Silver Mirror Test Essay

1038 Words3 Pages

Molisch test is one of the useful qualitative test for presence of carbohydrates in solution. The three glucose solutions all have a violet colored ring formed at the junction between the two layers. This showed that carbohydrates are present in these sugar solution. This test involved the addition of concentrated sulphuric acid which causes dehydration of all carbohydrates to give ‘furfural’ compound, where pentoses are dehydrated to furfural, and hexoses are dehydrated to 5-hydroxymethylfurfural (Molish's reagent, 2009). These compounds will later react with – naphtol which is Molisch reagent to give a purple colored complex. The test has to be carried out slowly as the violet colour formed at the surface of contact of concentrated sulfuric …show more content…

Tollens' reagent is an alkaline solution of ammoniacal silver nitrate. There is brown precipitate which is the silver hydroxide present and it will dissolves in aqueous ammonia, forming the diamminesilver(I) ion, [Ag(NH3)2]+. Aldehydes is then reduce the diamminesilver(I) ion to metallic silver and is oxidised to a salt of the corresponding carboxylic acid. Similar to Fehling’s test and Benedict’s test, Tollen’s reagent test will not react with ketones (Baker, 2007). The wall of glucose and fructose solution are deposited by layer of silver mirror while maltose solution has grey precipitate of silver within the solution. Even grey precipitate is shown in maltose solution instead of silver mirror layer, the result is still positive. This is because glucose, fructose and maltose are reducing sugar that contain of free aldehyde sugar which can reduce the diamminesilver(I) ion to metallic silver(Clark, 2004). The electron-half-equation for the reduction of of the diamminesilver(I) ions to silver …show more content…

Starch consist of mainly two parts, which are amylose and amylopectin. Amylose in starch is the main cause for the formation of a deep blue color in the presence of iodine as the iodine molecule will slips inside of the amylose coil. Iodide molecules is dissolving in water with potassium iodide because iodide is not so soluble in water. This makes a linear triiodide ion complex with is soluble that slips into the coil of the starch causing an intense blue-black colour. In this test, blue-black colour is shown and this indicates the presence of starch. Starch amylopectin does not give the blue-black colour, nor does cellulose, nor do disaccharides such as sucrose in sugar (Orhardt,

More about Silver Mirror Test Essay

Open Document