datavorous/entrance-exam-dataset huggingface dataset:
Consider the following reaction:

3PbCl2+2NH43PO4→Pb3PO42+6NH4Cl

If 72 mmol PbCl2 is mixed with 50 mmol of NH43PO4, then amount of Pb3PO42 formed in mmol. (nearest integer)

Options:

The correct answer is:
24

Solution:
The reaction is given as,

3PbCl2+2NH43PO4→Pb3PO42+6NH4Cl

The products formed in the above reaction is Pb3PO42 and NH4Cl .

According to the above stoichiometric equation, three mole of each lead chloride and two moles of NH43PO4 is required to make one mole of Pb3PO42 and 6 moles of NH4Cl

Here the Limiting Reagent is PbCl2

mmol of Pb3(PO4)2 formed

=mmol of PbCl2 reacted 3=723

=24 m mol

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2024 (01 Feb Shift 1)
Consider the following reaction :

A+alkali→B MajorProduct

If B is an oxoacid of phosphorus with no P-H bond, then A is

Options:
A White P4
B Red P
C H3PO3
D P2O3

The correct answer is:
Red P

Solution:
P4+3NaOH+3H2O→PH3+3NaH2PO2

Red P + Alkali →H4P2O6

White Phosphorous react with alkali metal hydroxide to form phosphine and a salt of H3PO2-. While red phosphorous react with alkali to give oxoacid of phosphorus with no P-H bonds.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2022 (26 Jun Shift 1)
Consider the following reaction and identify (B) .

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Consider the following reaction and statements:
CoNH34Br2++Br-→CoNH33Br3+NH3
(i) Two isomers are produced if the reactant complex ion is a cis-isomer.
(ii) Two isomers are produced if the reactant complex ion is a trans-isomer.
(iii) Only one isomer is produced if the reactant complex ion is a trans-isomer.
(iv) Only one isomer is produced if the reactant complex ion is a cis-isomer.
The correct statements are

Options:
A (ii) and (iv)
B (i) and (ii)
C (i) and (iii)
D (iii) and (iv)

The correct answer is:
(i) and (iii)

Solution:
From the below image it is clear that two isomers are produced if the reactant complexion is a cis-isomer, as well as only one isomer, is produced if the reactant complex is a trans-isomer.

Two isomers are produced if the reactant complex ion is a cis-isomer because the cis isomer can further form facial and meridional isomers.

Only one isomer is produced if the reactant complex ion is a trans-isomer because it cannot form any isomeric isomers.

Trans → only meridonial
Cis → fac and mer

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2018 (08 Apr)
Consider the following reaction at 298 K.

32O2g⇌O3g.Kp=2.47×10-29

∆rG0 for the reaction is _________ kJ. (Given R=8.314 JK–1 mol–1)

Round off your answer to the nearest integer.

Options:

The correct answer is:
163

Solution:
Given R=8.314 JK–1 mol–1

32O2g⇌O3g. Kp=2.47×10-29

Standard Gibbs free energy of the formation of a compound is basically the change of Gibbs free energy that is followed by the formation of one mole of that substance from its component element available at their standard states or the most stable form of the element which is at 25°C and 100 kPa. Its symbol is ΔfG˚.

∆rG0=-RT ln Kp

=-8.314×10–3×298×ln (2.47×10–29)

=–8.314×10–3×298×(–65.87)

=163.19 kJ

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2024 (31 Jan Shift 1)
Consider the following reaction,
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{Cl}+\mathrm{AgCN} \stackrel{\mathrm{EtOH} / \mathrm{H}_2 \mathrm{O}}{\longrightarrow} \underline{X} \text { (major) }$
Which one of the following statements is true for $\underline{X} ?$
(I) It gives propionic acid on hydrolysis
(II) It has an ester functional group
(III) It has a nitrogen linked to ethyl carbon
(IV) It has a cyanide group

Options:
A IV
B III
C II
D I

The correct answer is:
III

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
$$
\text { Consider the following reaction }
$$
The product $Z$ is :

