datavorous/entrance-exam-dataset huggingface dataset:
An observer while going on scooter hears sound of two sirens of same frequencies from two opposite directions. If he travels along the direction of one of the siren, then he

Options:
A Listens resonance
B Listens beats
C Will not listen sound due to destructive interference
D Will listen intensive sound due to constructive interference

The correct answer is:
Listens beats

Solution:
Correct option is 2. Listens beats
Let the two waves produced due to the two sirens have frequency and velocity as $\nu_1, v_1$ and $\nu_2, v_2$
Since, both are sound waves, we have $v_1=v_2=v$
For one wave the traveler is moving towards the source,
$\therefore \nu^{\prime}=\nu\left(\frac{v-v_o}{v}\right)$
For the other the traveler is moving away from the source
$\therefore \nu^{\prime}=\nu\left(\frac{v+v_o}{v}\right)$
Seeing this, we can conclude that the waves are in phase with each other and are fluctuating repeatedly.
Since, they are sound waves they are periodic in nature. Hence, we got two waves that are periodic and are repeatedly fluctuating. This is the condition of beats. So, beats will be produced and the traveler will listen to beats. Option (2) is the correct answer.

Topic: Physics, Waves and Sound, JEE Main
An observer whose least distance of distinct vision is d, views his own face in a convex mirror of radius of curvature r. The magnification produced can not exceed?

Options:
A r d + r 2 + d 2
B r d + d 2 - r 2
C r d - r + d
D r d + d + r

The correct answer is:
r d + r 2 + d 2

Solution:
Magnification, â m = f f - u â m = f f + x
m = r / 2 r / 2 + x â m = r r + 2 x ------- 1
from mirror formula
1 v + 1 u = 1 f â 1 ( d - x ) + 1 - x = 1 f â 1 d - x - 1 x = 2 r
+ x - ( d - x ) x ( d - x ) = 2 r â 2 x - d r = 2 x d - x
2rx - rd = 2xd - 2x2 â 2x2 + 2(r - d)x - rd = 0
2 x 2 + 2 r - d x - r d = 0 â x = - 2 r - d Â± 4 r - d 2 + 8 r d 4
x = 2 d - r Â± 4 r 2 + d 2 4 â x = d - r Â± r 2 + d 2 2
2 x = d - r Â± r 2 + d 2
putting the value of x in equation (1)
m = r r + [ d - r Â± r 2 + d 2 ]
For m to be maximum, x should be minimum
m min = r d + r 2 + d 2

Topic: Physics, Ray Optics, JEE Main
An octahedral complex is formed when hybrid orbitals of the following type are involved

Options:
A $s p^{3}$
B $d s p^{2}$
C $d^{2} s p^{3}$
D $s p^{2} d^{2}$

The correct answer is:
$d^{2} s p^{3}$

Solution:
An octahedral complex is formed when hybrid orbital of $s p^{3} d^{2}$ (outer orbital) or $d^{2} s p^{3}$ (inner orbital) types are involved.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
An octahedral complex of Co3+ is diamagnetic. The hybridisation involved in the formation of the complex is :

Options:
A dsp2
B d2sp3
C dsp3d
D sp3d2

The correct answer is:
d2sp3

Solution:
2 7 Co 3 + has 24 Electrons i.e 6 electrons in inner 3d orbital of central metal ion for diamagnetic behaviour these d6 Electrons have to be paired and due to pairing two inner d-orbitals get empty and hybridisation is d2sp3.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2014 (09 Apr Online)
An octahedral complex with molecular composition M.5NH3.Cl.SO4 has two isomers, A and B. The solution of A gives a white precipitate with AgNO3 solution and the solution of B gives white precipitate with BaCl2 solution .The type of isomerism exhibited by the complex is :

Options:
A Coordinate isomerism
B Geometrical isomerism
C Ionisation isomerism
D Linkage isomerism

