An alcohol ‘A' having molecular formula `C_3H_8O` gives iodoform test. ‘A’ on treatment with copper at 573k gives B. 'B' on treatment with sodium hydroxide 'C' obtained .Identify A,B andC.

An organic compound 'A' has the molecular formula C_3H_6O , it undergoes iodoform test. Wheir saturated with dil. HCl it gives 'B' of molecular formula C_9H_(14)O . A and B respectively are

Compounds 'A','B' and 'C' have the same molecular formula C_(7)H_(8)O . Compound 'A' and 'B' liberate hydrogen gas with sodium metal. When treated with sodium hydroxide, compound 'B' alone dissolves. Compound 'C' is inert towards both sodium metal and sodium hydroxide. Compounds 'A', 'B' and 'C' are respectively

An organic compound A on treatment with CHCl_3 and KOH gives two compounds B and C . Both B an C give the same product (D) when distilled with zinc dust. Oxidation of D gives E . having molecular formula C_7 H_6 O_2 . The sodium salt of E on heating with soda lime gives F which may also be obtained by distilling A with zinc dust. Identify A to F ,

Molecular formula: C_3H_8O , Functional group: OH, What is the difference between them?

A compóund A with molecular formula C_4 H_10O . on oxidation forms compound B. The compound B gives positive iodoform test. Compound B on reaction with CH_3 MgBr followed:by hydrolysis gives: C. Identify A, Band C and give the sequence of reactions.