Hi,

To test your sample you need to prepare fresh Fehling's reagent each time. Generally, the positive test with Fehling's reagent is red precipitate.

IN SHORT THE ANSWER IS:

"However, variations in the color change from Orange to brick red have been noted. The reason for this could be different amounts of reducing sugar. 1% will give red color. So you may have different variation in the colors depending on amount of sugar."

DETAIL:

You need to understand little chemistry behind this:

Fehling's solution is actually composed of Fehling's A and Fehling's B solution. 

Fehling's A is a  aqueous solution (blue) of copper(II) sulfate pentahydrate and Fehling's B is a colorless solution of aqueous potassium sodium tartrate (Rochelle salt) and a strong alkali like NaOH.

When the equal volumes of A and B are mixed, the bistartratocuprate(II) complex (i.e. Cu2+) is formed which is Fehling's reagent (Deep blue in color).

This reagent is an oxidizing agent which when mixed with aldehyde containing compounds such as reducing sugars, it oxidizes aldehyde to carboxylate ions. During this process, the Cu2+ (Divalent) of bistartratocuprate(II) complex gets reduce to Copper I i.e.Cu+ (Monovalent). Due to this, red copper (I) oxide is precipitate out of the reaction. 

non reducing sugar generally do not react with this.

Hope this help....

Thanks for all details 

what is green color with fehling reagent ? 

its a clear green color without precipitate.

Hi,

initially the supernatant is blue in color, then it turns to form red colored ppt (copper oxide) as end point. This determination of end point is rather difficult.

So, to answer the green color suspension, the following reaction may be of some hint, 

Glucose + Cupric Oxide (2CuO) ----→ C6H11O5.COOH (Gluconic acid [Yellow color]) + Cu2O (Red ppt.)

From this reaction I guess, that the clean yellow color of Gluconic acid and the bluish color of the remaining suspension (Because end point is hard to determine) may be giving the green color appearance.

However, this is theoretical answer......i am also eagerly waiting for correct answer....