datavorous/entrance-exam-dataset huggingface dataset:
The magnetic moment of an iron bar is $M$. It is now bent in such a way that it forms an arc section of a circle subtending an angle of $60^{\circ}$ at the centre. The magnetic moment of this arc section is

Options:
A $\frac{3 M}{\pi}$
B $\frac{4 M}{\pi}$
C $\frac{M}{\pi}$
D $\frac{2 M}{\pi}$

The correct answer is:
$\frac{3 M}{\pi}$

Solution:
$M=m \cdot L$
$R \theta=L$
$\frac{R \pi}{3}=L$
$R=\frac{3 L}{\pi}$
$M^{\prime}=m(2 R) \sin 30^{\circ}$
$=m(2) \frac{3 L}{\pi} \times \frac{1}{2}=\frac{3}{\pi} m L=\frac{3 M}{\pi}$

Topic: Physics, Magnetic Properties of Matter, NEET, NEET 2024 (Re-NEET)
The magnetic moment of electron due to orbital motion is proportional to (n = principal quantum numbers)

Options:
A 1n2
B 1n
C n2
D n

The correct answer is:
n

Solution:
Magnetic moment M0=e2m0×L0
∴ L0∝n
∴ M0∝n

Topic: Physics, Magnetic Properties of Matter, NEET
The magnetic moment of [MnCl₄]²⁻ is 5.92 BM. Explain giving a reason.

Options:
A The manganese ion has a d⁵ configuration, leading to a magnetic moment of 5.92 BM.
B The manganese ion has a d³ configuration, resulting in a magnetic moment of 5.92 BM.
C The manganese ion has a d⁴ configuration, causing a magnetic moment of 5.92 BM.
D The manganese ion has a d⁶ configuration, contributing to a magnetic moment of 5.92 BM.

The correct answer is:
The manganese ion has a d⁴ configuration, causing a magnetic moment of 5.92 BM.

Solution:
Correct Option is : (C)
The manganese ion has a d⁴ configuration, causing a magnetic moment of 5.92 BM.

Topic: Chemistry, Coordination Compounds, NEET
The magnetic moments of complex given below are in order
I. [Ni(CO)4]
II. [Mn(CN)6]4-
III. [Cr(NH3)6]3+
IV. [CoF6]3-

Options:
A I > II > III > IV
B I < II < III < IV
C IV > II > I > III
D IV < II < I < III

The correct answer is:
I < II < III < IV

Solution:
Magnetic moment μ=n(n+2)B.M.
n = unpaired electrons.
I. [Ni(CO)4] n = 0, so μ→=0 BM
II. [Mn(CN)6]4- n = 1, so μ→=3 BM
III. [Cr(NH3)6]3+ n = 3, so μ→=15 BM
IV. [CoF6]3- n = 4, so μ→=24 BM
Order of magnetic moment is I < II < III < IV.

Topic: Chemistry, Coordination Compounds, NEET
The magnetic needle of a vibration magnetometer makes 12 oscillations per minute in the horizontal component of earth's magnetic field. When an external short bar magnet is placed at some distance along the axis of the needle in the same line, it makes 15 oscillations per minute. If the poles of the bar magnet are interchanged, the number of oscillations it will make per minute is

Options:
A 61
B 63
C 65
D 67

The correct answer is:
63

Solution:
Magnetic field=BH, T=6012=5 s.
Magnetic field=BH + B, T=6015=4 s.
Let after reversing magnetic needle makes x oscillation/s.
âMagnetic field=BH-B, T=60x s.
T Î±1Magnetic fieldâ 54=BH+BBH
âBBH=916â 560/x=BH-BBH
â x=63

Topic: Physics, Magnetic Properties of Matter, NEET
The magnetic needle of a vibration magnetometer makes 12 oscillations per minute in the horizontal component of Earth's magnetic field. When an external short bar magnet is placed at some distance along the axis of the needle in the same line, it makes 15 oscillations per minute. If the poles of the bar magnet are interchanged, the number of oscillations it makes per minute is