The correct answer is:
Ionisation isomerism

Solution:
The central metal atom M is in +3 oxidation state and ionisation isomers due to exchange of ligands outside & inside the co-ordination sphere are
[M(NH3)5SO4][A]Cl→AgNO3AgCl(s)↓white ppt
[M (N H 3 ) 5 Cl] [B] S O 4 → BaC l 2 BaS O 4 (s) ↓ white  ppt
[A] and [B] are ionisation isomers

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2014 (19 Apr Online)
An octahedral complex with the formula \(\mathrm{CoCl}_3 \cdot \mathrm{nNH}_3\) upon reaction with excess of \(\mathrm{AgNO}_3\) solution gives 2 moles of \(\mathrm{AgCl}\). Consider the oxidation state of \(\mathrm{Co}\) in the complex is ' \(x\) '. The value of " \(x+n\) " is ______

Options:
A 6
B 8
C 3
D 5

The correct answer is:
8

Solution:
$\begin{aligned} & \mathrm{x}+0-1-2=0 \\ & \mathrm{x}=+3 \\ & \mathrm{n}=5 \\ & \therefore \mathrm{x}+\mathrm{n}=8\end{aligned}$

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2024 (08 Apr Shift 1)
An oil drop carrying a charge q has a mass m kg. It is falling freely in air with terminal speed v. The electric field required to make the drop move upwards with the same speed is

Options:
A mgq
B 2mgq
C mgvq2
D 2mgvq

The correct answer is:
2mgq

Solution:
QV â = 4 3 π r 3 ρ g
When the oil drop is falling freely under the effect of gravity is a viscous medium with terminal speed v, then
mg=6π η r v ...(i)
To move the oil drop upward with terminal velocity v if E is the electric field intensity applied, the
Eq=mg+6πηrv=mg+mg=2mg

Topic: Physics, Electromagnetic Waves, JEE Main
An oil drop of the radius 2 mm with a density 3 g cm-3 is held stationary under a constant electric field 3.55×105 V m-1 in the Millikan's oil drop experiment. What is the number of excess electrons that the oil drop will possess? consider g=9.81 m s-2.

Options:
A 48.8×1011
B 1.73×1010
C 17.3×1010
D 1.73×1012

The correct answer is:
1.73×1010

Solution:
qE=Mg⇒neE=ρ43πr3×g
⇒n×1.6×10-19×3.55×105=3×103×43×π×2×10-33×9.81
⇒n=173×10(3-9-5+19)⇒n=1.73×1010.

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2021 (18 Mar Shift 1)
An open beaker of water in equilibrium with water vapour is in a sealed container. When a few grams of glucose are added to the beaker of water, the rate at which water molecules :

Options:
A leaves the vapour increases
B leaves the solution increases
C leaves the solution decreases
D leaves the vapour decreases

The correct answer is:
leaves the vapour increases

Solution:
The vapour pressure of solution will be less than vapour pressure of pure solvent, so some vapour molecules will get condensed to maintain new equilibrium.

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2020 (02 Sep Shift 1)
An open cylindrical vessel filled with water is rotated together with water at angular speed 1 rad s-1. At this angular speed, the centre of the bottom is just exposed. If the angular speed is increased to 32 rad s-1, how much area (in cm2) of the bottom of the vessel is now exposed to surrounding? [Area of the cross-section of cylinder is 27 cm2]

Options:

The correct answer is:
9

Solution:
Let ω1 be initial angular velocity
ω12R22g=h
For angular velocity ω2 let radius of exposed surface be x
ω222gR2-x2=h
Putting the value of ω1 and ω2
πx2=9
Since πR2=27

Topic: Physics, Mechanical Properties of Solids, JEE Main
An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional 46 cm. What will be length of the air column above mercury in the tube now ?
(Atmospheric pressure = 76 cm of Hg)

Options:
A 16 cm
B 22 cm
C 38 cm
D 6 cm

The correct answer is:
16 cm

Solution:
Using isothermal process conditions
⇒ P 1 V 1 = P 2 V 2 ⇒ P 1 A ℓ 1 = P 2 A ℓ 2
8 p 0 = xp ⇒ p = 8 p 0 x (here p0 = 76 cm of Hg)
8 p 0 x + ρ m g 54 - x = ρ m g (76)
8 x + 54 - x 76 = 1
x = 16 cm

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2014 (06 Apr)
An open manometer attached to a flask containing ammonia gas have no difference in mercury level initially as shown in diagram. After sparking into the flask, ammonia is partially dissociated as 2 NH 3 g ⟶ N 2 g + 3 H 2 g . Now it have difference of 18 cm in mercury level in two columns, what is partial pressure of H 2 g at equilibrium ?

