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Organic Chemistry Some Basic Principles and Techniques Test - 72

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Organic Chemistry Some Basic Principles and Techniques Test - 72
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Weekly Quiz Competition
  • Question 1
    1 / -0
    Number of $$\pi$$ electrons in conjugation for these compounds will be respectively:

  • Question 2
    1 / -0
    Which of these functional groups is not present in the  below compound given  

  • Question 3
    1 / -0
    Which among the following is not correctly matched with their colour? 
          $$\text {Compound}$$                            $$\text {Colour}$$
    Solution
    $$A)\underset{from\ sodium\ extracted}{Na_{2}S}+Na_{2}[Fe(CN)_{5}NO]\rightarrow \underset{purple\ colouration}{Na_{4}[Fe(CN)_{5}NOS]}$$


    $$B) Na_{4}[Fe(CN)_{6}]+Fe^{+3}\rightarrow \underset{Ferric\ ferrocynid\\ \\ (blue\ colour)}{Fe_{4}[Fe(CN)_{6}]_{3}}$$


    $$C)Fe^{+3}+3NaSCN \rightarrow \underset{\ \ Ferric\ thiocynate\\  \\   (Blood\ red\ colouration)}{Fe(SCN)_{3}}$$


    $$D)Na+Cl\xrightarrow[From\ o.c]{\Delta }NaCl\xrightarrow[]{AgNO_{3}}\underset{white\ ppt}{AgCl}+NaNO_{3}$$


    $$\therefore AgCl$$ is incorrect match with its colour

    $$\mathbf{Hence\ correct\ answer\ is\ option\ (D)}$$



  • Question 4
    1 / -0
    A person is suffering from Diabetes, which get confirmed by testing his urine with Benedict's reagent. The organic compound and functional group present in the urine of suffering person is :
  • Question 5
    1 / -0
    Chromyl chloride gives yellow ppt. when it reacts with respectively_________
    Solution
    $$CrO_2Cl_2+4NaOH\rightarrow Na_2CrO_4+2NaCl+2H_2O$$
    Chromyl chloride

    $$Na_2CrO_4+(CH_3COO)_2Pb\rightarrow PbCrO_4+2CH_3COONa$$
                                                            yellow ppt

    Option C is correct.
  • Question 6
    1 / -0
    Which functional group is there in methyl ethanoate?
  • Question 7
    1 / -0
    The correct order of +M effect of 'N' containing functional group on benzene ring amongst the given compounds is:

  • Question 8
    1 / -0
    Which compound cannot give a precipitate with an ammoniacal silver nitrate solution?
    Solution

  • Question 9
    1 / -0

    Directions For Questions

    An aromatic compound $$(A), C_{9}H_{12}O$$ was subjected to a series of test in the laboratory. It was found that this compound:
    (i) does not form silver mirror with Tollens' reagent.
    (ii) rotates the plane of polarised light.
    (iii) reacts with sodium metal to evolve hydrogen gas.
    (iv) does not decolourise pink colour of bromine water.
    (v) reacts with hot $$KMnO_{4}$$ to form monocarboxylic acid $$'B'$$ which on decarboxylation gives benzene.
    (vi) reacts with Lucas reagent in about $$5$$ min.
    (vii) reacts with $$I_{2}$$ and $$NaOH$$ to produce yellow coloured precipitate of the compound $$(C)$$.
    (viii) loses its optical activity due to formation of compound $$(D)$$ on reaction with Red $$P$$ and $$HI$$.
    Answer the following questions:

    ...view full instructions

    The functional groups present in the compound $$(A)$$ is
    Solution

  • Question 10
    1 / -0
    The reaction
    $$ (CH_3)_3 C-Br \overset { H_ 2O }{ \longrightarrow  } (CH_3)_3 C-OH $$  is
    Solution
    The reaction $$ (CH_3)_3 C-Br \overset { H_ 2O }{ \longrightarrow  } (CH_3)_3 C-OH $$  is an example of Substitution reaction.

    Hence, Option "B" is the correct answer.
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