Options:
A 61
B 63
C 65
D 67

The correct answer is:
63

Solution:
Magnetic field =BH
T=6012=5 sec.
Magnetic field =BH+B
T=6015=4 sec.
Let after reversing magnetic needle makes x oscillation per minute.
⇒ Magnetic field =BH-B,T=60xsec.
T∝1Magnetic field⇒ 54=BH+BBH
⇒BBH=916
⇒ 560x=BH-BBH
⇒ x=63

Topic: Physics, Magnetic Properties of Matter, NEET
The magnetic potential due to a magnetic dipole at a point on its axis distant 40 cm from its centre is found to be 2.4Ã10-5 J A m-1.The magnetic moment of the dipole will be

Options:
A 28.6 A m2
B 32.2 A m2
C 38.4 A m2
D None of these

The correct answer is:
38.4 A m2

Solution:
Here, r=40 cm=0.4 m
Î¸=0Â° (an axial line)
M=?
As V=Î¼04ÏM cosÎ¸r2
2.4Ã10-5=10-7MÃ10.42
M=38.4 Am2

Topic: Physics, Magnetic Effects of Current, NEET
The magnetic potential energy stored in a certain inductor is 25 mJ, when the current in the inductor is 60 mA . This inductor is of inductance

Options:
A 1.389 H
B 138.88 H
C 0.138 H
D 13.89 H

The correct answer is:
13.89 H

Solution:
For current I through an inductor coil having inductance L , Magnetic potential energy stored in the inductor is given by
U=12LI2

Given

U=25 mJ=25×10-3 J

I=60 mA =60×10-3 A
∴ 25×10-3=12L60×10-32
Solving L=13.89 H

Topic: Physics, Alternating Current, NEET, NEET 2018
The magnetic potential energy, when a magnetic bar of magnetic moment $\vec{m}$ is placed perpendicular to the magnetic field $\vec{B}$ is

Options:
A $-\frac{m B}{2}$
B Zero
C $-m B$
D $m B$

The correct answer is:
Zero

Solution:
Potential energy stored in external magnetic field is given by $U=-\vec{m} \cdot \vec{B}$
$\because$ Angle between $\vec{m}$ and $\vec{B}$ is $90^{\circ}$
$\therefore \quad U=-m B \cos 90^{\circ}$
$=0$

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2024 (Re-NEET)
The magnetic susceptibility is negative for:

Options:
A diamagnetic material only
B paramagnetic material only
C ferromagnetic material only
D paramagnetic and ferromagnetic materials

The correct answer is:
diamagnetic material only

Solution:
As we know the relation between the magnetic permeability and susceptibility of material is, μr=1+χm …(i)

∵ For diamagnetic substances, μr<1

So, according to Eq. (i), the magnetic susceptibility χm of diamagnetic substance will be negative. While in case of para and ferromagnetic substances subsceptibility is positive.

Topic: Physics, Magnetic Properties of Matter, NEET, NEET 2016 (Phase 1)
The magnetic susceptibility of a material of a rod is 499. Permeability of vacuum is 4Ï Ã 10-7 H m-1. Absolute permeability of the material of the rod in H m-1 is

Options:
A π×10-4
B 2ÏÃ10-4
C 3π×10-4
D 4π×10-4

The correct answer is:
2π×10-4

Solution:
We know

μ=μ01+X

⇒μ=500×4π×10-7

⇒μ=2π×10-4 H m-1

Topic: Physics, Magnetic Properties of Matter, NEET
The magnetic susceptibility of a paramagnetic substance at -73 oC is 0.0060, then its value at -173 oC will be

