datavorous/entrance-exam-dataset huggingface dataset:
$\mathrm{CH}_3 \mathrm{COOH} \stackrel{\mathrm{LiAlH}_4}{\longrightarrow} A$
$\mathrm{A}+\mathrm{CH}_3 \mathrm{COOH} \stackrel{\mathrm{H}_3 \mathrm{O}^{+}}{\longrightarrow} \mathrm{B}+\mathrm{H}_2 \mathrm{O}$
In the above reactions ' $A$ ' and ' $B$ ' respectively are

Options:
A $\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5, \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$
B $\mathrm{CH}_3 \mathrm{CHO}, \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$
C $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}, \mathrm{CH}_3 \mathrm{CHO}$
D $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}, \mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5$

The correct answer is:
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}, \mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5$

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
$\mathrm{CH}_3 \mathrm{COOH} \xrightarrow{\mathrm{LiAlH}_4}$$(A) \xrightarrow{I_2+\mathrm{Na}{OH}}(B) \xrightarrow{\text { Ag(Dust)}}(C)$
the final product $(C)$ is

Options:
A $\mathrm{C}_2 \mathrm{H}_5 I$
B $\mathrm{C}_2 \mathrm{H}_3 \mathrm{OH}$
C $\mathrm{C}_2 \mathrm{H}_2$
D $\mathrm{CH}_3 \mathrm{COCH}_3$

The correct answer is:
$\mathrm{C}_2 \mathrm{H}_2$

Solution:
$\mathrm{CH}_3 \mathrm{COOH} \xrightarrow{\mathrm{LiAlH}_4} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH} \xrightarrow{\mathrm{I}_2 / \mathrm{NaOH}} \mathrm{CHI}_3 \xrightarrow{\mathrm{Ag}} \mathrm{C}_2 \mathrm{H}_2$

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
$\mathrm{CH}_{3} \mathrm{COOH} \stackrel{\mathrm{LiAlH}_{4}}{\longrightarrow} \mathrm{X} \underset{300^{\circ} \mathrm{C}}{\stackrel{\mathrm{Cu}}{\longrightarrow}} \mathrm{Y} \underset{\mathrm{NaOH}}{\stackrel{\text { Dilute }}{\longrightarrow}} \mathrm{Z}$
In the above reaction $Z$ is

Options:
A butanol
B aldol
C ketol
D acetal

The correct answer is:
aldol

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main
$\mathrm{CH}_3 \mathrm{COOH} \xrightarrow[\mathrm{P}_2 \mathrm{O}_3]{\Delta} \mathrm{X}$, Identify X

Options:
A $\mathrm{CH}_3 \mathrm{COCH}_3$
B $\mathrm{CH}_3 \mathrm{CHO}$
C $\left(\mathrm{CH}_3 \mathrm{CO}\right)_2 \mathrm{O}$
D $\mathrm{CH}_4$

The correct answer is:
$\left(\mathrm{CH}_3 \mathrm{CO}\right)_2 \mathrm{O}$

Solution:
$\mathrm{CH}_3 \mathrm{COOH} \xrightarrow{\Delta / \mathrm{P}_2 \mathrm{O}_5}\left(\mathrm{CH}_3 \mathrm{CO}\right)_2 \mathrm{O}$

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
$\mathrm{CH}_3-\mathrm{Mg}-\mathrm{Br}$ is an organo metallic compound due to

Options:
A
$\mathrm{Mg}-\mathrm{Br}$ bond

B
$C$ - $Mg$ bond

C
$C$ - $B r$ bond

D
$\mathrm{C}-\mathrm{H}$ bond

The correct answer is:
$C$ - $Mg$ bond

Solution:
Bond between C of organic molecule and metal atom

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2002
$\mathrm{CH}_3 \mathrm{Mgl}$ is an organometallic compound due to

Options:
A $M g-I$ bond
B $C-I$ bond
C $C-M g$ bond
D $\mathrm{C}-\mathrm{H}$ bond

The correct answer is:
$C-M g$ bond

Solution:
Due to $C-M g$ bond.