Options:
A $\mathrm{CH}_3 \mathrm{CH}_2-\mathrm{OH}$
B $\mathrm{CH}_2=\mathrm{CH}_2$
C $\mathrm{CH}_3 \mathrm{CH}_2-\mathrm{O}-\mathrm{CH}_2 \mathrm{CH}_3$
D $\mathrm{CH}_3 \mathrm{CH}_2-\mathrm{O}-\mathrm{SO}_3 \mathrm{H}$

The correct answer is:
$\mathrm{CH}_3 \mathrm{CH}_2-\mathrm{OH}$

Solution:

Topic: Chemistry, Hydrocarbons, NEET, NEET 2009 (Mains)
Consider the following reaction
\(\mathrm{MnO}_2+\mathrm{KOH}+\mathrm{O}_2 \rightarrow \mathrm{A}+\mathrm{H}_2 \mathrm{O}\).
Product ' \(\mathrm{A}\) ' in neutral or acidic medium disproportionate to give products ' \(\mathrm{B}\) ' and ' \(\mathrm{C}\) ' along with water. The sum of spin-only magnetic moment values of \(\mathrm{B}\) and \(\mathrm{C}\) is \(\qquad\) BM. (nearest integer) (Given atomic number of \(\mathrm{Mn}\) is 25)

Options:

The correct answer is:
4

Solution:
$\mathrm{MnO}_2+\mathrm{KOH}+\mathrm{O}_2 \rightarrow \mathrm{K}_2 \mathrm{MnO}_4+\mathrm{H}_2 \mathrm{O}$
(A)
$\begin{aligned}
& \mathrm{K}_2 \mathrm{MnO}_4 \xrightarrow{\text { Neutral/acidic solution }} \mathrm{KMnO}_4+\mathrm{MnO}_2 \\
& \mathrm{Mn}^{+4}:-[\mathrm{Ar}] 3 \mathrm{~d}^3 \\
& \mathrm{n}=3, \mu=\sqrt{3(3+2)}=3.87 \text { B.M. }
\end{aligned}$
Nearest integer is (4)

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2024 (04 Apr Shift 1)
Consider the following reaction

Minor by product(s) is/are

Options:
A 2-chloro propane
B Propene
C Isopropyl nitrite
D All of these

The correct answer is:
All of these

Solution:
Here the reaction is as follows

Topic: Chemistry, Amines, JEE Main
Consider the following reaction

MnO4-+8H++5e-→Mn+2+4H2O ,E°=1.51 V

The quantity of electricity required in Faraday to reduce five moles of MnO4- is

Options:

The correct answer is:
25

Solution:
For this balanced reaction, the change in oxidation number of the metal ion is Mn+7 →Mn+2.

Therefore one mole Mn required 5 Faraday

Hence, for five mole Mn required 25 Faraday.

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2021 (26 Feb Shift 1)
Consider the following reaction:
N2O4(g)=2NO2(g): ΔH0=+58k
For each of the following cases (a, b), the direction in which the equilibrium shifts is:
(a) Temperature is decreased.
(b) Pressure is increased by adding N2 at constant T.

Options:
A (a) towards product, (b) towards reactant
B (a) towards reactant, (b) towards product
C (a) towards reactant, (b) no change
D (a) towards product, (b) no change

The correct answer is:
(a) towards reactant, (b) no change

Solution:
(i) As reaction is endothermic so on decrease in temperature equilibrium shift in reactant side.
(ii) On increase in pressure by adding inert gas at same temperature, no shifting will take place

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2020 (05 Sep Shift 1)
Consider the following reaction, Phenol $\stackrel{\mathrm{Zn} \text { dust }}{\longrightarrow} \mathrm{X} \underset{\text { Anhydrous } \mathrm{AlCl}_3}{\stackrel{\mathrm{CH}_3 \mathrm{Cl}}{\longrightarrow}} \mathrm{Y} \stackrel{\begin{array}{l}\text { Alkaline } \\ \mathrm{KMnO}\end{array}}{\longrightarrow} \mathrm{Z}$, the product, $Z$, is

