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Chemistry Test - 26

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Chemistry Test - 26
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Weekly Quiz Competition
  • Question 1
    1 / -0

    Chlorobenzene on heating with NaOH at 300oC and high pressure followed by reaction with dil. HCl gives-

    Solution


    Concepts :
    Main Concept :
    Preparation of phenols from Dow process The Dow process is the electrolytic method of bromine extraction from brine, and was Herbert Henry Dow's second revolutionary process for generating bromine commercially.

    This process was patented in 1891. In the original invention, bromide-containing brines are treated with sulfuric acid and bleaching powder to oxidize bromide to bromine, which remains dissolved in the water. The aqueous solution is dripped onto burlap, and water is blown through causing bromine to volatilize. Bromine is trapped with iron turnings to give a solution of ferric bromide. Treatment with more iron metal converted the ferric bromide to ferrous bromide via comproportionation. Where desired, free bromine may be obtained by thermal decomposition of ferrous bromide.
    Before Dow got into the bromine business, brine was evaporated by heating with wood scraps and then crystallized sodium chloride was removed. An oxidizing agent was added, and bromine was formed in the solution. Then bromine was distilled. This was a very complicated and costly process.
    Dow's Process may also refer to the hydrolysis of chlorobenzene in the preparation of phenol. Benzene can be easily converted to chlorobenzene by electrophilic aromatic substitution. It is treated with dilute sodium hydroxide at 350oC and 300 bar to convert it to sodium phenoxide, which yields phenol upon acidification. This reaction is quickened manifold in the presence of electron withdrawing groups (such as -NO2) ortho and/or para to the halogen group.

    Hence, the correct option is (A)

  • Question 2
    1 / -0

    Consider a compound 'A' with formula C3H6O. It forms phenyl hydrazone and gives negative Tollen's test. Compound 'A' on reduction gives propane. Then, identify compound 'A'.

    Solution

    Compound 'A' forms phenyl hydrazone, so it must be a carbonyl compound. Since, it also gives negative Tollen's test, so it can't be an aldehyde.

    'A' on reduction gives propane. So, it must be acetone.

    Ketone reacts with phenyl hydrazine to form phenyl hydrazone but gives negative Tollen's test.
    Hence, the correct option is (C)

  • Question 3
    1 / -0

    Which of the following does not react with diethyl ether?

    Solution

    Diethyl ether is a lewis base. It reacts with lewis acids such as HCl, BF3 and AlCl3. Sodium ethoxide contains strongly basic ethoxide ion, which does not react with diethyl ether, which also is a base.
    Hence, the correct option is (A)

  • Question 4
    1 / -0
    Identify \(z\) in the given reaction.
    \(H_{2} S O_{4}+H g S O_{4}+6 \sigma^{n} C\)
    Solution
    \(C H \equiv C H \stackrel{H_{2} S O_{4}+H g S O_{4}+6 \sigma^{\rho} C}{\longrightarrow} C_{2} H_{3} O H \rightarrow C H_{3} C H O\)
    \(C_{2} H_{3} O H\) formed will undergo keto-enol tautomerism.
    Hence, the correct option is (D)
  • Question 5
    1 / -0

    The reagent (A) is:

    Solution

    3-bromo cyclo hexanone reacts with LAH and AlCl3 to form cyclo hexene.

    Hence, the correct option is (A)

  • Question 6
    1 / -0

    C6H5OH+CHCl3+NaOH→C6H4(CHO)OH

    In the above reaction, the electrophile involved is:

    Solution

    \(H_{2} O+C C l_{3} \rightarrow C C l_{2}+C l\)

    Here, elecrophile is dichlorocarbene.
    Hence, the correct option is (B)

  • Question 7
    1 / -0

    In Riemer-Tiemann reaction, reactants are:

    Solution

    In Riemer-Tiemann reaction, reactants are both Phenol, CHCl3 and alkali and phenol, CCl4 and alkali.

    When phenol, CHCl3 and alkali are used, the product is salicylaldehyde.

    When phenol, CCl4 and alkali are used, the product is salicylic acid.
    Hence, the correct option is (D)

  • Question 8
    1 / -0

    \(C H_{3} \equiv C-C H_{3} \frac{{ }^{40 \% H_{2} S O_{4}}}{{ }_{185 H g S O_{4}}} B\)

    Then, \(B\) is :

    Solution

    Hydration of alkynes is carried out in presence of warm 40% sulfuric acid in the presence of 1% mercuric sulphate.

    In this reaction, first a molecule of water is added to form enol which then tautomerizes to keto form. Thus, B is acetone.
    Hence, the correct option is (A)

  • Question 9
    1 / -0

    IUPAC name of the tertiary butyl alcohol is:

    Solution

    IUPAC name of tertiary butyl alcohol is 2-methyl-2-propanol.
    Hence, the correct option is (D)

  • Question 10
    1 / -0

    Which one of the following is a secondary alcohol?

    Solution


    Hence, the correct option is (D)

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