Haloalkanes and Haloarenes Test

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Haloalkanes and Haloarenes Test - 24

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Weekly Quiz Competition

Question 1
1 / -0

The compound $$C_{7}H_{8}$$ undergoes the following reactions:
$$C_{7}H_{8}\xrightarrow {3Cl_{2}/\triangle}A \xrightarrow {Br_{2}/ Fe} B \xrightarrow {Zn/HCl}C$$. The product $$C$$ is:

A
$$3-$$bromo$$-2,4,6-$$trichlorotoluene

B
$$m-$$bromotoluene

C
$$p-$$bromotoluene

D
$$o-$$bromotoluene

Solution

Option B is correct
Question 2
1 / -0

Which of the following is the most correct electron displacement for a nucleophilic reaction to take place?

A

B

C

D

Solution

The most correct electron displacement for a nucleophilic reaction to take place is as given above.
Question 3
1 / -0

Which of the following alkane cannot be made in good yield by Wurtz reaction?

A
2,3-Dimethylbutane

B
n-Heptane

C
n-Butane

D
n-Hexane

Solution

Wurtz reaction is a reaction in which alkyl halide is converted to alkane by Na/ether. It is limited to the synthesis of symmetrical alkanes.

As we know that n-Heptane involves the odd number of carbon hence it won't be produced by this method. Rest molecules can be formed by the Wurtz reaction.

Option B is correct.
Question 4
1 / -0

2-phenylethylbromide when heated with $$NaOEt$$, elimination takes place. No deuterium exchange takes place when the reaction is carried out in $${C}_{2}{H}_{5}OD$$ solvent. The mechanism will be:

A
$$E1$$ elimination

B
$$E2$$ elimination

C
$$E1cB$$ elimination

D
$$E2$$ or $${E1}cB$$

Solution

It is a primary bromide. So it will undergo elimination either by $$E2$$ or $$E1cB$$. Since there is no deuterium exchange in $${C}_{2}{H}_{5}OD$$ solvent, $$C-H$$ bond is not broken to form carbanion.

Hence, the actual mechanism is $$E2$$ only.
Question 5
1 / -0

Which one of the following undergoes nucleophilic substitution exclusively by $$S_N1$$ mechanism?

A
Benzyl chloride

B
Ethyl chloride

C
Chlorobenzene

D
Isopropyl chloride

Solution

Aliphatic $$S_N1$$ reaction is carried out in two steps. In first step carbocation is formed and its formation is based on the stability of carbocation.

$$C_6H_5\overset{+}{C}H_2 > CH_3\overset{+}{C}H CH_3> CH_3 \overset{+}{C}H_2$$

In the second step, the nucleophile is attracted to carbocation to give final products. Hence, an order of $$S_N1$$ reaction is
$$C_6H_5\overset{+}{C}H_2 >CH_3 CH-CH_3 > CH_3 CH_2$$

The aryl halides e.g., chlorobenzene are less reactive as compared to alkyl halides towards nucleophilic reagents in either $$S_N2$$ or $$S_N1$$ reaction because carbon-halogen bond in the aryl halide is strong (due to its double bond character).

So, the correct option is $$A$$
Question 6
1 / -0

On commercial scale, phenol is obtained from chlorobenzene. The chlorobenzene needed for the purpose is prepared by:

A

B

C

D

Solution

In Raschig's process a mixture of benzene vapours, air and hydrogen chloride is passed over heated $$Cu{Cl}_{2}$$ at $$500K$$.
$${ C }_{ 6 }{ H }_{ 6 }+CHCl_3+{ O }_{ 2 }(air)\xrightarrow [ 500K ]{ Cu{ Cl }_{ 2 } } 2{ C }_{ 6 }{ H }_{ 6 }Cl+2{ H }_{ 2 }O$$

Phenol is obtained from chlorobenzene as shown in the above reaction
Hence, the correct option is D
Question 7
1 / -0

Which of the following halides undergoes nucleophilic substitution most readily?

A
$$p- H_3CC_6H_4Cl$$

B
$$O- MeOC_6H_4Cl$$

C
$$p-ClC_6H_4Cl$$

D
$$C_6H_5CH(Cl)CH_3$$

Solution

$$\displaystyle C_6H_5CH(Cl)CH_3 $$ undergoes nucleophilic substitution most readily as Cl is attached to aliphatic C atom.

In $$\displaystyle p- H_3CC_6H_4Cl, O- MeOC_6H_4Cl, p-ClC_6H_4Cl$$, Cl is attached to aromatic carbon atom. The nucleophilic substitution of aryl chlorides requires drastic conditions.
Question 8
1 / -0

$$C_7H_8\overset{3Cl_2/ heat}{\rightarrow}A\overset{Fe/Br}{\rightarrow}B\overset{Zn/HCl}{\rightarrow}C$$. Here, the compound C is:

A
$$3$$-bromo $$2,4,6$$-trichlorotoluene

B
o-bromotoluene

C
P-bromotoluene

D
m-bromotoluene

Solution

The compound C is m-bromotoluene. First, the methyl group of toluene is chlorinated. Due to this, ortho para directing methyl group is converted into meta directing trichloromethyl group. Then aromatic ring is brominated. This is followed by the conversion of trichloromethyl group to methyl group. The overall reaction is toluene to m-bromotoluene.
Question 9
1 / -0

Isopropyl bromide on Wurtz reaction gives:

A
hexane

B
propane

C
$$2, 3$$-dimethyl butane

D
neo-hexane

Solution
Question 10
1 / -0

Product $$(B)$$ is :

A

B

C

D
None of these

Solution

(Refer to Image 1) on reaction with $$NBS (N-Bromosuccinimide)$$, followed by reaction with sodium methanethiolate gives (Refer to Image 2)
$$NBS$$ is a chemical agent which is used in radical substitution, electrophilic substitution. It is a source of bromine radical. Also, $$CH_3SNa$$ is the conjugate base of methanethiol. It is commercially available as white solid. It is a powerful nucleophile. Its hydrides in moist air produces methanethiol.