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Chemistry Test - 32

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Chemistry Test - 32
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Weekly Quiz Competition
  • Question 1
    1 / -0

    Aldol condensation between which of the following two compounds followed by dehydration gives methyl vinyl ketone?

    Solution
    Aldol condensation of formaldehyde and acetone gives \(C H_{2} O+C H_{3} O C H_{3} \longrightarrow H O C H_{2}-C H_{2}-C O-C H_{3}\)
    Heating the product gives of methyl vinyl ketone. \(H O C H_{2}-C H_{2}-C O-C H_{3} \longrightarrow C H_{2}=C H-C O-C H_{3}\)
    Hence, the correct option is (A)
  • Question 2
    1 / -0

    Of the following, the oxime of which compound shows geometrical isomerism?

    Solution
    For benzaldehyde, \(C_{6} H_{5} C H O+N H_{2} O H \longrightarrow C_{6} H_{5} C N(H)(O H)(a n t i)+C_{6} H_{5} C N(H)(O H)(s y n)\)
    gives syn and anti benzaldehyde oximes. For other cases, only one product is formed. For acetone, \(\left(C H_{3}\right)_{2} C O+H_{2} N O H \longrightarrow\left(C H_{3}\right)_{2} C N O H+H_{2} O\)
    For formaldehyde, \(H C H O+N H_{2} O H \longrightarrow H_{2} C N O H\)
    For diethyl ketone, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}_{2} \mathrm{CH}_{3}+\mathrm{NH}_{2} \mathrm{OH} \longrightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CN}(\mathrm{OH}) \mathrm{CH}_{3}\)
    Hence, the correct option is (D)
  • Question 3
    1 / -0

    Which one of the following is a tertiary alcohol?

    Solution

    \(\left(C H_{3}\right)_{3} C-O H\) is a tertiary alcohol, as the -OH group is attached to a tertiary carbon
    atom.
    Hence, the correct option is (D)
  • Question 4
    1 / -0

    Phenol on treatment with conc. HNO3 gives:

    Solution

    When phenol is heated with a mixture of conc. nitric acid and conc. sulphuric acid, 2,4,6 trinitrophenol is obtained which is known as picric acid.
    Hence, the correct option is (A)
  • Question 5
    1 / -0

    The IUPAC name of cinnamaldehyde is:

    Solution

    The structure of cinnamaldehyde is as follows:
    The IUPAC name is \((2 E)-3\)
    phenylprop- - -enal. (The notation \(E\) is used as the two
    groups on the double bond are on the opposite side).
    Hence, the correct option is (A)
  • Question 6
    1 / -0

    Which of the following gives ethyl alcohol by the action of methyl magnesium halide followed by hydrolysis?

    Solution

    4HCHO + CH3MgBr / H2O ⟶ CH3CH2OH + MgXOH
    Hence, the correct option is (A)

  • Question 7
    1 / -0

    In the presence of manganese acetate, ethyne produces:

    Solution

    Manganese acetate is an organometallic compound. Its action is like that of the Grignard reagent which first of all takes active hydrogen. Active hydrogen is the hydrogen attached to highly electronegative element like oxygen, nitrogen, chlorine. The electronegativity of sp hybridised carbon lies between that of oxygen and nitrogen and so, hydrogen attached to it is the active hydrogen.

    The reaction is as follows:

    Mn (CH3COO)2+ HC ≡ CH → 2CH3COOH + Mn(C2)
    Hence, the correct option is (A)

  • Question 8
    1 / -0

    The correct IUPAC name of the compound is:

    Solution

    The correct IUPAC name of the given compound is 3-cyclopentyl-3-methyl butanal.

    The parent compound contains 4 carbon atoms and an aldehyde group. Hence, it is named as butanal. Two substituents are present at third carbon atoms. One is methyl group and the other is cyclopentyl group.

    In the names given in option A and C, the parent hydrocarbon contains 3 carbon atoms. This is not the longest chain. Hence, these names are incorrect.

    In the name given in option D, two ethyl and one cyclobutyl groups are named, but are not present in the given structure. Hence, this name is also incorrect.
    Hence, the correct option is (B)

  • Question 9
    1 / -0

    2- phenyl propene on acidic hydration gives :

    Solution

    Hence, option A is correct.

  • Question 10
    1 / -0

    An alcohol (A) on heating with concentrated H2SO4 gives alkene (B), which shows geometrical isomerism. The compound (A) is:

    Solution

    since, alkene shows geometrical isomerism. So, the groups should be different on both sides of double bond. When 2 -butanol is heated with conc. sulfuric acid, it forms 2 -butene, which shows geometrical isomerism in the form of cis and trans isomers.
    \(C H_{3} C H_{2} C H(O H) C H_{3} \frac{d \text { ehydration }}{\operatorname{conc} H_{2} S O_{4}} C H_{3} C H=C H C H_{3}\)
    Hence, the correct option is (C)
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