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

What is the number of mono-chloro derivatives of Ethyl cyclohexane possible?

A
5
B
4
C
6
D
3
2
COMEDK 2024 Morning Shift
MCQ (Single Correct Answer)
+1
-0

Which one of the following is the correct IUPAC name of the given compound?

COMEDK 2024 Morning Shift Chemistry - IUPAC Nomenclatures Question 1 English

A
5-Bromo-2, 3-dimethylheptanoyl chloride
B
5,6-Dimethyl-3-bromohexanoyl chloride
C
1-Chloro-5-bromo-2, 3-dimethyl-1-oxoheptane
D
3-Bromo-5, 6-dimethylhexanoyl chloride
3
COMEDK 2024 Morning Shift
MCQ (Single Correct Answer)
+1
-0

An aromatic hydrocarbon $$[\mathrm{A}]\left(\mathrm{Molecular}\right.$$ formula $$\left.\mathrm{C}_9 \mathrm{H}_{12}\right)$$, undergoes air oxidation followed by reaction with dil. $$\mathrm{HCl}$$ to give [B] as one of the products

Compound [B] forms a white precipitate on treatment with $$\mathrm{Br}_2(\mathrm{aq})$$

Sodium salt of [B] reacts with $$\mathrm{CO}_2$$ at $$400 \mathrm{~K}$$ under high pressure followed by acid hydrolysis to give [C]

Compound [C] reacts with acetic anhydride / conc. $$\mathrm{H}_2 \mathrm{SO}_4$$ to form [D] Identify compound $$[\mathrm{D}]$$

A
COMEDK 2024 Morning Shift Chemistry - Hydrocarbons Question 4 English Option 1
B
COMEDK 2024 Morning Shift Chemistry - Hydrocarbons Question 4 English Option 2
C
COMEDK 2024 Morning Shift Chemistry - Hydrocarbons Question 4 English Option 3
D
COMEDK 2024 Morning Shift Chemistry - Hydrocarbons Question 4 English Option 4
4
COMEDK 2024 Morning Shift
MCQ (Single Correct Answer)
+1
-0

A given chemical reaction is represented by the following stoichiometric equation.

$$3 X+2 Y+\frac{5}{2} Z \rightarrow P_1+P_2+P_3$$

The rate of reaction can be expressed as _________.

A
$$\frac{2}{3}\left(\frac{d[X]}{d t}\right)=\left(\frac{d[Y]}{d t}\right)=\frac{4}{5}\left(\frac{d[Z]}{d t}\right)$$
B
$$\frac{3}{2}\left(\frac{d[X]}{d t}\right)=\frac{1}{4}\left(\frac{d[Y]}{d t}\right)=\frac{8}{5}\left(\frac{d[Z]}{d t}\right)$$
C
$$\frac{4}{15}\left(\frac{d[X]}{d t}\right)=\frac{4}{5}\left(\frac{d[Y]}{d t}\right)=\frac{8}{15}\left(\frac{d[Z]}{d t}\right)$$
D
$$\left(\frac{d[X]}{d t}\right)=\frac{3}{2}\left(\frac{d[Y]}{d t}\right)=\frac{15}{4}\left(\frac{d[Z]}{d t}\right)$$
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