Options:
A 0.0030
B 0.0120
C 0.0180
D 0.0045

The correct answer is:
0.0120

Solution:
χm∝1T
⇒χm1χm2=T2T1
⇒χm1χm2=100200
⇒χm2=2χm1
⇒χm2=2×0.0060=0.0120

Topic: Physics, Magnetic Properties of Matter, NEET
The magnification of an image by a convex lens is positive only when the object is placed

Options:
A At its focus F
B Between F and 2F
C At 2F
D Between F and optical centre

The correct answer is:
Between F and optical centre

Solution:
The magnification of an image by the convex lens is positive only when the object is placed between focal point F and optical centre because image formed by the convex lens for this location of the object is virtual and magnified and for real image m=-ve
For virtual image, m=+ve

Topic: Physics, Ray Optics, NEET
The magnifying power of a telescope is 9. When it is adjusted for parallel rays the distance between the objective and eyepiece is $20 \mathrm{~cm}$. The focal length of lenses are

Options:
A $10 \mathrm{~cm}, 10 \mathrm{~cm}$
B $15 \mathrm{~cm}, 5 \mathrm{~cm}$
C $18 \mathrm{~cm}, 2 \mathrm{~cm}$
D $11 \mathrm{~cm}, 9 \mathrm{~cm}$

The correct answer is:
$18 \mathrm{~cm}, 2 \mathrm{~cm}$

Solution:
Given, $M=\frac{f_0}{f_e}=9$ and $f_0+f_e=20$
$f_o=9 f_e$
So, $9 f_e+f_e=20$
$\begin{aligned}
& \therefore f_e=2 \mathrm{~cm} \\
& f_o=9 \times 2 \\
& f_0=18 \mathrm{~cm}
\end{aligned}$

Topic: Physics, Ray Optics, NEET, NEET 2012 (Screening)
The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the objective and the eye-piece is found to be 20 cm. The focal lengths of the lenses are

Options:
A 18 cm, 2 cm
B 11 cm, 9 cm
C 10 cm, 10 cm
D 15 cm, 5 cm

The correct answer is:
18 cm, 2 cm

Solution:
M=f0fe
9=f0fe or f0=9fe
Also, L=fo+fe or 20=fo+fe
Or 20=9fe+fe or 20=10 fe
Or fe=2 cm
∴fo=9×2 cm=18 cm

Topic: Physics, Ray Optics, NEET
The magnifying power of a telescope is 10 and length of telescope is 1.1 m for normal adjustment. The magnification when image is formed at least distance of distinct vision is

Options:
A 14
B 6
C 16
D 18

The correct answer is:
14

Solution:
Magnifying power of a telescope is f0fe=10
Length of telescope is f0+fe=1.1 m
∴ f0=1 m & fe=0.1 m
When final image is at D; then magnification is
m=f01D+1fe=100125+110
m=14

Topic: Physics, Ray Optics, NEET
The magnitude and direction of the current in the following circuit is

Options:
A 1.5 A from B to A through E
B 0.2 A from B to A through E
C 0.5 A from A to B through E
D 59 A from A to B through E

The correct answer is:
0.5 A from A to B through E

Solution:
Let the direction of electric current in the circuit be i from A to B. Using Kirchhoff's law in clockwise direction, we get

-i×2+10-5-i×1-i×7=0⇒5=10i⇒i=510=0.5 A

Topic: Physics, Current Electricity, NEET, NEET 2023 (All India)
The magnitude and direction of the electric field at point P can be best represented by

Options:
A 3kqx2(i^+j^)
B 2kqx2(i^+j^)
C 32kqx2(-i^-j^)
D 32kqx2(i^+j^)

The correct answer is:
32kqx2(i^+j^)

Solution:
EâP=Eâ²cos45Â°i^+Eâ²sin45Â°j^
= Eâ²i^2+j^2 = k3qx22i^2+j^2
= 6kqx2i^2+j^2 = 32kqx2(i^+j^)

Topic: Physics, Electrostatics, NEET
The magnitude of binding energy of the satellite is E and kinetic energy is K. The ratio E/K is

