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Test: Preparation of Alkyl Halides - NEET MCQ


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19 Questions MCQ Test Chemistry Class 12 - Test: Preparation of Alkyl Halides

Test: Preparation of Alkyl Halides for NEET 2024 is part of Chemistry Class 12 preparation. The Test: Preparation of Alkyl Halides questions and answers have been prepared according to the NEET exam syllabus.The Test: Preparation of Alkyl Halides MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Preparation of Alkyl Halides below.
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*Multiple options can be correct
Test: Preparation of Alkyl Halides - Question 1

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

Detailed Solution for Test: Preparation of Alkyl Halides - Question 1

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 options B and C, the configuration at the second carbon atom is R. Hence, they cannot be formed.

Test: Preparation of Alkyl Halides - Question 2

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

Detailed Solution for Test: Preparation of Alkyl Halides - Question 2

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.

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Test: Preparation of Alkyl Halides - Question 3

Which is incorrect about Hunsdiecker's reaction?

Detailed Solution for Test: Preparation of Alkyl Halides - Question 3

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

Test: Preparation of Alkyl Halides - Question 4

The major product of the following reaction is 

Detailed Solution for Test: Preparation of Alkyl Halides - Question 4

Free radical bromination occur. Preferably at highest degree carbon where most stable free radical is formed.

Test: Preparation of Alkyl Halides - Question 5

Racemic mixture is obtained due to the halogenation of

Detailed Solution for Test: Preparation of Alkyl Halides - Question 5

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.


Test: Preparation of Alkyl Halides - Question 6

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

Detailed Solution for Test: Preparation of Alkyl Halides - Question 6


The gaseous byproducts escape out on its own continuously driving the reaction in forward direction.

Test: Preparation of Alkyl Halides - Question 7

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

Detailed Solution for Test: Preparation of Alkyl Halides - Question 7



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.

Test: Preparation of Alkyl Halides - Question 8

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?

Detailed Solution for Test: Preparation of Alkyl Halides - Question 8

Both SO2CI2 and Cl2 undergo homolytic bond fission when heated or irradiated with light.

*Multiple options can be correct
Test: Preparation of Alkyl Halides - Question 9

What is the order of SNreaction of the alkyl halide?

Detailed Solution for Test: Preparation of Alkyl Halides - Question 9

Answer: (b) The order of SNreaction of the alkyl halide is RI > RBr > RCl > RF.

Explanation: Iodine is a good nucleophile and a good leaving group. Thus, it eliminates easily from an alkyl halide favouring SNelimination reaction

*Multiple options can be correct
Test: Preparation of Alkyl Halides - Question 10

Consider the following reaction,

Q. 

The expected product(s) is/are

Detailed Solution for Test: Preparation of Alkyl Halides - Question 10

NBS in CCI4 brings about allylic brom ination by free radical mechanism:





*Multiple options can be correct
Test: Preparation of Alkyl Halides - Question 11

Consider the following reaction,

Q.

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

Detailed Solution for Test: Preparation of Alkyl Halides - Question 11


*Multiple options can be correct
Test: Preparation of Alkyl Halides - Question 12

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

Detailed Solution for Test: Preparation of Alkyl Halides - Question 12

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.

Test: Preparation of Alkyl Halides - Question 13

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?

Detailed Solution for Test: Preparation of Alkyl Halides - Question 13

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.

Test: Preparation of Alkyl Halides - Question 14

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? 

Detailed Solution for Test: Preparation of Alkyl Halides - Question 14

ZnCI2 is a Lewis acid, helps in the form ation of carbocation intermediate, hence catalyse the reaction.

Test: Preparation of Alkyl Halides - Question 15

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? 

 

      

 

    

Detailed Solution for Test: Preparation of Alkyl Halides - Question 15

The order of stability of carbocation intermediates formed follows the order of reactivity.

*Answer can only contain numeric values
Test: Preparation of Alkyl Halides - Question 16

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?


Detailed Solution for Test: Preparation of Alkyl Halides - Question 16

*Answer can only contain numeric values
Test: Preparation of Alkyl Halides - Question 17

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


Detailed Solution for Test: Preparation of Alkyl Halides - Question 17




Only three pairs of enantiomers are formed.

*Answer can only contain numeric values
Test: Preparation of Alkyl Halides - Question 18

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


Detailed Solution for Test: Preparation of Alkyl Halides - Question 18

*Answer can only contain numeric values
Test: Preparation of Alkyl Halides - Question 19

Consider the following reaction,

Q. 

How many different monobromo derivatives would be produced?


Detailed Solution for Test: Preparation of Alkyl Halides - Question 19


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

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