Haloalkanes and Haloarenes Test

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Haloalkanes and Haloarenes Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

For the reaction, $$C_2H_5OH+HX\overset{ZnX_2}{\longrightarrow} C_2H_5X$$, the order of reactivity is:

A
$$\;HI>HCl>HBr$$

B
$$\;HI>HBr>HCl$$

C
$$HCl>HBr>HI$$

D
$$\;HBr>HI>HCl$$

Solution

Because the reaction follows $$S_N1$$ mechanism and hence the halide formed is stable when it is of a larger size and size increases down the group. $$I>Br>Cl$$
Question 2
1 / -0

$$1,\:3$$- Dibromopropane reacts with metallic zinc to form:

A
propene

B
cyclopropane

C
propane

D
hexane

Solution

1,3- Dibromopropane reacts with zinc metal to give Wurtz reaction. Wurtz reaction is also useful in preparing cyclic compounds as given in this case.
Question 3
1 / -0

Secondary butyl chloride undergo alkaline hydrolysis in the polar solvent by ___________ mechanism.

A
$$\:S_{N}2$$

B
$$\;S_{N}1$$

C
$$\;S_{N}1$$ and $$S_{N}2$$

D
$$\;E\ 2$$

Solution

In polar medium sec. alkyl halide undergo $$S_N1$$ mechanism.

The carbocation intermediate formed during the $$S_N1$$ mechanism is very stable in polar solvents.

So, option B is correct.
Question 4
1 / -0

$$1-phenyl-2-chloropropane$$ on treating with alc. $$KOH$$ gives mainly:

A
$$1$$-phenylpropene

B
$$2$$-phenylpropene

C
$$1$$-phenylpropane-$$2$$-ol

D
$$1$$-phenylpropan-$$1$$-ol

Solution

$$1$$−phenyl−$$2$$−chloropropane on treating with alc. $$KOH$$ gives mainly $$1$$−phenylpropene. This is dehydrochlorination reaction.
The double bond in the product is in conjugation with benzene ring.$$1$$−phenyl−$$2$$−chloropropane.
Question 5
1 / -0

$$1$$-Chlorobutane on reaction with alcoholic potash gives:

A
$$1$$-Butene

B
$$1$$-Butanol

C
$$2$$-Butene

D
$$2$$-Butanol

Solution

$${1}^{o}RX$$ undergoes $$E_2$$ elimination with alc.$$KOH$$ in which no rearrangement takes place.
Question 6
1 / -0

Allyl chloride is a .............. compound while vinyl chloride is inert towards nucleophilic substitution.

A
reactive

B
mildly reactive

C
not reactive

D
netural

Solution

Allyl Chloride forms a stable carbocation due to resonating structure of the ion formed. Now a nucleophile can attack on the electron deficient carbon atom leading to nucleohilic substitution of chloride ion.
Question 7
1 / -0

The product $$(D)$$ of the reaction will be:
$$C{ H }_{ 3 }Cl\xrightarrow { \ KCN\ } \left( A \right) \xrightarrow { \ { H }_{ 3 }{ O }^{ + }\ } \left( B \right) \xrightarrow { \ N{ H }_{ 3 }\ } \left( C \right) \xrightarrow { \Delta \ } \left( D \right) $$

A
$$C{ H }_{ 3 }C{ H }_{ 2 }N{ H }_{ 2 }$$

B
$$C{ H }_{ 3 }CN$$

C
$$HCON{ H }_{ 2 }$$

D
$$C{ H }_{ 3 }CON{ H }_{ 2 }$$

Solution

Reaction: $$C{ H }_{ 3 }-Cl\xrightarrow { KCN\ } C{ H }_{ 3 }-CN\xrightarrow { { H }_{ 3 }{ O }^{ \oplus } } C{ H }_{ 3 }-COOH \xrightarrow { N{ H }_{ 3 } } C{ H }_{ 3 }-CO\overset { \ominus }{ O } \overset { \oplus }{ N } { H }_{ 4 }\xrightarrow { \Delta } C{ H }_{ 3 }-CON{ H }_{ 2 }$$

Option D is correct.
Question 8
1 / -0

The most reactive towards $$S_{ N }{ 1 }$$ is:

A
$$PhC{ H }_{ 2 }Cl$$

B
$$PhCl$$

C
$$PhCH(Cl)C{ H }_{ 3 } $$

D
$$p-N{ O }_{ 2 }PhC{ H }_{ 2 }Cl$$

Solution

The rate of $$S_{N}{1 }$$ reaction depends upon the stability of the carbocation formed.