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

The correct answer is:
1

Solution:
Kinetic energy of satellite (E)
KEk=12mv2
=12mGMR2 âµv=GMR
[ m= mass of satellite]
=12mGMR
[Symbols have usual meanings]
Potential energy of satellite,
U=-mGMR
â´ Binding energy of the satellite,
Eb=KE+U
=12mGMR+-mGMR
=12mGMR-mGMR
=-GMm2R
â´ Magnitude of the binding energy E=Eb=GMm2R
The ratio of EK=GMm2RÃ2RGMm=1

Topic: Physics, Gravitation, NEET
The magnitude of displacement vector of a particle which is moving in a circle of radius a with constant angular velocity ω as a function of time is

Options:
A 2a sin ωt
B 2a sinÏt2
C 2a cos ωt
D 2a cosωt2

The correct answer is:
2a sinÏt2

Solution:
In time t the particle has rotated through an angle Î¸=Ït

This means the magnitude of the displacement of the particle is
sâ¡ = PQ = QR 2 + PR 2
= a sin Ï t 2 + a - a cos Ï t 2
sâ¡=2a sinÏt2

Topic: Physics, Laws of Motion, NEET
The magnitude of electric field intensity at point B (2, 0, 0) due to a dipole of dipole moment pâ=i^+3j^ kept at origin is (assume that the point B is at a large distance from the dipole and K=14ÏÎµ0)

Options:
A 13K8
B 13K4
C 7K8
D 7K4

The correct answer is:
7K8

Solution:
Let p→=i^+3j^=p1→+p2→
Therefore p1→=i^ and p2→=3j^
Field due to p1→ at point B ,
E1=2×K23=K4
Field due to p2→ at point B ,
E2=K×38
Enet=E12+E22
Enet=K216+K2×364
Enet=7K8

Topic: Physics, Electrostatics, NEET
The magnitude of force acting on a body varies with time as shown in the figure. The magnitude of total impulse of the force on the body from t= 0 to t=16 Î¼s is-

Options:
A 6×10-3 N s
B 5.8×10-3 N s
C 8.4×10-3 N s
D 7.6×10-3 N s

The correct answer is:
5.8×10-3 N s

Solution:
Impulse =∫Fdt = area under curve
Area =200×4+12×2×200+800+12×10×800×10-6
Area =800+1000+4000×10-6
Area =5800×10-6=5.8×10-3Ns

Topic: Physics, Center of Mass Momentum and Collision, NEET
The magnitude of resistance X in the circuit shown in the given figure, when no current flows through the 5 Ω resistor is

Options:
A

3â Î©

B

6 Î©

C

9 Î©

D

12 Î©

The correct answer is:

6 Î©

Solution:
Since no current flows through the 5 Î© resistor, the circuit represents a balanced Wheatstone bridge. Therefore,

X 18 = 2 6

X= 2 6 Ã18=6âÎ©

Topic: Physics, Current Electricity, NEET
The magnitude of the magnetic field at O (centre of the circular part) due to the current-carrying coil as shown is:

Options:
A μ0i4π4πa+2b
B μ0i2π3π2a+2b
C μ0i2ππ3a+32b
D μ0i4π3πa+2b

The correct answer is:
μ0i4π3πa+2b

Solution:
Magnetic field due to the circular segment =34.μ0i2a
Due to one straight wire segment =μ0i4πb( sin 45°+ sin 0°)=μ0i42πb
Net field
=3μ0i8a+2 ×μ0i42πb=μ0i4π 3 π a + 2 b

Topic: Physics, Magnetic Effects of Current, NEET
The magnitude of the maximum acceleration is π times that of the maximum velocity of a simple harmonic oscillator. The time period of the oscillator in second is

Options:
A 4
B 2
C 1
D 0.5

The correct answer is:
2

Solution:
Maximum acceleration = Maximum velocity×π
i.e., ω2A=πωA
where A is amplitude and ω is angular velocity.
⇒ ω=π
⇒ 2πT=π
⇒ T=2 s

Topic: Physics, Oscillations, NEET
The magnitude of x and y components of A→ are 7 and 6 respectively. Also, the magnitudes of x and y components of A→+B→ are 11 and 9 respectively. Calculate the magnitude of vector B→.

