Question

CHCl3 and H2O form two separate immiscible layers Br2 (orange) and I2 (violet) are soluble in organic layer due to a
greater extent. In test tube A (KI + CHCl3 + H2O) and test tube B (KI + KBr + CHCl3 + H2O) Cl2 gas is
passed then
organic layer in A appear violet
organic layer in B appear orange
organic layer in B appear black
organic layer in B first appears violet then orange​

Answer 1

Explanation:

CHCl3 and H2O form two separate immiscible layers Br2 (orange) and I2 (violet) are soluble in organic layer due to a

greater extent. In test tube A (KI + CHCl3 + H2O) and test tube B (KI + KBr + CHCl3 + H2O) Cl2 gas is

passed then

organic layer in A appear violet

organic layer in B appear orange

organic layer in B appear black

organic layer in B first appears violet then orange

Answer 2

$Br_{2}$ is orange in color.

$I_{2}$ is violet in color.

They both are soluble in organic layer.

In test tube A, the following reaction takes place.

$KI+CHCl_{3} +H_{2}O$

The chlorine($Cl$) would displace iodine($I$) from potassium iodidie($KI$).

The violet-colored iodine would dissolve in the organic layer.

Hence, the organic layer in test tube would appear violet in color.

In the test tube B, the following reaction takes place.

$KBr+CHCl_{3} +H_{2}O$

The chlorine($Cl$) would displace bromine($Br$)  from potassium bromide($KBr$).

The orange-colored bromide would dissolve in the organic layer.

Hence, the organic layer in the test tube would appear orange in color.