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Compound (A) C(4)H(10)O, is found to be...

Compound (A) `C_(4)H_(10)O`, is found to be soluble in sulphuric acid . (A) does not react with sodium or potassium permanganate. When (A) is heated with excess of HI, it is converted into a single alkyl halide. What is (A) ?

Text Solution

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(i) Since compound `A(C_(4)H_(10)O)` does not reat with Na metal or `KMnO_(4)`, it cannot be an alcohol.
(ii) Since compound A dissolve in conc. `H_(2)SO_(4)`, it may be an ether.
(iii) Since ether A on heating with excess of HI gives a single alkyl halide, therefore, ether (A) must be symmetrical. now the only symmetrical ether having M.F. `C_(4)H_(10)O` is diethyl ether `(CH_(3)CH_(2)OCH_(2)CH_(3))`.
(iv) All the reactions cann now be explained as follows.
`underset("Diethyl etehr (A) M.F. "C_(4)H_(10)O)((CH_(3)CH_(2)-O-CH_(2)CH_(3)) overset("conc. "H_(2)SO_(4))hArr underset("Soluble ozonium salt")([CH_(3)CH_(2)-underset(H)underset(|)overset(+)(O)-CH_(2)CH_(3)])HSO_(4)^(-)`
`underset("Diethyl ether (A)")(CH_(3)CH_(2)-O-CH_(2)CH_(3)) underset(Delta)overset("HI (excess)")to underset("Ethyl iodide")(2CH_(3)CH_(2)-I)+H_(2)O`
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Knowledge Check

  • When conc. H_(2)SO_(4) is heated with P_(2)O_(5), the acid is converted to

    A
    sulphur trioxide
    B
    sulphur dioxide
    C
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    C
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    D
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  • An organic compound of molecular formula C_4H_10O does not react with sodium. With excess of HI, it gives only one type of alkyl halide. The compound is

    A
    Ethoxy ethane
    B
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    C
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