Options:
A 10
B 5
C 6
D 3

The correct answer is:
5

Solution:
Given, A→=7i^+6j^
Let B→=b1i^+b2j^ ......(i)
Given that, A→+B→=11i^+9j^
Or 7+b1i^+6+b2j^=11i^+9j^
Comparing the scalar component of i^ and j^
7+b1=11 and 6+b2=9
⇒b1=4 and b2=3
∴ From Equation. (i),
B→=4i^+3j^
B→=42+32=25=5

Topic: Physics, Mathematics in Physics, NEET
The major androgen present in humans is

Options:
A testosterone
B estrogen
C androsterone
D progesterone

The correct answer is:
testosterone

Solution:
Testosterone is the major androgen in humans and is secreted by testis. Other androgens are secreted by zona reticulata of adrenal cortex.

Topic: Biology, Human Reproduction, NEET
The major drawback of the two-kingdom classification system was that it did not distinguish between

Options:
A eukaryotes and prokaryotes.
B photosynthetic and non-photosynthetic plants.
C unicellular and multicellular organisms.
D all of the above

The correct answer is:
all of the above

Solution:
The Two Kingdom system of classification with Plantae and Animalia kingdoms was developed that included all plants and animals respectively. This system did not distinguish between the eukaryotes and prokaryotes, unicellular and multicellular organisms and photosynthetic (green algae) and non-photosynthetic (fungi) organisms. Classification of organisms into plants and animals was easily done and was easy to understand, but, a large number of organisms did not fall into either category. Hence the two kingdom classification used for a long time was found inadequate.

Topic: Biology, Biological Classification, NEET
The major organic product formed from the following reaction:

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Amines, NEET, NEET 2005
The major organic product in the reaction,
$\mathrm{CH}_3-\mathrm{O}-\mathrm{CH}\left(\mathrm{CH}_3\right)_2+\mathrm{HI} \rightarrow$ products is:

Options:
A $\mathrm{CH}_3 \mathrm{I}+\left(\mathrm{CH}_3\right)_2 \mathrm{CHOH}$
B $\mathrm{CH}_3 \mathrm{OH}+\left(\mathrm{CH}_3\right)_2 \mathrm{CHI}$
C $\mathrm{ICH}_2 \mathrm{OCH}\left(\mathrm{CH}_3\right)_2$
D \(\begin{array}{c}\mathrm{CH}_3 \mathrm{OC}\left(\mathrm{CH}_3\right)_2 \\ | \\ \mathrm{I} \end{array}\)

The correct answer is:
$\mathrm{CH}_3 \mathrm{I}+\left(\mathrm{CH}_3\right)_2 \mathrm{CHOH}$

Solution:
In the case of an unsymmetrical ether, the alkyl halide is always formed from the smaller alkyl group. This happens because the $\mathrm{I}^{-}$ ion being larger in sizc approaches the smaller alkyl group to avoid steric hindrance.
Caution
$H I$ is an acid and when it is hot and concentrated it means we are having an excess of $\mathrm{HI}$. Iodide ion produced from $\mathrm{HI}$ in this reaction will act as a nucleophile.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET, NEET 2006
The major part of the organic matter in the sewage water is decomposed in:

Options:
A Primary treatment
B Aeration tank/oxidation pond
C Anaerobic sludge digester
D Tertiary treatment

The correct answer is:
Anaerobic sludge digester

Solution:
In sewage treatment plant aerobic microbes cause breakdown of organic matter present in water and forms flocs. These flocs are anaerobically degraded in anaerobic sludge digester by methanogens forming mixture of gase consisting of methane, hydrogen etc.

