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Chemistry Test - 13

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Chemistry Test - 13
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  • Question 1
    3 / -1

    Which of the following compound on heating does not produce metal oxide?

    Solution

    Here \(\mathrm{K}_{2} \mathrm{CO}_{3}\) on heating does not produce its oxide as it is thermally stable.

    \(\mathrm{MgCl}_{2} .6 \mathrm{H}_{2} \mathrm{O} \stackrel{\Delta}{\rightarrow} \mathrm{MgO}+2 \mathrm{HCl}+5 \mathrm{H}_{2} \mathrm{O}\)

    \(2 \mathrm{~K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \stackrel{\Delta}{\rightarrow} 2 \mathrm{~K}_{2} \mathrm{CrO}_{4}+\mathrm{Cr}_{2} \mathrm{O}_{3}+\frac{3}{2} \mathrm{O}_{2}\)

    \(\mathrm{K}_{2} \mathrm{CO}_{3} \stackrel{\Delta}{\rightarrow}\) does not decompose but melts

    \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2} \stackrel{\Delta}{\rightarrow} \mathrm{CuO}+2 \mathrm{NO}_{2}+\frac{1}{2} \mathrm{O}_{2}\)

  • Question 2
    3 / -1

    What is the total number of isomers, stereoisomers included, designated by the dichlorocyclopentane?

    Solution

    3 structural isomers, but if you are considering chirality/diastereomers, there are two chiral carbons, so the maximum number of stereoisomers is 22=4, from which cis-1,2-dichlorocyclopentane and cis-1,3-dichlorocyclopentane will have same structure in their individual 4 isomers as shown below, so total isomers will be 7.

  • Question 3
    3 / -1

    Which of the following can be used as the halide component for Friedel-Crafts reaction?

    Solution

    Friedel-Craft reaction is carried out by the generation of a carbocation but in chlorobenzene, bromobenzene, chloroethene, lone pair of halogen are delocalized with π bonds to attain double bond character. According to the Friedel-Crafts reaction, Isopropyl chloride can be used as the halide component.

  • Question 4
    3 / -1

    What is the paper strip developed in partition chromatography called?

    Solution

    The paper strip so developed by retaining different components according to their different partitions in the two phases is called a chromatogram. The spots of the separated colored components are visible at different lengths from the location of the initial point on the chromatogram.

  • Question 5
    3 / -1

    The correct order of basicity of the following compounds is:

    Solution

    The correct order of basicity of the given compounds is 2 < 1 < 3 < 4.

    According to the structure given in the question:

    (1) ; one lone pair

    (2) ; one lone pair

    (3) ; This donation helps the sp2- hybridized nitrogen atoms to donate lone pair of its own with greater effort.

    (4)  Much more donation here than the previous case

    We know that, 

    (i) Donation capacity of nitrogen increases if electron donating group(s) (+I or +R-effect containing) is/are attached with it.

    (ii) More nitrogen atoms (or donor atoms) means more basicity (in general).

    (iii) sp2-hybridized nitrogen is less capable of donating lone pair than sp3-hybridized.

    From the second point, we know that 4>3 and both will be greater than 1 and 2. Out of 1 and 2, 1 has an sp3 hybridized N which is bonded to electron releasing methyl group and therefore is more basic than 2.

  • Question 6
    3 / -1

    A catalyst is added to a reaction to increase the rate of reaction. Which of the following statements is correct?

    Solution

    A catalyst works by providing an alternative reaction pathway to the reaction product. The rate of the reaction is increased as this alternative route has lower activation energy than the reaction route not mediated by the catalyst.

  • Question 7
    3 / -1

    Which type of chromatography involves the separation of a mixture over a column of adsorbent packed in a glass tube?

    Solution

    Column chromatography involves the separation of a mixture over a column of adsorbent packed in a glass tube. The mixture adsorbed on the adsorbent is placed at the top of the column and the mobile phase is allowed to flow through the column slowly.

  • Question 8
    3 / -1

    The half-life of a substance A following first-order kinetics is \(5\) days. Starting with \(100\) g of A, the amount left after \(15\) days is:

    Solution
    The initial concentration of the substance, \(A_{0}=100\) g, Half-life, \(\mathrm{t}_{1 / 2}=5\) days, Total time, \(t=15\) days
    We have to find out the amount of radioactive substance left after \(15\) days.
    We know that, Radioactive decay follows first order kinetics. Thus, the amount of substance left after \(n\) half-lives is given by the relation:
    \(A_{n}=A_{0} \times\left(\frac{1}{2}\right)^{n} \quad \ldots\) (ii)
    Substituting the values of \(t\) and \(t _{1 / 2}\) in equation (ii), we get \(n=\frac{15}{5}\)\(=3\)
    Now, Substituting the values of \(n\) and \(A_{0}\) in equation (i), we get \({An}=100×\left(\frac{1}{2}\right)^{3}\)\(=\frac{100}{8}=12.5\) g
  • Question 9
    3 / -1

    Gallium remains liquid up to __________ Kelvin.

    Solution

    Gallium remains liquid up to \(2276\) Kelvin.

    Low melting point of gallium is due to the fact that it consists of \(Ga_{2}\) molecules and gallium remains liquid up to \(2276\) \(k\). Hence it is used in high-temperature thermometer. Gallium as a chemical symbol that is \(Ga\) and its atomic number is given as \(31\).

  • Question 10
    3 / -1

    Gabriel’s phthalimide synthesis is used for the preparation of which amine?

    Solution

    This reaction is used for transforming primary alkyl halide into primary amines.

    Secondary and tertiary amines are not formed by this synthesis, so pure primary amines are formed which is the reason why this reaction is preferred for formation of primary amine.This is a 3 step reaction:

    Step 1:

    Phthalimide is treated with potassium hydroxide (KOH) initiating an acid-base reaction. Imide is deprotonated by the hydroxide ion. This forms a potassium salt of phthalimide (imide ion) which is nucleophilic in nature.

    Step 2:

    Now, the obtained imide ion is treated with alkyl halide (R−X). The electrophilic carbon of the alkyl halide is attacked by the nucleophilic imide. Halogen atom of the alkyl halide is removed and replaced by the nitrogen atom which bonds with the carbon itself. This results in the formation of N-Alkyl Phthalimide.

    Step 3:

    N-Alkyl Phthalimide then undergoes alkaline hydrolysis by reaction with aqueous Sodium Hydroxide (NaON(aq)). The hydroxide ion from Sodium hydroxide reacts with the carbon atom from the alkyl group causing the cleavage of N-Alkyl Phthalimide and the cation attaches itself to the oxygen atom. When the oxygen atom replaces the nitrogen in phthalimide the ejected hydrogen ions get attached to the nitrogen which was already attached to the alkyl (R) group.

    Only aliphatic primary amines are formed and not aromatic because aryl halides do not undergo the nucleophilic substitution reaction with the anion formed by phthalimide.

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