Options:
A toluene
B benzaldehyde
C benzoic acid
D benzene

The correct answer is:
benzoic acid

Solution:
Key Idea (i) $Z n$ dust converts $-\mathrm{OH}$ group into- $\mathrm{H}$.
(ii) Reaction with $\mathrm{CH}_3 \mathrm{Cl}$ in presence of anthy. $\mathrm{AlCl}_3$ is called Friedel-Craft's alkylation.
(iii) Alkaline $\mathrm{KMnO}_4$ converts complete carbon chain which is directly attached with benzene ring, into - $\mathrm{COOH}$ group.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Consider the following reaction

Propanal+Methanal=→(ii) Δ(iii) NaCN(iv) H3O+(i) dil. NaOHProduct B(C5H8O3)

The correct statement for product B is. It is

Options:
A optically active and adds one mole of bromine
B racemic mixture and is neutral
C racemic mixture and gives a gas with saturated NaHCO3 solution
D optically active alcohol and is neutrall

The correct answer is:
racemic mixture and gives a gas with saturated NaHCO3 solution

Solution:
Propanal and methanal undergo aldol condensation reaction with sodium hydroxide. The α,β-unsaturated compound undergo 1,4-conjugated addition with cyanide ion gives a racemic mixture. The cyanide undergo acidic hydrolysis gives carboxylic acid.

Carboxylic acid will give CO2 gas, with NaHCO3 solution

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2023 (31 Jan Shift 1)
Consider the following reaction,

Select the incorrect statement.

Options:
A It is not a disproportionation reaction
B It is intramolecular redoc reaction
C OH- is a reducing as well as oxidising agent
D is a reducing as well as oxidising agent

The correct answer is:
OH- is a reducing as well as oxidising agent

Solution:
One CHO is oxidised to COO- and one CHO is reduced to CH2OH .
Thus, it is not a disproportionation reaction. It is intramolecular redox reaction. Thus, option A and B are true and is reducing as well as oxidising agent.
Thus, option D is also true and option C is incorrect.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Consider the following reaction sequence,
Isomers are

Options:
A C and E
B C and D
C D and E
D $\mathrm{C}, \mathrm{D}$ and $\mathrm{E}$

The correct answer is:
D and E

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Consider the following reaction sequence

âââââââ

The product 'B' is

Options:
A

B

C

D

The correct answer is:

Solution:
Diisobutylaluminium hydride, AlH(i-Bu)2 also known as DIBAL or DIBAL-H, is an exceedingly useful and versatile reducing agent. It is an electrophilic reducing agent, usually employed in selective reductions of esters or nitriles to aldehydes

So, in first step nitrile group will be reduced to aldehyde group.

In second step cross aldol condensation reaction will take place which will give mixture of products. Out of which one of the product will be option B.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2022 (29 Jul Shift 1)
Consider the following reaction that goes from A to B in three steps as shown below:

Choose the correct option

Number of
Intermediates	Number of
Activated
Complexes	Rate
determining
step

1	2	3	I

2	2	3	III

3	2	3	II

4	3	2	II

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:
The curve contain three activated complex as shown below in the curve.In the curve P and Q are the intermediates. The step with the highest activation energy is the rate determining step. Hence, in the given reaction, two intermediates, three activated complexes are involved and step-II is the rate determining step.

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2023 (06 Apr Shift 2)
Consider the following reaction
The product ' $A$ ' is

Options:
A $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CHO}$
B $\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}$
C $\mathrm{C}_6 \mathrm{H}_5 \mathrm{COCH}_3$
D $\mathrm{C}_6 \mathrm{H}_5 \mathrm{Cl}$

The correct answer is:
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{CHO}$

Solution:
In Rosenmund reaction, acid chloride reacts with $\mathrm{H}_2$ in the presence of $\mathrm{Pd} / \mathrm{BaSO}_4$ to yield aldehyde.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Consider the following reaction
The product ' $A$ ' is

Options:
A $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CHO}$
B $\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}$
C $\mathrm{C}_6 \mathrm{H}_5 \mathrm{COCH}_3$
D $\mathrm{C}_6 \mathrm{H}_5 \mathrm{Cl}$

The correct answer is:
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{CHO}$

Solution:
In Rosenmund reaction, acid chloride reacts with $\mathrm{H}_2$ in the presence of $\mathrm{Pd} / \mathrm{BaSO}_4$ to yield aldehyde.

