# Can iproniazid be prepared by reacting isoniazid with 2-chloropropane?

I'm asking as I fear that the 2-chloropropane might react with the pyridine ring to form k-isopropylpridine-4-carbohydrazidine where k=2,3,5. Or that no reaction will take place.

Figure 1: Isoniazid

Figure 2: Iproniazid

Figure 3: 2-chloropropane

• Could you elaborate on why you think that the reaction with the pyridine ring will be preferential? – Michiel Apr 15 '13 at 18:23
• Because the only difference between a pyridine and a benzene is a C instead of N and aromatic compounds react with alkyl halides like 2-chloropropane to create products in of which the benzene and alkyl halides have reacted to give the corresponding alkylbenzene. – BH2017 Apr 15 '13 at 19:09
• @BrentonHorne - Pyridine is deactivated relative to benzene, and isoniazid is likely more so. When I have time, this will become an answer, unless someone else beats me too it. – Ben Norris Apr 15 '13 at 20:53
• Wouldn't the chloropropane react with both the nitrogens in the side chain of isoniazid? – kaliaden Apr 17 '13 at 3:41
• @kaliaden The amide NH is much less active, so no, it won't. However, it will readily react with $NH2$ group twice unless there are sterical issues, what I doubt. – permeakra Jun 16 '13 at 18:22

The $\ce{NH_2}$ group in given compounds is very similar to regular amine fragment and is more active than acylated $\ce{NH}$. However, alkylated amines are more active than non-substituted ones. So, I'd expect a very significant amount of double-alkylated product.
To avoid purification issues I recommend to consider synthesis of acetone hydrazone followed by reduction with $\ce{Na[BH_3(CN)]}$ or similiar low-active hydrogen donor (hi-active ones will reduce amide fragment)