Options:
A 9âcm of Hg
B 18 cm of Hg
C 27âcm of Hg
D None of these

The correct answer is:
27 cm of Hg

Solution:
2 NH 3 g â¶ N 2 g + 3 H 2 7 6 - 2 x x 3 x
7 6 + 2 x = 7 6 + 1 8
2 x = 1 8
â´ x = 9
â´ p H 2 = 3 Ã 9
= 27 cm of Hg

Topic: Chemistry, States of Matter, JEE Main
An open pipe is in resonance in 2nd harmonic with frequency f1. Now one end of the tube is closed and frequency is increased to f2 such that the resonance again occurs in nth harmonic. Choose the correct option.

Options:
A n = 3 , fâ¡ 2 = 3 4 fâ¡ 1
B n = 3 , fâ¡ 2 = 5 4 fâ¡ 1
C n = 5 , fâ¡ 2 = 5 4 fâ¡ 1
D n = 5 , fâ¡ 2 = 3 4 fâ¡ 1

The correct answer is:
n = 5 , fâ¡ 2 = 5 4 fâ¡ 1

Solution:
fâ¡ 1 = vâ¡ lâ¡ (2nd harmonic of open pipe)
fâ¡ 2 = n vâ¡ 4 lâ¡ (nth harmonic of closed pipe)
Here, n is odd and fâ¡ 2 > fâ¡ 1
It is possible when n = 5
because with n = 5
fâ¡ 2 = 5 4 vâ¡ lâ¡ = 5 4 fâ¡ 1

Topic: Physics, Waves and Sound, JEE Main
An open pipe of length / vibrates in fundamental mode. The pressure variation is maximum at

Options:
A $1 / 4$ from ends
B The middle of pipe
C The ends of pipe
D At $1 / 8$ from ends of pipe middle of the pipe

The correct answer is:
The middle of pipe

Solution:
The pressure variation is maximum at a distance $1 / 2$ because the displacement node is pressure antinode

Topic: Physics, Waves and Sound, JEE Main
An open pipe of sufficient length is dipping in water with a speed v vertically. If at any instant l is the length of the tube above water. Then the rate at which fundamental frequency of pipe changes, is (speed of sound = c)

Options:
A cv2l2

B cv4l2
C c2v2l2

D c4v2l2

The correct answer is:
cv4l2

Solution:
f=c4l
∴ dfdt=c4l2dldt=cv4l2

Topic: Physics, Waves and Sound, JEE Main
An open pipe, when closed at one end, can resonate in its third harmonic with a frequency which is 100 Hz more than its fundamental frequency as an open pipe. If the fundamental frequency of the pipe when it is open at both ends is 50n Hz, then the value of n is

Options:

The correct answer is:
4

Solution:
Length of the organ pipe is same in both the cases. Fundamental frequency of open pipe is f 1 = v 2 l Frequency of third harmonic of closed pipe will be.
fâ¡ 2 = 3 vâ¡ 4 lâ¡
Given that, fâ¡ 2 = fâ¡ 1 + 1 0 0
or fâ¡ 2 - fâ¡ 1 = 1 0 0
or 3 4 vâ¡ lâ¡ - 1 2 vâ¡ lâ¡ = 1 0 0 â vâ¡ 4 lâ¡ = 1 0 0
â´ vâ¡ 2 lâ¡ or fâ¡ 1 = 2 0 0 Hz 50.ânâHzân = 4
Therefore, fundamental frequency of the open pipe is 200 Hz. and n = 4

Topic: Physics, Waves and Sound, JEE Main
An open U-tube contains mercury. When $11.2 \mathrm{~cm}$ of water is poured into one of the arms of the tube, how high does the mercury rise in the other arm from its initial unit?

