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Aldehydes, Ketones and Carboxylic Acids Test 9
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Aldehydes, Ketones and Carboxylic Acids Test 9

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  • Question 1/12
    1 / -0

    Only One Option Correct Type

    Direction (Q. Nos. 1-10) This section contains 10 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct.

    Q. 

    The major organic product in the following reaction is

    Solutions

  • Question 2/12
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    Consider the following reaction,

    Solutions

  • Question 3/12
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    Which is the best hydride (H-) donor in the key step of Cannizaro reaction? 

    Solutions

    Dianion is better hydride donor. Also the electron donating group (CH3O) increases hydride (H-) donating a bility .

  • Question 4/12
    1 / -0

    Consider the following reaction,

    Q.

    The major organic product is

    Solutions

    It is a simple Tischenko reaction

  • Question 5/12
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    What is the major organic product in the following reaction?

    Solutions


  • Question 6/12
    1 / -0

    Predict the major organic product in the following reaction.

    Solutions

    Intramolecular Cannizaro reaction occur.

  • Question 7/12
    1 / -0

    Whta is the major organic product formed in the follwing reaction ?

    Solutions


  • Question 8/12
    1 / -0

    For a Cannizaro reaction,


    Rate law is derived as : Rate = k [RCHO]2 [HO-]2
    From the above rate law, it can be concluded that

    Solutions

    For formation of dianion hydride donor, an additional mole of NaOH is consumed

    Hence, rate becomes 2nd order with respect to HO- and fourth order overall.

  • Question 9/12
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    Consider the following reaction,


    Q. 

    How the product Z can be prepared selectively using X and Y and other reagents?

    Solutions

  • Question 10/12
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    What is the major product formed in the following intramolecular Cannizaro reaction?

    Solutions

    —CHO para to electron withdrawing —NO2 group is better hydride (H-) acceptor.

  • Question 11/12
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    Matching List Type

    Direction (Q. Nos. 11 and 12) Choices for the correct combination of elements from Column I and Column II are given as options (a), (b), (c) and (d), out of which one is correct.

    Q. 

    Consider the reactions of Column I and match with the products of Column II. Mark the correct option from the codes given below.

     

    Solutions

    (i)  is better H- acceptor, reduced to FCH2OH.
    (ii) CH2O is oxidised and C6H5CHO is reduced in cross Cannizaro reaction.
    (iii) Electron withdrawing —NO2 makes —CHO a better hydride acceptor, hence p-nitrobenzaldehyde is reduced and benzaldehyde is oxidised.
    (iv) p-nitrobenzaldehyde is reduced, p-methoxy benzaldehyde becomes better hydride donor after attack by HO-, hence oxidised.

  • Question 12/12
    1 / -0

    Match the reactions of Column I with the type of reactions from Column II. Mark the correct option from the codes given below.

     

    Solutions

    (i) Initially, aldol reaction followed by Cannizaro reaction giving C(CH2OH)4 + HCOOH.
    (ii) F—CHO undergo Cannizaro reaction due to absence of α -H.
    (iii) It has difficulty in aldol condensation, hence undergo Cannizaro reaction predominantly.
    (iv) All aldol, Cannizaro and Claisen reaction occur.

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