Topic: Chemistry, Aldehydes and Ketones, NEET, NEET 2012 (Mains)
Consider the following reaction:

The product'P'gives positive ceric ammonium nitrate test. This is because of the presence of which of these–OH group(s)?

Options:
A (b) only
B (c) and (d)
C (d) only
D (b) and (d)

The correct answer is:
(b) only

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2020 (03 Sep Shift 2)
Consider the following reaction:
The product 'X' is used:

Options:
A in protein estimation as an alternative to ninhydrin
B in acid base titration as an indicator
C as food grade colourant
D in laboratory test for phenols

The correct answer is:
in acid base titration as an indicator

Solution:
Methyl orange is used as indicator in acid base titration.

Azo coupling is the most widely used industrial reaction in the production of dyes, lakes and pigments. Aromatic diazonium ions acts as electrophiles in coupling reactions with activated aromatics such as anilines or phenols.

Topic: Chemistry, Amines, JEE Main, JEE Main 2020 (07 Jan Shift 1)
Consider the following reaction
What is $Y$ ?

Options:
A Acetanilide
B Benzanilide
C Azo-benzene
D Hydrazo-benzene

The correct answer is:
Benzanilide

Solution:
$\mathrm{C}_6 \mathrm{H}_6 \mathrm{NO}_2+6(\mathrm{H}) \underset{\mathrm{HCl}}{\stackrel{\mathrm{Sn}}{\mathrm{HCl}}} \underset{\mathrm{C}}{\mathrm{C}} \mathrm{H}_8 \mathrm{NH}_2+2 \mathrm{H}_2 \mathrm{O}$

Topic: Chemistry, Amines, JEE Main
Consider the following reaction

Which of these reactions are possible?

Options:
A b , c and d
B b and d
C a and d
D a and b

The correct answer is:
b and d

Solution:
Vinyl halides and Aryl halides do not give Friedel Craft Reaction due to partial double bond character which is reso stabilised .

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Consider the following reaction

[X] and [Y] respectively be

Options:
A 2,3-dimethyl-2-butanol and 3,3-dimethyl-2-butanol
B 3,3-dimethyl-2-butanol and 3,3-dimethyl-2-butanol
C 3,3-dimethyl-2-butanol and 3,3-dimethyl-3-butanol
D 2,3-dimethyl-2-butanol and 2,3-dimethyl-2-butanol

The correct answer is:
2,3-dimethyl-2-butanol and 3,3-dimethyl-2-butanol

Solution:
Here [X] and [Y] are 2,3-dimethyl-2-butanol and 3,3-dimethyl-2-butanol respectively

Topic: Chemistry, Hydrocarbons, JEE Main
Consider the following reaction

xMnO4-+yC2O42-+zH+→xMn2++2yCO2+z2H2O

The values of x, y and z in the reaction are, respectively

Options:
A 5, 2 and 8
B 2, 5 and 8
C 2, 5 and 16
D 5, 2 and 16

The correct answer is:
2, 5 and 16

Solution:
2MnO4-+5C2O42-+16H+→2Mn2++10CO2+8H2O
Here x = 2, y = 5 and z = 16.