Options:
A $0.56 \mathrm{~cm}$
B $1.35 \mathrm{~cm}$
C $0.41 \mathrm{~cm}$
D $2.32 \mathrm{~cm}$

The correct answer is:
$0.41 \mathrm{~cm}$

Solution:
On pouring water on left side, mercury rises $x$ $\mathrm{cm}$ (say) from its previous level in the right limb of U-tube creating a difference of levels of mercury by $2 x \mathrm{~cm} .$ Equating pressure at $A$ and $B$, we get
$$
\begin{array}{l}
\quad p_{A}=p_{B} \\
\therefore 11.2 \times 10^{-2} \times \rho_{\text {water }} \times g=2 x \times \rho_{\mathrm{Hg}} \times g \\
11.2 \times 10^{-2} \times 1000 \mathrm{~kg} / \mathrm{m}^{3} \\
=2 x \times 13600 \mathrm{~kg} / \mathrm{m}^{3} \\
x=\frac{11.2 \times 10^{-2} \times 1000}{2 \times 13600} \mathrm{~m} \\
=0.41 \mathrm{~cm}
\end{array}
$$

Topic: Physics, Kinetic Theory of Gases, JEE Main
An open vessel at 27°C is heated until three fifth of the air has been expelled. Assuming the volume of the vessel remains constant the temperature to which vessel has to be heated is

Options:
A 1500 K
B 75 K
C 750 K
D None

The correct answer is:
750 K

Solution:
From question, pressure and volume are constant
T1=273+27=300K
n1=1,
n2=1-35=25
n1n2=T2T1
So T2=n1T1n2=5×3002=750K .

Topic: Chemistry, States of Matter, NEET
An open vessel at 27o C is heated until 3/8 th of the air in it has been expelled. Assuming that the volume remains constant, calculate the temperature at which the vessel was heated

Options:
A 307 o  C
B 107 o  C
C 480 o  C
D 207 o  C

The correct answer is:
207 o  C

Solution:
If 3th8 of the air is expelled out then remaining air =58
In the open vessel volume and pressure are both constant but moles are decreasing
â´â n 1 T 1 â=â n 2 T 2
â´ n 1 âÃâ300â= 5 8 n 1 T 2
â´â T 2 â=â 8 5 âÃâ T 1 â=â 8 5 âÃâ300â=â480âK
=480â273= 207 o C

Topic: Chemistry, States of Matter, JEE Main
An open vessel at 27oC is heated until two fifth of the air (assumed as an ideal gas) in it has escaped from the vessel. Assuming that the volume of the vessel remains constant, the temperature to which the vessel has heated is:

Options:
A 750oC
B 500oC
C 750 K
D 500 K

The correct answer is:
500 K

Solution:
n1T1=n2T2
ânÃ300=n-2n5T2
â300=35T2
âT2=500 K

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2019 (12 Jan Shift 2)
An open vessel at $300 \mathrm{~K}$ is heated till $2 / 5^{\text {th }}$ of the air in it is expelled. Assuming that the volume of the vessel remains constant, the temperature to which the vessel is heated, is

Options:
A
$1500 \mathrm{~K}$

B
$400 \mathrm{~K}$

C
$500 \mathrm{~K}$

D
$750 \mathrm{~K}$

The correct answer is:
$500 \mathrm{~K}$

Solution:
No Solution Available

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2012 (19 May Online)
An ordered pair α, β for which the system of linear equations
1+αx+βy+z=2
αx+1+βy+z=3
αx+βy+2z=2 has a unique solution, is :

Options:
A -3,Â 1
B 1, -3
C 2,Â 4
D -4,Â 2

The correct answer is:
2, 4

Solution:
For unique solution 1+αβ1α1+β1αβ2≠0

Applying C1→C1+C2+C3, we get

α+β+2β1α+β+21+β1α+β+2β2≠0

⇒α+β+21β111+β11β2≠0

⇒α+β+2≠0

Clearly, point 2, 4 satisfying the given condition.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2019 (12 Jan Shift 1)
An ordinary dice is rolled for a certain number of times. If the probability of getting an odd number 2 times is equal to the probability of getting an even number 3 times, then the probability of getting an odd number for odd number of times is:

