1
JEE Main 2026 (Online) 22nd January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The dibromo compound $[\mathrm{P}]$ (molecular formula: $\mathrm{C}_9 \mathrm{H}_{10} \mathrm{Br}_2$ ) when heated with excess sodamide followed by treatment with dilute HCl gives [Q]. On warming [Q] with mercuric sulphate and dilute sulphuric acid yield $[R]$ which gives positive Iodoform test but negative Tollen's test. The compound $[\mathrm{P}]$ is :

A
JEE Main 2026 (Online) 22nd January Evening Shift Chemistry - Hydrocarbons Question 15 English Option 1
B
JEE Main 2026 (Online) 22nd January Evening Shift Chemistry - Hydrocarbons Question 15 English Option 2
C
JEE Main 2026 (Online) 22nd January Evening Shift Chemistry - Hydrocarbons Question 15 English Option 3
D
JEE Main 2026 (Online) 22nd January Evening Shift Chemistry - Hydrocarbons Question 15 English Option 4
2
JEE Main 2026 (Online) 22nd January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Which of the following mixture gives a buffer solution with $\mathrm{pH}=9.25$ ?

Given : $\mathrm{pK}_{\mathrm{b}}\left(\mathrm{NH}_4 \mathrm{OH}\right)=4.75$

A

$0.5 \mathrm{M} \mathrm{NH}_4 \mathrm{OH}(0.2 \mathrm{~L})+0.2 \mathrm{M} \mathrm{HCl}(0.5 \mathrm{~L})$

B

$0.2 \mathrm{M} \mathrm{NH}_4 \mathrm{OH}(0.5 \mathrm{~L})+0.1 \mathrm{M} \mathrm{HCl}(0.5 \mathrm{~L})$

C

$0.2 \mathrm{M} \mathrm{NH}_4 \mathrm{OH}(0.4 \mathrm{~L})+0.1 \mathrm{M} \mathrm{HCl}(1 \mathrm{~L})$

D

$0.4 \mathrm{M} \mathrm{NH}_4 \mathrm{OH}(1 \mathrm{~L})+0.1 \mathrm{M} \mathrm{HCl}(1 \mathrm{~L})$

3
JEE Main 2026 (Online) 22nd January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

When 1 g of compound $(\mathrm{X})$ is subjected to Kjeldahl's method for estimation of nitrogen, 15 mL 1 M $\mathrm{H}_2 \mathrm{SO}_4$ was neutralized by ammonia evolved. The percentage of nitrogen in compound $(\mathrm{X})$ is :

A

0.21

B

21

C

42

D

0.42

4
JEE Main 2026 (Online) 22nd January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The energy of first (lowest) Balmer line of H atom is $x \mathrm{~J}$. The energy (in J) of second Balmer line of H atom is :

A

$\frac{x}{1.35}$

B

$1.35 x$

C

$x^2$

D

$2 x$

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