Topic: Chemistry, Redox Reactions, JEE Main
Consider the following reaction:
x MnO4-+y C2O42-+zH+⟶x Mn2++2y CO2+z2H2O
The values of x, y and z in the reaction are, respectively:

Options:
A 2,5 and 16
B 5,2 and 8
C 5,2 and 16
D 2,5 and 8

The correct answer is:
2,5 and 16

Solution:
C 2 O 4 2 - ⟶ 2 CO 2 + 2 e - ] × 5 MnO 4 - + 8 H + + 5 e- ⟶ Mn 2 + + 4 H 2 O ] × 2
-----------------------------------------------------------------------------------
2 MnO 4 - + 1 6 H + + 5 C 2 O 4 2 - ⟶ 1 0 CO 2 + 2 Mn 2 + + 8 H 2 O
MnO 4 - ⇒ x = 2
C 2 O 4 2 - ⇒ y = 5
H + ⇒ z = 1 6

Topic: Chemistry, Redox Reactions, JEE Main
Consider the following reaction:
x MnO4-+y C2O42-+zH+⟶x Mn2++2y CO2+z2H2O
The values of x, y and z in the reaction are, respectively:

Options:
A 2,5 and 16
B 5,2 and 8
C 5,2 and 16
D 2,5 and 8

The correct answer is:
2,5 and 16

Solution:
C 2 O 4 2 - ⟶ 2 CO 2 + 2 e - ] × 5 MnO 4 - + 8 H + + 5 e- ⟶ Mn 2 + + 4 H 2 O ] × 2
-----------------------------------------------------------------------------------
2 MnO 4 - + 1 6 H + + 5 C 2 O 4 2 - ⟶ 1 0 CO 2 + 2 Mn 2 + + 8 H 2 O
MnO 4 - ⇒ x = 2
C 2 O 4 2 - ⇒ y = 5
H + ⇒ z = 1 6

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2013 (07 Apr)
$$
\text { Consider the following reaction : }
$$

Options:
A Benzene
B Toluene
C Benzaldehyde
D Benzoic acid

The correct answer is:
Benzoic acid

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, NEET, NEET 2009 (Mains)
Consider the following reactions

(1) (CH 3 ) 3 CCH(OH)CH 3 → conc .H 2 SO 4

(2) (CH 3 ) 2 CHCH(Br)CH 3 → conc.KOH

(3) ( CH 3 ) 2 CHCH(Br)CH 3 → (CH 3 ) 3 CO ⊖ K ⊕

(4)

Which of these reactions(s) will produce Saytzeff product?

Options:
A (1), (2) and (4)
B (3) only
C (4) only
D (2) and (4)

The correct answer is:
(1), (2) and (4)

Solution:
Bulky base like (CH3)3O⊖K⊕ always prefers Hoffmann product (less stable alkene) in case of 1, 2 and 4 elimination product are according to Saytzeff Rule (more stable alkene)

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Consider the following reactions,
$A$ and $B$ are

Options:
A $A=\mathrm{C}_2 \mathrm{H}_4, B=\mathrm{C}_6 \mathrm{H}_6$
B $A=\mathrm{C}_2 \mathrm{H}_6, B=\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2$
C $A=\mathrm{C}_2 \mathrm{H}_4, B=\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2$
D $A=\mathrm{C}_6 \mathrm{H}_6, B=\mathrm{C}_6 \mathrm{H}_5 \mathrm{NO}_2$

The correct answer is:
$A=\mathrm{C}_6 \mathrm{H}_6, B=\mathrm{C}_6 \mathrm{H}_5 \mathrm{NO}_2$

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
Consider the following reactions:
a CH33CCHOHCH3→conc.H2SO4
b CH32CHCHBrCH3→alc.KOH
c CH32CHCHBrCH3→CH33O⊖K⊕
d

Which of the reaction(s) will not produce Saytzeff product?

Options:
A a,c and d
B d only
C c only
D b and d

The correct answer is:
c only

Solution:
We have to know that a Saytzeff product is basically an alkene compound which has undergone many substitution reactions to form the alkene product. From Saytzeff rule, a stable alkene is formed when removal of hydrogen from beta carbon has a low number of hydrogen atoms.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2020 (07 Jan Shift 1)
Consider the following reactions in which all the reactants and the products are in gaseous state.
2PQ⇌P2+Q2;K1=3.6×105
PQ+1/2R2⇌PQR;K2=6×10-3
The value of K3 for the equilibrium
1/2P2+1/2Q2+1/2R2⇌PQR is