Topic: Biology, Microbes in Human Welfare, NEET
The major portion of the dry weight of plants comprises of:

Options:
A nitrogen, phosphorus and potassium
B calcium, magnesium and sulphur
C carbon, nitrogen and hydrogen
D carbon, hydrogen and oxygen

The correct answer is:
carbon, hydrogen and oxygen

Solution:
The four most common elements in living organisms are $\mathrm{H}, \mathrm{C}, \mathrm{O}, \mathrm{N}$. These are also known as framework element. Carbon, hydrogen and oxygen together constitute about $94 \%$ of the total dry weight of the plant.

Topic: Biology, Biomolecules (B), NEET, NEET 2003
The major product A and B for the following reaction are, respectively

Options:
A
B
C
D

The correct answer is:

Solution:
The reaction occurs as follows
Step 1 is nucleophilic substitution reaction.
Step 2 is catalytic reduction which reduces carbonyl group to alcohol and nitriles to amines.

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
The major product C in the below mentioned reaction is:
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br} \xrightarrow[\Delta]{\text { alc. } \mathrm{KOH}} \mathrm{A} \xrightarrow{\mathrm{HBr}} \mathrm{B} \xrightarrow[\Delta]{\text { aq. } \mathrm{KOH}} \mathrm{C}$

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

The correct answer is:
Propan-2-ol

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, NEET, NEET 2024 (Re-NEET)
The major product D formed in the following reaction sequence is:

Options:
A

B

C $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}$
D $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Cl}$

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

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, NEET, NEET 2024 (Re-NEET)
The major product during hydroboration-oxidation of 1-methylcyclopentene is

Options:
A
B
C
D

The correct answer is:

Solution:
During hydroboration-oxidation, addition of H2O across the double bond occurs anti to Markownikoff's rule and since the stereochemistry of addition cis, therefore trans-2-methylcyclopentanol is formed

Topic: Chemistry, Hydrocarbons, NEET
The major product expected from the following reaction is:

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
The major product form by the monobromination of phenyl benzoate is

Options:
A
B
C
D

The correct answer is:

Solution:
In phenyl benzoate the 1string gets deactivated, while 2nd ring gets activated due to lone pairs on oxygen atom. So it can give ortho or para bromo product. But the ortho position is sterically hindered, so it will give para compound as a major product.

Topic: Chemistry, Carboxylic Acid Derivatives, NEET
The major product formed by the acid catalyzed hydration of

is

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, NEET
The major product formed in the dehydrohalogenation reaction of 2 -Bromopentane is Pent 2-ene. This product formation is based on?

Options:
A Saytzeff's Rule
B Hund's Rule
C Hofmann Rule
D Huckel's Rule

The correct answer is:
Saytzeff's Rule

Solution:
2 -bromopentane gives 2 -pentene as the major product according to Saytzeff's rule as it is more substituted alkene

CH3-CHBr-CH2-CH2-CH3→alc.KOHCH2=CH-CH2-CH2-CH3 (Hofmann product, least substituted alkene)

CH3-CHBr-CH2-CH2-CH3→alc.KOHCH3-CH=CH-CH2-CH3 (Satzeff's product, more substituted alkene)

Topic: Chemistry, Hydrocarbons, NEET, NEET 2021
The major product formed in the following conversion is

Options:
A

B

C

D

The correct answer is:

Solution:
Major product formed will be a conjugated alkene.

Topic: Chemistry, Aldehydes and Ketones, NEET, NEET 2023 (Manipur)
The major product formed in the following reaction is

Options:
A
B
C
D

The correct answer is:

Solution:
An Aldehyde reacts at a faster rate than a ketone during aldol condensation and sterically less hindered anion is a better nucleophile.

