}^{-},\,3\,\text{bonds},\,1\,\text{LP on centre})

Step 5:Pair E:

\text{CrO}_4^{2-}

has Cr in

+6

(

d^{0}

); the dot structure has two Cr

=

O and two Cr

-O^{-}

in equivalent resonance, no d-electron on metal.

\text{MnO}_4^{2-}

has Mn in

+6

formal oxidation state but Mn is one position to the right of Cr, so the species carries one extra electron at the metal centre (

d^{1}

). The dot patterns therefore differ in electron count on Mn.

\text{CrO}_4^{2-}\ (d^{0}) \ne \text{MnO}_4^{2-}\ (d^{1})

Step 6:Collect the dissimilar pairs.

\text{Dissimilar} = \{A,\,E\}

Final answer: A and E

Q55Single correctChemical Thermodynamics

Arrange the following isothermal processes in order of the magnitude of the work

(p-V)

involved between states 1 and 2.
A. Expansion in single stage

w_{A}

.
B. Expansion in multi stages

w_{B}

.
C. Compression in single stage

w_{C}

.
D. Compression in multi stages

w_{D}

.
Choose the correct option.

(A)

\displaystyle |w_{B}| > |w_{A}| > |w_{C}| > |w_{D}|

(B)

\displaystyle |w_{C}| > |w_{D}| > |w_{A}| > |w_{B}|

(C)

\displaystyle |w_{C}| > |w_{D}| > |w_{B}| > |w_{A}|

(D)

\displaystyle |w_{B}| > |w_{A}| > |w_{D}| > |w_{C}|

Q56Single correctSolutions

When 0.25 moles of a non-volatile, non-ionizable solute was dissolved in 1 mole of a solvent the vapour pressure of solution was

x

% of vapour pressure of pure solvent. What is

x

% ?

(A)

\displaystyle 50\%

(B)

\displaystyle 60\%

(C)

\displaystyle 70\%

(D)

\displaystyle 80\%

Q57Single correctEquilibrium

One mole each of He and A(g) are taken in a 10 L closed flask and heated to 400 K to establish the following equilibrium.

A(g) \rightleftharpoons B(g)

K_{c}

for this reaction at 400 K is 4.0. The partial pressures (in atm) of He and B(g) are respectively (at equilibrium) (Assume He, A(g) and B(g) behave as ideal gases) (Given :

R = 0.082\,\text{L atm K}^{-1}\,\text{mol}^{-1}

)

(A)

3.28, 2.624

(B)

2.624, 3.28

(C)

3.28, 0.656

(D)

0.656, 6.56

Q58Single correctRedox Reactions and Electrochemistry

Consider the following data.
Electrolyte |

\Lambda_{m}^{\circ}

(in

\text{S cm}^{2}\,\text{mol}^{-1}

) |

\text{BaCl}_2

x_{1}

\text{H}_2\text{SO}_4

x_{2}

\text{HCl}

x_{3}

\text{BaSO}_4

is sparingly soluble in water. If the conductivity of the saturated

\text{BaSO}_4

solution is

x\,\text{S cm}^{-1}

then the solubility product of

\text{BaSO}_4

can be given as (Here

\Lambda_{m} = \Lambda_{m}^{\circ}

)

(A)

\displaystyle \dfrac{10^{6}x^{2}}{\alpha^{2}\,(x_{1}+x_{2}-2x_{3})^{2}}

(B)

\displaystyle \dfrac{x^{2}}{(x_{1}+x_{2}-2x_{3})^{2}}

(C)

\displaystyle \dfrac{\alpha^{2}\,(x_{1}+x_{2}-2x_{3})^{2}}{10^{6}x^{2}}

(D)

\displaystyle \dfrac{x^{2}}{(x_{1}+x_{2}+2x_{3})^{2}}

SolutionAnswer: Option 1

\dfrac{10^{6}x^{2}}{\alpha^{2}\,(x_{1}+x_{2}-2x_{3})^{2}}

Approach:

Build

\Lambda_{m}^{\circ}

\text{BaSO}_{4}

by combining the three tabulated electrolytes via Kohlrausch's law (cancelling

H^{+}

and C

l^{-}

ion conductivities). Convert the saturated-solution conductivity to molar solubility using

S=1000\,\kappa/(\alpha\,\Lambda_{m}^{\circ})

, then square because BaS

O_{4}

is a 1:1 salt.

