An example of a disproportionation reaction is

First, we recall the definition of a disproportionation reaction. A single species undergoes simultaneous oxidation as well as reduction, producing two different products that contain the same element in two different oxidation states. In symbolic terms:

$$\text{A}^{n+} \;\longrightarrow\; \text{A}^{(n+m)+} \;+\; \text{A}^{(n-k)+}$$

Here one portion of A loses electrons (oxidation, oxidation state increases by $$m$$) while another portion gains electrons (reduction, oxidation state decreases by $$k$$).

We now analyse each option by writing the oxidation state of the key element in reactants and products.

Option A: $$2KMnO_4 \;\longrightarrow\; K_2MnO_4 + MnO_2 + O_2$$

In $$KMnO_4$$, manganese has oxidation state $$+7$$ (because $$K$$ is $$+1$$ and each $$O$$ is $$-2$$: $$+1 + x + 4(-2)=0 \implies x=+7$$). In $$K_2MnO_4$$, manganese is $$+6$$ ($$2(+1)+x+4(-2)=0 \implies x=+6$$). In $$MnO_2$$, manganese is $$+4$$ ($$x+2(-2)=0 \implies x=+4$$).

Both changes $$+7 \to +6$$ and $$+7 \to +4$$ are reductions; there is no simultaneous oxidation. Hence this is not disproportionation.

Option B: $$2MnO_4^- + 10I^- + 16H^+ \;\longrightarrow\; 2Mn^{2+} + 5I_2 + 8H_2O$$

$$Mn$$ goes $$+7 \to +2$$ (reduction) while $$I$$ goes $$-1 \to 0$$ (oxidation). Two different elements change oxidation state, so again it is not disproportionation.

Option C: $$2\,CuBr \;\longrightarrow\; CuBr_2 + Cu$$

Let us calculate the oxidation states step by step.

• In $$CuBr$$ the bromide ion is $$-1$$. Setting the sum to zero: $$x+(-1)=0 \implies x=+1$$. So copper is $$+1$$.

• In $$CuBr_2$$ we have $$x+2(-1)=0 \implies x=+2$$. Thus copper becomes $$+2$$ — this is oxidation because the oxidation state increases from $$+1$$ to $$+2$$ (loss of one electron).

• In metallic $$Cu$$ the oxidation state is $$0$$. The change $$+1 \to 0$$ is reduction because the oxidation state decreases (gain of one electron).

We see the same species $$Cu^{+}$$ has simultaneously undergone oxidation ($$+1 \to +2$$) and reduction ($$+1 \to 0$$). This exactly fits the definition of disproportionation.

Option D: $$2 NaBr + Cl_2 \;\longrightarrow\; 2NaCl + Br_2$$

Here $$Cl$$ goes $$0 \to -1$$ (reduction) and $$Br$$ goes $$-1 \to 0$$ (oxidation). Two different elements again; so it is not disproportionation.

Only Option C satisfies the criterion that one element in a single oxidation state splits into two different oxidation states, evidencing both oxidation and reduction of that element.

Hence, the correct answer is Option C.