Options:
A 132
B 516
C 316
D 12

The correct answer is:
12

Solution:
C2n12n=C3n12n⇒C2n=C3n

⇒ n=5

Probability of getting an odd number for odd number of times is

C15125+C35125+C55125=1255+10+1

=12

Topic: Mathematics, Probability, JEE Main, JEE Main 2021 (24 Feb Shift 1)
An organ pipe of length L is open at one end and closed at other end. The wavelengths of the three lowest resonating frequencies that can be produced by this pipe are

Options:
A 4L, 2L, L
B 2L, L, L/2
C 2L, L, 2L/3
D 4L, 4L/3, 4L/5

The correct answer is:
4L, 4L/3, 4L/5

Solution:
L=2n-1λ4
λ=4L2n-1
λ1=4L21-1=4L
λ2=4L4-1=4L3
λ3=4L6-1=4L5

Topic: Physics, Waves and Sound, JEE Main
An organ pipe open at one end and closed at other, has a length L. The wavelengths of the three lowest resonating frequencies that can be produced by this pipe are

Options:
A 4L, 2L, L
B 2L, L, L/2
C 2L, L, 2L/3
D 4L, 4L/3, 4L/5

The correct answer is:
4L, 4L/3, 4L/5

Solution:
L=(2n−1)λ4
λ=4L(2n−1)

Hence,
λ1=4L2(1)−1=4L

λ2=4L4−1=4L3
λ3=4L6−1=4L5

Topic: Physics, Waves and Sound, JEE Main
An organ pipe open at one end is vibrating in first overtone and is in resonance with another pipe open at both ends and vibrating in third harmonic. The ratio of length of two pipes is

Options:
A 1:2
B 4:1
C 8:3
D 3:8

The correct answer is:
1:2

Solution:
If 3V 4 l 1 = 3V 2 l 2
l 1 l 2 = 1 2

Topic: Physics, Waves and Sound, JEE Main
An organ pipe $P_1$ closed at one end vibrating in its first overtone and another pipe $P_2$ open at both ends vibrating in its third overtone are in resonance with a given tuning fork. The ratio of lengths of $P_1$ and $P_2$ is

Options:
A 1:2
B 1:3
C 3:8
D 3:4

The correct answer is:
1:2

Solution:
An organ pipe P1 closed at one end and vibrating in its first overtone pipe P2 open at booth ends vibrating in its third overtone are in resonance with a given tuning fork. The ratio of the lengths of $\mathrm{P} 1$ to that of $\mathrm{P} 2$ is. $1 / 2$.

Topic: Physics, Waves and Sound, JEE Main
An organic acid A reacts with conc. H2SO4 to give a neutral oxide B, acidic oxide C and a diatomic oxide D. When D reacts with chlorine gas, a poisonous gas E is evolved. This gas with NH3 gives an organic compound F. The compound A and F are

Options:
A H2C2O4,NH2CONH2
B CH3COOH,NH2CONH2
C CHCl3,CH3CONH2
D CCl4,CH3CONH2

The correct answer is:
H2C2O4,NH2CONH2

Solution:
H 2 C 2 O 4 A Oxalic acid  +  H 2 SO 4 → H 2 O B Neutral  +  CO 2 C Acidic  +  CO D Diatomic oxide
CO+Cl2→COCl2(E)
COCl 2  +  2NH 3 → NH 2 CONH 2 F Urea  + 2HCl

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
An organic compound 'A' C4H8on treatment with KMnO4/H+ yields compound 'B' C3H6O. Compound 'A' also yields compound 'B' an ozonolysis. Compound 'A' is :

