datavorous/entrance-exam-dataset huggingface dataset:
The main reduction product of the following compound with $\mathrm{NaBH}_4$ in methanol is :

Options:
A

B

C

D

The correct answer is:

Solution:
Sodium borohydride reduces ketonic group to alcohol, but not the amide group and $\mathrm{C}=\mathrm{C}$ double bond.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
The main role of bacteria in the carbon cycle involves

Options:
A assimilation of nitrogenous compound
B photosynthesis
C chemosynthesis
D digestion or breakdown of organic compounds

The correct answer is:
digestion or breakdown of organic compounds

Solution:
Bacteria are essential decomposers in nature. They play a vital role in the recycling of nutrients by causing the breakdown of dead organic matter; thus, they play a critical role in the carbon cycle. The most important bacterium involved in the carbon cycle is Pseudomonas denitrificans.

Topic: Biology, Ecosystem, NEET
The main role of DNA polymerase III is

Options:
A addition and polymerization of nucleotides from 5' end to 3' end.
B as major repair enzyme from 5' end to 3' end.
C removal of DNA primer from 5' end.
D addition and polymerization of nucleotides from 3' end to 5' end.

The correct answer is:
addition and polymerization of nucleotides from 5' end to 3' end.

Solution:
DNA polymerase III is an important enzyme that can polymerase in 5' to 3'. Also, it has 3' to 5' exonuclease proofreading activities. It plays an important role in DNA replication by adding and polymerizing nucleotide bases.
DNA polymerase II has a role in DNA repair.

Topic: Biology, Molecular Basis of Inheritance, NEET
The main scale of a vernier calliper has $\mathrm{n}$ divisions/cm. n divisions of the vernier scale coincide with $(n-1)$ divisions of main scale. The least count of the vernier callipers is

Options:
A $\frac{1}{(n+1)(n-1)} \mathrm{cm}$
B $\frac{1}{n} \mathrm{~cm}$
C $\frac{1}{n^2} \mathrm{~cm}$
D $\frac{1}{n(n+1)} \mathrm{cm}$

The correct answer is:
$\frac{1}{n^2} \mathrm{~cm}$

Solution:
As it is given that $\mathrm{n}$ divisions of vernier scale coincide with $(n-1)$ divisions of main scale i.e.
$$
\begin{aligned}
& \mathrm{n}(\mathrm{VSD})=(\mathrm{n}-\mathrm{l}) \text { MSD } \\
& \Rightarrow \quad \text { IVSD }=\frac{(\mathrm{n}-\mathrm{l})}{\mathrm{n}} \mathrm{MSD} \\
&
\end{aligned}
$$
The least count is the difference between one main scale division (MSD) and one vernier scale division (VSD).
$$
\begin{aligned}
& \therefore \text { Least Count }(\mathrm{LC})=1 \mathrm{MSD}-1 \mathrm{lVSD} \\
& =1 \mathrm{MSD}-\frac{(\mathrm{n}-\mathrm{l})}{\mathrm{n}} \mathrm{MSD} \\
& =\left(1-\frac{(\mathrm{n}-\mathrm{l})}{\mathrm{n}}\right) \mathrm{MSD}=\frac{1}{\mathrm{n}} \text { MSD } \\
& \text { Here, } 1 \mathrm{MSD}=\frac{1}{\mathrm{n}} \mathrm{cm} \\
& \Rightarrow \quad \mathrm{LC}=\frac{\mathrm{l}}{\mathrm{n}} \times \frac{\mathrm{l}}{\mathrm{n}} \mathrm{cm}=\frac{\mathrm{l}}{\mathrm{n}^2} \mathrm{~cm} \\
&
\end{aligned}
$$

Topic: Physics, Experimental Physics, NEET, NEET 2019 (Odisha)
The main scale of a vernier calliper reads 10 mm in 10 divisions. 10 divisions of vernier scale coincide with 9 divisions of the main scale. When the two jaws of the callipers touch each other, the fifth division of the vernier coincides with a main scale division and the zero of the vernier is to the right of zero of the main scale. When a cylinder is tightly placed between the two jaws, the zero of vernier scale lies slightly to the left of 3.2 cm and the fourth vernier division coincides with the main scale division. The diameter of the cylinder is

