datavorous/entrance-exam-dataset huggingface dataset:
The process that is NOT endothermic in nature is :

Options:
A Ar(g)+e-→Arg-
B H(g)+e-→Hg-
C O(g)-+e-→O(g)2-
D Na(g)→Nag++e-

The correct answer is:
H(g)+e-→Hg-

Solution:
H(g)+e- →extremicH-(g)	ΔHeg=-72KJ/mol

O-(g)+e-⟶ →endeterm c O2-(g)	ΔHeg=+744 KJ/mol

Ar(g)+e- →endeterm c Ar-(g)	ΔHeg=+96 KJ/mol

Na(g)⟶ endotherm c Na+(g)+e-	IE =495.8KJ/mol

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2020 (04 Sep Shift 2)
The process that mainly place in stars to produce energy

Options:
A nuclear fission
B nuclear fusion
C ionization
D annihilation

The correct answer is:
nuclear fusion

Solution:
In deep inside the core of stars, enough protons collide into each other with enough speed and form a helium nucleus generating a tremendous amount of energy at the same time. This process is called nuclear fusion.

Topic: Physics, Nuclear Physics, AP EAMCET, AP EAMCET 2023 (19 May Shift 1)
The process to increase the rate of crystallisation is known as

Options:
A filtration
B re-crystallisation
C seeding
D distillation

The correct answer is:
seeding

Solution:
Crystallization is defined as how a substance is converted from a liquid solution into a solid crystalline state. In crystallization, seeding is one of the most straightforward methods used to control supersaturation. During seeding, a small mass of crystals is added to a solution to start or increase the crystallization rate at the desired level. It provides sufficient surface area to ensure supersaturation is consumed in a controlled manner.

Topic: Chemistry, General Organic Chemistry, JEE Main
The process with negative entropy change is:

Options:
A Synthesis of ammonia from N2 and H2.
B Dissolution of iodine in water.
C Dissociation of CaSO4 s to CaO s and SO3 g.
D Sublimation of dry ice.

The correct answer is:
Synthesis of ammonia from N2 and H2.

Solution:
If a process leads to the arrangement in a system, it will have a negative entropy, whereas if disorder increases, the entropy will be positive. When the number of moles of a gas increases, or ice melts, the disorder increases. Freezing of water, crystallization, combination reactions etc, leads to an arrangement.
CO2 s→CO2 g; ∆S=+ve ∆ng>0
N2 (g)+3H2 (g)→2NH3(g); ∆S=-ve ∆ng=-ve
CaCO3(s)→CaO(s)+CO2(g); ∆S=+ve ∆ng>0
Dissolution of I2 in H2O; ΔS=+ve.

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2019 (10 Jan Shift 2)
The processes of calcination and roasting in metallurgical industries, respectively, can lead to :

Options:
A Global warming and photochemical smog
B Global warming and acid rain
C Photochemical smog and ozone layer depletion
D Photochemical smog and global warming

The correct answer is:
Global warming and acid rain

Solution:
In Calcination and roasting CO2 and SO2 are released which are responsible for Global warning and acid rain.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2020 (04 Sep Shift 2)
The product 214∙4116∙8148∙161128∙.... to ∞ is equal to:

Options:
A 212
B 214
C 1
D 2

The correct answer is:
212

Solution:
214+216+348+……….∞
S=214+18+116+…..∞ ....i

∴14+18+116+...∞is an infinite G.P with a=14, r=12

S∞=a1-r=12

Using equation i, S=2

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2020 (09 Jan Shift 1)
The product "A" and "B" formed in above reactions are

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2021 (16 Mar Shift 1)
The product "A" in the above reaction is:

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2021 (17 Mar Shift 1)
The product A is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The product ' $\mathrm{A}^{\prime}$ is

Options:
A succinic acid
B melonic acid
C oxalic acid
D maleic acid

The correct answer is:
succinic acid

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The product (A) of the following reaction

Options:
A
B
C
D

The correct answer is:

Solution:
The reaction produced as

β - lactone do not exist but can only be made by special method.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The product B in the above reaction is

Options:
A
B
C Mixture of
and
D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
The product 'B' is

Options:
A
B
C
D

The correct answer is:

Solution:
Reaction of alkyl halides with metals (i.e., Mg) will produce Grignard reagent, which on hydrolysis produce
hydrocarbons.
\( \mathrm{Mg}(\mathrm{OH}) \mathrm{Br} \)
Cyclohexane
(B)

