Haloalkanes and Haloarenes Test

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Haloalkanes and Haloarenes Test - 58

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Question 1
1 / -0

The main difference in C - X bond of a haloalkane and a haloarene is

A
C - X bond in haloalkanes is shorter than haloarenes.

B
In haloalkanes the C attached to halogen in C - X bond is $$sp^{3}$$ hybridised while in haloarenes it is $$sp^{2}$$ hybridised.

C
C - X bond in haloarenes acquires a double bond character due to higher electronegativity of X than haloalkanes.

D
haloalkanes are less reactive than haloarenes due to difficulty in C - X cleavage in haloalkanes.

Solution

The main difference in C - X bond of a haloalkane and a haloarene is that in haloalkanes carbon of C - X is $$sp^{2}$$ hybridised while in haloarenes it is $$sp^{3}$$ hybridised.
Hence, Option "B" is the correct answer.
Question 2
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An alkyl halide on reaction with sodium in the presence of ether gives 2,2,5,5-tetramethylhexane. The alkyl halide is:

A
1-chloro pentane

B
1-chloro-2,2-dimethyl propane

C
3-chloro-2,2-dimethyl butane

D
2-chloro-2-methyl-butane

Solution

Alkyl halide on reaction with sodium in presence of ether to give alkane is called Wurtz reaction.
The reaction takes place as:
$$2ClCH_2-C(CH_3)_2-CH_3+2Na\rightarrow CH_3-C(CH_3)_2-CH_2-CH_2-C(CH_3)_2-CH_3+ 2NaCl$$
Question 3
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Statement 1: $$S_{N}2$$ reaction is carried out in the presence of polar aprotic solvents.
Statement 2: Polar aprotic solvents does not contain acidic hydrogen.

A
Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation for Statement 1.

B
Statement 1 and Statement 2 are correct and Statement 2 is not the correct explanation for Statement 1.

C
Statement 1 is correct and statement 2 is wrong.

D
Statement 1 is wrong and statement 2 is correct.
Question 4
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$$CH_3-CH_3\xrightarrow[]{Br_2/hv}A\xrightarrow[]{LiAlH_4}B\xrightarrow[]{Br_2/hv}C\xrightarrow[]{Na/ether}D$$

The compound $$D$$ is:

A
$$CH_{3}-CH_{2}-CH_{2}-CH_{3}$$

B
$$CH_{3}-CH_{2}-O-CH_{2}-CH_{3}$$

C
$$CH_{3}-CH(CH_{3})-CH(CH_{3})-CH_{3}$$

D
$$CH_{3}-CH_{2}-CH_{3}$$

Solution

By this process, first alkyl halide forms which reduced by $$LiAlH_4$$ to form alkane and again bromination takes place to give alkyl halide which in presence of $$Na/ dry \:ether$$ forms alkane as product (Wurtz reaction).

$$CH_3CH_3\xrightarrow[]{Br_2/hv}CH_3CH_2Br\xrightarrow[]{LiAlH_4}CH_3CH_3\xrightarrow []{Br_2/hv}CH_3-CH_2-Br\xrightarrow[]{Na/dry\,ether}CH_3-CH_2-CH_2-CH_3$$

Option A is correct.
Question 5
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One of the iosmers of $$C_{4}H_{9}Cl$$ by Wurtz reaction forms $$2,2,3,3$$- tetramethylbutane. The structure of that isomer of $$C_{4}H_{9}Cl$$:

A
$$1$$-Chloro - $$2$$- methylpropane

B
$$2$$-Chloro - $$2$$- methylpropane

C
$$2$$-Chloro butane

D
$$1$$-Chlorobutane

Solution

Wurtz reaction: In this process alkyl halide forms alkane in presence of sodium and dry ether.
Reaction:
$$CH_3C(CH_3)ClCH_3+2Na+CH_3C(CH_3)ClCH_3\rightarrow CH_3C(CH_3)_2C(CH_3)_2CH_3$$
$$2$$-Chloro - $$2$$- methylpropane is the answer.
Question 6
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The compound $$Y$$ is:

A
1, 3 - dichloro benzene

B
benzyl chloride

C
1, 3, 5 - trichloro benzene

D
mixture of 1, 2 and 1, 4 dichlorobenzene

Solution

$$C_6H_6 \xrightarrow {Cl_2,Fe} C_6H_5Cl \xrightarrow {Cl_2}$$ $$o-$$ and $$p-$$ dichlorobenzene

In first reaction chlorobenzene forms as a product and as $$Cl-$$ group is ortho-, para- directing so ortho, para dichloro benzene forms as product.

Second step is chlorination of chlorobenzene.

Option D is correct.
Question 7
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$$CH_3-CH(Br)-CH_3\xrightarrow[]{alc.\,KOH} A\xrightarrow[peroxide]{HBr}B\xrightarrow[acetone]{NaI}C$$

The compound $$C$$ is:

A
$$CH_{3}-CH_{2}-CH_{2}-I$$

B
$$CH_{3}-CH(I)-CH_{3}$$

C
$$CH_{3}-CHI-CH_{2}-I$$

D
$$CH_{3}-CH=CH-I$$

Solution

This process is used to form alkyl iodide as a product.

Alcohol $$KOH$$ is dehydrohalogenating agent and peroxide is used to form anti-markovnikov product.

$$CH_3-CHBr-CH_3\xrightarrow[]{alc.KOH}CH_3-CH=CH_2\xrightarrow[peroxide]{HBr} CH_3CH_2CH_2Br\xrightarrow[acetone]{Nal}CH_3CH_2CH_2I$$

Option A is correct.
Question 8
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Assertion: Aryl halides undergo electrophilic substitution more readily than benzene.
Reason: Aryl halide gives a mixture of $$o -, p -$$ products when they undergo electrophilic substitution.

A
Assertion and Reason are correct and Reason is the correct explanation of Assertion

B
Assertion and Reason are correct but Reason is not the correct explanation of Assertion

C
Assertion is true and Reason is false

D
Assertion is false and Reason is true

Solution

$$X$$- (halogen group) has two effects$$ -I$$ & $$+M$$.

Due to $$- I$$ effect of $$X$$- group, it reduces electron density of benzene so halobenzene is less reactive than benzene for electrophilic substitution reaction.

But due to $$+M$$ effect, it is ortho, para directing and in halobenzene further substitution takes place at ortho and para positions.
Question 9
1 / -0

Which one of the following occurs by an addition elimination mechanism?

A
Toluene, Br$$_2$$, FeBr$$_3 \rightarrow$$ $$\rho$$-bromotoluene

B
Benzene, HNO$$_3$$, H$$_2$$SO$$_4 \rightarrow$$ nitrobenzene

C
Benzene, CO, HCl, AlCl$$_3$$, CuCl $$\rightarrow$$ benzaldehyde

D
2, 4-dinitrochlorobenzene, NaOH, heat, then H$$^+ \rightarrow$$ 2, 4-dinitrophenol

Solution

The reaction in option D proceeds by addition elimination reaction or nucleophilic substitution reaction. First the nuclophile (hydroxide ion) is added and then the chloride ion is eliminated. This reaction is possible due to presence of two electron withdrawing nitro groups.
Question 10
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The major product of the above shown reaction is :

A

B

C

D

Solution

Both bromine and methoxy group are ortho and para directing groups.

But, bromine deactivates the aromatic ring and methoxy group activates the ring.

Hence, the effect of methoxy group is more pronounced.

Therefore, the substitution occurs in the position indicated in option C.

Option C is correct.