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What exactly is the mechanism by which Benedict’s test works for reducing sugars? As I understand it the free hemiacetal groups on reducing sugars opens up and an aldehyde is formed, this is then oxidised by Cu2+ which then results in a red copper(I) oxide.

RCHO + 2 Cu2++ 5 OH- → RCOO- + Cu2O + 3 H2O.

The equation above isn’t really clear to me, what is exactly going on?

Additionally why are alkaline conditions required for the test? And if a red precipitate is formed how come there is such a wealth of colours observed (green, yellow, orange) for varying concentrations of reducing sugars?

Perhaps these are a couple of questions too many :D

Hashim Ishfaq
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  • In addition to the post linked above you can look at https://chemistry.stackexchange.com/questions/62352/why-is-the-reduction-by-sugars-more-efficient-in-basic-solutions-than-in-acidic – Buck Thorn Oct 20 '20 at 09:10
  • See also: https://chemistry.stackexchange.com/questions/40127/why-does-fructose-reduce-tollens-reagent-and-fehlings-solution – Buck Thorn Oct 20 '20 at 09:14
  • And: https://chemistry.stackexchange.com/questions/82677/why-are-different-colors-according-to-sugar-concentration-observed-in-benedict – Buck Thorn Oct 20 '20 at 09:31

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