Step 1:Combine the three electrolytes so that the ion conductivities of

H^{+}

and C

l^{-}

cancel:

\text{BaCl}_{2} + \text{H}_{2}\text{SO}_{4} - 2\,\text{HCl}

has the net composition

\text{Ba}^{2+}+\text{SO}_{4}^{2-}=\text{BaSO}_{4}

\Lambda_{m}^{\circ}(\text{BaSO}_{4}) = x_{1} + x_{2} - 2x_{3}\ \text{S cm}^{2}\,\text{mol}^{-1}

Step 2:Convert the conductivity

\kappa = x

S c

m^{-1}

into molar solubility. The factor of 1000 arises because

\kappa

uses c

m^{-3}

for volume whereas solubility is in mol

L^{-1}

(1 L = 1000 c

m^{3}

S = \dfrac{1000\,x}{\alpha\,(x_{1}+x_{2}-2x_{3})}\,\text{mol L}^{-1}

Step 3:BaS

O_{4}

dissociates 1:1 into B

a^{2+}

and SO

_{4}^{2-}

, so

K_{sp} = [\text{Ba}^{2+}][\text{SO}_{4}^{2-}] = S \cdot S = S^{2}

K_{sp} = \left[\dfrac{1000\,x}{\alpha\,(x_{1}+x_{2}-2x_{3})}\right]^{2}

Step 4:Expand the square.

K_{sp} = \dfrac{10^{6}\,x^{2}}{\alpha^{2}\,(x_{1}+x_{2}-2x_{3})^{2}}

Final answer:

\dfrac{10^{6}\,x^{2}}{\alpha^{2}\,(x_{1}+x_{2}-2x_{3})^{2}}

Q59Single correctp-Block Elements

Given below are two statements: Statement I: Aluminium is more electropositive than thallium as the standard electrode potential value of

E^{\circ}_{\text{Al}^{3+}|\text{Al}}

is negative and

E^{\circ}_{\text{Tl}^{3+}/\text{Tl}}

is positive. Statement II: The sum of first three ionization enthalpies of boron is very high when compared to that of aluminium. Due to this reason boron forms covalent compounds only and aluminium forms

\text{Al}^{3+}

ion. In the light of the above statements, choose the correct answer from the options given below

(A)

Both Statement I and Statement II are true

(B)

Both Statement I and Statement II are false

(C)

Statement I is true but Statement II is false

(D)

Statement I is false but Statement II is true

SolutionAnswer: Option 1Both Statement I and Statement II are true

Approach:

Statement I is checked against the standard electrode potentials of A

l^{3+}

/Al and T

l^{3+}

/Tl and the rule that the more negative

E^{\circ}

corresponds to the more electropositive metal. Statement II is checked against the sum of the first three ionisation enthalpies of B and Al and the resulting Fajan / energetics argument for covalent vs ionic behaviour.

Step 1:Standard electrode potentials:

E^{\circ}_{\text{Al}^{3+}/\text{Al}} = -1.66

V (negative) and

E^{\circ}_{\text{Tl}^{3+}/\text{Tl}} = +0.72

V (positive). The positive value for T

l^{3+}

/Tl follows from the inert-pair effect, which destabilises T

l^{3+}

relative to T

l^{+}

E^{\circ}_{\text{Al}^{3+}/\text{Al}} = -1.66\,\text{V} < 0 < +0.72\,\text{V} = E^{\circ}_{\text{Tl}^{3+}/\text{Tl}}

Step 2:Therefore Statement I, which asserts both signs and the electropositive ordering, is correct.

\text{Statement I: True}

Step 3:Standard sums of the first three ionisation enthalpies (kJ mo

l^{-1}

): B

= 800+2427+3660 \approx 6887

, Al

= 577+1816+2744 \approx 5137

. B's much smaller size and absence of inner d-shielding push its sum well above Al's.

