Explain whether oxidation, reductio, or neither occurs in the following reactions :
(a) `H_2 C= CH_2 rarr H_3 C - CH_3`
(b) `HC -= CH rarr H_3 C- CH = O`
( c) `CH_4 rarr CH_3 OH`
(d) `H_2 CB r_2 rarr H_2 C = O`
II. Give the structural formula for the simplest hydrocarbon in which `C` has a zero `O.N`.

Explanation

(a) Reduction, because `O.N`. Of `C` changes from `-2` to `-3`
`H_2 C = CH_2 (C_2 H_4) : 2x + 4 = 0, x = - 2`
`H_3 C - CH_3(C_2 H_6) : 2 x + 6 = 0, x = - 3`
(b) Neither, because `O.N` of `C` in both reactant and products is `-1`, i.e., no change in `O.N`.
`CH -= CH , (C_2 H_2) : 2 x + 2 = 0, x = -1`
`CH_3 OH(CH_4 O) = 2x + 4 - 2 = 0, x= -1`
( c) Oxidation, because `O.N` of `C` changes from `-4` to `-2`
`CH_4 = x + 4 = 0. x =-4`
`CH_3 OH(CH_4 O) = x + 4 - 2 = 0, x = -2`
(d) Neither, because `O.N` of `C` in both reactant and product is zero, i.e., no change in `O.N`.
`H_2 CB r_2 -= x + 2 - 2 = 0, x = 0`
`CH_2 O = x + 2 - 2 = 0, x = 0`
II. The simplest hydrocarbon is `(HC -= CH)`. The number of `H` atoms and `C` atoms must be same so that their oxidation number with opposite signs would cancel.