1. Identify what disproportionation means

A disproportionation reaction is one in which the same species undergoes both:

• oxidation, and
• reduction

So, one element in a single oxidation state changes into two different oxidation states.

2. Check each option

Option A

$2NaBr + Cl_2 \to 2NaCl + Br_2$

• In $NaBr$, $Br$ is $-1$
• In $Br_2$, $Br$ is $0$ → bromide is oxidized
• In $Cl_2$, $Cl$ is $0$
• In $NaCl$, $Cl$ is $-1$ → chlorine is reduced

Here, two different species are involved:

• $Br^-$ is oxidized
• $Cl_2$ is reduced

So this is a redox displacement reaction, not disproportionation.

Option B

Given reaction is intended as thermal decomposition of permanganate: $2KMnO_4 \to K_2MnO_4 + MnO_2 + O_2$

Now check oxidation states of Mn:

• In $KMnO_4$, $Mn = +7$
• In $K_2MnO_4$, $Mn = +6$ → reduction
• In $MnO_2$, $Mn = +4$ → reduction

Oxygen:

• In reactant oxide form, $O = -2$
• In $O_2$, $O = 0$ → oxidation

Thus the same element does not undergo both oxidation and reduction. Manganese only gets reduced, while oxygen gets oxidized.

Hence this is not disproportionation.

Option C

$2CuBr \to CuBr_2 + Cu$

Find oxidation state of Cu in each species:

• In $CuBr$, since $Br = -1$, $Cu = +1$
• In $CuBr_2$, $Cu = +2$
• In $Cu$, oxidation state $= 0$

So copper in $+1$ state changes into:

• $+2$ (oxidation)
• $0$ (reduction)

The same species $Cu^+$ undergoes both oxidation and reduction.

Therefore, this is a disproportionation reaction.

Option D

$2MnO_4^- + 10I^- + 16H^+ \to 2Mn^{2+} + 5I_2 + 8H_2O$

• $Mn$ in $MnO_4^-$ is $+7$
• $Mn^{2+}$ is $+2$ → manganese is reduced
• $I^-$ is $-1$
• $I_2$ is $0$ → iodide is oxidized

Again, two different species are involved. This is a normal redox reaction, not disproportionation.

3. Conclusion

Only Option C is a disproportionation reaction.

$\boxed{C}$