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The product B is

Options:
A Benzyl alcohol
B 2-Phenylethanol
C 1-Phenylethanol
D Quinol

The correct answer is:
2-Phenylethanol

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The product (C) in the below mentioned reaction is:

CH3-CH2-CH2-Br→∆KOHalcA→HBrB→KOHaq∆C

Options:
A Propan-1-ol
B Propene
C Propyne
D Propan-2-ol

The correct answer is:
Propan-2-ol

Solution:
When haloalkane or alkyl halide with a β hydrogen is heated with alcoholic solution of KOH, elimination of a hydrogen atom from β carbon and halogen atom from α-carbon occurs as a result, alkene is formed as product. Since β hydrogen atom is involved in the elimination reaction, it is often called β elimination. An alkene is formed. This alkene then on reaction with HBr undergoes addition reaction to give bromopropane, which on reaction with aqueous KOH gives alcohol, propan-2-ol.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2024 (31 Jan Shift 1)
The product $C$ in the following sequence of chemical reaction is
$\mathrm{CH}_{3}-\mathrm{COOH} \stackrel{\mathrm{Ca}(\mathrm{OH})_{2}}{\longrightarrow} \mathrm{\longrightarrow} \stackrel{\text { Heat }}{\longrightarrow} B \stackrel{\mathrm{NH}_{2} \mathrm{OH}}{\longrightarrow} \mathrm{C}$

Options:
A acetaldoxime
B formaldoxime
C ethane nitrile
D acetoxime

The correct answer is:
acetoxime

Solution:
The complete reaction is as follows

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The product (C) in the following sequence of reactions has ______ \(\pi\) bonds.

Options:

The correct answer is:
4

Solution:
$\pi$ bonds $=4$

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2024 (05 Apr Shift 2)
The product $D$ of the reaction
$\mathrm{CH}_3 \mathrm{Cl} \xrightarrow{\mathrm{KON}}(A) \xrightarrow{\mathrm{H}_2 \mathrm{O}}(B) \xrightarrow{\mathrm{NH}_3}(C) \xrightarrow{\Delta}(D)$ is

Options:
A $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{NH}_2$
B $\mathrm{CH}_3 \mathrm{CN}$
C $\mathrm{HCONH}_2$
D $\mathrm{CH}_3 \mathrm{CONH}_2$

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

Solution:
$\mathrm{CH}_3 \mathrm{Cl} \xrightarrow[-\mathrm{KCl}]{\mathrm{KCN}} \mathrm{CH}_3 \mathrm{CN} \xrightarrow{\mathrm{H}_2 \mathrm{O}} \mathrm{CH}_3 \mathrm{COOH} \xrightarrow{\mathrm{NH}_3} \mathrm{CH}_3 \mathrm{COONH}_4 \xrightarrow{\Delta} \mathrm{CH}_3 \mathrm{CONH}_2$

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The product formed from the following reaction sequence is:

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The product formed in the first step of the reaction of with excess Mg/Et2OEt=C2H5 is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2021 (24 Feb Shift 1)
The product formed in the following reaction.

is:

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2022 (25 Jun Shift 1)
The product formed when toluene is heated in light with $\mathrm{Cl}_2$ and in absence of halogen carrier is

Options:
A Benzotrichloride
B Gammexane
C Chlorobenzene
D None of these

The correct answer is:
Benzotrichloride

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
The product \( (\mathrm{X}) \) obtained in this reaction is

Options:
A \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} \)
B \( \left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHOH} \)
C \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} \)
D \( \mathrm{HO}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{OH} \)

The correct answer is:
\( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} \)

Solution:
The nucleophilic addition of Grignard reagent at the epoxide group gives primary alcohol.

X = CH 2 OH | - CH 2 - CH 3

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The product obtained by the reaction of \( \mathrm{HBr} \) with phenol is

Options:
A
B
C
D There is no reaction

The correct answer is:
There is no reaction

Solution:
Phenol reacts with PCl5 to form chlorobenzene. Halogenation of phenol does not take place with HX.