Topic: Chemistry, Aldehydes and Ketones, NEET
The major product formed in the reaction given below will be:

Options:
A
B
C
D

The correct answer is:

Solution:
Mechanism:

Topic: Chemistry, Amines, NEET
The major product formed in the reaction is
CH3CHClCH2CH2OH→KOH/H2O

Options:
A CH3CH=CHCH2OH
B CH2=CHCH2CH2OH
C
D CH3CHCH2CH2OH

The correct answer is:

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
The major product formed on nitration of N, N-dimethylaniline with conc. H2SO4,HNO3 mixture is

Options:
A
B
C
D

The correct answer is:

Solution:
Nitration of N,N-dimethyl aniline with conc. H2SO4,HNO3 mixture give a mixture of meta and para isomers. In which meta isomer is the major product. In presence of strong acid H2SO4 , -N(CH3)2 group changes into -N+H(CH3)2 by accepting proton so, it is a meta directing group.

Topic: Chemistry, Amines, NEET
The major product formed when 3-methyl-2-pentene reacts with chlorine water is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, General Organic Chemistry, NEET
The major product obtained in the following reaction is :

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Amines, NEET
The major product obtained in the following reaction is

Options:
A
B
C
D

The correct answer is:

Solution:
The reaction occurs as follows

Topic: Chemistry, Amines, NEET
The major product of following reaction is
R—C≡N→(1)ALH(i-Bu2)(2)H2O?

Options:
A RCHO
B RCOOH
C RCH2NH2
D RCONH2

The correct answer is:
RCHO

Solution:
Here aldehyde is formed as follows
R-C≡N→(1)AlH(i-Bu2)R-CH=NH→H2OR-CH=O

Topic: Chemistry, Amines, NEET
The major product of the following chemical reaction is:

Options:
A
B
C
D

The correct answer is:

Solution:
Unsymmetrical alkenes undergo anti-markovnikov's addition reaction with HBr in the presence of peroxide.

Topic: Chemistry, Hydrocarbons, NEET, NEET 2021
The major product of the following reaction is

Options:
A P
B Q
C R
D S

The correct answer is:
Q

Solution:
In phenol -OH-group is highly ring activating due to strong +R effect so if –COOH- or - SO3H - groups are present at ortho and para positions they can be substituted by electrophile like Br+ . It is important to use Br2 in water so that Br+ can be easily generated in good amount.

Topic: Chemistry, Hydrocarbons, NEET
The major product of the following reaction is

Options:
A
B
C
D

The correct answer is:

Solution:
This conversion is possible in following steps as follows
(I) Nucleophilic substitution reaction
(II) Oxidation
(III) Loss of H2O resulting in cyclization.

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
The major product of the following reaction is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
The major product of the following reaction is

Options:
A
B
C
D

The correct answer is:

Solution:
-OH group is highly ring activating so groups like –COOH & -SO3H present at ortho or para positions are also substituted by electrophile.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
The major product of the following reaction is

Options:
A
B
C
D

The correct answer is:

Solution:
This reaction occurs by the following mechanism given here.

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
The major product of the following reaction is

Options:
A
B
C
D

The correct answer is:

Solution:
This conversion can be done as follows

Topic: Chemistry, Carboxylic Acid Derivatives, NEET
The major product of the following reaction is

Options:
A An ester
B An ether
C An acetal
D A hemiacetal

The correct answer is:
An acetal

Solution:
Here an acetal is formed as follows.

Topic: Chemistry, Aldehydes and Ketones, NEET
The major product of the following reaction is

Options:
A
B
C
D

The correct answer is:

Solution:
It follows E2 mechanism

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
The major product of the following reaction is:

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, NEET
The major product of the following reaction is:

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, NEET, NEET 2019 (All India)
The major product of the following reactions

Options:
A
B
C
D

The correct answer is:

Solution:
This conversion occurs as follows

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
The major product of the reaction is

is

Options:
A

​

B

​

C

​

D

​

The correct answer is:

​

Solution:

Topic: Chemistry, Amines, NEET