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In the above reaction, I do know that $\ce{SnCl2/HCl}$ reduces $\ce{-NO2}$ group to $\ce{-NH2}$ but which one of the two $\ce{-NO2}$ will be reduced first by $\ce{SnCl2/HCl}$? Why so?

And how does $\ce{NH4HS}$ react and again which one of the two $\ce{-NO2}$ will be reduced first by $\ce{NH4HS}$ and why so?

NOTE: 1eq represents 1 equivalent.

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Hyperconjugation accounts for the somewhat greater negativity of the 4-nitro-group compared with the 2-nitro-group in 2:4-dinitrotoluene; in consequence, selective reduction of the former by alkaline sulphides produces mainly 2-nitro-4-aminotoluene, although some 4-nitro-2-aminotoluene is also formed. In acid medium the 2-nitro-group becomes relatively negative in comparison with the 4-nitro-group, owing to the inductive (+I) effect having the chief role, with resultant preferential salt formation to give the very positive $\ce{-NO2H}$ group at 2-position and preferential reduction of it by acid stannous chloride.

Quoted from Coloration Technology 62 (1946) 114

As for how ammonium sulphide reacts see this answer: Selective nitro reduction of poly nitro compounds

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