\Sigma\,\text{IE}_{1-3}(\text{B}) \approx 6887 \gg 5137 \approx \Sigma\,\text{IE}_{1-3}(\text{Al})

Step 4:The energy released on hydration/lattice formation by A

l^{3+}

(high charge density) can offset the 5137 kJ mo

l^{-1}

ionisation cost, so A

l^{3+}

is observed in ionic solids such as Al

F_{3}

and in aqueous A

l^{3+}

. For B the 6887 kJ mo

l^{-1}

cost is too large for hydration/lattice energy to recover, so B is restricted to covalent compounds (B

F_{3}

, BC

l_{3}

B_{2}H_{6}

\Delta H_{\text{hyd/lat}}(\text{B}^{3+}) < \Sigma\,\text{IE}(\text{B}),\quad \Delta H_{\text{hyd/lat}}(\text{Al}^{3+}) \sim \Sigma\,\text{IE}(\text{Al})

Step 5:Both statements are independently true; the correct option is the one stating so.

\text{I: True;\ II: True}

Final answer: Both Statement I and Statement II are true

Q60Single correctd- and f-Block Elements

The correct statements among the following are. A. Basic vanadium oxide is used in the manufacture of

\text{H}_2\text{SO}_4

. B. The spin-only magnetic moment value of the transition metal halide employed in Ziegler-Natta polymerization is 2.84 BM . C. The p-block metal compound employed in Ziegler-Natta polymerization has the metal in

+3

oxidation state. D. The number of electrons present in the outer most ' d ' orbital of metal halide employed in Wacker process is 8 . Choose the correct answer from the options given below:

(A)

A and B only

(B)

A, C and D only

(C)

C and D Only

(D)

B, C and D only

Q61Single correctCoordination Compounds

Match the LIST-I with LIST-II

List-I (Electronic configuration of tetrahedral metal ion)	List-II (Crystal Field Stabilization Energy $(\Delta_{t})$ )
A. $d^{2}$	I. $-0.6$
B. $d^{4}$	II. $-0.8$
C. $d^{5}$	III. $-1.2$
D. $d^{8}$	IV. $-0.4$

(A)

A-III, B-IV, C-II, D-I

(B)

A-III, B-I, C-IV, D-II

(C)

A-III, B-IV, C-I, D-II

(D)

A-II, B-I, C-IV, D-III

Q62Single correctCoordination Compounds

Which of the following are true about the energy of the given d-orbitals of a tetrahedral complex? A.

d_{xy} = d_{xz} > d_{x^{2}-y^{2}}

. B.

d_{xy} = d_{yz} > d_{z^{2}}

. C.

d_{x^{2}-y^{2}} > d_{z^{2}} > d_{xz}

. D.

d_{x^{2}-y^{2}} = d_{z^{2}} < d_{xz}

. Choose the correct answer from the given below:

(A)

A, B and D only

(B)

A and B only

(C)

B and D only

(D)

B, C and D only

SolutionAnswer: Option 1A, B and D only

Approach:

In a tetrahedral field the d-orbitals split into a lower e set (

d_{z^{2}}

and

d_{x^{2}-y^{2}}

, degenerate at

-0.6\,\Delta_{t}

) and an upper

t_{2}

set (

d_{xy}

d_{xz}

and

d_{yz}

, degenerate at

+0.4\,\Delta_{t}

). Test each statement against this ordering.

