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Aldehydes Ketones and Carboxylic Acids Test - 74

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Aldehydes Ketones and Carboxylic Acids Test - 74
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Weekly Quiz Competition
  • Question 1
    1 / -0
    Addition of alcohols to aldehydes in presence of anhydrous acid yield
    Solution
    Option D is correct.

  • Question 2
    1 / -0
    $$CH_3CONH_2\xrightarrow[]{NaNO_2IHCI}X$$
    Solution
    Amide, on treating with $$HNO_2$$ give acids. 

    $$CH_3CONH_2\xrightarrow[(h\nu)]{NaNO_2/HCl}CH_3COOH+N_2+H_2O$$
                                                    acetic acid
  • Question 3
    1 / -0
    Addition of alcohols to aldehydes in presence of anhydrous acid yield:
    Solution
    Alcohols are added to aldehydes and ketones in the presence of dry hydrogen chloride to form hemiacetals. This reaction can continue by adding another alcohol to form an acetal. 

    Hence, option $$D$$ is correct.

  • Question 4
    1 / -0

    Directions For Questions

    Four isomeric optically active compounds $$\xrightarrow [  ]{ NaH{ CO }_{ 3 } } { CO }_{ 2 }(g)$$
    $$(A,B,C,D)({C}_{4}{H}_{8}{O}_{3})$$
    Compound $$(A)\xrightarrow [  ]{ LAH } (E)$$ (Achiral compound)
    compound $$(B\xrightarrow [ or\\ Cr{ O }_{ 3 } ]{ KMn{ O }_{ 4 } }  $$ Inert to oxidation
    Compound $$(C)\xrightarrow [  ]{ NaOI/{ H }_{ 3 }{ O }^{ \oplus  } } CH{ I }_{ 3 }$$ (Yellow ppt.)+Compound $$(F)$$

    ...view full instructions

    Compound $$(A)$$ is:
    Solution

  • Question 5
    1 / -0

    Directions For Questions

    In above reaction

    ...view full instructions

    Carboxylic acid $$(A)$$ is:
    Solution
    Formation of $$(E)$$ suggests monosubstituted acid and reaction of 1 mol of $${H}_{2}$$ with (A) and (B) shows one $$(C=C)$$ bond. There are 6 D.U (4 D.U for $$(Ph)$$ group, 1 D.U for $$(COOH)$$ and 1 D.U for $$(C=C)$$ in both (A) and (B)
    Chiral compound with nine $$C$$ atoms would be (given figure)
    Here $$(C)$$ and $$(D)$$ can be distinguished quantitatively by H.V.Z reaction. $$(C)$$ has one $$\alpha$$- H atoms and would give dichlorinated product. The chlorine content in the chlorinated derivaties of $$(C)$$ and $$(D)$$ is determined by quantitative methods

  • Question 6
    1 / -0
    $${CH}_{3}CH=CHCHO$$ is oxidised to $${CH}_{3}CH=CHCOOH$$ using:
    Solution
    Since alkaline $$KMnO_4$$ are powerful oxidizing agents, they can oxidize double bond also, to protect double bond, we use a mild oxidizing agent as $$Tollen's\  [Ag(NH_3​)_2​]+OH−$$ oxidizes $$CHO$$ to $$COOH$$ without affecting the $$(C=C)$$ bond. and $$Ag(0)$$ (mirror) is precipitated.

  • Question 7
    1 / -0
    Which of the following statements are correct?
    Solution

  • Question 8
    1 / -0
    The decreasing order of $$ pK_a $$ value of the following is:

    Solution

    Stabler the $$C_B$$, stronger is the acidic strength (lower $$pK_a$$).

    Therefore, the decreasing $$pK_a$$ value, (II) > (I) > (III). {Option (B) is correct}

  • Question 9
    1 / -0
    Five isomeric p-substituted aromatic compounds (A) to (E) with molecular formula $$C_8H_8O_2$$ are given for identification. Based on the following observation, give the structures of the compounds.
    i. Both (A) and (B) form a silver mirror with Tollens reagent. (B) also gives a positive test with neutral $$FeCl_3$$ solution.
    ii. (C) gives positive iodoform test.
    iii. (D) is readily extracted in aqueous $$NaHCO_3$$ solution.
    iv. (E) on acid hydrolysis gives 1,4 - dihydroxy benzene.
    Compound (D) is:
    Solution
    Hence, Option "B" is the correct answer.

  • Question 10
    1 / -0
    Select the correct order in sequence in order to get the final product (E).

    Solution
    Hence, Option "C" is the correct answer.

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