# Selective reduction of nitro group to amine, in benzene ring containing nitrile?

I would like to know if there is a possible route of synthesis that may reduce the shown compound A to the following product B.

I would normally reduce a nitro to an amine using $$\ce{Pd/C}$$ and $$\ce{H2}$$, but since the nitrile is sensitive to this method, I need an alternative.

• Clarification: If there are easier methods to arrive at product "B" through alternative reactions, besides reduction from "A", then I would gladly hear them. – Philip Dec 3 '19 at 16:32
• I think it would also works using Fe/HCl – Arnau Dec 3 '19 at 17:30
• So would Sn/HCl or Zn/HCl – Waylander Dec 3 '19 at 17:32
• Sure! There are some "active metals" that would work – Arnau Dec 3 '19 at 17:35
• I just did a Chem Abstracts search of this reaction with 3031 hits! Many of them involve hydrogenations. Hope you have access. – user55119 Dec 3 '19 at 17:36

My answer is intended to support Waylander's suggestion elsewhere. I thought it'd be better to give OP some insight of this reduction mentioned (Ref.1). To my knowledge, after doing thorough literature search, the aromatic nitro group reduction by $$\ce{SnCl2.2H2O}$$ is the best method so far in the presence of other sensitive groups such as nitrile group on the aromatic nucleus. The abstract of the reference 1 stated that:

Aromatic nitro compounds are readily reduced by $$\ce{SnCl2.2H2O}$$ in alcohol or ethyl acetate or by anhydrous $$\ce{SnCl2}$$ in alcohol where other reducible or acid sensitive groups such as aldehyde, ketone, ester, cyano, halogen, and O-benzyl remain unaffected.

This point is depicted in following diagram as two examples given in Ref.1:

However, as shown in the separate box in the diagram (under the title, Stephen reduction of nitriles), if you use $$\ce{HCl}$$ (hydrogen chloride gas) with $$\ce{SnCl2}$$ as other suggestions, the result would be different (Ref.2). $$\ce{HCl}$$ is added to the nitrile group to give the imidoyl chloride salt first, which is then reduced by the tin(II) chloride. The initial product is the complex with $$\ce{SnCl4}$$ (which is often crystalline) that is then hydrolyzed to the corresponding aldehyde (see the insert in the diagram).

For the convenience of OP and other readers, I'd include a typical reduction procedure in Ref.1:

The reduction of p-nitrobenzoic acid: A mixture of $$\pu{1.67 g}$$ ($$\pu{0.01 mol}$$) of p-nitrobenzoic acid and $$\pu{11.275 g}$$ ($$\pu{0.05 mol}$$) of $$\ce{SnCl2.2H2O}$$ in $$\pu{20 mL}$$ of absolute ethanol is heated at $$\pu{70 ^\circ C}$$ under nitrogen. After $$\pu{30 min}$$ the starting material has disappeared and the solution is allowed to cool down and then poured into ice. The $$\mathrm{pH}$$ is made slightly basic ($$\mathrm{pH}\ 7-8$$) by addition of 5% aqueous sodium bicarbonate before being extracted with ethyl acetate. The organic phase is thoroughly washed with brine, treated with charcoal and dried over sodium sulfate. Evaporation of the solvent leaves $$\pu{1.5 g}$$ ($$94.5\%$$) of p-aminobenzoic acid, which gives one spot on TLC and melts over $$\pu{300 ^\circ C}$$.

References:

1. F. D. Bellamy, K. Ou, “Selective reduction of aromatic nitro compounds with stannous chloride in non acidic and non aqueous medium,” Tetrahedron Letters 1984, 25(8), 839-842 (https://doi.org/10.1016/S0040-4039(01)80041-1).
2. James A. Knight, Harry D. Zook, “Reduction of Aliphatic Nitriles by the Stephen Reaction,” J. Am. Chem. Soc. 1952, 74(18), 4560-4562 (https://doi.org/10.1021/ja01138a031).
• I prefer basifying with 50% NaOH. The initial basification produces a persistent emulsion of Sn(IV) salts. If you add enough NaOH, Sn(OH)4 forms which goes back into solution and the extraction becomes straightforward. – Waylander Dec 3 '19 at 20:55
• @Philip: You should accept Waylander's answer since it is the one prompt me to do some research. And also, it is the best method so far. – Mathew Mahindaratne Dec 3 '19 at 21:07
• @MathewMahindaratne: Thank you for your very expounding answer <3 – Philip Dec 4 '19 at 20:04
• @Philip: I'd appreciate if you accept Waylander's answer, because my answer initiated by his reference. :-) – Mathew Mahindaratne Dec 4 '19 at 20:12

$$\ce{SnCl2.2H2O}$$ in refluxing $$\ce{EtOAc}$$ as reported by Bellamy & Ou here will do this very cleanly. Work up into ice then add concentrated $$\ce{NaOH}$$ until you get phase separation. I have done this transformation.