Options:
A 3.19 cm
B 3.14 cm
C 3.04 cm
D none of these

The correct answer is:
3.19 cm

Solution:
Zero error = 0.5 mm = 0.05 cm
observed reading of cylinder = 3.2 cm + (4) (0.01 cm) = 3.24 cm
actual thickness of cylinder = (3.24) – (0.05) = 3.19 cm

Topic: Physics, Units and Dimensions, NEET
The main scale of a vernier calliper reads 10 mm in 10 divisions. 10 divisions of vernier scale coincide with 9 divisions of the main scale. When the two jaws of the callipers touch each other, the fifth division of the vernier coincides with a main scale division and the zero of the vernier is to the right of zero of the main scale. When a cylinder is tightly placed between the two jaws, the zero of vernier scale lies slightly to the left of 3.2 cm and the fourth vernier division coincides with a main scale division. The diameter of the cylinder is

Options:
A 3.19 cm
B 3.14 cm
C 3.04 cm
D none of these

The correct answer is:
3.19 cm

Solution:
Zero error = 0.5 mm = 0.05 cm
observed reading of cylinder = 3.2 cm + (4) (0.01 cm) = 3.24 cm
actual thickness of cylinder = (3.24) – (0.05) = 3.19 cm.

Topic: Physics, Experimental Physics, JEE Main
The main scale of a vernier calliper reads in mm and its vernier scale is divided into 8 divisions, which coincides with 7 divisions of the main scale. It was also observed that, when the two jaws are brought in contact, the zero of the vernier scale coincided with the zero of the main scale. The edge length of a cube is measured using this vernier calliper. The main scale reads 12 mm and the 2nd division of the vernier scale coincides with the main scale. The edge length (in mm) of the cube is

Options:

The correct answer is:
12.25

Solution:
8 VSD = 7 MSD

⇒1 VSD = 78 MSD

LC = 1 MSD - 1 VSD

LC = 1 MSD - 78 MSD = 18 MSD = 18 mm

l = 12 + 2×18=12.25 mm

Topic: Physics, Experimental Physics, JEE Main
The main scale of a Vernier Callipers reads in millimeter and its vernier is divided into 10 divisions which coincides with 9 divisions of the main scale. The length of the object for situation is found to be mm . Find the value of x .

Options:

The correct answer is:
5.20

Solution:
List count =0.1 mm
Negative zero error =LC×10-9=0.1 mm
Measured value =5.0+LC×1+ negative
Zero error =5.20

Topic: Physics, Units and Dimensions, JEE Main
The main structural feature of proteins is

Options:
A The ester linkage
B The ether linkage
C The peptide linkage
D All of these

The correct answer is:
The peptide linkage

Solution:
The main structural feature of proteins is the peptide linkage.

Topic: Chemistry, Biomolecules, JEE Main
The maintenance of favourable internal conditions, by self-regulated mechanisms, even though there are changes in the environment, is known as

Options:
A

Homeostasis

B

Entropy

C

Enthalpy

D

Steady state

The correct answer is:

Homeostasis

Solution:
For metabolic processes to continue, the living things need to remain in a steady state maintained by a self-regulatory mechanism called homeostasis.

Topic: Biology, Organisms and Populations, NEET
The major component of which of the following ore is sulphide based mineral?

Options:
A Calamine
B Siderite
C Sphalerite
D Malachite

The correct answer is:
Sphalerite

Solution:
Calamine, Siderite and Malachite are carbonate containing ores. Sphalerite is a sulphide ore.

Calamine: ZnCO3

Siderite: FeCO3

Sphalerite : ZnS

Malachite : CuCO3·CuOH2

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2023 (29 Jan Shift 2)
The major component/ingredient of Portland Cement is :

Options:
A dicalcium silicate
B tricalcium aluminate
C dicalcium aluminate
D tricalcium silicate

The correct answer is:
tricalcium silicate

Solution:
Portland cement contain
(i) dicalcium silicate =26%,
(ii) Tricalcium silicate =51%,
(iii) Tricalcium aluminate =11%

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2021 (31 Aug Shift 1)
The major components in "Gun Metal" are:

Options:
A Cu,Sn and Zn
B Al,Cu,Mg and Mn
C Cu,Ni and Fe
D Cu,Zn and Ni

The correct answer is:
Cu,Sn and Zn

Solution:
Gun metal is used for making guns and is also known as red brass. Gun metals are resistant to corrosion that is caused by steam and salt water. That’s why their composition includes metals which provides resistance to corrosion. Usually, gun metal is a type of bronze which is an alloy of copper, tin and zinc.

