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Chemistry Test - 26

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Chemistry Test - 26
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  • Question 1
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    Glucose is oxidized by nitric acid and produce:

    Solution

    Let's understand glucose in detail. It is also known by the name dextrose. It is the most abundant organic compound on earth. In honey and sweet fruits, glucose is present. In ripe grapes also, it is found in large amounts. The structure of glucose is,

    Now, come to the question. We are asked to identify the product of the reaction of glucose with nitric acid. Nitric acid oxidizes glucose to saccharic acid. The reaction is,

  • Question 2
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    Identify the product formed in the given reaction.

    Solution

    All the alkylated benzenes will oxidize to benzoic acid. t-butyl benzene is inert to \(KMnO _{4}\). tert. Butylbenzene has no benzylic hydrogens and hence does not give benzoic acid on oxidation. But as the substituted tert. Butyl group get seperated from the benzene ring due to steric effect and forms tert.-butanoic acid.

  • Question 3
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    Which of the following method cannot be used for preparation of aromatic amine?

    Solution

    Aromatic primary amines cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.

  • Question 4
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    Natural rubber is a polymer of isoprene. It is made to market quality through a process called vulcunisation. Name the new bonds introduced that stiffens rubber.

    Solution

    Sulphur cross links stiffens rubber.

    To improve upon the physical properties of rubber, a process of vulcanisation is carried out. This process consists of heating a mixture of raw rubber with sulphur and an appropriate additive at a temperature range between 373 K to 415 K. On vulcanisation, sulphur forms cross links at the reactive sites of double bonds and thus, the rubber gets stiffened.

  • Question 5
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    Molecules: General formula. (n represents the number of carbon atoms in chain)

    1. Aldehydes: \(C _{ n } H _{2 n } O\)

    2. Carboxylic acids: \(C _{ n } H _{2 n } O _{2}\)

    3. Cycloalkanes: \(C _{ n } H _{2 n }\)

    4. Alkanes: \(C _{ n } H _{2 n +2}\)

    Which of the above matching is correct? Choose from the options given below:

    Solution

    All the given options are correctly matched.

    Aldehyde has \(C _{n} H _{2} O\) general formula. Example: Ethanal \(- C _{2} H _{4} O , n =2\)

    Carboxylic acid has \(C _{n} H _{2 n} O _{2}\) general formula. Example: Propionic acid \(- C _{3} H _{6} O _{2}, n =3\)

    Cycloalkane has \(C _{ n } H _{2 n }\) general formula. Example: Cyclobutane \(- C _{4} H _{8}, n =4\)

    Alkane also has \(C _{n} H _{2 n +2}\) general formula. Example: Ethane \(- C _{2} H _{6}, n =2\)

  • Question 6
    1 / -0

    Which of these polymers are co-polymers?

    (i) Neoprene

    (ii) Buna-N

    (iii) Nylon-6,6

    Solution

    Buna-N and Nylon-6,6 are co-polymers.

    When two or more different monomers combine, they form co-polymer.

    Neoprene is an addition homo-polymer of chloroprene.

    Buna-N is formed by co-polymerization of 1,3-Butadiene and Acrylonitrile.

    Nylon-6,6 is formed by co-polymerization of adipic acid and hexamethylene diamine by elimination of water.

  • Question 7
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    Which one of the following is the strongest base in aqueous solution?

    Solution

    Comparision of Basic nature in aliphatic amines in aqueous phase. -

    The basic strength of an amine in aqueous medium is determined not only on the basic of +1 effect of the alkyl groups attached to the nitrogen atom but also the solvation of alkyl ammonium ion formed by uptake of proton.

    The solvation of alkyl ammonium ion follows the above order. Both the +1 effect and solvation of amine make basic order of aniline in aqueous medium as

    General order of basicity -

    Aliphatic Amines > Ammonia > Aromatic Amines

    In aqueous solution the order of basic strength is -

  • Question 8
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    Which type of 'defect' has the presence of cations in the interstitial sites?

    Solution

    Frenkel defect: It is a stoichiometric defect. Frenkel defect arises when an ion leaves its normal site and occupies an interstitial site. A cation is much smaller than an anion, thus, this defect is caused by the displacement of cations from original sites to interstitial sites.

  • Question 9
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    The given reaction is \(a\):

    Solution

    The given reaction is \(a 1,4\) addition reaction.

