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Practical Organic Chemistry Test - 9
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Practical Organic Chemistry Test - 9
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  • Question 1/10
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    The structure of 4-methylpent-2-en-1-ol is

    Solutions

    The structure of 4-methylpent-2-en-1-ol is (CH3)2CHCH=CHCH2OH.
    It is evident from the IUPAC name that -OH is attached to the first carbon in the chain and the double bond is at the second position.

     

  • Question 2/10
    1 / -0

    The IUPAC name of the above given compound is

    Solutions

    According to the IUPAC rules, all the three carboxylic acid groups are treated as equal. So, the correct IUPAC name of the above given compound is pentane-1,3,5-tricarboxylic acid.

     

  • Question 3/10
    1 / -0

    The following question has four choices, out of which ONLY ONE is correct.

    The total number of contributing structures for hyperconjugation in CH3CH=CH2 is

    Solutions

    Propene has three hyperconjugated C-H bonds. The hyperconjugating structures of propene are as follows:

     

  • Question 4/10
    1 / -0

    Which of the following structures does not exhibit tautomerism?

    Solutions

    The structure  does not exhibit tautomerism because this structure is extra stable due to the presence of conjugated double bond system.

     

  • Question 5/10
    1 / -0

    Which of the following statements is/are correct about resonance energy?

    Solutions

    Only statement (2) is correct.
    Stability ∝ Resonance energy

    Whereas, statements 1 and 3 are incorrect.
    Resonance energy ∝ Number of canonical structures
    Resonance energy = (Expected heat of hydrogenation) - (Actual heat of hydrogenation)

     

  • Question 6/10
    1 / -0

    Arrange the following species in increasing order of their stability:

    Solutions

    I, II, III and IV have 6, 5, 3 and 2 hyperconjugating H-atoms, respectively. The more are the hyperconjugating atoms, the more is the stability.

     

  • Question 7/10
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    Soda-lime test is used to detect _________ element of organic compound.

    Solutions

    Soda-lime test is used to detect nitrogen element of organic compound.

     

  • Question 8/10
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    The reactions given above form the basis of direct estimation of which of the following elements of organic compound?

    Solutions

    These reactions are used for the direct estimation of oxygen in organic compound from the mass of carbon dioxide produced.

     

  • Question 9/10
    1 / -0

    In Kjeldahl's method for estimation of nitrogen present in a soil sample, ammonia evolved from 0.75 g of sample neutralised 10 ml of 1 M H2SO4. The percentage of nitrogen in the soil was

    Solutions

    ∵ M × V (ml) = m mol
    number of mili moles of H2SO4 = 10
    H2SO4 + 2NH3 → (NH4)2SO4
    According to the balanced equation:
    10 m mol of H2SO4 reacts with 20 m mol of NH3
    1 mol NH3 contains 14 g of nitrogen.
    20 × 10-3 mol NH3 contains 14 × 20 × 10-3 g of nitrogen.
    Percentage of nitrogen in 0.75 g of the sample = × 100 = 37.33%

     

  • Question 10/10
    1 / -0

    In Kjeldahl's method, the nitrogen present in an organic compound is quantitatively converted into

    Solutions

    A known mass of an organic compound is heated with conc. H2SO4 when nitrogen present in the organic compound is quantitatively converted into ammonium sulphate. Ammonium sulphate thus obtained is boiled with conc. NaOH solution. The NH3 thus liberated is absorbed in a known volume of standard acid solution (HCl or H2SO4).

     

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