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Haloalkanes and Haloarenes Test - 55

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Haloalkanes and Haloarenes Test - 55
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Weekly Quiz Competition
  • Question 1
    1 / -0
    Halogenation of alkanes is
    Solution
    The reaction of a halogen with an alkane in the presence of ultraviolet (UV) light or heat leads to the formation of a haloalkane (alkyl halide).This is a oxidative process
  • Question 2
    1 / -0
    Stilbene $$(PhCH=CHPh)$$ can exist in two diastereomeric forms $$(X)$$ and $$(Y)$$, and $$(X)$$ is found to be more soluble in water than $$(Y)$$. Predict which of the following statement is correct. 
    Solution

  • Question 3
    1 / -0
    Benzene reacts with $$CH_{3}Cl$$ in the presence of anhydrous $$AlCl_{3}$$ to form:
    Solution
    Benzene reacts with $$CH_3Cl$$ in presence of anhydrous $$AlCl_3$$ to give toulene.

    This is Friedel crafts Alkylation and an electrophilic substitution reaction.

  • Question 4
    1 / -0
    In the following reaction, the major product $$'X'$$ is

    Solution

  • Question 5
    1 / -0
    Identify $$'B'$$ in the following reaction

    Solution
    Answer is C, as :

    $$CH_3-C_6H_5-Cl \xrightarrow{KMnO_4} COOH-C_6H_5-Cl \xrightarrow{C_2H_5OH} COOC_2H_5-C_6H_5-Cl$$
  • Question 6
    1 / -0
    Toluene on treatment with $$CH_{3}Cl$$ and $$AlCl_{3}$$ at $$80^{\circ}C$$ gives the major product as
    Solution
    Friedel-Craft's alkylation of toluene by reaction with methyl chloride in the presence of $$AlCl_3$$ and heat produces meta-isomer that is m-xylene as major product
  • Question 7
    1 / -0
    The product of the following reaction $$C_{6}H_{6} + Cl_{2} \xrightarrow {Sunlight} ?$$; is
    Solution

  • Question 8
    1 / -0

    The final product of the given reaction sequence is:

    Solution

  • Question 9
    1 / -0
    If the light waves pass through a nicol prism then all the oscillations occur only in one plane such beam of light is called as
    Solution
    Correct option is $$B$$.
  • Question 10
    1 / -0
    What is the major product of the following reaction $$CH_{3}C\equiv C-CH_{2}-CH_{3}\xrightarrow[]{1 \,mole \,of \,Cl_{2}}$$
    Solution
    $$CH_{3}C\equiv C-CH_{2}-CH_{3}\xrightarrow[]{1 \,mole \,of \,Cl_{2}}$$$$CH_{3}-\overset{Cl}{\overset{|}{\underset{Cl}{\underset{|}{C}}}}-\overset{Cl}{\overset{|}{\underset{Cl}{\underset{|}{C}}}}-CH_{2}CH_{3}$$
    Option D is a correct answer. 

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