3
$\begingroup$

Chlorine is an oxidizing agent, so could it replace oxygen? If so, what would be the products, and how much energy would such reactions release? Do any other carbohydrates burn in chlorine? I cannot find any information on this.

Clarification: How does glucose interact with chlorine chemically?

Improve this question
$\endgroup$
6
  • $\begingroup$ related chemistry.stackexchange.com/questions/64541/… chemistry.stackexchange.com/questions/19883/… $\endgroup$
    – Mithoron
    Commented Apr 30, 2018 at 22:52
  • $\begingroup$ It shouldn't work for same reason as why PVC is difficult to put on fire, IMO. $\endgroup$
    – Mithoron
    Commented Apr 30, 2018 at 22:56
  • 2
    $\begingroup$ "Burn" is typically used only to described reactions with oxygen. $\endgroup$
    – MaxW
    Commented Apr 30, 2018 at 23:28
  • $\begingroup$ @Mithoron I don't quite see the analogy. ;-) The reaction yields HCl, and will definitely work. Ok, sugar even doesn't burn very well in oxygen. $\endgroup$
    – Karl
    Commented May 1, 2018 at 13:43
  • $\begingroup$ @MaxW Not necessarily, you can make flame with various other oxidants, fluorine would work even in cases when oxygen isn't strong enough. $\endgroup$
    – Mithoron
    Commented May 1, 2018 at 14:14

1 Answer 1

Reset to default
2
$\begingroup$

I have not heard of glucose being burned in chlorine, but I do know that glucose reacts with chlorine in aqueous-buffer medium. See reference 1. The full paper is behind a paywall.

It says that oxidation of glucose by bromine water was studied previously and the reaction of glucose and aqueous chlorine also happens in a similar fashion although the reaction is slow and products yield is less. Nonetheless, major product of the reaction is D-gluconic acid in both acidic and alkaline medium but unbuffered chlorine may further react with D-gluconic acid to various products like 5-keto-D-gluconic acid and D-glucaric acid.

Reference

  1. J. Am. Chem. Soc., 1955, 77 (7), 1875–1880. DOI: 10.1021/ja01612a050.
Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.