Haloalkanes and Haloarenes Test

Result

Haloalkanes and Haloarenes Test - 5

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Question 1
4 / -1

Only One Option Correct Type

Direction (Q. Nos. 1-7) This section contains 7 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.

Q. When (S)-2-bromopentane is brominated, several 2,3-dibromopentane are formed, which of the following is not formed?

A

B

C
None of these

D
All of the Above

Solution

During bromination of (S)-2-bromopentane, the configuration at second carbon is not affected. It remains R configuration. The configuration at the third carbon atom can be either S or R.

In option B the configuration at the second carbon atom is R. Hence, they cannot be formed.
Question 2
4 / -1

The yield of alkyl bromide obtained as a result of heating the dry silver salt of carboxylic acid with bromine in CCI₄ is

A
1°> 3°> 2° bromides

B
3 ° > 2° > 1°bromides

C
1°> 2° > 3° bromides

D
3 ° > 1°> 2 ° bromides

Solution

The reaction follows free radical mechanism and alkyl free radical is formed in the propagation step as

Hence, stability of alkyl free radical (3° > 2° > 1°) determine the reactivity.
Question 3
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Which is incorrect about Hunsdiecker's reaction?

A
Only Cl₂ can give alkyl halide.

B
I₂ will give ester when treated with RCOOAg.

C
The reaction proceeds through free radical.

D
F₂ cannot give alkyl halide.

Solution

Except F₂, almost all halogens react with RCOOAg giving alkyl halide via Hunsdiecker reaction.
With l₂ if RCOOAg is in excess, R— I formed in the first step reacts further with unreacted salt to give ester as
R—COOAg + R—I → R—COOR + AgI
Question 4
4 / -1

The major product of the following reaction is

A

B

C

D

Solution

Free radical bromination occur. Preferably at highest degree carbon where most stable free radical is formed.
Question 5
4 / -1

Racemic mixture is obtained due to the halogenation of

A
n-pentane

B
isopentane

C
neopentane

D
Both (a) and (b)

Solution

If free radical halogenation generate a chiral carbon, racemic mixture of halides are always formed due to equal probability of halogenation from both sides of planar free radical.
Question 6
4 / -1

The reaction of SOCI₂ on alkanols to form alkyl chlorides gives good yields because

A
alkyl chlorides are immiscible with SOCI₂

B
the other products of the reaction are gaseous and escape out

C
alcohol and SOCI₂ are soluble in water

D
the reaction does not occur via intermediate formation of an alkyi chlorosulphite

Solution

The gaseous byproducts escape out on its own continuously driving the reaction in forward direction.
Question 7
4 / -1

Addition of bromine on propene in the presence of brine yields a mixture of

A
CH₃CHCICH₂Br and CH₃CHBrCH₂CI

B
CH₃CHCICH₂CI and CH₃CHBrCH₂Br

C
CH₃CHCICH₂Br and CH₃CHBrCH₂Br

D
CH₃CHCICH₂Cl and CH₃CHBrCH₂Cl

Solution

Nucleophilic attack in step-ll occur at the carbon atom which can better accommodate the positive charge. Hence, attack of Br^- or Cl^- in second step occur at 2° carbon rather that at 1° carbon.
Question 8
4 / -1

One or More than One Options Correct Type
Direction (Q. Nos. 8-12) This section contains 5 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or MORE THAN ONE are correct.
Q. Which of the following reagents can bring about free radical chlorination of propane?

A
SOCI₂

B
SO₂CI₂

C
none

D
PCI₃

Solution

Both SO₂CI₂ and Cl₂ undergo homolytic bond fission when heated or irradiated with light.
Question 9
4 / -1

In which of the following reaction(s), the major product is a primary alkyl bromide?

