Synthesize 1,5-diazocane using only benzene as a source of carbon atoms, and ammonia as the only source of nitrogen atoms.

My attempt: Noticing that the target compound has split the benzene ring, I attempted to split the benzene ring as well. I thought of two options:

  1. H2/Pt at high temperature and pressue

  2. Creation of phenol and subsequent oxidation

As the first process would create cyclohexane and be difficult to manipulate, I went with the second option. To synthesize phenol from benzene, I used a nitration, followed by reduction to create aniline, then used nitrous acid to create the benzenediazonium salt, and decomposed the salt in the presence of water to create phenol. After the oxidation of phenol I got 1,4-benzoquinone.

Now, realizing that I would need to create 1,3-propandial in order to use $\ce{NH3/NaBH3CN}$ to create 1,5-diazocane, I attempted to make 1,3-propandial from 1,4-benzoquinone. Ozonolysis of 1,4-benzoquinone produces 2-oxo,1,3-propandial, which has an extra carbonyl group. Because of the many carbonyl groups involved in this compound, I was thinking of removing the carbonyl groups directly from benzoquinone, and doing an ozonelysis from there. However, I was not sure if any standard carbonyl reactions, such as the Wolff-Krishner or the Clemmensen reaction could be applied to remove the carbonyls, seeing as they are heavily resonance stabilized.

Therefore, I was wondering if these reactions do proceed on benzoquinone, and if not, then what other plausible ways of synthesizing 1,5-diazocane exist?

  • $\begingroup$ I think the idea with benzoquinone is good, you're overcomplicating it, though. $\endgroup$
    – Mithoron
    Aug 18, 2017 at 13:33
  • $\begingroup$ This is my approach, start with reaction of benzene and ozone in the presence of zinc and water to form glyoxal. React two equivalents of glyoxal with two equivalents of ammonia to form two equivalents of the monoimine. Next convert some glyoxal into formyl chloride, react two equivalents of formyl chloride with two equivalents of imine, make the product undergo Pinacol Coupling (Magnesium, high temperature), and then dehydrate and reduce. $\endgroup$
    – AS_1000
    Aug 19, 2017 at 12:43
  • 2
    $\begingroup$ Think double Beckmann rearrangement. $\endgroup$
    – ron
    Apr 3, 2018 at 1:55
  • $\begingroup$ @ron: In doing some research on my answer, I found an attempted double Beckmann on the bis-oxime of 1,4-cyclohexanedione (arduously prepared from benzene via 1,4-dimethoxybenzene?). Beckmann (29%) and Schmidt (HN3) on the diketone (40%). Rothe, M.; Timler, R. Chemische Berichte, (1962), 95, 783. $\endgroup$
    – user55119
    Apr 3, 2018 at 15:42
  • $\begingroup$ @user55119, Yes, that's the reference. The double Beckmann from 1,3-cyclohexanedione has also been reported to yield only the desired precursor bis-amide, 1,5‐diazocine‐2,4‐dione. The preparation of 1,3-cyclohexanedione and 1,4-cyclohexanedione seem reasonably straightforward. $\endgroup$
    – ron
    Apr 3, 2018 at 17:11

2 Answers 2


Here is a relatively straightforward synthesis of 1,5 diazocane that uses the Beckmann rearrangement as a key step. The Beckmann rearrangement is a useful synthetic tool for converting ketones to amides. If the carbonyl is in a ring, then conversion to the amide will enlarge the ring by one nitrogen atom.

Beckmann Rearrangement

The novel twist in the following synthesis is to perform a double Beckmann rearrangement in the penultimate step.

