1
COMEDK 2024 Morning Shift
MCQ (Single Correct Answer)
+1
-0

Which one of the following shows the correct increasing order of basic nature of the given compounds?

A: Phenylmethanamine

B: N-Ethylethanamine

C:N, N-Dimethylaniline

D : N, N-Dimethylmethanamine

A
$$ \mathrm{B}<\mathrm{C}<\mathrm{A}<\mathrm{D} $$
B
A < D < B < C
C
D < B < A < C
D
C < A < D < B
2
COMEDK 2024 Morning Shift
MCQ (Single Correct Answer)
+1
-0

Given: $$\Delta \mathrm{G}^0{ }_{\mathrm{f}}$$ of $$\mathrm{C}_2 \mathrm{H}_2$$ is $$2.09 \times 10^5 \mathrm{~J} / \mathrm{mol}$$ and $$\Delta \mathrm{G}_{\mathrm{f}}^0$$ of $$\mathrm{C}_6 \mathrm{H}_6$$ is $$1.24 \times 10^5 \mathrm{~J} / \mathrm{mol}$$. Calculate the equilibrium constant for the cyclic polymerisation of Ethyne to Benzene at $$27^{\circ} \mathrm{C}$$. $$(\mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1})$$

A
$$1.1134\times10^{10}$$
B
0.1113
C
$$1.429\times10^{88}$$
D
88.15
3
COMEDK 2024 Morning Shift
MCQ (Single Correct Answer)
+1
-0

A Carbonyl compound $$\mathrm{X}$$ when reacted with Methyl magnesium bromide followed by hydrolysis gave product $$\mathrm{Y}$$. When $$\mathrm{Y}$$ is heated with $$\mathrm{Cu}$$ at $$573 \mathrm{~K}$$, it gave $$\mathrm{Z}$$ (2- Methylbut-2-ene) as one of the products. Identify $$\mathrm{X}$$.

A
Pentan-2-one
B
Butanal
C
Butanone
D
Pentan-3-one
4
COMEDK 2024 Morning Shift
MCQ (Single Correct Answer)
+1
-0

Identify the Reagents I and II to be used in the course of the given reactions.

COMEDK 2024 Morning Shift Chemistry - Aldehyde and Ketone Question 6 English

A
$$ \text { Reagent I: DIBAL- } \mathrm{H} / \mathrm{H}_2 \mathrm{O} \quad \text { Reagent II: conc. } \mathrm{KOH} / \text { heat } $$
B
$$ \text { Reagent I: } \mathrm{Zn} / \mathrm{Hg} \text { in conc. } \mathrm{HCl} \quad \text { Reagent II: }\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+} $$
C
$$ \text { Reagent I: } \mathrm{NH}_2 \mathrm{NH}_2 / \mathrm{KOH} \quad \text { Reagent II: } \mathrm{Cr}_2 \mathrm{O}_7{ }^{2-} / \mathrm{H}^{+} $$
D
$$ \text { Reagent I: } \mathrm{H}_2 / \mathrm{Pd}-\mathrm{BaSO}_4 \quad \text { Reagent II: } \mathrm{CrO}_3 /\left(\mathrm{CH}_3 \mathrm{CO}\right)_2 $$
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