Question

Which one of the following turns acidified potassium dichromate from orange to green?
A. $C{H_3} - C{H_2} - OH$
B. ${C_{6{\text{ }}}}{H_{14}}$
C. $C{H_{3{\text{ }}}} - COOH$
D. $C{H_{3{\text{ }}}} - {\text{ }}C{H_{2{\text{ }}}} - {\text{ }}C{H_{2{\text{ }}}} - {\text{ }}Br$

Answer 1

When an organic compound reacts with an acidified potassium dichromate solution, the potassium dichromate gets reduced to its sulfate by the reaction. This is a green colored solution formed from an orange dichromate solution.

Complete step by step answer:
The chemical reaction between acidified potassium dichromate and primary alcohol is characterized by color change, from the orange of dichromate to green of chromium sulfate. Primary alcohol reduces acidified potassium dichromate from $C{r_2}O_7^{2 - }$ (orange) to $C{r^{3 + }}$ (green) ions without leaving a residue and that primary alcohol get oxidized into a carboxylic acid, as follows,
${K_2}C{r_2}{O_7} + {H_2}S{O_4} + C{H_3} - C{H_2} - OH \to {K_2}S{O_4} + C{r_2}{(S{O_4})_3} + {H_2}O + C{H_3} - COOH$
SO, the correct answer is A.
Here, the oxidizing agent is potassium dichromate while the reducing agent is a hydrocarbon. This results in a redox reaction. The orange-colored dichromate solution will turn green due to the formation of chromium sulfate, $C{r_2}{(S{O_4})_3}$. Sulphuric acid is used as a dehydrating agent here.

Note:
The chemical reaction between acidified potassium dichromate and sulfur dioxide gas is characterized by color change, from the orange of dichromate to green of chromium sulfate. Sulfur dioxide gas reduces acidified potassium dichromate from $C{r_2}O_7^{2 - }$ (orange) to $C{r^{3 + }}$ (green) ions without leaving a residue and itself oxidized from $SO_3^{2 - }$ ions in sulphuric acid to $SO_4^{2 - }$ ions in the same acid.
${K_2}C{r_2}{O_7} + {H_2}S{O_4} + 3S{O_2} \to {K_2}S{O_4} + C{r_2}{(S{O_4})_3} + {H_2}O$