1,5-diazocane synthesis

  • heating benzene with concentrated sulfuric acid containing excess sulfur trioxide (oleum) will lead to the disulfonated product, benzene-1,3-disulfonic acid
  • heating the disulfonic acid with potassium carbonate produces resorcinol (see here)
  • treating resorcinol with hydrogen gas at low pressure over a rhodium catalyst yields 1,3-cyclohexanedione (see here)
  • hydroxylamine can be prepared by oxidizing ammonia with hydrogen peroxide to produce ammonium nitrite which can then be converted by a number of methods to hydroxylamine
  • the hydroxylamine is used in standard fashion to afford the bis-oxime (various oxime stereoisomers are likely produced, but this does not matter, all oxime stereoisomers will undergo the Beckmann rearrangement in the next step)
  • the bis-oxime is converted to the bis-amide through a double Beckmann rearrangement (see here for this specific conversion; note that the double Beckmann could afford isomers but only the desired isomer is produced; see the preceding link for the authors explanation as to why this is so)
  • finally, standard reduction of the bis-amide with $\ce{LiAlH4}$ and aqueous workup gives rise to the target molecule

The double Beckmann could also be run starting with 1,4-cyclohexanedione (readily prepared from benzene, see here). 1,4-cyclohexanedione has the advantage of symmetry - the first Beckmann rearrangement can produce only one isomer. Although the second Beckmann rearrangement could produce 2 isomers, only the desired 1,5-diazocane-2,4-dione was isolated (see here).


Birch reduction of benzene provides 1,4-cyclohexadiene (2), which yields malonic acid upon ozonolysis and subsequent oxidation with peroxide and formic acid to form malonic acid 3 [2 --> 3 was found via Chem. Abstr.]. The half acid chloride of malonic ester 5 is well-known. In addition, 1,3-dibromopropane (6) is also accessible from malonic acid (3), or for that matter, from diethyl malonate (4) or the acid chloride 5. Moreover, 1,3-diaminopropane (8) is available from malonic acid (3) in addition to treating dibromide 6 with an excess of NH4OH.

At this juncture the temptation would be to alkylate diamine 8 with dibromide 6 to synthesize 1,5-diazacane (13). But this type of reaction is fraught with problems such as polymerization and overalkylation. Boc protected hydrazine 7 has been shown1 to provide pyrazolidine (9). Although hydrazine can be prepared from ammonia via chloramine and ammonia via the Olin Raushig process, it is unlikely to be conducive to the academic laboratory. However, diamine 8 has been transformed into pyrazolidine (9) with bleach via a more facile intramolecular process.2 To assure the use of ammonia, the dibromide 6 can be treated with excess ammonia to provide diamine 8.

Pyrazolidine (9) can be acylated with malonyl acid halide 5 followed by reduction to alcohol 11. Ring closure to tetrahydro-1H,5H-pyrazolo[1,2-a]pyrazole (12) involves conversion of alcohol 11 to the bromide, neutralization of the intermediate hydrobromide salt with base and heating to effect facile intramolecular alkylation. A second option is to alkylate hydrazine 9 with dibromide 6, a process that is not as susceptible to polymerization as the previously discussed process. Finally, reduction of the N-N bond in hydrazine 12 is accomplished by hydrogenation3 to afford the target molecule 1,5-diazacane (13).

ADDENDUM: If ester 10 were to cyclize, LiAlH4 reduction would form hydrazine 12.

enter image description here

1) Watanabe, H.; Kuwata, S.; Koyama, S. Bull. Chem. Soc. Japan, 1963, 143.
2) El Hajj, A.; Bougrine, A. J.; Le, D. M.; Pasquet, V.; Delalu, H. Reaction Kinetics, Mechanisms and Catalysis, 2016, 117, 429.
3) Stetter, H.; Spangenberger, H. Chemische Berichte, 1958, 91, 1982.

  • $\begingroup$ Tempting, but no. Only benzene and ammonia are supposed to provide the carbon and nitrogen atoms you use. $\endgroup$ Apr 3, 2018 at 1:26
  • 1
    $\begingroup$ Good catch, OL. Forgot to add that dibromide 6 gives diamine 8 with ammonia. See new last sentence in second paragraph. Benzene provides all of the carbons. Overall, a sensible approach. Clearly, one wouldn't use benzene to make the product. An academic exercise. $\endgroup$
    – user55119
    Apr 3, 2018 at 2:03
  • 1
    $\begingroup$ @Oscar Lanza; See above reply and 3 --> 8, which uses ammonium hydroxide. $\endgroup$
    – user55119
    Apr 3, 2018 at 2:23
  • $\begingroup$ Other than "ammonium hydroxide" being a misnomer ($\ce{NH3(aq)}$ would be more accurate), now it's all good. $\endgroup$ Apr 3, 2018 at 9:00

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