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Haloalkanes & Haloarenes Test - 12
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Haloalkanes & Haloarenes Test - 12

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  • Question 1/9
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    Consider the following reaction,

    Q.

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

     

    Solutions

     

     


     

     

  • Question 2/9
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    Choose the correct statement(s) from the following regarding free radical chlorination and bromination reaction of alkane.

     

    Solutions

     

     

    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 3/9
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    Comprehension Type

    Direction (Q. Nos. 3 - 5) 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?

     

    Solutions

     

     

    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 4/9
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    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? 

     

    Solutions

     

     

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

     

     

  • Question 5/9
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    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? 

     

        

     

       

     

    Solutions

     

     

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

     

     

  • Question 6/9
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    One Integer Value Correct Type

    Direction (Q, Nos. 6 - 9) 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?

     

    Solutions

     

     

     

     

  • Question 7/9
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    On free radical chlorination reaction of butane, how many different, optically active, dichloroalkanes would be formed ?

     

    Solutions

     

     




    Only three pairs of enantiomers are formed.

     

     

  • Question 8/9
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    If 1, 3-butadiene is treated with excess of bromine in CCI4 , how many different tetrabromides would be formed?

     

    Solutions

     

     

     

     

  • Question 9/9
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    Consider the following reaction,

    Q. 

    How many different monobromo derivatives would be produced?

     

    Solutions

     

     


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

     

     

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