Free Objective Test 02 Practice Test - 11th and 12th

Question 1

Conversion of $C H_{4}$ to $C H_{3} C l$ is an example of which of the following reaction?

Electrophilic substitution

Free radical addition

Free radical substituion

SOLUTION

Solution : D

When methane gas is treated with chlorine in the presence of sunlight, onE hydrogen of methane is replaced by the chlorine atom and forms methyl chloride.

The mechanism involved in this reaction is the free radical mechanism.

So it is an example of free radical substitution reaction.

Question 2

The addition of $H C l$ to vinyl chloride gives 1,1-dichloroethane because of?

Mesomeric effect of $C l$

Inductive effect of $C l$

Restricted rotation around double bond

None of these

SOLUTION

Solution : D

$C H_{2} | | C H | C l + H C l$ $\to$ $C H_{3} {| C H}_{╲ C l}^{C l ╱}$

(1,1-dichloroethane)

It is an addition reaction which is according to Markownikoff rule.

Question 3

The nitration of a compound is due to the?

$N O_{2}$

$N O_{3}$

$N O$

$N O_{2}^{+}$

SOLUTION

Solution : D

The process of nitration takes place as below

$H O N O_{2} + 2 H_{2} S O_{4} ⇌$ $H_{3} O^{+} + 2 H S O_{4}^{-} + N O_{2}^{+}$

(nitronium ion)

The electrophile responsible for nitration is

$N O_{2}^{+}$ ion.

Question 4

Which of the following can't be used in Friedal Craft's reactions?

$F e C l_{3}$

$F e B r_{2}$

$A l C l_{3}$

$N a C l$

SOLUTION

Solution : D

All cations are expected to act as Lewis acid since they are electron deficient in nature.

However, cations such as $N a^{+}$ , $K^{+}$ etc. (Inert gas configuration) have a very little tendency to accept electrons.

Therefore they do not acts as lewis acids in Friedel Craft's reaction.

Question 5

Which of the following is a polar compound?

$C_{2} H_{6}$

$C C l_{4}$

$H C l$

$C H_{4}$

SOLUTION

Solution : C

We know that there is more difference in the electronegativities of hydrogen and chlorine. Therefore $H C l$ is a polar compound.

Question 6

Homolytic fission of C - C bond in ethane gives an intermediate in which carbon is?

$s p^{3}$ hybridised

$s p^{2}$ hybridised

sp hybridised

$s p^{2} d$ hybridised

SOLUTION

Solution : B

Question 7

The total number of pi-bonds in $C H_{2} = C H - C H = C H - C \equiv C H$ are ______.

A. two

B. three

C. four

D. five

SOLUTION

Solution : C

$\begin{matrix} Nature of bond & Number of sigma bonds & Number of pi-bonds Single bond & 1 & 0 Double bond & 1 & 1 Triple bond & 1 & 2 \end{matrix}$ In the compound $C H_{2} = C H - C H = C H - C \equiv C H$ , there are (2 $\times$ 1) + (1 $\times$ 2) = 4 pi-bonds in total.

Question 8

Which of the following has a bond formed by overlap of $s p - s p^{3}$ hybrid orbitals?

$C H_{3} - C \equiv C - H$

$C H_{2} = C H - C H = C H_{2}$

$H C \equiv C H$

SOLUTION

Solution : A

$C s p^{3} H_{3} - C s p \equiv C s p - H$

Question 9

The bond between carbon atom (1) and carbon atom (2) in compound $N \equiv C - C H = C H_{2}$ involves the hybridised carbon as?

$s p^{2}$ and $s p^{2}$

$s p^{3}$ and $s p$

$s p$ and $s p^{2}$

$s p$ and $s p$

SOLUTION

Solution : A

sp and $s p^{2}$

$N \equiv s p C 1 - s p^{2} C H 2 = C 3 H_{2}$

Question 10

Shape of the methane molecule is ________.

A. square planar

B. tetrahedral

C. trigonal planar

D. pyramidal

SOLUTION

Solution : B

Carbon has four electrons in its valence shell. In methane molecule, the carbon atom forms covalent bonds with four different hydrogen atoms. There are no lone pairs on carbon atom. Therefore, the four hydrogen atoms are arranged at four corners of tetrahedron and carbon at the centre of the tetrahedron. The structure of methane molecule is tetrahedral.

Question 11

Dehydrohalogenation of an alkyl halide is a/an?

Nucleophilic substitution reaction

Elimination reaction

Both nucleophilic substitution and elimination reaction

Rearrangement

SOLUTION

Solution : B

Dehydrohalogenation of an alkyl halide in the presence of alcoholic potash is an example of elimination reaction.

$R - C H_{2} C H_{2} - C l + K O H a l c Δ \to$

$R C H = C H_{2} + K C l + H_{2} O$

Question 12

Among the following compounds, the one which can be dehydrated very easily is?

$C H_{3} - C H_{2} - C H_{3} | C | O H - C H_{2} - C H_{3}$

$C H_{3} - C H_{2} - C H_{2} - O H | C H - C H_{3}$

$C H_{3} - C H_{2} - C H_{2} - C H_{2} - C H_{2} - O H$

$C H_{3} - C H_{2} - C H | C H_{3} - C H_{2} - C H_{2} - O H$

SOLUTION

Solution : A

$C H_{3} - C H_{2} - C H_{3} | C | O H - C H_{2} - C H_{3} H^{+} - - \to$

$C H_{3} - C H_{2} - C H_{3} | C | \oplus - C H_{2} - C H_{3}$

Higher the stability of the carbocation, more easily it will be dehydrated.

Question 13

Number of $^{π}$ electrons in cyclobutadienyl anion $(C_{4} H_{4})^{- 2}$ is?

SOLUTION

Solution : D

Question 14

Which one of the following orders is correct regarding the inductive effect of the substituents?

$- N R_{2} < - O R > - F$

$- N R_{2} > - O R > - F$

$- N R_{2} < - O R < - F$

$- N R_{2} > - O R < - F$

SOLUTION

Solution : C

Relative inductive effects have been experimentally measured with reference to hydrogen, in decreasing order of -I effect or increasing order of +I effect, as follows:
–NH₃⁺ > –NO₂ > –SO₂R > –CN > –SO₃H > –CHO > –CO > –COOH > –COCl> –CONH₂ >
–F > –Cl > –Br > –I > –OR > -OH > –NH₂ > –C₆H₅ > –CH=CH₂ > –H

From this, you can get the required answer.

Question 15

All bonds in benzene are equal due to?

Tautomerism

Inductive effect

Resonance

Isomerism

SOLUTION

Solution : C

All the bonds (C-C) are equal in benzene. The

C-C bond length is 1.39 $\circ A$ which is in between

C-C bond (1.54 $\circ A$ ) abd C=C (1.34 $\circ A$ ).

Free Objective Test 02 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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