Options:
A 2-Methylpropene
B 1-Methylcyclopropane
C But-2-ene
D Cyclobutane

The correct answer is:
2-Methylpropene

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2021 (25 Jul Shift 1)
An organic compound $\mathrm{A}, \mathrm{C}_5 \mathrm{H}_8 \mathrm{O}$; reacts with $\mathrm{H}_2 \mathrm{O}, \mathrm{NH}_3$ and $\mathrm{CH}_3 \mathrm{COOH}$ as described below:

Options:
A

B

C

D

The correct answer is:

Solution:
Given compound $\mathrm{A}$ is
$$
\text { Reactions given are as following : }
$$

Topic: Chemistry, Aldehydes and Ketones, JEE Main
An organic compound $\mathrm{A}, \mathrm{C}_5 \mathrm{H}_8 \mathrm{O}$; reacts with $\mathrm{H}_2 \mathrm{O}, \mathrm{NH}_3$ and $\mathrm{CH}_3 \mathrm{COOH}$ as described below:

Options:
A

B

C

D

The correct answer is:

Solution:
Given compound $\mathrm{A}$ is
$$
\text { Reactions given are as following : }
$$

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2014 (11 Apr Online)
An organic compound A C 5 H 8 on hydrogenation gives compound B C 5 H 12 compound A on ozonolysis give methanal and 2-ketopropanal. The compound A is

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
An organic compound AC6H6O gives dark green colouration with ferric chloride. On treatment with CHCl3 and KOH, followed by acidification gives compound B. Compound B can also be obtained from compound C on reaction with pyridinium chlorochromate (PCC). Identify A, B and C.

Options:
A

B

C

D

The correct answer is:

Solution:
The reaction with ferric chloride is test for phenol. The reaction of A with CHCl3 and KOH to give B is Riemer-Tiemann reaction. The conversion C to B is an oxidation reaction.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2021 (22 Jul Shift 1)
An organic compound A containing C, H and O has a pleasant odour with boiling point of 78°C . On boiling A with concentrateed H2SO4 , a colourless gas is produced which decolourises bromine water and alkaline KMnO4. The organic liquid A is

Options:
A C2H5Cl
B C2H5COOCH3
C C2H5OH
D C2H6

The correct answer is:
C2H5OH

Solution:
The organic liquid A is C2H5OH .
(i) Ethyl alcohol is a colourless liquid with a characteristic pleasant smell, having boiling point 78.1Â°C .
(ii) C 2 H 5 OH → conc.  H 2 SO 4 CH 2  =  CH 2 Decolourises  Br 2  water   and alkaline  KMnO 4

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
An organic compound ‘A’ is oxidized with Na2O2 followed by boiling with HNO3. The resultant solution is then treated with ammonium molybdate to yield a yellow percipitate. Based on a above observation, the element present in the given compound is:

Options:
A Fluorine
B Phosphorus
C Nitrogen
D Sulphur

The correct answer is:
Phosphorus

Solution:
The phosphorus containing organic compounds are detected by 'Lassaigne's test' by heated with an oxidizing agent (sodium peroxide). The phosphorus present in the compound is oxidised to phosphate.

The solution is boiled with nitric acid and then treated with ammonium molybdate to produced canary yellow precipitate.

Phosphorus gets oxidised when it is treated with fuming conc. HNO3.
Na3PO4+3HNO3→H3PO4+3NaNO3

Phosphoric acid is used along with molybdenum compound for the identification of Magnesium.

Topic: Chemistry, Practical Chemistry, JEE Main, JEE Main 2019 (12 Apr Shift 1)
An organic compound (A) (molecular formula C6H12O2) was hydrolysed with dil. H2SO4 to give a carboxylic acid (B) and an alcohol (C). ‘C ’ gives white turbidity immediately when treated with anhydrous ZnCl2 and conc. HCl. The organic compound (A) is:

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2020 (04 Sep Shift 1)
An organic compound [A], molecular formula C10H20O2 was hydrolyzed with dilute sulphuric acid to given a carboxylic acid [B] and an alcohol [C]. Oxidation of [C] with CrO3-H2SO4 produced [B]. Which of the following structures are not possible for [A]

Options:
A
B
C CH3CH2CH2COOCH2CH2CH2CH3
D CH33-C-COOCH2CCH33

The correct answer is:

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2020 (03 Sep Shift 1)
An organic compound 'A' on reaction with NH3 followed by heating gives compound B. Which on further strong heating gives compound C C8H5NO2. Compound C on sequential reaction with ethanolic KOH, alkyl chloride and hydrolysis with alkali gives a primary amine. The compound A is

Options:
A

B

C

D

The correct answer is:

Solution:
The carboxylic acids on heating with ammonia give amides. A diamide on strong heating give an imide.