The major components in "Gun Metal" are

Cu:87%

Zn:3%

Sn:10%

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2021 (24 Feb Shift 1)
The major components of German Silver are :

Options:
A Zn,Ni and Ag
B Cu,Zn and Ni
C Ge,Cu and Ag
D Cu,Zn and Ag

The correct answer is:
Cu,Zn and Ni

Solution:
German silver is silvery-white in appearance. Hence, it was named German silver. But nowadays, if an alloy does not contain silver, then it cannot be named as silver. German silver is also called Nickel silver.

Major components of German silver are:

Cu,Zn,Ni50%30%20%

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2021 (25 Feb Shift 2)
The major organic compound formed by the reaction of 1, 1, 1 - trichloroethane with silver powder is :

Options:
A Acetylene
B Ethene
C 2 - Butyne
D 2 - Butene

The correct answer is:
2 - Butyne

Solution:
The 1, 1, 1 - trichloroethane form â 2-butyne when reacted with silver powder.
CH3CCl3+ 6Ag →CH3C≡CCH3+6AgCl

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2014 (06 Apr)
The major organic compound formed by the reaction of 1, 1, 1 - trichloroethane with silver powder is :

Options:
A Acetylene
B Ethene
C 2 - Butyne
D 2 - Butene

The correct answer is:
2 - Butyne

Solution:
â
The molecule dimerizes to give 2 butyne
CH3 - C ≡ C - CH3

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
The major product (A) formed in the reaction given below is :

Options:
A
B
C
D

The correct answer is:

Solution:
Major product will be formed by elimination reaction. Product having conjugated double bonds is more stable.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2021 (27 Aug Shift 2)
The major product 'A' of the following given reaction has sp2 hybridized carbon atoms.

2,7-Dimethyl-2,6-octadiene ⟶H+AMajor Product

Options:

The correct answer is:
2

Solution:
So, there are total of two sp2 hybrid carbon in the final product.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2022 (24 Jun Shift 1)
The major product B formed in the following reaction sequence is:

Options:
A
B
C
D

The correct answer is:

Solution:
Initially aldehyde reacts with Grignard reagent to form a secondary alcohol. Which undergo nucleophilic substitution reaction.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2018 (16 Apr Online)
The major product [B] in the following reaction is :

Options:
A CH2=CH2
B
C
D CH3-CH2CH=CH-CH3

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2020 (04 Sep Shift 2)
The major product [B] in the following sequence of reaction is having how many sp3 hybridized Carbon atoms?

Options:

The correct answer is:
6

Solution:
All single bonded C-atoms are sp3 hybridised here in major product.

Topic: Chemistry, Hydrocarbons, JEE Main
The major product B in the following sequence of reactions is:

Options:
A
B
C
D

The correct answer is:

Solution:
Alcohol upon reaction with protic acids tends to lose a molecule of water to form alkenes. These reactions are known as dehydrogenation or dehydration of alcohols. It is an example of an elimination reaction.The order of the ease of dehydration of alcohols is: tertiary > secondary > primary.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2020 (08 Jan Shift 2)
The major product [C] of the following reaction sequence will be :

Options:
A
B
C
D

The correct answer is:

Solution:
Markovnikov's addition states that, in hydrohalogenation of an unsymmetrical alkene, the hydrogen atom in the hydrogen halide forms a bond with the doubly bonded carbon atom in the alkene, bearing the greater number of hydrogen atoms.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2020 (04 Sep Shift 2)
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, JEE Main, JEE Main 2017 (08 Apr Online)
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, JEE Main
The major product formed during the 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 is cis, therefore trans-2-methylcyclopentanol is formed

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The major product formed in the following is

Options:
A

B

C

D

The correct answer is:

Solution:
The major product in the following is
$\because$ In the reaction $\mathrm{C}=\mathrm{C}$ remain as such and $-\mathrm{CH}$ changes to - CHO.

