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Hydrocarbons Test - 10
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Hydrocarbons Test - 10
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  • Question 1/10
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    The alkene which on reaction with HBr follows Markovnikov rule is:

    Solutions

    The addition in unsymmetrical alkenes follows the Markovnikov rule, according to which the negative part of the addendum goes to the carbon having lesser number of hydrogens.

     

  • Question 2/10
    1 / -0

    Identify 'A' and 'X' in the following conversion:

    Solutions

    The step-1 involves ozonolysis to form hexan-1,6-dione.
    The step-2 involves aldol condensation in the presence of a base to form α, β-unsaturated aldehydes.

     

  • Question 3/10
    1 / -0

    Identify the alkene which on ozonolysis gives the following products:

    Solutions

    4-Methyl-penta-1,3-diene on ozonolysis yields methanal, glyoxal and acetone.

     

  • Question 4/10
    1 / -0

    Identify 'X' in the given reaction:

    Solutions

    The given reaction is an example of dehydrohalogenation of alkyl halides in the presence of alc. KOH.

     

  • Question 5/10
    1 / -0

    Which of the following statements account(s) for the acidic nature of acetylene?

    Solutions

    The acidic nature of acetylene is due to the greater sigma electron density of C-H bond of acetylene towards carbon, which is sp hybridised and has 50% of s-character.
    Hence, the electron pair of C-H bond gets displaced towards the carbon atom; and as a result, abstraction of H+ ion becomes easier.

     

  • Question 6/10
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    Identify 'X' in the given reaction:

    Solutions

    Since the alkyne is reduced to an alkene, and cis product is formed, the reducing agent is H2/Pd, BaSO4 (Lindlar's reagent).

     

  • Question 7/10
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    But-1-yne reacts with cold KMnOto give

    Solutions

     

  • Question 8/10
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    Consider the following conversion:

    The 'Reagent I' and 'Reagent 'II' respectively are:

    Solutions

    Reagent I is a Lewis acid that catalyses the electrophilic substitution (acylation) of the benzene ring.
    Reagent II is Zn-Hg,HCl that brings Clemmensen reduction of carbonyl compounds to form hydrocarbons.

     

  • Question 9/10
    1 / -0

    The aromatic compounds among the following are:

    Solutions

    (A), (C) and (D) are aromatic as there is complete conjugation and they obey Huckel's rule of (4n + 2) π-electrons.

     

  • Question 10/10
    1 / -0

    During the monobromination of 1-methoxy-3-methylbenzene with (Br2/FeBr3), the major product obtained is:

    Solutions

    Both methyl group (–CH3) and methoxy group (-OCH3) are ortho and para directing, and activating groups.
    Since the effect of methoxy group is stronger, the major product will have bromine substituted at the para position with respect to the methoxy group.

     

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