Options:
A 3×103
B 6×103
C 1.0×10-5
D 5×10-3

The correct answer is:
1.0×10-5

Solution:
Given , 2PQ⇌P2+Q2
K1=P2Q2[PQ]2=3.6×105
PQ+1/2R2⇌PQR;
K2=[PQR]PQR21/2
=6×10-3
For the required equilibrium
1/2P2+1/2Q2+1/2R2⇌PQR;
K3=[PQR]P21/2Q21/2R21/2
K3=K2×1K1
=6×10-3×13.6×105=6×10-30.6×103
=1×10-5

Topic: Chemistry, Chemical Equilibrium, JEE Main
Consider the following reactions
\(\mathrm{NiS}+\mathrm{HNO}_3+\mathrm{HCl} \rightarrow \mathrm{A}+\mathrm{NO}+\mathrm{S}+\mathrm{H}_2 \mathrm{O}\)
\(\begin{aligned} \mathrm{A}+\mathrm{NH}_4 \mathrm{OH}+\mathrm{H}_3 \mathrm{C}-\mathrm{C} & =\mathrm{N}-\mathrm{OH} \\ | ~\\ \mathrm{H}_3 \mathrm{C}-\mathrm{C} & =\mathrm{N}-\mathrm{OH}\end{aligned} \rightarrow \mathrm{B}+\mathrm{NH}_4 \mathrm{Cl}+\mathrm{H}_2 \mathrm{O}\)
The number of protons that do not involve in hydrogen bonding in the product \(B\) is ______.

Options:

The correct answer is:
12

Solution:

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2024 (06 Apr Shift 2)
Consider the following reactions
NaCl+K2Cr2O7+Conc.H2SO4âA+side products
A+NaOHâB+ side products
B+dilute H2SO4+H2O2âC+ Side products
The sum of atoms in one molecule each of (A),â(B) and (C) is

Options:

The correct answer is:
18

Solution:
The reaction is as follows
NaCl → K 2 Cr 2 O 7 /H ⊕ CrO 2 Cl 2 A → NaOH Na 2 CrO 4 B → H ⊕ H 2 O 2 CrO 5 C
CrO2Cl2→5
Na2CrO4→7
CrO 5 →6, Total = 18

Topic: Chemistry, d and f Block Elements, JEE Main
Consider the following reactions:
NaCl+K2Cr2O7+H2SO4Conc.→A+ side products
A+NaOH→B+ side products
B+H2SO4dilute+H2O2→C+ side products
The sum of the total number of atoms in one molecule each of A,B and C is ________

Options:

The correct answer is:
18

Solution:
A=CrO2Cl2
B=Na2CrO4
C=CrO5

Chromyl chloride CrO2Cl2 can be prepared by the reaction of potassium chromate or potassium dichromate with hydrogen chloride in the presence of sulfuric acid, followed by distillation. The sulfuric acid serves as the dehydration agent.

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2020 (07 Jan Shift 2)
Consider the following reactions,
The compound P is:

Options:
A
B
C
D

The correct answer is:

Solution:
When meta methyl aniline reacts with sodium nitrite in presence of an acidic medium then diazonium salt is formed then on reaction with β-naphthol a colored compound is formed.

Meta methyl aniline also undergoes for the halogentaion reaction and desired compound is formed.

Topic: Chemistry, Amines, JEE Main, JEE Main 2020 (09 Jan Shift 2)
Consider the following reactions,

The end product 'S' is

Options:
A
B
C
D

The correct answer is:

Solution:
Here the conversion occurs as follows

Topic: Chemistry, Amines, JEE Main
Consider the following reactions,
$$
\text { What are the structures of } Y \text { and } Z \text { ? }
$$

Options:
A

B

C

D

The correct answer is:

Solution:
$\mathrm{KMnO}_4$ (alk.) oxidise alkyl group $(R)$ directly attached to benzene ring, irrespective of length of alkyl group (if has at least one $\alpha-H$ ).

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
Consider the following reactions:
Which of these reactions are possible?

