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Give distinguishing test for the following : (a) ethane and ethyne (Alkane and alkyne RC-=CH) (b) Ethene and Ethyne (Alkene and Alkyne RC-=CH) (c) Dimethyl ethyne and ethyne (d) Ethane and Ethene (Alkane and alkene) |
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Answer» Solution :Ethane and ethyne : (x) in ethyne `(HC-=CH)`, H is acidic in nature and so it reacts with sodium or sodamide while ethane does not give such reaction. (i) `UNDERSET("Ethyne")(HC-=CH)underset(or NanH_(2))overset(Na(NH_(3)))rarr underset("Disodium ethynide")(NaC-=CNa)` (Eq. (i)) (ii)On reaction of ethyne with ammonium silver nitrite it gives precipitates of silver ethynide. while ethane does not give such reaction. `HC-=CH+ 2Ag(NH_(3))_(2)NHO_(3) rarr underset("Disilver ethynide")(AgC-=C Ag_(s)) + 2NH_(4)NO_(3) + 2NH_(3)` (Eq. (ii)) (iii) `CH_(3)-CH_(3)+2Ag(NH_(3))NO_(3) rarr "No reaction"` OR Alkane `+ 2Ag(NH_(3))NO_(3) rarr "No reaction"` (b) Ethene is alkene in nature and H-atom present in them are not acidic in nature. So ethene, means all alkenes does not react with Na, `NaNH_(2)` and `Ag(NH_(3))_(2)NO_(3)`. But ethyne can react according to reaction (i) and (ii). (c) When dimethyl ethyne `(CH_(3)C-=C CH_(3))` and ethyne `(CH-=CH)` both REACTED with `Na(NH_(3)), NaNH_(2)` and ammonium silvernitrate, then ethyne shows the reaction while dimethyl ethyne does not shows the reaction. `underset("(Dimethyl ethyne)")underset("But-2-yne")(H_(3)C C-=C-CH_(3))underset(underset(or Ag(NH_(3))_(2)NO_(3))(or NaNH_(2)))overset(Na(NH_(3)))rarr"No reaction"` The hydrogen of ethye `(HC-=CH)` acidic in nature and so it react with Na and `NaNH_(2)`, according to answer of (a) and reaction (i) and (ii). (d)Distinguishing reaction of ethane and ethene :Ethane is alkane and ethene is alkene. Ethene is unsaturated in nature and so it gives following two-reaction of unsaturation. (i) Bayer test : Alkene (Ethene) remove color of `KMnO_(4)`. `underset("Ethene")(CH_(2))=CH_(2)+H_(2)O + O underset(KMnO_(4))overset("Cold, Dilute")rarr underset("Glycol")(CH_(2)OH-CH_(2)OH)` (ii) Ethene react with `Br_(2)` in `C Cl_(4)` and decolorize RED color of `Br_(2)`. `underset("Ethene")(CH_(2))=CH_(2)+underset("Red liquid")(Br_(2)) overset(C Cl_(4))rarr CH_(2)Br CH_(2)Br` Ethane is saturated in nature, it does not gives above reaction and so can be distinguished. |
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