Topic: Chemistry, Aldehydes and Ketones, TS EAMCET, TS EAMCET 2020 (10 Sep Shift 2)
The major product formed in the following reaction is:

Options:
A
B
C
D

The correct answer is:

Solution:
In the first step hydrogen iodide attacks the ether according to SN2 mechanism and in the second step hydrogen iodide attacks the ether according to SN1 mechanism.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2018 (08 Apr)
The major product formed in the following reaction is:

Options:
A

B

C

D

The correct answer is:

Solution:
PCC oxidizes primary alcohols to aldehydes and secondary alcohols to ketones. In the above reaction, $-\mathrm{OCOCH}_3$ group is hydrolyzed to secondary alcohol which is then oxidised (with PCC) to ketone.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
The major product formed in the following reaction is:

Options:
A

B

C

D

The correct answer is:

Solution:
Note: In the given reaction a molecule of $\mathrm{HCl}$ is lost and $\mathrm{C}=\mathrm{C}$ double bond is formed. Thus it is dehydrohalogenation reaction.
Nitro group is electron withdrawing group. Hence increases the acidity of $\mathrm{H}$ atom (attached to $\mathrm{C}$ atom bearing nitro group) which is removed easily. Further the newly formed double bond is in conjugation with nitro group.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
The major product formed in the following reaction is:

Options:
A

B

C

D

The correct answer is:

Solution:
This reaction is cross aldol condensation reaction.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2019 (09 Jan Shift 2)
The major product formed in the following reaction is:

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2021 (26 Aug Shift 1)
The major product formed in the following reaction is:

Options:
A
B
C
D

The correct answer is:

Solution:
At 25°C thermodynamicaly controlled Product is formed as major product.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2021 (26 Aug Shift 1)
The major product formed in the following reaction is

Options:
A

B

C

D

The correct answer is:

Solution:
Lithium borohydride is commonly used for the selective reduction of esters and lactones to the corresponding alcohols in the presence of carboxylic acids, tertiary amides, and nitriles.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2023 (08 Apr Shift 1)
The major product formed in the following reaction is

Options:
A

B

C

D

The correct answer is:

Solution:
Alkyl halide is $1^{\circ}$.
Keep in mind $1^{\circ}$ halide give product by $\mathrm{S}_{\mathrm{N}} 2 / \mathrm{E}-2$ mechanism and $1^{\circ}$ halide always gives substitution reaction except when strongly hindered base is used.
e.g., with
$$
\text { it gives mainly elimination. }
$$
$$
\text { The reaction involves carbocation intermediate. }
$$
i.e.
but as it is a primary carbocation it will rearrange to give a tertiary carbocation, which completes the reaction
Stability of carbocation : $3^{\circ}>2^{\circ}>1^{\circ}>\stackrel{\oplus}{\mathrm{CH}_3}$
It is because the stability of a charged system is increased by dispersal of the charge. The more stable the carbocation, the faster it is formed.
N.B. - Rearrangement can be done in two ways.

Topic: Chemistry, Alcohols Phenols and Ethers, AIIMS, AIIMS 2005
The major product formed in the following reaction is :

CH2CH=CHCHCH32→HBr

Options:
A CH3CH2CH(Br)CHCH32
B CH3CH(Br)CH2CHCH32
C BrCH23CHCH32
D CH3CH2CH2C(Br)CH32

The correct answer is:
CH3CH2CH2C(Br)CH32

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2020 (05 Sep Shift 2)
The major product formed in the following reaction is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The major product formed in the following reaction sequence is

Options:
A
B
C
D

The correct answer is:

Solution:
N-acylation of aniline will take place and acetanilide is form.

Now nitration of acetanilide will take place and para nitro acetanilide is formed as major product.

Now this para nitro acetanilide on hydrolysis gives para nitro aniline as major product.

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main
The major product formed in the following reactions is

Options:
A

B

C

D

The correct answer is:

Solution:
The given chain reaction give the final product as follows :
(i) with $\mathrm{KI}$, we get $\mathrm{CH}_3 \mathrm{I}$, which on treatment with $\mathrm{Mg}$, dry ether give ethane $\left(\mathrm{CH}_3-\mathrm{CH}_3\right)$.
(ii) on treatment with
followed by $\mathrm{H}_2 \mathrm{O}$ and $20 \% \mathrm{H}_3 \mathrm{PO}_4$ at $358 \mathrm{~K}$, we get the final product.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
The major product formed in the Friedel-Craft acylation of chlorobenzene is

Options:
A

B

C

D

The correct answer is:

Solution:
Friedel-Crafts acylation is a reaction in which the addition of an acyl group to an aromatic ring takes place. The reaction will lead to the aromatic ring being transformed into a ketone.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2023 (15 Apr Shift 1)
The major product formed in the given reaction is:

Options:
A
B
C
D

The correct answer is:

Solution:
The Birch reduction is an organic reaction where aromatic compounds which have a benzenoid ring are converted into 1,4-cyclohexadiene which have two hydrogen atoms attached at opposite ends of the molecule. The solvated electrons (the free electrons in the solution of sodium in liquid ammonia which are responsible for the intense blue colour) add to the aromatic ring, giving a radical anion. This radical anion is supplied with a proton by the alcohol. The alcohol also supplies another proton to the penultimate carbanion. Now, with the alcohol present, the products 1,4– cyclohexadiene and an alkoxide ion are formed.