A

B

C

D

Solution
Question 10
4 / -1

Consider the following reaction,
Q.
The expected product(s) is/are

A

B

C

D

Solution

NBS in CCI₄ brings about allylic brom ination by free radical mechanism:
Question 11
4 / -1

Consider the following reaction,
Q.
When a pure enantiomer of X is taken in the above reaction, correct completion regarding the reaction is/are

A
Four different dichlorocyclohexane are formed as significant products

B
a pair of enantiomers is formed

C
two pairs of diastereomers are formed

D
product mixture has zero specific rotation

Solution
Question 12
4 / -1

Choose the correct statement(s) from the following regarding free radical chlorination and bromination reaction of alkane.

A
The first propagation step is exothermic in bromination reaction and endothermic in chlorination reaction

B
The first propagation step is endothermic in bromination reaction and exothermic in chlorination reaction

C
Chlorination reaction is more reactive process but very less selective, reverse is true for bromination

D
Chlorination is more selective but less reactive process and its reverse is true for bromination

Solution

In free radical halogenation of alkane, the first step of propagation is exothermic when Cl₂ is used while it is endothermic when Br₂ is used. Also chlorination occur at very fast rate, hence very less selective while bromination occur at very slow rate, occurs selectively where most stable free radical is formed.
Question 13
4 / -1

Comprehension Type
Direction (Q. Nos. 13-15) This section contains a paragraph, describing theory, experiments, data, etc.
Three questions related to the paragraph have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d).
Passage
An alcohol (R — OH) can be converted into alkyl chloride by the treatm ent with HCI. Reaction involves protonation of alcohol followed by the formation of carbocation intermediate. Carbocation intermediate in the final step undergo nucleophilic attack by Cl^- ion as :
Q.
Which of the following alcohol reacts most easily?

A

B

C

D

Solution

As mentioned in mechanism, reaction proceed via carbocation intermediate. Hence, alcohol forming most stable carbocation reacts most easily. Alcohol (c) forms aromatic (highly stable) carbocation, hence most reactive.
Question 14
4 / -1

An alcohol (R — OH) can be converted into alkyl chloride by the treatm ent with HCI. Reaction involves protonation of alcohol followed by the formation of carbocation intermediate. Carbocation intermediate in the final step undergo nucleophilic attack by Cl^- ion as :
Q.
Which of the following can catalyse the above reaction?

A
AgNO₃

B
ZnCl₂

C
CCl₄

D
H₃PO₄

Solution

ZnCI₂ is a Lewis acid, helps in the form ation of carbocation intermediate, hence catalyse the reaction.
Question 15
4 / -1

An alcohol (R — OH) can be converted into alkyl chloride by the treatm ent with HCI. Reaction involves protonation of alcohol followed by the formation of carbocation intermediate. Carbocation intermediate in the final step undergo nucleophilic attack by Cl^- ion as :
Q.
What is the correct order of reactivty of the followings with HCl?

A
I < II < III < IV

B
IV < II< III< I

C
III < II < I < IV

D
IV < I < II < III

Solution

The order of stability of carbocation intermediates formed follows the order of reactivity.
Question 16
4 / -1

One Integer Value Correct Type
Direction (Q, Nos. 16-19) This section contains 4 questions. When worked out will result in an integer from 0 to 9 (both inclusive).
Q.
If 2,4-dimethyl pentane is subjected to free radical chlorination reaction, how many different monochlorinated products would be formed?

A
4

B
344

C
5423

D
1427

Solution
Question 17
4 / -1

On free radical chlorination reaction of butane, how many different, optically active, dichloroalkanes would be formed ?

A
6

B
346

C
5623

D
1627

Solution

Only three pairs of enantiomers are formed.
Question 18
4 / -1

If 1, 3-butadiene is treated with excess of bromine in CCI₄ , how many different tetrabromides would be formed?

A
3

B
343

C
5323

D
1327

Solution
Question 19
4 / -1

Consider the following reaction,
Q.
How many different monobromo derivatives would be produced?

A
4

B
344

C
5423

D
1427

Solution

(I) has two optically active enantiomers and (II) has two geometrical isomers.