Topic: Chemistry, Coordination Compounds, JEE Main
$\mathrm{CH} \equiv \mathrm{CH} \frac{\mathrm{HCl}}{\text { Excess }} A \frac{\text { Hydrolysis }}{\Delta} B \frac{\text { Reduction }}{\mathrm{Na}-\mathrm{Hg} / \mathrm{H}_{2} \mathrm{O}} C$
Compound $C$ is

Options:
A ethanal
B propenal
C ethanol
D 2-methyl butan-1-ol

The correct answer is:
ethanol

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
$\mathrm{CH} \equiv \mathrm{CH} \xrightarrow[\mathrm{H}_2 \mathrm{SO}_4]{\mathrm{HgSO}_4} \xrightarrow[\mathrm{H}_2 \mathrm{O}]{\mathrm{CH}_3 \mathrm{MgBr}^2} \xrightarrow{\mathrm{P} / \mathrm{Br}_2}$

Options:
A $\mathrm{CH}_3 \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_3$
B $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}$
C $\mathrm{CH}_2=\mathrm{CH}-\mathrm{Br}$
D $\mathrm{BrCH}=\mathrm{CH}-\mathrm{CH}_3$

The correct answer is:
$\mathrm{CH}_3 \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_3$

Solution:
$\mathrm{CH} \equiv \mathrm{CH} \xrightarrow[\mathrm{H}_2 \mathrm{SO}_4]{\mathrm{HgSO}_4} \mathrm{CH}_3 \mathrm{CHO} \xrightarrow{\mathrm{CH}_3 \mathrm{MgBr}}$

Topic: Chemistry, Hydrocarbons, JEE Main
Chalcogen group elements are :

Options:
A Se, Te and Po.
B O, Ti and Po.
C Se, Tb and Pu.
D S, Te and Pm.

The correct answer is:
Se, Te and Po.

Solution:
Oxygen family also called as chalcogen family So, O, S, Se, Te, Po are chalcogen family member. These are ore forming elements.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2021 (26 Aug Shift 2)
Change in temperature of the medium changes

Options:
A
frequency of sound waves

B
amplitude of sound waves

C
wavelength of sound waves

D
loudness of sound waves

The correct answer is:
wavelength of sound waves

Solution:
As we know that,
$$
v_t=v_0(1+0.61 t)
$$
Speed of sound wave in a medium $v \propto \sqrt{T}$
(where $T$ is temperature of the medium)
Clearly, when temperature increase then the speed also increase as frequency does not change during propogation of wave by formula.
As, $v=v \lambda$
where $v$ is frequency and $\lambda$ is wavelength.
Frequency $(v)$ remains fixed,
$v \propto \lambda$ or $\lambda \propto v$
So velocity $v$ and wavelength $(\lambda)$ both increases.

Topic: Physics, Waves and Sound
Change in volume of the system does not alter the number of moles in which of the following equilibria ?

Options:
A
$\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})$

B
$\mathrm{PCl}_5(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})$

C
$\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g})$

D
$\mathrm{SO}_2 \mathrm{Cl}_2(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_2(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})$

The correct answer is:
$\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})$

Solution:
In this reaction the ratio of number of moles of reactants to products is same i.e. 2 : 2, hence change in volume will not alter the number of moles.

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2002
Characteristic feature of alkali metals is

Options:
A Good conductor of heat and electricity
B High melting points
C Low oxidation potentials
D High ionization potentials

The correct answer is:
Good conductor of heat and electricity

Solution:
Alkali metal are good conductor of heat and electricity.

