Organic Chemistry Some Basic Principles and Techniques Test

Result

Organic Chemistry Some Basic Principles and Techniques Test - 12

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The compound that does not produce nitrogen has by the thermal decomposition is:

A
$$NH_4NO_2$$

B
$$(NH_4)_2SO_4$$

C
$$Ba(N_3)_2$$

D
$$(NH_4)_2Cr_2O_7$$

Solution

$$NH_4NO_2\rightarrow2H_2O+N_2$$

$$2(NH_4)_2SO_4\rightarrow2NH_3+H_2SO_4$$

$$Ba(N_3)_2\rightarrow Ba+3N_2$$

$$(NH_4)_2Cr_2O_7\rightarrow Cr_2O_3+N_2+4H_2O$$

So, correct answer is $$(B)$$.
Question 2
1 / -0

The ratio of mass percent of $$C$$ and $$H$$ of an organic compound $$(C_XH_YO_Z)$$ is $$6:1$$. If one molecule of the above compound $$(C_XH_YO_Z)$$ contain half as much oxygen as required to burn one molecule of compound $$C_XH_Y$$ completely to $$CO_2$$ and $$H_2O$$. The empirical formula of compound $$C_XH_YO_Z$$ is :

A
$$C_3H_4O_2$$

B
$$C_2H_4O_3$$

C
$$C_3H_6O_3$$

D
$$C_2H_4O$$

Solution

$$C_xH_yO_z$$

$$C_xH_y+nO_2\rightarrow xCO_2+\frac {y}{2} H_2O$$

$$n=\dfrac {4x+y}{4}$$ and amount of $$O$$ required, $$m=n*2$$

Given that $$z=m/2=n$$

and mass percent ratio of $$C$$:$$H$$ $$::6:1=$$ $$12:2$$

therefore for each $$C$$ two $$H$$ are present in the molecule.

For $$x=1$$,$$y=2$$ $$\therefore z=\dfrac {4\times 1+2}{4}$$
$$z=1.5$$

Emperical formula is $$C_2H_4O_3$$

Option $$B$$ is correct.
Question 3
1 / -0

The product of acid hydrolysis of P and Q can be distinguished by:

A
Lucas Reagent

B
2,4-DNP

C
Fehling's Solution

D
$$\mathrm{N}\mathrm{a}\mathrm{H}\mathrm{S}\mathrm{O}_3$$

Solution

The product of acid hydrolysis of P and Q can be distinguished by Fehlings Solution.
The acid hydrolysis of P gives an enol which tautomerizes to a ketone.
The acid hydrolysis of Q gives an enol which tautomerizes to an aldehyde.
Aldehydes reduce Fehlings Solution but ketones do not reduce Fehlings Solution.
Question 4
1 / -0

The total number of $$\pi $$ electrons in the given structure are :

A
12

B
16

C
4

D
8

Solution

The given compound is tetraene. In other words, it contains 4 C=C double bonds.

Each bond contains 2 electrons.

Hence, the number of bond electrons is $$4 \times 2 = 8$$.

Therefore, the correct option is D.
Question 5
1 / -0

Suitable reagent for following conversion.

A
$$CH_3MgBr, H_3O^+, I_2/NaOH, H-Br/R_2O_2$$

B
$$KMnO_4, NaOH, HBr/R_2O_2$$

C
$$CH_3MgBr, KMnO_4, HBr$$

D
$$CH_3MgBr, H_3O^+, H-Br, I_2/NaOH$$

Solution

Solution:- (A) $$C{H}_{3}MgBr, \; {{H}_{3}O}^{+}, \; {I}_{2}/NaOH, \;l H-Br/{R}_{2}{O}_{2}$$
Suitable reagent for the given conversion will be-
$$C{H}_{3}MgBr, \; {{H}_{3}O}^{+}, \; {I}_{2}/NaOH, \;l H-Br/{R}_{2}{O}_{2} \left( \text{Peroxide} \right)$$
Question 6
1 / -0

Marsh gas is:

A
$$C_6H_6$$

B
$$CH_4$$

C
$$C_2H_2$$

D
$$CO$$

Solution

Marsh gas is methane, $$CH_4$$. It is called so because it is often obtained from marshy places.
Question 7
1 / -0

Methyl group is:

A
$$- CH_{2}$$

B
$$- CH_{2}OH$$

C
$$- CH_{2}Cl$$

D
$$- CH_{3}$$

Solution

$$-CH_3$$ group is called methyl group, not $$-CH_2$$.
Question 8
1 / -0

How many pi bonds are present in phosgene?

A
1

B
2

C
3

D
4

Solution

From the image we can see that there is only 1 pi-bond present in phosgene (cabonyl chrolide).
Question 9
1 / -0

Correct IUPAC name for the given compound is:

A
1-Ethoxy-1-aminobutane

B
1-Amino-1-ethoxy-1-methylpropane

C
1-Ethoxy-2-propanol

D
2-Ethoxybutan-2-amine

Solution

The functional group on first priority in the given compound is an amine .
So we will number the parent chain so as to give amine the lowest number.
Thus the carbon-containing amine will be numbered (2).
So , the correct IUPAC name is 2-Ethoxybutan-2-amine .
Question 10
1 / -0

Which of the following is called ethanoic acid :

A
$$HCOOH$$

B
$$CH_{3}COOH$$

C
$$CH_{3}CH_{2}COOH$$

D
$$CH_{3}CH_{2}CH_{2}COOH$$

Solution

Methanoic acid :$$ H COOH$$
ethanoic acid : $$CH_3 COOH$$