Step 1:Group the d-orbitals into the two sets.

e = \{d_{z^{2}},\,d_{x^{2}-y^{2}}\}\ \text{(lower)};\quad t_{2} = \{d_{xy},\,d_{xz},\,d_{yz}\}\ \text{(upper)}

Step 2:Statement A asserts

d_{xy} = d_{xz} > d_{x^{2}-y^{2}}

. Both

d_{xy}

and

d_{xz}

are in

t_{2}

(equal) and

d_{x^{2}-y^{2}}

is in e (lower). Both parts hold.

t_{2}: d_{xy} = d_{xz}\ \text{and}\ t_{2} > e \ni d_{x^{2}-y^{2}}

Step 3:Statement B asserts

d_{xy} = d_{yz} > d_{z^{2}}

. Both

d_{xy}

and

d_{yz}

are in

t_{2}

(equal) and

d_{z^{2}}

is in e (lower). Both parts hold.

t_{2}: d_{xy} = d_{yz}\ \text{and}\ t_{2} > e \ni d_{z^{2}}

Step 4:Statement C asserts

d_{x^{2}-y^{2}} > d_{z^{2}} > d_{xz}

. Since

d_{x^{2}-y^{2}}

and

d_{z^{2}}

are degenerate e members, the strict inequality

d_{x^{2}-y^{2}} > d_{z^{2}}

is false. Also

d_{xz}

t_{2}

member) is higher in energy than either e member, so the second inequality is reversed.

d_{x^{2}-y^{2}} = d_{z^{2}} < d_{xz},\ \text{not}\ d_{x^{2}-y^{2}} > d_{z^{2}} > d_{xz}

Step 5:Statement D asserts

d_{x^{2}-y^{2}} = d_{z^{2}} < d_{xz}

. Both e members are degenerate and lower than every

t_{2}

member; this is exactly the tetrahedral splitting.

d_{x^{2}-y^{2}} = d_{z^{2}}\ (\text{both in } e),\ d_{xz}\in t_{2}\Longrightarrow e < t_{2}

Step 6:Collect the true statements.

\{A,\,B,\,D\}

Final answer: A, B and D only

Q63Single correctPurification and Characterisation of Organic Compounds

R_{f}

value for 2-methylpropene in a solvent system (Ethyl acetate + ether) is 0.42 . 2-methylpropene is treated with dilute

\text{H}_2\text{SO}_4

to give major organic product (X).

R_{f}

value for (X) in the same solvent system under identical condition will be:

(A)

0.42

(B)

0.82

(C)

0.62

(D)

0.12

Q64Single correctSome Basic Principles of Organic Chemistry

Given below are two statements:

Statement I: 2,6-diethylcyclohexanone and 6-methyl-2-n-propylcyclohexanone are metamers.

Statement II: 2,2,6,6 - tetramethylcyclohexanone exhibits keto-enol tautomerism.

In the light of the above statements, choose the correct answer from the options given below

(A)

Both Statement I and Statement II are true

(B)

Both Statement I and Statement II are false

(C)

Statement I is true but Statement II is false

(D)

Statement I is false but Statement II is true

SolutionAnswer: Option 3Statement I is true but Statement II is false

Approach:

For Statement I, derive the molecular formulae of the two ketones and verify that they share the carbonyl functional group with different alkyl distributions around it (definition of metamerism). For Statement II, identify the

\alpha

-carbons of 2,2,6,6-tetramethylcyclohexanone and check whether any

\alpha

-H remains for enolisation.

Step 1:Build the molecular formula of 2,6-diethylcyclohexanone. The cyclohexanone ring is

\text{C}_{6}\text{H}_{10}\text{O}

; adding two ethyl groups (net

+\text{C}_{4}\text{H}_{8}

after removing two ring H) gives

\text{C}_{10}\text{H}_{18}\text{O}

\text{2,6-diethylcyclohexanone}:\ \text{C}_{10}\text{H}_{18}\text{O}

Step 2:Build the molecular formula of 6-methyl-2-n-propylcyclohexanone. Methyl + n-propyl add

\text{C}_{4}\text{H}_{8}

\text{C}_{6}\text{H}_{10}\text{O}

\text{6-methyl-2-n-propylcyclohexanone}:\ \text{C}_{10}\text{H}_{18}\text{O}

Step 3:Both are ketones with the same C=O functional group, with the two

\alpha

-positions carrying different alkyl groups (Et/Et versus Me/n-Pr), satisfying the metamerism criterion.