Topic: Chemistry, s Block Elements, JEE Main
Charge is distributed within a sphere of radius R with a volume charge density ρr=Ar2e-2ra, where A and a are constants. If Q is the total charge of this charge distribution, the radius R is:

Options:
A a2log⁡11-Q2πaA
B alog⁡11-Q2πaA
C alog⁡1-Q2πaA
D a2log⁡1-Q2πaA

The correct answer is:
a2log⁡11-Q2πaA

Solution:
ρr=Ar2 e-2ra
Charge enclosed between r and r+dr is
dq=ρr 4πr2dr
To get total charge Q
Q=∫0Rdq=∫0RAr2e-2r/a 4πr2dr=-4πAa2[e-2r/a]0R
Q=-4π Aa2e-2R/a-1
R=-ln⁡1-Q24πAaa2=a2ln⁡11-Q2πaA

Topic: Physics, Electrostatics, JEE Main, JEE Main 2019 (09 Jan Shift 2)
Charge \( Q \), is divided into two parts which are then kept some distance apart. The force between them will be maximum if the two parts are having the charge -

Options:
A \( Q / 2 \) each
B \( Q / 4 \) and \( 3 Q / 4 \)
C \( Q / 3 \) and \( 2 Q / 3 \)
D \( e \) and \( (Q-e) \), where \( e= \) electron charge

The correct answer is:
\( Q / 2 \) each

Solution:
Let charges are q and ( Qâq ).
force F= kq( Q–q ) r 2 = k r 2 ( qQ− q 2 )
For maximum force
dF dq =0
∴ k r 2 [ Q−2q ]=0
∴ q= Q 2
â´ Charge of other part
=Q−q
=Q− Q 2 = Q 2

Topic: Physics, Electrostatics, JEE Main
Charge $q$ is uniformly spread on a thin ring of radius $R$. The ring rotates about its axis with a uniform frequency $f \mathrm{~Hz}$. The magnitude of magnetic induction at the centre of the ring is

Options:
A $\frac{\mu_0 q f}{2 R}$
B $\frac{\mu_0 q}{2 f R}$
C $\frac{\mu_0 q}{2 \pi f R}$
D $\frac{\mu_0 q f}{2 \pi R}$

The correct answer is:
$\frac{\mu_0 q f}{2 R}$

Solution:
We know
$$
\begin{array}{ll}
B=\frac{\mu_0 i}{2 R} \\
q=i t \Rightarrow i=\frac{q}{t}=q f \quad\left[\because f=\frac{1}{t}\right] \\
B=\frac{\mu_0 q f}{2 R}
\end{array}
$$

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2011 (Mains)
Charges are placed at the vertices of a square as shown in the diagram. If charges at A and B are interchanged with C and D respectively, then,

Options:
A Only magnitude of electric field will change at the centre
B Both magnitude and direction of electric field will change at the centre
C Only direction of electric field at centre will change
D Both magnitude and direction of electric field will remain unchanged

The correct answer is:
Only direction of electric field at centre will change

Solution:

Topic: Physics, Electrostatics, NEET
Charges are placed at the vertices of square as shown in diagram. If charge at A and B are interchanged with C and D respectively then

Options:
A only magnitude of electric field will change at the centre
B both magnitude and direction of electric field will change at the centre
C only direction of electric field at centre will change
D both magnitude and direction of electric field will remain unchanged

The correct answer is:
only direction of electric field at centre will change

Solution:
Direction of electric field will be as shown

Therefore only direction of electric field at centre will change

Topic: Physics, Electrostatics, NEET
Charges are placed on the vertices of a square as shown. Let $\mathrm{E}$ be the electric field and $\mathrm{V}$ the potential at the centre. If the charges on $\mathrm{A}$ and $\mathrm{B}$ are interchanged with those on $\mathrm{D}$ and $\mathrm{C}$ respectively, then

Options:
A
$\mathrm{\vec{E}}$ remains unchanged, $\mathrm{V}$ changes

B
Both $\mathrm{\vec{E}}$ and $\mathrm{V}$ change

C
$\mathrm{\vec{E}}$ and $\mathrm{V}$ remains unchanged

D
$\mathrm{\vec{E}}$ changes, $\mathrm{V}$ remains unchanged

The correct answer is:
$\mathrm{\vec{E}}$ changes, $\mathrm{V}$ remains unchanged

