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Haloalkanes and Haloarenes Test - 40

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Haloalkanes and Haloarenes Test - 40
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  • Question 1
    1 / -0
    Bond pair of electrons in $$C-X$$ bond is nearer to:
    Solution
    Halogens are more electronegative than carbon. Halogens are electron withdrawing group which pulls a bond pair of electrons towards it.
                                  $$C- X$$
    Thus, Bond pair of electrons in $$C-X$$ is near to X
  • Question 2
    1 / -0
    $$C-X$$ bond is polar because:
    Solution
    $$C-X$$ bond is polar because X is more E.N than C. X pulls electron cloud towards itself and X gains negative charge on it and carbon gains slight positive charge.
  • Question 3
    1 / -0
    Which is the correct increasing order of bond dissociation enthalpy of $$C-X$$ bond?
    Solution
    Due to high bond length C-I has less bond strength and has less bond dissociation enthalpy. Because of short bond length C-F has high bond dissociation enthalpy so that increasing order of bond dissociation enthalpies is    $$C-I<C-Br<C-Cl<C-F$$.
  • Question 4
    1 / -0
    Which of the following has least polar $$C-X$$ bond?
    Solution

    strong electro withdrawing group like $$NO_2$$ when attach to the compound containing $$C-Cl$$ bond, it decreases the polarity of $$C-Cl$$ bond due to the opposite pull of sheared pair of electrons.

  • Question 5
    1 / -0
    Which is the correct increasing order of bond length of $$C-X$$ bond?
    Solution
    Increase in size of the atom increases the bond length because of largely sized atom formation of bond takes not easily.
    Hence $$C-F<C-Cl<C-Br<C-I$$ is the increasing order. 
  • Question 6
    1 / -0
    Which is the correct increasing order of dipole moment of $$C-X$$ bond?
    Solution
    $$\begin{array}{l}\text { Dipole moment-  it's occur when there is a } \\\text { separation of charge. They can occur between } \\\text { two ions in ionic bond. The dipole moment is a } \\\text { measure of the polarity of the molecule. }\end{array}$$
    $$C-I>C-Br>C-Cl>C-F$$
  • Question 7
    1 / -0
    $$CH_3 - CH_2 - Br \xrightarrow[\Delta]{Alco. KOH} B \xrightarrow{HBr} C\xrightarrow{Na/ether}D$$ the compound D is
    Solution
    The compound D is $$\displaystyle n-butane$$.
    $$\displaystyle CH_3 - CH_2 - Br \xrightarrow[\Delta]{Alco. KOH} \underset {B}{CH_2 = CH_2 } \xrightarrow{HBr} \underset {C}{CH_3 - CH_2 -Br} \xrightarrow{Na/ether}\underset {D}{CH_3 - CH_2 -CH_2-CH_3}$$

    When $$\displaystyle ethyl \: bromide$$ is heated with $$\displaystyle alc. KOH$$, it undergoes dehydrohalogenation to form $$\displaystyle ethylene$$. Addition of a molecule of $$\displaystyle HBr$$ to $$\displaystyle C=C$$ double  bond gives $$\displaystyle ethyl \: bromide$$. In presence of $$\displaystyle Na/ether$$, two molecules of $$\displaystyle ethyl \: bromide$$ couple to form $$\displaystyle n-butane$$. In this reaction, $$\displaystyle C-C$$ single bond is formed.
  • Question 8
    1 / -0
    Which alky halide has maximum density?
    Solution
    $${ CH }_{ 3 }I$$ has the highest molecular weight hence, it is alkyl halide with maximum density.

    $$(Mol. \ wt.\ of\ { CH }_{ 3 }I)=(Mol.\ wt.\ of\ C)+3\times (Mol.\ wt.\ of\ H)+(Mol.\ wt.\ of\ I)$$

    $$=12+(3\times 1)+127$$

    $$=142$$

    So, the correct option is $$D$$
  • Question 9
    1 / -0
    Consider the reactions
    (i) $$(CH_{3})_{2}CH - CH_{2}Br \xrightarrow {C_{2}H_{5}OH} (CH_{3})_{2} CH - CH_{2}OC_{2}H_{5} + HBr$$
    (ii) $$(CH_{3})_{2}CH - CH_{2}Br \xrightarrow {C_{2}H_{5}O^{-}} (CH_{3})_{2} CH - CH_{2}OC_{2}H_{5} + Br^{-}$$
    The mechanisms of reactions (i) and (ii) are respectively:
    Solution
    $$C_{2}H_{5}OH$$ being a weaker nucleopbile, when used as a solvent in case of hindered $$1^{\circ}$$ halide, favours $$S_{N}1$$ mechanism while $$C_{2}H_{5}O^{-}$$ being a strong nucleophile in this reaction favours $$S_{N}2$$ mechanism.
  • Question 10
    1 / -0
    Which of the following does not give white precipitate when boiled with alcoholic silver nitrate:
    Solution
    In chlorobenzene the electron lone pair of chlorine atom is in conjugation with benzene ring and hence chlorine atoms takes part in resonance and chlorobenzene does not show removal of chloride ion readily therefore it does not give white precipitate with alcoholic silver nitrate
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