Options:
A a and b
B a and d
C b,c and d
D b and d

The correct answer is:
b and d

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2020 (07 Jan Shift 2)
Consider the following reactions :
$Y$ can be conserted to $X$ on heating with at temperature :

Options:
A $\mathrm{Al}_2 \mathrm{O}_3, 350^{\circ} \mathrm{C}$
B $\mathrm{Cu}, 300^{\circ} \mathrm{C}$
C $\mathrm{Ca}(\mathrm{OH})_2+\mathrm{CaOCl}_2, 60^{\circ} \mathrm{C}$
D $\mathrm{NaOH} / \mathrm{I}_2, 60^{\circ} \mathrm{C}$

The correct answer is:
$\mathrm{Al}_2 \mathrm{O}_3, 350^{\circ} \mathrm{C}$

Solution:
$\underset{(Y)}{\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}} \underset{350^{\circ} \mathrm{C}}{\stackrel{\mathrm{Al}_2 \mathrm{O}_3}{\longrightarrow}} \underset{(X)}{\mathrm{C}_2 \mathrm{H}_4}+\mathrm{H}_2 \mathrm{O}$

Topic: Chemistry, Hydrocarbons, JEE Main
Consider the following recuction processes
Zn2++2e-→Zn(s);E°=-0.76V
Ca2++2e-→Ca(s);E°=-2.87V
Mg2++2e-→Mg(s);E°=-2.36V
Ni2++2e-→Ni(s);E°=-0.25V
The reduction power of the metals increases in the order

Options:
A Ca < Zn < Mg < Ni
B Ni < Zn < Mg < Ca
C Zn > Mg < Ni < Ca
D Ca < Mg < Zn < Ni

The correct answer is:
Ni < Zn < Mg < Ca

Solution:
Reducing power â Tendency to undergo oxidation
∝ 1 reduction potential values
so, reducing power decreases as Ca > Mg > Zn > Ni

Topic: Chemistry, Electrochemistry, JEE Main
Consider the following reduction processes

Zn2++2e-âZns; Eo=-0.76 V

Ca2++2e-âCas; Eo=-2.87 V

Mg2++2e-âMgs; Eo=-2.36 V

Ni2++2e-âNis; Eo=-0.25 V

The reduction power of the metals increases in the order

Options:
A Ca<Zn<Mg<Ni
B Ni<Zn<Mg<Ca
C Zn>Mg<Ni<Ca
D Ca<Mg<Zn<Ni

The correct answer is:
Ni<Zn<Mg<Ca

Solution:
Reducing power â Tendency to undergo oxidation
â1reduction potential values
So, reducing power decreases as Ca>Mg>Zn>Ni

Topic: Chemistry, Electrochemistry, JEE Main
Consider the following relations for emf of a electrochemical cell
(A) Emf of cell = (oxidation potential of anode) - (reduction potential of cathode)
(B) Emf of cell = (oxidation potential of anode) + (reduction potential of cathode)
(C) Emf of cell = (reduction potential of anode) + (reduction potential of cathode)
(D) Emf of cell = (oxidation potential of anode)-(Oxidation potential of cathode)
Which of the above relations are correct?

Options:
A (C) and (A)
B (A) and (B)
C (C) and (D)
D (B) and (D)

The correct answer is:
(B) and (D)

Solution:
$\begin{aligned}
& \mathrm{E}_{\text {cell }}=\underset{\text { cathode }}{D}-\underset{\text { (red) }}{\mathrm{E}_{\text {anode }}^{\circ}} \\
& \text {or } E_{\text {cell }}=\underset{\text { cred) }}{\mathrm{\mathrm{E}}_{\text {cathode }}^{\circ}}+\underset{\text { (oxide) }}{E_{\text {Anode }}^{\circ}} \\
& \text {or } \mathrm{E}_{\text {cell }}=\underset{\text { (oxide) }}{\mathrm{E}} \underset{\text { anode }}{\mathrm{E}}-\mathrm{E}_{\text {cathode }}^{\circ} \text { (oxide) }
\end{aligned}$

Topic: Chemistry, Electrochemistry, NEET, NEET 2010 (Mains)