The maximum number of moles of isopropyl alcohol that can be oxidized to acetone by one mole of `K_(2)Cr_(2)O_(7)` in the presence of `H_(2)SO_(4)` is

Options

A. four
B. one
C. two
D. three

Answer

D

Explanation

`CH_(3)-underset(OH)underset(|)overset(O)(CH)-CH_(3) rarr CH_(3)-underset(O)underset(||)overset(+2)(C)-CH_(3)`
`overset(+6)(C)underset((+12))(r_(2)O_(7)^(2-))rarr2underset((+6))overset(+3)(C)r^(3+)`
Since `1 ` mole of `K_(2)Cr_(2)O_(7)` can accept `6` moles of `e^(-)s` (as `O.N`. of `Cr`changes by `6` units from `+12` to `+6`), three moles of isopropyl alcohol can be oxidized to acetone as each mole loses `2e^(-)` (as `O.N. of C` changes by `2` units).