Solution:
As $\overrightarrow{\mathrm{E}}$ is a vector quantity

Topic: Physics, Electrostatics, JEE Main, JEE Main 2007
Charges of $+\frac{10}{3} \times 10^{-9} \mathrm{C}$ are placed at each of the four corners of a square of side $8 \mathrm{~cm}$. The potential at the intersection of the diagonals is

Options:
A $150 \sqrt{2}$ volt
B $1500 \sqrt{2}$ volt
C $900 \sqrt{2}$ volt
D $900$volt

The correct answer is:
$1500 \sqrt{2}$ volt

Solution:
Potential at the centre 0 , $V=4 \times \frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{a / \sqrt{2}}$
Where $Q=\frac{10}{3} \times 10^{-9} \mathrm{C}$ and $a=8 \mathrm{~cm}=8 \times 10^{-2} \mathrm{~m}$
So $V=5 \times 9 \times 10^9 \times \frac{\frac{10}{3} \times 10^{-9}}{\frac{8 \times 10^{-2}}{\sqrt{2}}}=1500 \sqrt{2}$ volt

Topic: Physics, Electrostatics, JEE Main
Charges Q1 and Q2 are at points A and B of a right-angled triangle OAB. The resultant electric field at point O is perpendicular to the hypotenuse, then Q1/Q2 is proportional to:

Options:
A x13x23
B x2x1
C x1x2
D x22x12

The correct answer is:
x1x2

Solution:
Using triangle properties

tanθ=x1x2tanθ=E2E1=KQ2/x22KQ1/x12⇒x1x2=KQ2/x22KQ1/x12

Q1Q2= x1x2

Topic: Physics, Electrostatics, JEE Main, JEE Main 2020 (06 Sep Shift 1)
Charges $q, 2 q, 3 q$ and $4 q$ are placed at the corners $A, B, C$ and $D$ of a square as shown in the following figure. The direction of electric field at the centre of the square is along

Options:
A $A B$
B $C B$
C $B D$
D $A C$

The correct answer is:
$C B$

Solution:
$E_A=E, E_B=2 E, E_C=3 E, E_D=4 E$

Topic: Physics, Electrostatics, JEE Main
Charges Q, 2Q and -Q are given to three concentric conducting spherical shells A, B and C respectively, as shown in the figure. The ratio of charges on the inner and outer surfaces of the shell C will be

Options:

The correct answer is:
-1.5

Solution:
Due to induction of charge on surface facing each other will be equal and opposite.
Ratio = - 3 2

Topic: Physics, Electrostatics, JEE Main
Charges Q, 2Q and -Q are given to three concentric conducting spherical shells A, B and C respectively, as shown in the figure. The ratio of charges on the inner and outer surfaces of the shell C will be ________. If your answer is K, mark absolute value of 2K as answer after rounding off to nearest integer.

Options:

The correct answer is:
3

Solution:
Due to induction of charge on surface facing each other will be equal and opposite.
Ratio = - 3 2

Topic: Physics, Electrostatics, JEE Main
Charges -q and +q, located at A and B, respectively, constitute an electric dipole. Distance AB=2a, O is the mid point of the dipole and OP is perpendicular to AB. A charge Q is placed at P where OP=y and y≫2a. The charge Q experiences an electrostatic force F. If Q is now moved along the equatorial line to P' such that OP'=y3 the force on Q will be close toy3≪2a

Options:
A 27F.
B F3.
C 3F.
D 9F.

The correct answer is:
27F.