Topic: Chemistry, Hydrocarbons, JEE Main
The product obtained from the electrolytic oxidation of acidified sulphate solutions, is :

Options:
A HSO4-
B HO3SOOSO3H
C HO2SOSO2H
D HO3SOSO3H

The correct answer is:
HO3SOOSO3H

Solution:
Electrolysis of concentrated solution of acidified sulphate solution yields H2S2O8

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2021 (27 Jul Shift 1)
The product obtained when acetic acid is treated with phosphorus trichloride is

Options:
A $\mathrm{CH}_3 \mathrm{COOPCl}_3$
B $\mathrm{CH}_3 \mathrm{COOCl}$
C $\mathrm{CH}_3 \mathrm{COCl}$
D $\mathrm{ClCH}_2 \mathrm{COOH}$

The correct answer is:
$\mathrm{CH}_3 \mathrm{COCl}$

Solution:
$3 \mathrm{CH}_3 \mathrm{COOH}+\mathrm{PCl}_3 \rightarrow \mathrm{H}_3 \mathrm{PO}_3+3 \mathrm{CH}_3 \mathrm{COCl}$

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The product of a 9Ã4 matrix and a 4Ã9 matrix contains a variable x in exactly two places. If Dx is the determinant of the matrix product such that D0=1, D-1=1 and D2=7, then D-2 is equal to

Options:

The correct answer is:
3

Solution:
Let, the order of matrix A is 9×4 and matrix B is 4×9
Dx=AB= a polynomial of degree 2 as it has x at 2 places.
Let, Dx=ax2+bx+c
D0=c=1
D-1=a-b+c=1⇒a-b=0
D2=4a+2b+c=7⇒4a+2b=6
⇒2a+b=3
Solving these two equations, we get,

a=1,b=1
⇒Dx=x2+x+1
⇒D-2=4-2+1=3

Topic: Mathematics, Determinants, JEE Main
The product of all the values of Î», such that the lines x+2y-3=0, 3x-y-1=0 and Î»x+y-2=0 cannot form a triangle, is equal to

Options:

The correct answer is:
1.8

Solution:
Case I: x+2y-3=0 and Î»x+y-2=0 are parallel
âÎ»1=12âÎ»=0.5
Case II: 3x-y-1=0 and Î»x+y-2=0 are parallel
âÎ»3=1-1âÎ»=3
Case III: Lines x+2y-3=0, 3x-y-1=0 and 2Î»x+y-2=0 are concurrent
âÎ»1-212-33-1-1=0â-5Î»-8+14=0âÎ»=1.2
Hence, the product of all the values of Î»=0.5Ã3Ã1.2=1.8

Topic: Mathematics, Straight Lines, JEE Main
The product of following reaction is

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
The product of oxidation of $\mathrm{I}^{-}$with $\mathrm{MnO}_4^{-}$in alkaline medium is

Options:
A $10_3^{-}$
B $I_2$
C $10^{-}$
D $10_4^{-}$

The correct answer is:
$10_3^{-}$

Solution:
$6 \mathrm{MnO} 4+1+6 \mathrm{OH}-----\gt6 \mathrm{MnO}+2 / \mathrm{O} 3+3 \mathrm{H} 2 \mathrm{O}$

Topic: Chemistry, Redox Reactions, JEE Main
The product of the above reaction is

Options:
A
B
C
D

The correct answer is:

Solution:
It is an example of intramolecular Cannizzaro reaction.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The product of the following reaction is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Amines, JEE Main
The product of the following reaction is P.

The number of hydroxyl groups present in the product P is________.

Options:

The correct answer is:
0

Solution:
The reaction proceeds via addition of Grignard's reagent (PhMgBr).

A Grignard reagent is an organomagnesium compound which can be described by the chemical formula ‘R-Mg-X’ where R refers to an alkyl or aryl group and X refers to a halogen.

When C6H4CHOOH reacts with grignard's reagent benzene along with an addition product is formed, which on further reaction with aq ammonium chloride gives C6H4CHOOH.

Here Ph- is a strong base which is extracting proton from the reactant to give benzene.

The reaction is as follows,

Product P is benzene and it has zero hydroxyl group.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2024 (31 Jan Shift 1)
The product of the lengths of perpendiculars drawn from any point on the hyperbola $x^2-2 y^2-2=0$ to its asymptotes is

Options:
A $1/2$
B $2/3$
C $3/2$
D $2$

The correct answer is:
$2/3$

Solution:
Given equation is $\frac{x^2}{2}-\frac{y^2}{1}=1$ ....(i)
Product of length of perpendiculars drawn from any point on the hyperbola (i) to the asymptotes is $\frac{a^2 b^2}{a^2+b^2}=\frac{2 \times 1}{2+1}=\frac{2}{3}$

Topic: Mathematics, Hyperbola, JEE Main
The product of the lengths of perpendiculars drawn from any point on the hyperbola $x^2-2 y^2-2=0$ to its asymptotes is :

