1
COMEDK 2023 Evening Shift
MCQ (Single Correct Answer)
+1
-0

The conversion of Propyne to Benzene can be brought out in 4 steps.

Choose the reagents to be used, in the proper sequential order, to bring out the conversion.

Reagents provided: alc. $$\mathrm{KOH}, \mathrm{H}_2 / \mathrm{Pd}, \mathrm{Na} /$$ dry ether, $$\mathrm{HBr}, \mathrm{HBr} /\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CO}\right)_2 \mathrm{O}_2$$, Soda-lime, $$\mathrm{Cr}_2 \mathrm{O}_3$$ (high T and P), Conc. $$\mathrm{H}_2 \mathrm{SO}_4$$

A
$$\mathrm{Na} /$$ Ether, $$\mathrm{Cr}_2 \mathrm{O}_3$$ (high $$\mathrm{T}$$ and $$\mathrm{P}$$), $$\mathrm{HBr}, \mathrm{H}_2 / \mathrm{Pd}$$
B
$$\mathrm{HBr}$$, alc. $$\mathrm{KOH}, \mathrm{HBr} /\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CO}\right)_2 \mathrm{O}_2, \mathrm{Cr}_2 \mathrm{O}_3$$ (high $$\mathrm{T}$$ and $$\mathrm{P}$$)
C
$$\mathrm{Cr}_2 \mathrm{O}_3$$ (high $$\mathrm{T}$$ and $$\mathrm{P}$$), $$\mathrm{Na} /$$ dry ether, $$\mathrm{HBr}$$, Conc. $$\mathrm{H}_2 \mathrm{SO}_4$$
D
$$\mathrm{H}_2 / \mathrm{Pd}, \mathrm{HBr} /\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CO}\right)_2 \mathrm{O}_2, \mathrm{Na} /$$ dry ether, $$\mathrm{Cr}_2 \mathrm{O}_3$$ (high $$\mathrm{T}$$ and $$\mathrm{P}$$)
2
COMEDK 2023 Evening Shift
MCQ (Single Correct Answer)
+1
-0

Para and ortho hydrogen differ in:

A
Atomic number
B
Number of neutrons
C
Size
D
Spins of protons
3
COMEDK 2023 Evening Shift
MCQ (Single Correct Answer)
+1
-0

What is the mole fraction of solute in a $$5 \mathrm{~m}$$ aqueous solution?

A
0.038
B
0.593
C
0.082
D
0.751
4
COMEDK 2023 Evening Shift
MCQ (Single Correct Answer)
+1
-0

Identify the reagents $$\mathrm{X}, \mathrm{Y}$$ and $$\mathrm{Z}$$ used to bring out the following reactions.

COMEDK 2023 Evening Shift Chemistry - Hydrocarbons Question 6 English

A
$$ \mathrm{X}=\text { Alkaline } \mathrm{KMnO}_4 \quad \mathrm{Y}=\text { dil } \mathrm{HCl} \quad \mathrm{Z}=\mathrm{CrO}_3 $$
B
$$ \mathrm{X}=\text { Air } \quad \mathrm{Y}=\text { dil. } \mathrm{H}_2 \mathrm{SO}_4 \quad \mathrm{Z}=\mathrm{Cr}_2 \mathrm{O}_7{ }^{2-} / \mathrm{H}^{+} $$
C
$$ \mathrm{X}=\mathrm{H}_2 \mathrm{O}_2 \quad \mathrm{Y}=\text { Conc. } \mathrm{HCl} \quad \mathrm{Z}=\mathrm{O}_3 $$
D
$$ \mathrm{X}=\text { DIBAL- } \mathrm{H} \quad \mathrm{Y}=\text { Conc. } \mathrm{HNO}_3 \quad \mathrm{Z}=\mathrm{CrO}_2 \mathrm{Cl}_2 $$
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