Solution:
Electric field intensity at the distance y along the perpendicular bisector is
E1=KPy3
So, force at distance y on Q,
F=Q(E1)=QKPy3 ...(1)
F'=QKPy33 ...(2)

Dividing 1 by 2, we get,

F'F=33
∴ F'=27F

Topic: Physics, Electrostatics, JEE Main, JEE Main 2019 (10 Jan Shift 2)
Charges Q are placed at the ends of a diagonal of a square and charges q are placed at the other two corners. The condition for the net electric force on Q to be zero is

Options:
A Q=-22q
B Q=-q2
C Q=-8 q
D Q=-2 q

The correct answer is:
Q=-22q

Solution:
F'+2F=0
F'=-2F
14ÏÎµ0 .Q22a2=-214ÏÎµ0 .qQa2
Q2=-2q
Q=-22q

Topic: Physics, Electrostatics, JEE Main
Check the injectivity and surjectivity of the following functions:
(i) $\mathrm{f}: \mathrm{N} \rightarrow \mathrm{N}$ given by $f(x)=x^2$
(ii) $f: Z \rightarrow Z$ given by $f(x)=x^2$
(iii) $f: R \rightarrow R$ given by $f(x)=x^2$
(iv) $f: N \rightarrow N$ given by $f(x)=x^3$
(v) $f: Z \rightarrow Z$ given by $f(x)=x^3$

Options:

None

Solution:
(i) $f: N \rightarrow N$ given by $f(x)=x^2$
(a) $f\left(x_1\right) \Rightarrow f\left(x_2\right) \Rightarrow x_1{ }^2=x_2{ }^2 \Rightarrow x_1=x_2$ $\therefore \mathrm{f}$ is one-one i.e. it is injective.
(b) There are such member of codomain which have no image in domain $\mathrm{N}$.
e.g. $3 \in$ codomain N. But there is no pre-image in domain of $\mathrm{f}$.
$\Rightarrow \mathrm{f}$ is not onto i.e. not surjective.
(ii) $f: z \rightarrow z$ given by $f(x)=x^2$
(a) $f(-1)=f(1)=1 \Rightarrow-1$ and 1 have the same image.
$\therefore \mathrm{f}$ is not one-one i.e. not injective.
(b) There are many such elements belonging to codomain have no pre-image in its codomain z.
e.g. $3 \in$ codomain $\mathrm{z}$ but $\sqrt{3} \notin$ domain $\mathrm{z}$ of $\mathrm{f}$,
$\therefore \mathrm{f}$ is not onto i.e. not surjective
(iii) $f: R \rightarrow R$, given by $f(x)=x^2$
(a) $f$ is not one-one since $f(-1)=f(1)=1$
$-1$ and 1 have the same image i.e., fis not injective
(b) $-2 \in$ codomain $\mathrm{R}$ of f but $\sqrt{-2}$ does not belong to domain $\mathrm{R}$ of $\mathrm{f}$.
$\Rightarrow \mathrm{f}$ is not onto i.e. $\mathrm{f}$ is not surjective.
(iv) Injective but not surjective.
(v) Injective but not surjective.

Topic: Mathematics, Relations and Functions (Class 12)
Check up the only correct statement in the following

Options:
A a body has a constant velocity and still it can have a varying speed
B a body has a constant speed but it can have a varying velocity
C a body having constant speed cannot have any acceleration
D a body in motion under a force acting upon it must always have work done upon it

The correct answer is:
a body has a constant speed but it can have a varying velocity

Solution:
If velocity is constant then both magnitude i.e. speed and direction remains constant and acceleration is zero.
But if speed is constant then direction of velocity may change and therefore it may have acceleration.