Gabriel Phthalimide reaction: The hydrogen of the imide is acidic which can be extracted with base. Now, the nucleophile formed undergo bimolecular nucleophilic substitution reaction with alkyl chloride. Now, the alkyl imide undergo hydrolysis to give primary amine.

Topic: Chemistry, Amines, JEE Main, JEE Main 2022 (25 Jul Shift 1)
An organic compound A on treatment with benzene sulfonyl chloride gives compound B. B is soluble in dil. NaOH solution. Compound A is :

Options:
A C6H5-NCH32
B C6H5-NHCH2CH3

C C6H5-CH2NHCH3
D

The correct answer is:

Solution:
Hinsberg reagent (Benzene sulfonyl chloride) gives reaction product with 1° amine, and it is soluble in dil. NaOH.

Topic: Chemistry, Amines, JEE Main, JEE Main 2021 (18 Mar Shift 2)
An organic compound ' $A$ ' on treatment with $\mathrm{NH}_3$ gives ' $B$ ' which on heating gives ' $C$ ', ' $C$ ' when treated with $\mathrm{Br}_2$ in the presence of $\mathrm{KOH}$ produces ethylamine. Compound ' $A$ ' is

Options:
A $\mathrm{CH}_3 \mathrm{COOH}$
B $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{COOH}$
C

D $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH}$

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

Solution:
The organic compound ' $A$ ' is propanoic acid $\left(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH}\right)$.

Topic: Chemistry, Amines, JEE Main
An organic compound ' $A$ ' on treatment with $\mathrm{NH}_3$ gives ' $B$ ' which on heating gives ' $C$ ', ' $C$ ' when treated with $\mathrm{Br}_2$ in the presence of $\mathrm{KOH}$ produces ethylamine. Compound ' $A$ ' is

Options:
A $\mathrm{CH}_3 \mathrm{COOH}$
B $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{COOH}$
C

D $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH}$

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

Solution:
The organic compound ' $A$ ' is propanoic acid $\left(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH}\right)$.

Topic: Chemistry, Amines, NEET, NEET 2011 (Mains)
An organic compound $A$ reacts with sodium metal and forms $B$. On heating with conc. $\mathrm{H}_2 \mathrm{SO}_4, A$ gives diethyl ether. $A$ and $B$ are

Options:
A $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$ and $\mathrm{C}_2 \mathrm{H}_5 \mathrm{ONa}$
B $\mathrm{C}_3 \mathrm{H}_7 \mathrm{OH}$ and $\mathrm{CH}_3 \mathrm{ONa}$
C $\mathrm{CH}_3 \mathrm{OH}$ and $\mathrm{CH}_3 \mathrm{ONa}$
D $\mathrm{C}_4 \mathrm{H}_9 \mathrm{OH}$ and $\mathrm{C}_4 \mathrm{H}_9 \mathrm{ONa}$

The correct answer is:
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$ and $\mathrm{C}_2 \mathrm{H}_5 \mathrm{ONa}$

Solution:
$\begin{aligned} & \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}+\mathrm{Na} \rightarrow \mathrm{C}_2 \mathrm{H}_5 \mathrm{ONa} \\ & 2 \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} \xrightarrow[\mathrm{H}_2 \mathrm{SO}_4]{\text { conc. }} \mathrm{C}_2 \mathrm{H}_5-\mathrm{O}-\mathrm{C}_2 \mathrm{H}_5+\mathrm{H}_2 \mathrm{O}\end{aligned}$

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main