\text{Same molecular formula and -CO-, different R/R'}

Step 4:Identify

\alpha

-carbons in 2,2,6,6-tetramethylcyclohexanone:

C_{2}

and

C_{6}

(the two ring carbons adjacent to the carbonyl). Each bears two methyl groups and is bonded to the ring on the remaining two positions, leaving zero hydrogen on either

\alpha

-carbon.

\text{C2: bonded to (C=O), 2\,CH}_3,\,\text{C3 — no H};\ \text{C6: bonded to (C=O), 2\,CH}_3,\,\text{C5 — no H}

Step 5:Keto-enol tautomerism requires removal of an

\alpha

-H to form the enol

\text{C=C-OH}

. With zero

\alpha

-H, enolisation cannot occur.

\text{No } \alpha\text{-H} \Longrightarrow \text{no keto-enol tautomerism}

Step 6:Combine: Statement I true, Statement II false.

\text{I: T,\ II: F}

Final answer: Statement I is true but Statement II is false

Q65Single correctHydrocarbons

Given below are two statements:

Statement I: Methane can be prepared by decarboxylation of sodium ethanoate, Kolbe's electrolysis of sodium acetate and reaction of

\text{CH}_3\text{MgBr}

with water.

Statement II: Methane cannot be prepared from unsaturated hydrocarbons and by Wurtz reaction.

In the light of the above statements, choose the correct answer from the options given below

(A)

Both Statement I and Statement II are true

(B)

Both Statement I and Statement II are false

(C)

Statement I is true but Statement II is false

(D)

Statement I is false but Statement II is true

SolutionAnswer: Option 4Statement I is false but Statement II is true

Approach:

Write the product of each named reaction on sodium acetate / methyl-Grignard / methyl halide and decide whether each preparation listed in the two statements yields methane.

Step 1:Test the first sub-claim of Statement I — soda-lime decarboxylation of sodium acetate. The product is methane.

\text{CH}_{3}\text{COONa} \xrightarrow{\text{NaOH/CaO}} \text{CH}_{4}

Step 2:Test the second sub-claim — Kolbe's electrolysis of sodium acetate. Two methyl radicals couple at the anode, giving ethane rather than methane.

2\,\text{CH}_{3}\text{COO}^{-} \xrightarrow{\text{electrolysis}} \text{CH}_{3}\text{-CH}_{3}\ (\text{ethane}) + 2\,\text{CO}_{2}

Step 3:The presence of one false sub-claim makes Statement I false even though the third sub-claim (

\text{CH}_{3}\text{MgBr}+\text{H}_{2}\text{O}\to\text{CH}_{4}

) is valid.

\text{Statement I: False}

Step 4:For Statement II, reduction of unsaturated hydrocarbons preserves the carbon skeleton, so the smallest possible alkane product has

n\ge 2

; methane (n=1) cannot be obtained this way.

\text{C}_{n}\text{H}_{2n}\ \text{or}\ \text{C}_{n}\text{H}_{2n-2} + \text{H}_{2} \to \text{C}_{n}\text{H}_{2n+2}\ (n \ge 2)

Step 5:The Wurtz reaction couples two alkyl halides, so the smallest alkane it can produce (from

\text{CH}_{3}\text{X}

) is ethane. Methane cannot be made by Wurtz.

\text{Wurtz on } \text{CH}_{3}\text{X} \to \text{C}_{2}\text{H}_{6}

Step 6:Combine the verdicts: Statement I false, Statement II true.

\text{I: F,\ II: T}

Final answer: Statement I is false but Statement II is true

Q66Single correctHydrocarbons

Given below are two statements:

Statement I: 3-phenylpropene reacts with HBr and gives secondary alkyl bromide having a chiral carbon atom as the major product.

Statement II: Aryl chlorides and aryl cyanides can be prepared by Sandmeyer reaction as well as Gattermann reaction.