Topic: Physics, Motion In Two Dimensions, NEET
Chemical name of washing soda is:

Options:
A Sodium chloride
B Sodium hydrogen carbonate
C Sodium carbonate
D Sodium hydroxide

The correct answer is:
Sodium carbonate

Solution:
Washing soda is a chemical compound with the formula $\mathrm{Na}_2 \mathrm{CO}_3$, known as sodium carbonate, and it's a salt of carbonic acid.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
Chemically heroin is

Options:
A
morphine monoacetate

B
morphine dibenzoate

C
morphine diacetate

D
morphine monobenzoate

The correct answer is:
morphine diacetate

Solution:
No Solution Available

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2012 (12 May Online)
Chemisorption

Options:
A Increases with temperature
B Decreases with temperature
C Remains unaffected by change of temperature
D Either increases or decreases with temperature

The correct answer is:
Either increases or decreases with temperature

Solution:
Chemisorption initially increases then decreases with rise in temperature. The initial increase is due to the fact that heat supplied acts as activation energy. The decrease afterward is due to the exothermic nature of adsorption equilibrium.

Topic: Chemistry, Surface Chemistry, JEE Main
Chemisorption

Options:
A involves the weak attractive interactions between adsorbent and adsorbate
B is irreversible in nature
C decreases with increase of temperature
D involves multilayer formation of adsorbent on adsorbate

The correct answer is:
is irreversible in nature

Solution:
Chemical adsorption is irreversible due to formation of new bonds.

Topic: Chemistry, Surface Chemistry, JEE Main
Options:
A Liebermann's reaction
B Phthalein fusion test
C Reimer-Tiemann reaction
D Schotten-Baumann reaction

The correct answer is:
Liebermann's reaction

Solution:
The given reaction is Liebermann's reaction. It is test for phenol.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Options:
A
B
C CH3-CH=CH-CH3
D

The correct answer is:

Solution:
The reaction between 3-methylbutan-2-ol and HBr is explained below.
In the first step, −OH group is protonated. In the second step, a molecule of water is lost to form secondary carbocation. Third step is the rearrangement of less stable secondary carbocation to more stable tertiary carbocation through 1,2-hydride ion shift. Final step is the nucleophilic attack of bromide ion on tertiary carbocation to form 2-bromo-2-methylbutane as shown below.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Options:
A
B
C
D

The correct answer is:

Solution:
Here benzene is formed as follows

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Options:
A
B
C
D

The correct answer is:

Solution:
Aromatic ring is most stable than aliphatic ring

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Options:
A
B
C
D

The correct answer is:

Solution:
Ammoniacal silver nitrate solution is Tollens’ reagent. Tollens’ reagent can be used to distinguish aldehyde & ketone as aldehyde upon warming with Tollens’ reagent produces a silver mirror due to formation of silver metal in alkaline medium. Aldehyde is oxidised to corresponding carboxylate anion.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Options:
A
B
C
D

The correct answer is:

Solution:
Here intramolecular Aldol condensation occurs and elimination takes place via E1cb mechanism due to presence of EWG.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Options:
A
B
C
D

The correct answer is:

Solution:
Nitrobenzene is reduced to aniline by passing hydrogen gas in presence of finely divided palladium and also by reduction with metals in an acidic medium. When aniline reacts with acetic anhydride in the presence of pyridine to give acetanilide. Aniline acts as a weak base, while acetic anhydride acts as electrophile.

Topic: Chemistry, Amines, JEE Main, JEE Main 2023 (31 Jan Shift 1)
Options:
A
B
C
D

The correct answer is:

Solution:
This conversion occurs as follows

Topic: Chemistry, Amines, JEE Main
Options:
A
2-bromo-6-chlorocyclohex-1-ene

B
6-bromo-2-chlorocyclohexene

C
3-bromo-1-chlorocyclohexene

D
1-bromo-3-chlorocyclohexene

The correct answer is:
3-bromo-1-chlorocyclohexene

Solution:
No Solution Available

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2006
Options:
A
B
C
D

The correct answer is:

Solution:
The tautomeric form is as follows

Topic: Chemistry, General Organic Chemistry, JEE Main
Options:
A
B
C
D All

The correct answer is:

Solution:
This reaction is possible as follows

Topic: Chemistry, General Organic Chemistry, JEE Main
Options:
A resonating structures
B tautomers
C geometrical isomers
D optical isomers

The correct answer is:
resonating structures

Solution:
Structure I and structure II are the resonating forms because the position of atoms remains the same and only electron redistribution takes place.

