1
JEE Main 2022 (Online) 26th July Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The major product of the following reaction is

JEE Main 2022 (Online) 26th July Morning Shift Chemistry - Alcohols, Phenols and Ethers Question 63 English

A
JEE Main 2022 (Online) 26th July Morning Shift Chemistry - Alcohols, Phenols and Ethers Question 63 English Option 1
B
JEE Main 2022 (Online) 26th July Morning Shift Chemistry - Alcohols, Phenols and Ethers Question 63 English Option 2
C
JEE Main 2022 (Online) 26th July Morning Shift Chemistry - Alcohols, Phenols and Ethers Question 63 English Option 3
D
JEE Main 2022 (Online) 26th July Morning Shift Chemistry - Alcohols, Phenols and Ethers Question 63 English Option 4
2
JEE Main 2022 (Online) 26th July Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The correct stability order of the following diazonium salt is

JEE Main 2022 (Online) 26th July Morning Shift Chemistry - Compounds Containing Nitrogen Question 71 English

A
$$(\mathrm{A})>(\mathrm{B})>(\mathrm{C})>(\mathrm{D})$$
B
$$(\mathrm{A})>(\mathrm{C})>(\mathrm{D})>(\mathrm{B})$$
C
$$(\mathrm{C})>(\mathrm{A})>(\mathrm{D})>(\mathrm{B})$$
D
$$(\mathrm{C})>(\mathrm{D})>(\mathrm{B})>(\mathrm{A})$$
3
JEE Main 2022 (Online) 26th July Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Which one of the following is a reducing sugar?

A
JEE Main 2022 (Online) 26th July Morning Shift Chemistry - Biomolecules Question 54 English Option 1
B
JEE Main 2022 (Online) 26th July Morning Shift Chemistry - Biomolecules Question 54 English Option 2
C
JEE Main 2022 (Online) 26th July Morning Shift Chemistry - Biomolecules Question 54 English Option 3
D
JEE Main 2022 (Online) 26th July Morning Shift Chemistry - Biomolecules Question 54 English Option 4
4
JEE Main 2022 (Online) 26th July Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A) : Experimental reaction of $$\mathrm{CH}_{3} \mathrm{Cl}$$ with aniline and anhydrous $$\mathrm{AlCl}_{3}$$ does not give $$o$$ and $$p$$-methylaniline.

Reason (R): The $$-\mathrm{NH}_{2}$$ group of aniline becomes deactivating because of salt formation with anhydrous $$\mathrm{AlCl}_{3}$$ and hence yields $$m$$-methyl aniline as the product.

In the light of the above statements, choose the most appropriate answer from the options given below :

A
Both A and R are true and (R) is the correct explanation of (A).
B
Both A and R are true but (R) is not the correct explanation of (A).
C
(A) is true, but (R) is false.
D
(A) is false, but (R) is true.
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