The reaction with nitro anisole is given below.

Na+NH3→NaNH3++ammoniated electrons

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2022 (26 Jul Shift 1)
The major product formed in the reaction given below will be :

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Amines, JEE Main, JEE Main 2019 (10 Jan Shift 1)
The major product formed in the reaction is:

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
The major product formed in the reaction is:

Options:
A
B
C
D

The correct answer is:

Solution:
− CH 3 has 3∝−H
and is more activating towards electrophilic substitution. It is ortho-para directing in nature.

Topic: Chemistry, Hydrocarbons, JEE Main
The major product formed when 1, 1, 1- trichloropropane is treated with aqueous potassium hydroxide is :

Options:
A 2 - Propanol
B Propyne
C Propionic acid
D 1 - Propanol

The correct answer is:
Propionic acid

Solution:
The major product formed when 1, 1, 1-trichloro-propane is treated with aqueous potassium hydroxide is propionic acid.

Alc. KOH causes elimination in the molecule of haloalkane generally, while aq. KOH leads to substitution. Here aqueous KOH is used, substitution reaction is most preferable.

The initial reaction is the nucleophilic substitution, which will form a trihydroxy compound, since, three hydroxy group on one carbon is unstable and result in dehydration and form carboxylic acid.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2014 (19 Apr Online)
The major product in the dehydrohalogenation of 3-Bromo-2, 2-dimethylbutane is

Options:
A 3, 3-dimethyl but-1-ene
B 2, 3-dimethylbut-1-ene
C 2, 3-dimethyl but-2-ene
D 4-methylpent-2-ene

The correct answer is:
2, 3-dimethyl but-2-ene

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
The major product in the follo wing conversion is

Options:
A
B
C
D

The correct answer is:

Solution:
As HBr is in excess it converts ether to alcohol and alkenes undergo electrophilic addition with it.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The major product in the following conversion is

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2019 (12 Jan Shift 2)
The major product in the following reaction is:

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Amines, JEE Main, JEE Main 2019 (08 Apr Shift 2)
The major product in the following reaction is:

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2020 (03 Sep Shift 2)
The major product in the following reaction is :

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2020 (02 Sep Shift 1)
The major product in the following reaction is:

Options:
A
B
C
D

The correct answer is:

Solution:
(Aromatic)

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2020 (08 Jan Shift 2)
The major product in the following reaction is

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
The major product in the following reaction is

Options:
A

B

C

D

The correct answer is:

Solution:
In the given reaction, the final product is as follows :
Here, $-\mathrm{CH}_3$ will make bond with $-\mathrm{COO}$ of ring. Hence, we get the product as shown.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The major product in the following reaction is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The major product in the following reaction is:

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Amines, JEE Main
The major product in the following reaction is having how many pi electrons here?

Options:

The correct answer is:
2

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Main
The major product in the following reaction is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The major product in the following reaction

Options:
A

B

C

D

The correct answer is:

Solution:
No Solution Available

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2013 (25 Apr Online)
The major product in the following reactions is

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
The major product in the following reactions, is

Options:
A

B

C

D

The correct answer is:

Solution:
In the given reaction, the final product is as follows :
In step (i), we get diazonium product, which go for coupling reaction to give the final product as shown in the reaction.

Topic: Chemistry, Amines, JEE Main
The major product in the given reaction is

Options:
A

B

C

D

The correct answer is:

Solution:
In first step of the reaction H+ will attack on the lone pairs of electrons of the oxygen atom. Which will get dehydrated forming carbocation intermediate. On this carbocation pi bond will attack forming a five membered cyclic ring. In last step Br- will combine with carbocation forming final product according to the following reactions.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2022 (28 Jul Shift 2)
The major product in the reaction is

Options:
A
B
C
D Both B and C

The correct answer is:

Solution:
The product is obtained by Me shift as follows

Topic: Chemistry, Hydrocarbons, JEE Main