Topic: Chemistry, General Organic Chemistry, NEET, NEET 2002
Options:
A
B
C
D

The correct answer is:

Solution:
Cl¯ is –I > +M but ortho/para directing
–M group is at leaving group to ortho present so ArSN2 will takes place.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Options:
A
B
C
D

The correct answer is:

Solution:
âCl is the good leaving group in the compound and SHâ is the best nucleophile. So, S N 2 reaction is more favoured.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Options:
A
B
C
D

The correct answer is:

Solution:
The reaction sequence can be given as follows

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Options:
A $\mathrm{CH}_3-\mathrm{Br}$
B $\mathrm{CH}_3 \mathrm{COI}$
C $\mathrm{CH}_3 \mathrm{COOH}$
D None of these

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

Solution:
$\mathrm{CH}_3 \mathrm{COOAg}+\mathrm{Br}_2 \xrightarrow{\mathrm{CS}_2} \mathrm{CH}_3 \mathrm{Br}+\mathrm{AgBr}+\mathrm{CO}_2$

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Options:
A

B

C

D

The correct answer is:

Solution:
It is oxymercuration–demercuration reaction. In this an alkene is treated with mercury(II) acetate, and the product is treated with sodium borohydride. The net result is a Markovnikov addition product without rearrangement.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2022 (26 Jun Shift 2)
Options:
A
B
C
D

The correct answer is:

Solution:
This conversion occurs as follows

Topic: Chemistry, Hydrocarbons, JEE Main
Options:
A $\mathrm{I}_2, \mathrm{IO}_4^{-}$
B $\mathrm{I}_2, \mathrm{IO}_3^{-}$
C $\mathrm{IO}_3^{-}, \mathrm{IO}_3^{-}$
D $\mathrm{IO}_3^{-}, \mathrm{I}_2$

The correct answer is:
$\mathrm{I}_2, \mathrm{IO}_3^{-}$

Solution:
(a) $2 \mathrm{MnO}_4^{-}+10 \mathrm{I}^{-}+16 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Mn}^{2+}+5 \mathrm{I}_2+8 \mathrm{H}_2 \mathrm{O}$
(b) $2 \mathrm{MnO}_4^{-}+\mathrm{I}^{-}+\mathrm{H}_2 \mathrm{O} \longrightarrow \underset{\mathrm{Y}}{2} \underset{\mathrm{X}}{2 \mathrm{MnO}_2}+\mathrm{IO}_3^{-}+\underset{\mathrm{O}}{-}$ Hence, $(\mathrm{X})$ and $(\mathrm{Y})$ are respectively $\mathrm{I}_2$ and $\mathrm{IO}_3^{-}$.

Topic: Chemistry, Redox Reactions, JEE Main
Chemosensitive area of respiratory centre in medulla is affected by

Options:
A less $\mathrm{CO}_2$ and $\mathrm{H}^{+}$ions
B less $\mathrm{O}_2$ and $\mathrm{H}^{+}$ions
C excess $\mathrm{CO}_2$ and $\mathrm{H}^{+}$ions
D excess $\mathrm{O}_2$ and $\mathrm{H}^{+}$ions.

The correct answer is:
excess $\mathrm{CO}_2$ and $\mathrm{H}^{+}$ions

Solution:
Human beings have a significant ability to maintain and moderate the respiratory rhythm to suit the demands of the body tissues. A specialised centre present in the medulla region of the brain called respiratory rhythm centre is primarily responsible for this regulation. A chemosensitive area is situated adjacent to the rhythm centre which is highly sensitive to $\mathrm{CO}_2$ and hydrogen ions. Increase in these substances can activate this centre, which in turn can signal the rhythm centre to make necessary adjustments in the respiratory process.

Topic: Biology, Breathing and Exchange of Gases, AIIMS, AIIMS 2010