Options:
A $\frac{1}{2}$
B $\frac{2}{3}$
C $\frac{3}{2}$
D 2

The correct answer is:
$\frac{2}{3}$

Solution:
Equation of hyperbola is
$x^2-2 y^2=2$
$\Rightarrow \quad \frac{x^2}{2}-\frac{y^2}{1}=1$
Here, $a^2=2, b^2=1$
Equation of asymptotes to the hyperbola
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \text { is } \frac{x^2}{a^2}-\frac{y^2}{b^2}=0$
$\therefore \quad \frac{x}{a}-\frac{y}{b}=0$ and $\frac{x}{a}+\frac{y}{b}=0$
Let $P(a \sec \theta, b \tan \theta)$ be any point, then the product of length of perpendiculars
$=\frac{\left[\frac{a \sec \theta}{a}+\frac{b \tan \theta}{b}\right]}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}} \frac{\left[\frac{a \sec \theta}{a}+\frac{b \tan \theta}{b}\right]}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}$
$=\frac{\sec ^2 \theta-\tan ^2 \theta}{\frac{1}{a^2}+\frac{1}{b^2}}=\frac{1}{\frac{1}{2}+\frac{1}{1}}=\frac{2}{3}$

Topic: Mathematics, Hyperbola, JEE Main
The product of the perpendicular distances drawn from the points $(3,0)$ and $(-3,0)$ to the tangent of the ellipse $\frac{x^2}{36}+\frac{y^2}{27}=1$ at $\left(3, \frac{9}{2}\right)$ is

Options:
A 36
B 27
C 9
D 63

The correct answer is:
27

Solution:
Let $P(6 \cos \theta, 3 \sqrt{3} \sin \theta)$ be any point on the ellipse $\frac{x^2}{36}+\frac{y^2}{27}=1$. The equation of the tangent at $P(6 \cos \theta, 3 \sqrt{3} \sin \theta)$ is $\frac{x}{6} \cos \theta+\frac{y}{3 \sqrt{3}} \sin \theta=1$.
$$
\Rightarrow \quad 3 \sqrt{3} x \cos \theta+6 y \sin \theta-18 \sqrt{3}=0
$$
The product of the lengths of the perpendiculars from $(3,0)$ and $(-3,0)$ an Eq. (i) is given by
$$
\begin{aligned}
& P=\left|\frac{3 \times 3 \sqrt{3} \cos \theta-18 \sqrt{3}}{\sqrt{27 \cos ^2 \theta+36 \sin ^2 \theta}}\right|\left|\frac{3 \times 3 \sqrt{3} \cos \theta+18 \sqrt{3}}{\sqrt{27 \cos ^2 \theta+36 \sin ^2 \theta}}\right| \\
& =\frac{36 \times 27-9 \times 27 \cos ^2 \theta}{36 \sin ^2 \theta+27 \cos ^2 \theta} \\
& =\frac{9 \times 27\left(4-\cos ^2 \theta\right)}{36\left(1-\cos ^2 \theta\right)+27 \cos ^2 \theta} \\
& =\frac{9 \times 27\left(4-\cos ^2 \theta\right)}{36-9 \cos ^2 \theta} \\
& =\frac{9 \times 27\left(4-\cos ^2 \theta\right)}{9\left(4-\cos ^2 \theta\right)}=27 . \\
&
\end{aligned}
$$

Topic: Mathematics, Ellipse, JEE Main
The product of the reaction between ethyl benzene and N-bromosuccinamide is

Options:
A

B

C

D

The correct answer is:

Solution:
$$
\mathrm{Ag}>\mathrm{Hg}>\mathrm{Cu}
$$

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2012 (19 May Online)
The product of the reaction give below is:

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2016 (03 Apr)
The product of the reaction given below is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
The product of the reaction is

Options:
A
B
C
D

The correct answer is:

Solution:
It is a substitution reaction, so the product
formed is

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
The product of the roots of the equation 9x2-18|x|+5=0 is :

Options:
A 59
B 2581
C 527
D 259

The correct answer is:
2581

Solution:
∵x2=|x|2

9x2-18x+5=0

⇒(3x-1)(3x-5)=0

⇒|x|=13,53

∴x=13,-13,53,-53

product of roots =13-1353-53=2581

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2020 (05 Sep Shift 1)
The product of the roots of the equation whose roots are greater by unity than the roots of the equation x3-5x2+6x-3=0 is equal to