In the light of the above statements, choose the correct answer from the options given below

(A)

Both Statement I and Statement II are true

(B)

Both Statement I and Statement II are false

(C)

Statement I is true but Statement II is false

(D)

Statement I is false but Statement II is true

SolutionAnswer: Option 3Statement I is true but Statement II is false

Approach:

For Statement I, identify the major HBr-addition product of 3-phenylpropene by selecting the most stable carbocation intermediate (benzylic), then check whether the brominated carbon is secondary and chiral. For Statement II, compare the substrate scope of the Sandmeyer (CuCl/CuBr/CuCN) and Gattermann (Cu/HCl or Cu/HBr) reactions.

Step 1:Protonation of 3-phenylpropene

\text{Ph-CH}_{2}\text{-CH}=\text{CH}_{2}

at the terminal carbon gives a benzylic carbocation

\text{Ph-}\overset{+}{\text{C}}\text{H-CH}_{2}\text{-CH}_{3}

, which is far more stable than the alternative non-benzylic cation. Bromide attacks the benzylic carbon.

\text{Ph-CH}_{2}\text{-CH=CH}_{2} + \text{HBr} \to \text{Ph-CHBr-CH}_{2}\text{-CH}_{3}

Step 2:The brominated carbon is attached to Ph, Br,

\text{CH}_{2}\text{CH}_{3}

and H — four different substituents — so it is a chiral centre.

\text{C* bonded to: Ph,\ Br,\ Et,\ H}

Step 3:The same brominated carbon is bonded to two carbons (the phenyl ring carbon and the methylene of the ethyl group) plus one H, so it is a secondary carbon.

\text{Ph-}\underset{2^{\circ}}{\text{CHBr}}\text{-CH}_{2}\text{CH}_{3}

Step 4:Sandmeyer scope: reacting

\text{ArN}_{2}^{+}

with CuCl, CuBr or CuCN gives ArCl, ArBr or ArCN respectively, so the reaction covers both aryl halides and aryl cyanides.

\text{Sandmeyer} \to \text{ArCl,\ ArBr,\ ArCN}

Step 5:Gattermann scope: using Cu powder in HCl or HBr converts

\text{ArN}_{2}^{+}

to ArCl or ArBr only. There is no CuCN / HCN variant; aryl cyanides are not obtained by Gattermann.

\text{Gattermann} \to \text{ArCl,\ ArBr only}

Step 6:Because Gattermann fails for cyanides, Statement II is false.

\text{Gattermann}\ \nrightarrow\ \text{ArCN}

Step 7:Combine the verdicts: Statement I true, Statement II false.

\text{I: T,\ II: F}

Final answer: Statement I is true but Statement II is false

Q67Single correctSome Basic Principles of Organic Chemistry

Consider the following sequence of reactions.
The major product P is:

tert-Butyl alcohol undergoes (i) Cu/573 K dehydration to isobutylene, then (ii) H+ and benzoic acid (PhCOOH) addition giving product P.

(A)

4-substituted benzoic acid (para-X-C6H4-COOH).

(B)

3-substituted benzoic acid (meta-X-C6H4-COOH).

(C)

(D)

SolutionAnswer: Option 3

\text{C}_6\text{H}_5\text{COO}\text{C(CH}_3)_3

Approach:

tert-Butanol has no

\alpha

-hydrogen on the carbinol carbon, so Cu/573 K cannot dehydrogenate it; instead it dehydrates to isobutylene. Protonation of isobutylene in step (ii) regenerates the tertiary carbocation, which is captured by benzoate, yielding tert-butyl benzoate.

Step 1:Identify the substrate:

(\text{CH}_3)_3\text{C-OH}

is a tertiary alcohol with no hydrogen on the carbon bearing the OH group, so the standard Cu/573 K dehydrogenation pathway (which removes the

\alpha

-C-H and the O-H to give a carbonyl) is unavailable.

(\text{CH}_3)_3\text{C-OH} \xrightarrow{\text{Cu/573 K}} (\text{CH}_3)_2\text{C}=\text{CH}_2 + \text{H}_2\text{O}

Step 2:In step (ii), benzoic acid provides a proton. Markovnikov addition places the proton on the terminal

\text{CH}_2

of isobutylene, generating the very stable tertiary cation.

