The M.F. = C_(3)H_(8)O represents two compounds A and B. Both of them react with sodium to liberate H_(2) and on dehydration results same alkene C". In presence of H_(2)O_(2), the alkene 'C" ads of HBr giving 'D', which on treatment with aqueous KOH gives 'B'. Further A' responds to iodoform test. Identify 'A'.

The M.F. = C_(3)H_(8)O represents two compounds A and B. Both of them

1.	The M.F. = C_(3)H_(8)O represents two compounds A and B. Both of them react with sodium to liberate H_(2) and on dehydration results same alkene C". In presence of H_(2)O_(2), the alkene 'C" ads of HBr giving 'D', which on treatment with aqueous KOH gives 'B'. Further A' responds to iodoform test. Identify 'A'.
Answer» `CH_(3)CH_(2)CH_(2)OH` `CH_(3)CHO` `CH_(3)-overset(OH)overset(\|)CH-CH_(3)` `CH_(3)-overset(Br)overset(\|)CH-CH_(3)` Solution :`MF=C_(3)H_(2)O` represents ETHER and alcohol. As the compound A and B react with sodium to liberate `H_(2)` gas, so they must be alcohol. Dehydration of A and B results same alkene, HENCE the alkene C is, `C_(3)H_(6)`(propene). `B= UNDERSET("Propan-1-ol")(CH_(3)-CH_(2)-CH_(2)-OH), A= underset(Propan-2-ol)(CH_(3)-overset(OH)overset(\|)CH-CH_(3))` `underset("(C)")(CH_(3)-CH=CH_(2)) underset(H_(2)O)overset(HBr)to underset("(D)")(CH_(3)-CH_(2)-CH_(2)-Br)` `CH_(3)-CH_(2)-CH_(2)-Br+KOH_("(aq)")to underset("(B)")(CH_(3)- CH_(2)-CH_(2)-OH+ KBr)` `underset("(A)")(CH_(3)- underset(OH)underset(\|)CH-CH_(3)) overset(I_(2)//NaOH)to underset("iodoform")(CHI_(3))+CH_(3)-COONa+4HI`