Options:
A 3
B 12
C 15
D 18

The correct answer is:
15

Solution:
Let the roots of the equation x3-5x2+6x-3=0 are α, β, γ, then
α+β+ γ=5, αβ+βγ+γα=6, αβγ=3
Roots of the other equation are greater by unity, so it will be
α+1, β+1, γ+1
Now, α+1β+1γ+1
=αβγ+αβ+βγ+γα+α+β+γ+1
=3+6+5+1
=15

Topic: Mathematics, Quadratic Equation, JEE Main
The product of the slopes of the common tangents of the ellipse x2+4y2=16 and the parabola y2-4x-4=0 is

Options:
A -115
B -15
C -13
D -12

The correct answer is:
-15

Solution:
Given, ellipse: x216+y24=1 and parabola: y2=4x+1
Putting x+1=X and Y=y, we get,

Y2=4X

So, the equation of the tangent is Y=mX+1m=mx+1+1m
Y=mx+m+1mâ¦i
If i is also a tangent to x216+y24=1,

then, c2=a2m2+b2
âm+1m2=16m2+4
âm2+1m2+2=16m2+4
â15m2-1m2+2=0
Let, m2=t>0

â15t2+2t-1=0

ât=15,-13âm2=15

Hence, m=Â±15

Topic: Mathematics, Ellipse, JEE Main
The product of three consecutive terms of a G.P. is 512. If 4 is added to each of the first and the second of these terms, the three terms now form an A.P., then the sum of the original three terms of the given G.P. is :

Options:
A 28
B 24
C 32
D 36

The correct answer is:
28

Solution:
To minimize the calculation, 3 numbers in G.P. can be taken as ar, a, ar where, a is the first term and r is the common ratio.

Given product of three numbers is 512.

⇒ar.a.ar=512⇒a3=83⇒a=8

Also given, ar+4, a+4, ar are in

A.P.

⇒2a+4=ar+4+ar

Putting a=8, we get

212=8r+4+8r

⇒2r2-5r+2=0

⇒2r-1r-2=0

⇒r=2 or 12

Hence, numbers may be 4, 8, 16 or 16, 8, 4

In both the cases, sum =4+8+16=28.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2019 (12 Jan Shift 1)
The product of three geometric means between 4 and $\frac{1}{4}$ will be

Options:
A $4$
B $2$
C $-1$
D $1$

The correct answer is:
$1$

Solution:
We have
$4, g_1, g_2, g_3, \frac{1}{4}$ is a G.P.
Here $a=4$,
$g_1=a r=4 \times r, \quad g_2=a r^2, \quad g_3=a r^3$
and $g_4=a r^4=4 \times r^4=\frac{1}{4}$
$\Rightarrow r^4=\frac{1}{16}=\left(\frac{1}{2}\right)^4 \Rightarrow r=\frac{1}{2}$
Now product of three G.M. $g_1 \cdot g_2 \cdot g_3=a r \cdot a r^2 \cdot a r^3$ mon.poce
$=a^3 r^6=4^3 \times\left(\frac{1}{2}\right)^6=\frac{4^3}{4^3}=1$
Note : The product of ' $n$ 'geometric means between ' $a$ ' and $\frac{1}{a}$ is always equal to 1 .

Topic: Mathematics, Sequences and Series, JEE Main
The product of two complex numbers each of unit modulus is also a complex number, of

Options:
A Unit modulus
B Less than unit modulus
C Greater than unit modulus
D None of these

The correct answer is:
Unit modulus

Solution:
If $\left|z_1\right|=1$ and $\left|z_2\right|=1$, then $\left|z_1 z_2\right|=\left|z_1 \| z_2\right|=1.1=1$

Topic: Mathematics, Complex Number, JEE Main
The product (P) formed from the following multistep reaction is:

Options:
A

B
C
D

The correct answer is:

Solution:
The first step of the sequence is involved with electrophilic substitution reaction. Nitro group is deactivating group and methyl group is activating group, hence, the substitution reaction takes place with respect to methyl group. In the second step nitro group is reduced to amine group with hydrogen in the presence of palladium metal. The aniline undergo diazotisation in the third step.The diazonium salt loses nitrogen gas in the presence of H3PO2.

