MHTCET 20 Apr 2026 Shift 2 Paper

Concept:Bromine water is a mild oxidising agent.It selectively oxidises the aldehyde group of glucose without affecting other functional groups.Explanation:Glucose contains an aldehyde group ($-\mathrm{CHO}$) and a primary alcohol group ($-\mathrm{CH}_2\mathrm{OH}$).When glucose reacts with bromine water, only the $-\mathrm{CHO}$ group is oxidised to a carboxylic acid group ($-\mathrm{COOH}$).The primary alcohol group remains unchanged because bromine water is too mild to oxidise it.So glucose is converted into gluconic acid.The reaction can be written as:$$\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6 \xrightarrow{\mathrm{Br}_2/\mathrm{H}_2\mathrm{O}} \mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_7$$Here, the product $\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_7$ is gluconic acid.Saccharic acid is formed only when glucose is oxidised with a stronger oxidising agent like nitric acid.Thus, the correct option is gluconic acid.Answer:Option C: Gluconic acid.