(\text{CH}_3)_2\text{C}=\text{CH}_2 \xrightarrow{\text{H}^{+}} (\text{CH}_3)_3\text{C}^{+}

Step 3:The benzoate anion (

\text{PhCOO}^{-}

) attacks the tertiary carbocation at oxygen, forming the C-O bond of an ester.

(\text{CH}_3)_3\text{C}^{+} + \text{PhCOO}^{-} \rightarrow \text{PhCOO-C(CH}_3)_3

Step 4:Match P with the structural options. Options (1) and (2) are aromatic carboxylic acids that cannot arise from a tert-butanol+PhCOOH coupling. Option (4) (allyl benzoate) would require an allylic cation, which is not formed here. Only option (3),

\text{PhCOOC(CH}_3)_3

, matches the deduced structure.

P \equiv \text{PhCOOC(CH}_3)_3

Final answer:

\text{C}_6\text{H}_5\text{COO}\text{C(CH}_3)_3

(tert-butyl benzoate)

Q68Single correctSome Basic Principles of Organic Chemistry

Arrange the following compounds according to increasing order of boiling points.

\text{n}-\text{C}_4\text{H}_9\text{OH(A)}, \text{n}-\text{C}_4\text{H}_9\text{NH}_2\text{(B)}, \text{n}-\text{C}_4\text{H}_{10}\text{(C)} \text{ and } \text{C}_2\text{H}_5\text{NHC}_2\text{H}_5\text{(D)}

(A)

\displaystyle \text{C} < \text{B} < \text{A} < \text{D}

(B)

\displaystyle \text{D} < \text{C} < \text{B} < \text{A}

(C)

\displaystyle \text{C} < \text{D} < \text{B} < \text{A}

(D)

\displaystyle \text{D} < \text{B} < \text{A} < \text{C}

Q69Single correctSome Basic Principles of Organic Chemistry

Match the LIST-I with LIST-II

Choose the correct answer from the options given below:

List-I (Deficiency Disease)	List-II (Vitamin)
A. Scurvy	I. Pyridoxine
B. Convulsions	II. Vitamin A
C. Cheilosis	III. Ascorbic Acid
D. Xerophthalmia	IV. Riboflavin

(A)

A-I, B-III, C-II, D-IV

(B)

A-I, B-III, C-IV, D-II

(C)

A-III, B-I, C-IV, D-II

(D)

A-III, B-I, C-II, D-IV

Q70Single correctSome Basic Principles of Organic Chemistry

Match the LIST-I with LIST-II

Choose the correct answer from the options given below:

List-I (Amino acid)	List-II (Positive reaction/Test for functional group present in side chain of amino acid)
A. Glutamine	I. Hinsberg's test
B. Lysine	II. Neutral $\text{FeCl}_3$ test
C. Tyrosine	III. Ceric ammonium nitrate test
D. Serine	IV. Hoffman bromamide degradation

(A)

A-IV, B-II, C-I, D-III

(B)

A-IV, B-I, C-II, D-III

(C)

A-III, B-II, C-I, D-IV

(D)

A-IV, B-I, C-III, D-II

SolutionAnswer: Option 2A-IV, B-I, C-II, D-III

Approach:

Identify the dominant functional group in the side chain of each amino acid, then match it to the chemical test whose positive response is diagnostic of that group: primary amide

\to

Hoffmann bromamide; primary amine

\to

Hinsberg; phenol

\to

neutral FeC

l_{3}

; alcohol

\to

ceric ammonium nitrate (CAN).

Step 1:A — Glutamine has the side chain

-\text{CH}_{2}\text{CH}_{2}\text{CONH}_{2}

, which contains a primary amide. Hoffmann bromamide degradation acts on primary amides (

\text{RCONH}_{2}\to\text{RNH}_{2}

\text{Glutamine side chain: }-\text{CH}_{2}\text{CH}_{2}\text{CONH}_{2}\ (1^{\circ}\text{ amide})

Step 2:B — Lysine has the side chain

-(\text{CH}_{2})_{4}\text{NH}_{2}

, a primary amine. The Hinsberg test (PhS

O_{2}

Cl in aq. NaOH) is the classic test for primary amines, yielding an alkali-soluble sulfonamide.