Topic: Chemistry, Amines, JEE Main, JEE Main 2023 (08 Apr Shift 2)
The product Q and R in the following reactions respectively are H2+CH3C≡CH→ Lindlar's catalystP→Zn+H2OO3Q+R

Options:
A Ethanol. Methanoic Acid
B Ethanoic acid, Methanol
C Ethanal, Methanal
D Ethanoic acid, Methanoic acid

The correct answer is:
Ethanal, Methanal

Solution:
Lindlar's Catalyst transforms an alkyne to a alkene because the hydrogenation reaction is occurring on the surface of the poisned metal. The alkene reacts with ozone in the presence of Zn/H2O gives carbonyl compounds.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The product (s) of the following reaction sequence is (are)

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Amines, JEE Main
The product, which is not obtained during the electrolysis of brine solution is

Options:
A H2
B HCl
C NaOH
D Cl2

The correct answer is:
HCl

Solution:
The aqueous solution of sodium chloride is known as brine solution. On electrolysis of brine solution, the following reactions are observed.

Oxidation reaction takes place at anode and reduction reaction takes place at cathode. The reaction

Anode: 2OH-→12O2+H2O+2e⊝

Cathode: H2O+e⊝→12H2+OH⊝

Hydrogen and oxygen are electrode products. The remaining solution contain NaOH.

Hence, HCl is not obtained

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2023 (06 Apr Shift 2)
The product 'X' formed in above reaction is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The product (X) will be

Options:
A
B
C
D

The correct answer is:

Solution:
Here glyoxal undergoes intramolecular Cannizzaro reaction as follows:

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The products A and B are

Options:
A
B
C
D

The correct answer is:

Solution:
LAH ( LiAlH4 ) reduces (CHO) group to CH2OH and also reduces double bond when Ph group is attached to β -position of double bond, whereas NaBH4 reduces only -CHO group to CH2OH group.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The products formed in the following reaction
$\mathrm{C}_6 \mathrm{H}_5-\mathrm{O}-\mathrm{CH}_3+\mathrm{HI} \xrightarrow{\text { heat }}$ are

Options:
A $\mathrm{C}_6 \mathrm{H}_5-\mathrm{I}$ and $\mathrm{CH}_3-\mathrm{OH}$
B $\mathrm{C}_6 \mathrm{H}_5-\mathrm{OH}$ and $\mathrm{CH}_3-\mathrm{I}$
C $\mathrm{C}_6 \mathrm{H}_5-\mathrm{CH}_3$ and $\mathrm{HOl}$
D $\mathrm{C}_6 \mathrm{H}_6$ and $\mathrm{CH}_3 \mathrm{Ol}$

The correct answer is:
$\mathrm{C}_6 \mathrm{H}_5-\mathrm{OH}$ and $\mathrm{CH}_3-\mathrm{I}$

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The products obtained during treatment of hard water using Clark's method are

Options:
A CaCO3 and MgCO3
B CaOH2 and MgOH2
C CaCO3 and MgOH2
D CaOH2 and MgCO3

The correct answer is:
CaCO3 and MgOH2

Solution:
In this method calculated amount of lime is added to hard water. It precipitates out calcium carbonate and magnesium hydroxide which can be filtered off In Clark's method lime water is used.

It is because of low solubility product of Mg(OH)2 as compared to that of MgCO3.

CaHCO32+2CaOH2→2CaCO3+2H2O

MgHCO32+2CaOH2→2CaCO3+MgOH2+2H2O

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2022 (28 Jul Shift 2)
The products obtained from a reaction of hydrogen peroxide and acidified potassium permanganate are

Options:
A Mn4+,H2O only
B Mn2+,H2O only
C Mn4+,H2O,O2 only
D Mn2+,H2O,O2 only

The correct answer is:
Mn2+,H2O,O2 only

Solution:
Hydrogen peroxide acts as an oxidising as well as reducing agent in both acidic and alkaline media.

In this reaction MnO4- act as an oxidising agent and hydrogen peroxide acts as a reducing agent. The reaction is shown below,

2MnO4-+6H++5H2O2→2Mn2++5O2+8H2O

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2022 (26 Jul Shift 2)
The products obtained from the following chemical reactions are respectively

(i) NaOH ( hotââandââconc. ) + Cl 2 â

(ii) XeF 6 ( excess ) + H 2 Oâ

Options:
A NaOH and XeO 3
B HClO 3 and XeO 2 F 2
C NaClO 3 and XeO 3
D None of these

The correct answer is:
NaClO 3 and XeO 3

Solution:
The reactions are as follows:
(i) 6 NaOH ( hot  and  conc. ) +3 Cl 2 →5NaCl+ NaClO 3 +3 H 2 O
(ii) XeF 6 ( excess ) +3 H 2 O→ XeO 3 +6HF

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main