State the role of alkaline potassium permanganate in this reaction

I saw this experiment many times but still can't explain why. The experiment is first prepare two solutions. The first one is to dissolve potassium permanganate crystal into distilled water to form potassium permanganate (VII) while the second one is to put sodium hydroxide and glucose (sugar) into distilled water. Secondly, I pour the second solution into the first one. My observation is the purple solution will slowly turn into blue, then slowly turn into green and lastly turn into yellowish orange. So, what really happen in the solution? This is a redox reaction in which the permanganate ion is reduced and the glucose is oxidised. Potassium permanganate is usually used in acid solution and under these conditions is a very powerful oxidising agent in which the manganese is reduced from the oxidation state of +7 to +2 with a colour change from purple to colourless (actually extremely pale pink). In alkaline solution manganese is only reduced from +7 to +6 changing colour from purple to green. (The blue colour you mention is due to a mixture of purple and green). The green manganate(VI) ion is unstable and slowly disproportionates to manganate(VII) (purple) and manganese(IV) oxide which is brown. The manganate(VII) goes on to react as before until the brown manganese(IV) oxide is all that remains. In alkaline solution this tends to form a colloidal suspension which (if fairly dilute) can appear orange. The oxidation of the glucose forms no coloured productss.

Preparation of Baeyer's Reagent There are some conditions to be careful when baeyer's reagent is prepared for alkene oxidation to diol. • KMnO 4 solution should be dilute. • Solution should be cold. Steps of preparation of 1% alkaline potassium permanganate solution • Dissolve 1 gram of solid KMnO 4 in 100 ml of distilled water to produce a 1% potassium permanganate solution • Add 10 grams of anhydrous sodium carbonate (Na CO 3) to the KMnO 4 solution and the stoppered bottle is shaken until fully dissolved and mixed. Sodium carbonate is weak alkaline compound. Therefore, solution will become slightly alkaline. • Keep the solution in a dark, cool cupboard when not in use to maintain it fresh. Oxidation of alkene and alkyne compounds by baeyer's reagent Alkene compounds can be oxidized to diol and alkyne compounds can be oxidized to vicinal diketones or dialdehyde (Vic-diketones or 1,2-diketones) or, under more vigorous conditions, carboxylic acids. Alkene and baeyer's reagent Two -OH groups are attached to the carbon atoms in alkene group as in the figure. Alkynes and baeyer's reagent Alkyne compounds can be oxidized to vicinal diketones or dialdehyde (Vic-diketones or 1,2-diketones) or, under more vigorous conditions, carboxylic acids by baeyer's reagent. More vigorous conditions can be made by increasing the alkalinity of the solution by adding potassium hydroxide (KOH). Questions Related Tutorials

This question was asked in my book: To my knowledge, side chain oxidation with KMnO4 occurs only when a benzylic hydrogen is present. However this doesn't contain a benzylic hydrogen and hence according to me it wouldn't give any reaction. However the answer in my book states that Phenylacetylene will be oxidized to benzoic acid. How does this take place then? According to this source Alkynes, similar to alkenes, can be oxidized gently or strongly depending on the reaction environment. Since alkynes are less stable than alkenes, the reactions conditions can be gentler. For examples, alkynes form vicinal dicarbonyls in neutral permanganate solution. For the alkene reaction to vicinal dialcohols, the permanganate reaction requires a lightly basic environment for the reaction to occur. During strong oxidation with ozone or basic potassium permanganate, the alkyne is cleaved into two products. Brief explanation Hot alkaline solution of potassium permanganate oxidises a terminal alkyne according to the following process: Triple bond between the first and second carbon atoms (they are marked in the figure above) converts into two carbonyl groups $\ce$: References • Oxidative cleavage of alkynes: • Oxidation of alkynes: One possibility: the alkyne gets its $\alpha$ hydrogen through formation of a hydrate. The triple bonded carbon is electrophilic with its relatively high electronegativity, with the attack occurring on the remote carbon to form a stabilized (benzylic) carbanion: $...