Identify the reaction situation

The question refers to nitration of aniline (or a related ring containing the $-NH_2$ group), where along with the expected products, a meta-nitro product is also formed.

Usual directing effect of $-NH_2$

The amino group $(-NH_2)$ is normally a strong activating, ortho/para-directing group because its lone pair donates electron density into the benzene ring by resonance.

So, if free aniline undergoes electrophilic substitution, we expect mainly:

• ortho product
• para product

Therefore, option A is false because $-NH_2$ is not meta-directing.

What happens during nitration conditions?

Nitration is carried out in a strongly acidic medium, usually with a mixture of concentrated $HNO_3$ and concentrated $H_2SO_4$.

In this acidic medium, the amino group gets protonated:

$C_6H_5NH_2 + H^+ \rightarrow C_6H_5NH_3^+$

This species is called the anilinium ion.

Directing effect of anilinium ion

The group $-NH_3^+$ is electron-withdrawing and meta-directing.

Hence, once aniline is converted into anilinium ion under nitration conditions, some substitution occurs at the meta position, which explains the formation of the meta-nitro product.

Check other options

• B: "$-NO_2$ substitution always takes place at meta-position"
  False. The entering electrophile is the nitronium ion $NO_2^+$, and the position of substitution depends on the group already present on the ring.

• D: low temperature
  False. Low temperature is not the reason for meta substitution here.

Conclusion

The meta product is formed because the amino group is protonated in acidic medium to give anilinium ion, which is meta-directing.

Therefore, the correct option is:

$\boxed{C}$