An organic compound 'A' `(C_2H_6O)` reacts with sodium to form a compound 'B' with the evolution of hydrogen gas and gives a yellow compound 'C' when treated with iodine and NaOH. When heated with conc. `H_2SO_4` at `140^@C` it gives a compound 'D' `C_4H_(10)` which on treatment with HI at `100^@C` gives 'E'. 'B' when heated with 'E' also gives back 'D'. Identify 'A' to 'E' and write equations for the reaction involved.

Explanation

(i) The compound A is an alcohol `(CH_(3)CH_(2)OH)` as it reacts with Na to evolve hydrogen and forms compound B, `(CH_(3)CH_(2)ON a)` . It also reacts with `I_(2)` and NaOH to from `C(CHI_(3))` which is yellow.
`CH_(3)underset([B])(CH_(2))ONaoverset(Na)underset(-1//2H_(2))larrCH_(3)underset([B])(CH_(2))OHoverset(I_(2)//NaOH)rarrunderset([C])(CHI_(3))`
(ii) Upon heating A with concentrated `H_(2)SO_(4)` at 413 K , diethyl ether [D] is formed which reacts wiht HI to from E `(C_(2)H_(5)I)`
`2CH_(3)underset([A])(CH_(2))OHoverset("Conc".H_(2)SO_(4)//413 K)rarrCH_(3)CH_(2)-underset([D])(O)-CH_(2)CH_(3)overset(2HI)rarr2CH_(3)underset([E])(CH_(2)I)`
(iii) According to the data, `C_(2)H_(5)ONa` [B] reacts with `CH_(3)CH_(2)I` [E] to from `CH_(3)CH_(2)OCH_(2)CH_(3)[D]`
`CH_(3)CH_(2)underset
([B])(O)Na+ICH_(2)underset([E])(CH_(2))CH_(3)rarrCH_(3)CH_(2)underset([D])(-)O-CH_(2)CH_(3)`