    In the first step, a molecule of \(H B r\) is added to a conjugated diene. Both \(H\) atom and \(B r\) atom are added to terminal \(C\) atoms. This is \(1,4\) addition or conjugate addition. In the next step, the reaction with \(M g\) in presence of ether gives a Grignard reagent \((R-M g X)\) which then attacks a molecule of carbon dioxide. Acid hydrolysis gives carboxylic acid.

  • Question 10
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    The ions \(\mathrm{Mg}^{+2}\) and \(\mathrm{PO}_{4}^{-3}\) combine to form compounds:

    Solution

    The ions \(\mathrm{Mg}^{+2}\) and \(\mathrm{PO}_{4}^{-3}\) combine to form compound \(\mathrm{Mg}_{3}\left(\mathrm{PO}_{4}\right)_{2}\)

    Total charge on cation \(=3(2+)=+6\).

    Total charge on anion \(=2(-3)=-6\).

    Therefore,

    Total charge on cation \(=\) Total charge on anion.

  • Question 11
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    The solution of H2O2 can be stored for a long time at room temperature. However, bubbles of oxygen form as soon as a drop of bromine is added. The role of bromine for the reaction 2H2O2(aq) → 2H2O(l) + O2(g) is:

    Solution

    In the decomposition of hydrogen peroxide, two successive reactions of H2O2, with bromide and then with bromine, take place. Because these two reactions together serve as a catalytic pathway for hydrogen peroxide decomposition, both of them must have significantly lower activation energies than the uncatalyzed decomposition, as shown schematically in Figure.

    Bromine acts as negative catalyst in the reaction. Bromine forms an intermediate (Br−) on reacting with H2O2 following reduction of bromine and then to show oxidation of Br− to Br2 by H2O2.

  • Question 12
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    Identify the functional group in Product \(B\)?

    Solution

    The functional group in Product \(B\) is carboxylic group \(-C O O H .\)

    The first step is allylic bromination with \(N\)-bromosuccinimide (NBS) in which \(B r\) atom is added to allylic \(C\) atom of cyclohexene to give compound \(A\).

    In the next step, the reaction with \(M g\) in presence of ether gives a Grignard reagent \((R-M g X)\) which then attacks a molecule of carbon dioxide. Acid hydrolysis gives carboxylic acid.


  • Question 13
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    \({IP}_{1}\) and \({IP}_{2}\) of \(Mg\) are \(178\) and \(384\) kcal. The enthalpy required for the reaction \({Mg} \rightarrow {Mg}^{2+}+2 {e}^{-}\)
    Solution
    \(M g(g) \rightarrow M g^{+}(g)+e^{-}\) \(IP_{1}=178 ~kcal ~mol ^{-1}\)
    \(M g^{+}(g) \rightarrow M g^{2+}+e^{-} \ {IP}_{2}=384 ~kcal~ mol ^{-1}\)
    \(\overline{M g \rightarrow M g^{2+}+2 e^{-}} \quad \Delta H=?\)
    The successive ionisation enthalpies i.e. IP1, and IP\(_{2}\) are additive so that the overall enthalpy change can be given as:\(\Delta {H}={IP}_{1}+{IP}_{2}\)
    On substituting the values, we get\(\Delta {H}=(178+384) ~{kcal}~ {mol}^{-1}=+562~ {kcal} {~mol}^{-1}\)
    Therefore, the enthalpy required for the reaction \({Mg} \rightarrow {Mg}^{2+}+2 {e}^{-}\) is \(+562 ~{kcal}~ {mol}^{-1}\).
  • Question 14
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    Zwitter ion can be formed by:

    Solution

    We know that a molecule consisting of both positively charged and negatively charged ions is known as a zwitterion. The number of positively charged functional groups in a zwitterion is equal to the number of negatively charged functional groups. Thus, we can say that electrical neutrality is maintained within the molecule. Due to the presence of both positive and negative charges, they are known as dipolar ions.

    Thus, we can say that to form a Zwitterion a molecule must have an acidic group and a basic group. Of all the given compounds, only sulphanilic acid fulfils both the requirements. Sulphanilic acid contains a sulfonic acid group and an amino group. The sulfonate group is deprotonated forming an anion and the amino group is protonated forming a cation giving rise to a dipolar zwitterion.

    Sulphanilic acid is a crystalline, off- white coloured solid. It is generally used in the production of food dye and various pharmaceutical products.

  • Question 15
    1 / -0

    Natural rubber has:

    Solution

    Natural rubber has all cis-configuration.

    The repeating unit has the cis-configuration in natural rubber with chain extensions on the same side of the ethylene double bond. This is essential for elasticity. If the configuration is trans, the polymer is either a hard plastic or like gutta-percha.

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