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Chemistry Test - 27

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Chemistry Test - 27
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  • Question 1
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    Major product of the following reaction is

    qImage668eb663067317c1e5c67234

    Solution

    qImage669e61c7d522ac1660e16aa3

    - The given reaction involves a compound with both an ester \(\left(- COOCH _3\right)\) and a nitrile \((- CN )\) group, treated with excess \(CH _3 MgBr\) followed by hydrolysis.

    - Steps of the reaction:

    - Step 1: The ester group reacts with excess \(CH _3 MgBr\).

    - The first mole of \(CH _3 MgBr\) attacks the carbonyl carbon of the ester, forming an intermediate.

    - A second mole of \(CH _3 MgBr\) reacts with the intermediate to produce a tertiary alcohol after hydrolysis.

    - Step 2: The nitrile group reacts with \(CH _3 MgBr\).

    - The Grignard reagent adds to the nitrile carbon, forming an imine intermediate.

    - Upon hydrolysis, the imine is converted into a ketone.

    - The final product contains a tertiary alcohol and a ketone group on the benzene ring.

    A compound with a tertiary alcohol and a ketone group attached to the benzene ring.

  • Question 2
    1 / -0

    The element with atomic number 57 belongs to:

    Solution

    The element with atomic number 57 belongs to d-block

    The d-block elements are called transition metals and have valence electrons in the d-orbital that's why such metals named as d-block elements.

    The d-block elements are found in groups 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 of the periodic table

    Valence shell electronic configuration of d-block element is (n- 1)d1-10 ns1-2.

    There are three series of transition elements which are known as 3d series, 4d series, 5d series.

    3d series start from Scandium (Sc) to Zinc (Zn), 4d series start from Yttrium (Y) to Cadmium (Cd) and 5d series start from Lanthanum(La) to Mercury (Hg), excluding Cerium (Ce) to Lutetium (Lu).

  • Question 3
    1 / -0

    On oxidation with a mild oxidising agent like \(Br _{2} / H _{2} O\), the glucose is oxidised to:

    Solution

    On oxidation with a mild oxidising agent like \(Br _{2} / H _{2} O\), the glucose is oxidised to six carbon carboxylic acid (gluconic acid). This indicates that the carbonyl group is present as an aldehydic group.

    \(CHO -( CHOH )_{4} CH _{2} OH \stackrel{ Br _{2} \text { water }}{\longrightarrow} COOH -\underset{\text { Gluconic acid }}{( CHOH )_{4}}- CH _{2} OH\)

  • Question 4
    1 / -0

    He and Ar are monoatomic gases and their atomic weights are 4 and 40 respectively. Under similar conditions, He will diffuse through semi-permeable membrane:

    Solution
    According to Graham's law of diffusion of gases,
    \(\frac{r_{1}}{r_{2}}=\sqrt{\frac{M_{2}}{M_{1}}}\)where, \(r_{1}, r_{2}=\) Rates of diffusion of the two gases,\(\mathrm{M}_{1}, \mathrm{M}_{2}=\) Molar masses of the two gases
    For He and Ar,
    where, \(M_{A r}=40\)\(M_{H e}=4\)
    Substituting the values in equation (i), we get\(\frac{r_{H e}}{r_{A r}}=\sqrt{\frac{40}{4}}=\sqrt{\frac{10}{1}}=3.16\)
    Under similar conditions, He will diffuse through a semi-permeable membrane 3.16 times as fast as Ar.
  • Question 5
    1 / -0

    Which one of the following element has the highest ionization energy?

    Solution

    \( (\mathrm{Ne})3\mathrm{s}^{2} 3 \mathrm{p}^{3}\) is stable configuration because the outer sub-shell is half-filled. Half filled configurations are stable configurations.

    The one half-filled and one completely filled orbital gives stability. Thus, It will be difficult to remove an electron from it and will take more energy.

    So, \([\mathrm{Ne}] 3 \mathrm{s}^{2} 3 \mathrm{p}^{3}\) has the highest ionization energy among the given elements.

  • Question 6
    1 / -0

    Which one of the following ions has the highest value of ionic radius?

    Solution

    If we compare \(\mathrm{Li}^{+}\)and \(\mathrm{B}^{3+}\) then both of the elements have the same number of electrons that is 2 that is they are isoelectronic.

    And in the case of isoelectronic species more the +ve charge less is the ionic radius.

    So, \(\mathrm{B}^{3+}\) has more ionic radius than \(\mathrm{Li}^{+}\).

    In the case of \(\mathrm{O}^{2-}\) and \(\mathrm{F}^{-}\)both have a total of 10 electrons so they are isoelectronic species.

    So, the more the -ve charge more is the ionic radius of the element. Hence the order of ionic radius will be \(\mathrm{C}>\mathrm{D}>\mathrm{A}>\mathrm{B}\)

  • Question 7
    1 / -0
    Match the Column I and Column II and choose the correct code given below.
    Column I Column II
    (A) Peroxyacetyl nitrile (i) Waste incineration
    (B) Indigo (ii) Vat dye
    (C) IR active molecules (iii) Global warming
    (D) Dioxins (iv) Photochemical smog
    Solution
    Column I Column II
    (A) Peroxyacetyl nitrate (iv) Photochemical smog
    (B) Indigo (ii) Vat dye
    (C) IR active molecules (iii) Global warming
    (D) Dioxins (i) Waste incineration

    Peroxyacetyl nitrate (PAN) is used in the photochemical song.

    Indigo is a vat dye. Dyeing with indignation was carried out in the wooden vats in the form of water-soluble indigotin-white.

    IR active molecules are used in global warming.

    Dioxins are used to burn the waste into ashes.

  • Question 8
    1 / -0

    A cell converts:

    Solution

    A cell converts chemical energy into electrical energy.

    • It consists of two solid electrodes and this solid electrodes are placed in an electrolyte and connected together by an electrical conductor such as wire.
    • The two electrodes should be made up of two different metals and the electrolyte solution can be an alkaline, acidic or salt solution depending on the situation.
    • These electrodes can be depicted as two half cells set up in different containers and are connected through a porous or salt solution.
  • Question 9
    1 / -0

    A drug which acts as antipyretic as well as an analgesic is:

    Solution

    Acetamidophenol is an antipyretic as well as analgesic drug. It lowers body temperature as well as is used to get relief from pain. It is commonly known by the name of paracetamol.

    Chloroquine is a drug which is used in the treatment of malaria. Malaria is caused by a malaria protozoa carried by the female Aedes mosquito.

    Penicillin is a collection of antibiotics derived from penicillium mould. It was the first antibiotic discovered. It was discovered by Alexander Fleming.

    Chlordiazepoxide is a sedative drug. It is used to treat insomnia, anxiety and acute alcohol withdrawal. It belongs to the benzodiazepines class of drugs. These act on the central nervous system producing a calming sensation.

  • Question 10
    1 / -0

    Stephen's reduction converts ethane nitrile into:

    Solution

    Stephen's reduction converts ethane nitrile into ethanal or Acetaldehyde.

    When ethane nitrile is treated with Sn/HCl, it undergoes Stephen's reaction to form  Acetaldehyde.

    The reaction is:

    The reaction is more efficient when aromatic nitriles are used rather than aliphatic nitriles. The reaction is a redox reaction and electron-withdrawing substituents can improve the rate of the reaction.

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