The compound given in option C forms the most stable carbocation after removal of Cl. This carbocation is most stable due to resonance, hyperconjugation and $$+I$$ effect of methyl group so it will react easily.

$$Ph-\overset { \oplus }{ C } H-C{ H }_{ 3 }$$

Option C is correct.
Question 9
1 / -0

Aryl halide is less reactive than alkyl halide towards nucleophilic substitution because:

A
of less stable carbonium ion

B
due to large $$C-Cl$$ bond energy

C
of inductive effect

D
of resonance stabilization and $$s{p}^{2}$$ hybridization of $$C$$ attached to halide

Solution

Aryl halide is less reactive than alkyl halide towards nucleophilic substitution because of the resonance stabilization and $$sp^2$$ hybridization of $$C$$ attached to halide. Due to resonance, carbon-chlorine bond acquires partial double bond character, hence it becomes shorter and stronger, and thus, cannot be easily replaced by nucleophiles. This effect is further enhanced by the $$sp^2$$ nature of the $$C$$ atom to which chlorine is attached.
Question 10
1 / -0

Which of the following alkyl halides will undergo $${S}_{N}{1}$$ reaction most readily?

A
$${ \left( C{ H }_{ 3 } \right) }_{ 3 }C-F$$

B
$${ \left( C{ H }_{ 3 } \right) }_{ 3 }C-Cl$$

C
$${ \left( C{ H }_{ 3 } \right) }_{ 3 }C-Br$$

D
$${ \left( C{ H }_{ 3 } \right) }_{ 3 }C-I$$

Solution

Due to large size, iodide ion have less density of charge so it will be most stable and it will follow reaction most easily.

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Haloalkanes and Haloarenes Test - 34

Result

Haloalkanes and Haloarenes Test - 34

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Question 1

$$\;HI>HCl>HBr$$

$$\;HI>HBr>HCl$$

$$HCl>HBr>HI$$

$$\;HBr>HI>HCl$$

Solution

Question 2

propene

cyclopropane

propane

hexane

Solution

Question 3

$$\:S_{N}2$$

$$\;S_{N}1$$

$$\;S_{N}1$$ and $$S_{N}2$$

$$\;E\ 2$$

Solution

Question 4

$$1$$-phenylpropene

$$2$$-phenylpropene

$$1$$-phenylpropane-$$2$$-ol

$$1$$-phenylpropan-$$1$$-ol

Solution

Question 5

$$1$$-Butene

$$1$$-Butanol

$$2$$-Butene

$$2$$-Butanol

Solution

Question 6

reactive

mildly reactive

not reactive

netural

Solution

Question 7

$$C{ H }_{ 3 }C{ H }_{ 2 }N{ H }_{ 2 }$$

$$C{ H }_{ 3 }CN$$

$$HCON{ H }_{ 2 }$$

$$C{ H }_{ 3 }CON{ H }_{ 2 }$$

Solution

Question 8

$$PhC{ H }_{ 2 }Cl$$

$$PhCl$$

$$PhCH(Cl)C{ H }_{ 3 } $$

$$p-N{ O }_{ 2 }PhC{ H }_{ 2 }Cl$$

Solution

Question 9

of less stable carbonium ion

due to large $$C-Cl$$ bond energy

of inductive effect

of resonance stabilization and $$s{p}^{2}$$ hybridization of $$C$$ attached to halide

Solution

Question 10

$${ \left( C{ H }_{ 3 } \right) }_{ 3 }C-F$$

$${ \left( C{ H }_{ 3 } \right) }_{ 3 }C-Cl$$

$${ \left( C{ H }_{ 3 } \right) }_{ 3 }C-Br$$

$${ \left( C{ H }_{ 3 } \right) }_{ 3 }C-I$$

Solution

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