\text{Lysine side chain: }-(\text{CH}_{2})_{4}\text{NH}_{2}\ (1^{\circ}\text{ amine})

Step 3:C — Tyrosine carries a p-hydroxyphenyl side chain (

-\text{CH}_{2}\text{-C}_{6}\text{H}_{4}\text{-OH}

), i.e. a phenolic group. Neutral

\text{FeCl}_{3}

gives the characteristic violet/blue colour with phenols.

\text{Tyrosine side chain: }-\text{CH}_{2}\text{-C}_{6}\text{H}_{4}\text{-OH}\ (\text{phenol})

Step 4:D — Serine has the side chain

-\text{CH}_{2}\text{OH}

, a primary aliphatic alcohol. Ceric ammonium nitrate gives a red colouration with alcohols.

\text{Serine side chain: }-\text{CH}_{2}\text{OH}\ (1^{\circ}\text{ alcohol})

Step 5:Combine the four pairings.

A\to IV,\ B\to I,\ C\to II,\ D\to \text{III}

Final answer: A-IV, B-I, C-II, D-III

Q71NumericalClassification of Elements and Periodicity in Properties

First and second ionization enthalpies of lithium are

520\,\text{kJ}\,\text{mol}^{-1}

and

7297\,\text{kJ}\,\text{mol}^{-1}

respectively. Energy required to convert 3.5 mg lithium (g) into

\text{Li}^{2+}(g)

\left[\text{Li(g)} \to \text{Li}^{2+}(g)\right]

is ____ kJ. (nearest integer) [Molar mass of

\text{Li} = 7\,\text{g}\,\text{mol}^{-1}

]

Q72NumericalSome Basic Principles of Organic Chemistry

Consider the following sequence of reactions.

\text{C}_6\text{H}_5 - \text{N} = \text{N} - \text{NH} - \text{C}_6\text{H}_5 \xrightarrow[\substack{(\text{ii}) \text{ Aniline (in excess)} \\ (\text{iii}) \text{ Warm } 40-45\,^\circ\text{C} \\ (\text{iv}) \text{ Cool, acetic acid}}]{(\text{i}) \text{ HCl}}

Yellow Product (X).

The percentage of nitrogen in the yellow product (X) formed is ____ %. (Nearest Integer)

(Given Molar mass in

\text{g}\,\text{mol}^{-1}\, \text{H}:1, \text{C}:12, \text{N}:14

)

Structure of 1,3-diphenyltriazene (diazoaminobenzene): C6H5-N=N-NH-C6H5

Q73NumericalHydrocarbons

4.7 g of phenol is heated with Zn to give product X. If this reaction goes to

60

% completion then the number of moles of compound X formed will be ____

\times 10^{-2}

. (Nearest Integer)

(Given molar mass in

\text{g}\,\text{mol}^{-1}: \text{H}:1, \text{C}:12, \text{O}:16

)

Q74NumericalChemical Kinetics

Sucrose hydrolyses in acidic medium into glucose and fructose by first order rate law with

t_{1/2} = 3

hour. The percentage of sucrose remaining after 6 hours is ____ . (Nearest integer)

(Given :

\log 2 = 0.3010

and

\log 3 = 0.4771

)

Q75NumericalChemical Thermodynamics

Consider the reaction

X \rightleftharpoons Y

at 300 K. If

\Delta H^0

and K are

28.40\,\text{kJ}\,\text{mol}^{-1}

and

1.8 \times 10^{-7}

at the same temperature, then the magnitude of

\Delta S^0

for the reaction in

\text{J}\,\text{K}^{-1}\,\text{mol}^{-1}

is ____ . (Nearest integer)

(Given :

R = 8.3\,\text{J}\,\text{K}^{-1}\,\text{mol}^{-1}, \ln 10 = 2.3, \log 3 = 0.48, \log 2 = 0.30

)

JEE Main 2026 April 06, Shift 1 Question Paper with Solutions

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