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Hydrocarbons Test - 20

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Hydrocarbons Test - 20
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  • Question 1
    1 / -0
    Among the following compounds 2, 3-dimethylexane is :

    Solution
    Simple nomenclature of alkane.
  • Question 2
    1 / -0
    Methane with the Molecular formula $$"CH_4"$$ has
    Solution
    Structure of methane $$H-\overset{\displaystyle H}{\overset{|}{\underset{\displaystyle H}{\underset{|}{C}}}}-H$$
    It has $$4$$ covalent bonds.
  • Question 3
    1 / -0
    The boiling point will increase in the order of:
    Solution
    Straight chain isomers have higher boiling point than the branched ones. Thus boiling point of n-butane is higher than iso-butane. Option (D) is correct.
  • Question 4
    1 / -0
    The number of total product produced upon monochloriation of (+) 2-chlorobutane is ___________.
    Solution
    B. 4
    2-Chlorobutane as reactant will produce four$${ C }_{ 4 }{ H }_{ 8 }{ Cl }_{ 2 }$$.The products are-
    $${ ClCH }_{ 2 }{ CHClCH }_{ 2 }{ CH }_{ 3 }+{ CH }_{ 3 }{ CCl }_{ 2 }{ CH }_{ 2 }{ CH }_{ 3 }+\\ { CH }_{ 3 }{ CHClCHClCH }_{ 3 }+{ CH }_{ 3 }{ CHClCH }_{ 2 }{ CH }_{ 2 }Cl$$
  • Question 5
    1 / -0
    Addition of $$HBr$$ on propene produce:
    Solution
    The addition of $$HBr$$ to propene takes place according to Markownikoff's rule. According to this rule negative part of reagent adds to carbon atom having lesser number of hydrogen atoms.
    $$CH_3CH=CH_2+HBr\rightarrow CH_3-\underset{Br\,\,\,}{\underset{|}{C}H}-CH_3$$
  • Question 6
    1 / -0
    The hydrocarbon obtained by treating sodium ethanoate with soda lime is _________.
    Solution
    $${ CH }_{ 3 }COONa+NaOH\rightarrow { CH }_{ 4 }+{ Na }_{ 2 }{ CO }_{ 3 }$$

    $$CH_3COONa$$=Sodium ethanoate
    $$NaOH$$=Soda lime
    $$CH_4$$=Methane
    $$Na_2CO_3$$=Sodium carbonate
  • Question 7
    1 / -0
    The reaction of one mol of bromine with ethyne yields:
    Solution

    Halogens, especially chlorine and bromine on reaction with alkynes readily produces tetra-halogen derivatives of alkane. The reaction is carried out in inert solvent like carbon tetrachloride.

    Due to the presence of two n-bonds, each molecule of the alkyne can react with two molecules of the halogen.

    One mole of Bromine with ethyne gives 1,2-dibromoethene.

    $$HC\equiv CH + Br_2 \xrightarrow{CCl_4} H(Br)C = C(Br)H$$

  • Question 8
    1 / -0
    Ethylidene dichloride is obtained by the reaction of excess of HCl with.
    Solution
    $$\underset{acetylene}{CH\equiv CH}+HCl\rightarrow \underset{vinyl  chloride}{CH_2=CHCl}\overset{HCl}{\rightarrow} \underset{ethylidene dichloride}{CH_3CHCl_2}$$
    [Alkenes due to presence of only $$1^o$$ of unsaturation, form only monochloro derivative.]
  • Question 9
    1 / -0
    X compound reacts with Na to give $$CH_3CH_2CH_2CH_3$$. The compound X is:
    Solution
    $$\underset{ethyl chloride}{CH_3 CH_2} Cl+2Na + Cl CH_2 CH_3 \rightarrow CH_3 - CH_2 - CH_2 CH_3 + 2 NaCl$$
    This is Wurtz reaction.
  • Question 10
    1 / -0
    The compound formed when ethyl alcohol is treated with conc.$${H}_{2}{SO}_{4}$$ at $${170}^{o}C$$ is _______.
    Solution
    $$Ethanol$$ can be dehydrated to give $$ethene$$ by heating it with an excess of concentrated sulphuric acid at about 170C.
    $$CH_3CH_2-OH \overset {Conc. H_2SO_4 170^oC}{\rightarrow} CH_2=CH_2 + H_2O$$
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