Haloalkanes and Haloarenes Test

Result

Haloalkanes and Haloarenes Test - 59

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Weekly Quiz Competition

Question 1
1 / -0

Select the best combination of regents which will bring the following conversion.

A
$$Br_2/FeBr_3; \;HNO_3/H_2SO_4;\ NaOH; \;H_2/Pt$$

B
$$CH_3COCl/AlCl_3; \;Br_2/FeBr_3;\; NaBH_4/CH_3OH$$

C
$$CH_3Cl/AlCl_3; \;KMnO_4; \;CH_3Cl/AlCl_3; \;NBS/hv;\;LiAiH_4/H_2O;\;NaOH$$

D
$$CH_3Cl/AlCl_3; \;HNO_3/H_2SO_4; \;NBS/hv/NaOH$$

Solution

Benzene is first acetylated with acetyl chloride in the presence of anhydrous aluminum chloride to obtain acetophenone (Friedel-crafts acylation reaction).

Then, it is brominated with bromine in the presence of ferric bromide. Acetyl group is a meta directing group.

Hence, m-bromo acetophenone is obtained. The carbonyl group is then reduced to alcoholic group by using sodium borohydride in methanol.

Option B is correct.
Question 2
1 / -0

In comparison to $$C-Cl$$ bond in ethyl chloride, the $$C-Cl$$ bond in vinyl chloride is:

A
longer and weaker

B
shorter and weaker

C
longer and stronger

D
shorter and stronger

Solution

$$C-Cl$$ bond in vinyl chloride has partial double bond character.

Therefore, the $$C-Cl$$ bond in vinyl chloride is shorter and stronger as compound to $$C-Cl$$ bond in ethyl chloride.

Option D is correct.
Question 3
1 / -0

$$A + Na/\Delta \rightarrow$$ no gaseous product
$$A + Br_2/FeBr_3 \rightarrow C_7H_7OBr$$ (two isomers).
Based on the above reactions, identify A from the given options.

A

B

C

D

Solution

If phenolic $$OH$$ group was present, treatment with sodium would have liberated hydrogen gas. Since no gaseous product is obtained, phenolic $$OH$$ group is absent.
Question 4
1 / -0

Which of the following method would be better for the synthesis of 1-bromo-3-ethylbenzene?

A

B

C

D

Solution

Hence option C is correct.
Question 5
1 / -0

In which of the following cases $$C-C$$ bond length will be the longest?

A
$$CH_3-CF_3$$

B
$$FCH_2-CH_2F$$

C
$$F_2CH-CHF_2$$

D
$$CF_3-CF_3$$

Solution

Fluorine is the most electro-negative element. Hence, it will try to pull the carbon towards itself which will result in increasing the bond length between two carbons.
Question 6
1 / -0

The most stable conformation of the product of following reaction is:

A

B

C

D

Solution

Conformation $$D$$ is most stable. This conformer corresponds to anti-peri-planar orientation (conformation) of two largest substituents $$Br$$ in the system. Due to the steric size the antiperiplanar conformer is the most stable.
Question 7
1 / -0

Which of the following is fast debrominated?

A
IV

B
II

C
III

D
I

Solution

The debromination of compound $$(IV)$$ is fast as it forms benzene which is aromatic and highly stable.
Question 8
1 / -0

The alkene limonene has following structure, (Ref. image). Which product results from the reaction of Limonene and $$1$$ molar equivalent chlorine water?

A

B

C

D

Solution

Chlorine water $$HOCl$$ reacts as $$Cl^+OH^-$$. $$Cl^+$$ being strong electrophile attacks on electron rich $$\pi$$ bond.
Question 9
1 / -0

Action of alcoholic $$AgNO_3$$ on chlorobenzene is similar to the action on:

A
allyl chloride

B
vinyl chloride

C
isopropyl chloride

D
benzyl chloride

Solution

In chlorobenzene, the chlorine atom is directly attached to the unsaturated carbon atom.

In vinyl chloride also, the chlorine atom is directly attached to the unsaturated carbon atom.

Hence, the action of alcoholic $$AgNO_3$$ on chlorobenzene is similar to the action on vinyl chloride.

Both vinyl chloride and chlorobenzene give no precipitate with alcoholic $$AgNO_3$$ because both have chlorine atoms that are not reactive.

Option B is correct.
Question 10
1 / -0

An aromatic compound $$\displaystyle 'A' (C_{7}H_{6}Cl_{2}$$), gives $$AgCl$$ on boiling with alcoholic $$AgNO_3$$ solution, and yields $$\displaystyle C_{7}H_{7}OCl$$ on treatment with sodium hydroxide. 'A' on oxidation gives a mono chlorobenzoic acid which affords only one mononitro derivative. The compound A is :

A

B

C

D

Solution

$$SN^1$$ mechanism proceeds through carbocation formation, as stability of carbocation increases, reaction favours $$SN^1$$ mechanism.Benylic carbocation is most stable due to more resonating structures and allylic carbocations are more stable than aliphatic carbocation due to resonance so order of stability of carbocations is benzyll > allyl > $$3^0 > 2^0 > 1^0$$.
$$3^0$$ carbocation are more stable than others or stability order of carbocations is $$3^0 > 2^0 > 1^0$$ so order of alkyl halides towards $$SN^1$$ reaction is $$3^0 > 2^0 > 1^0$$.
So, order of reactivity for $$SN^1$$ mechanism is
benzyll > allyl > $$3^0 > 2^0 > 1^0$$.
Here, given compound is $$Cl-C_6H_4-CH_2-Cl$$, in presence of $$AgNO_3$$ its benylic chloride reacts and AgCl forms white ppt.
In presence of NaOH, benzylic chlorine has been replaced by OH- group and $$Cl-C_6H_4-CH_2-OH$$ forms as product, which forms acid on oxidation.
So, answer is option A.