$0.25 mol$ of hexane-2,5-dione, $CH X_{3} - CO - CH X_{2} - CH X_{2} - CO - CH X_{3}$ , is treated with excess alkaline iodine until both $CH X_{3} CO$ groups have undergone the iodoform reaction (each $CH X_{3} CO$ group consumes $3$ mol of $I X_{2}$ and gives $CHI X_{3}$ ). Find the total number of moles of $I X_{2}$ consumed (two decimal places).

Numerical answer type, answer in mol, tolerance 0.05.

Answer: 1.5 mol

Each

CH X_{3} CO

group uses 3 mol

I X_{2}

and 4 mol

NaOH

(the haloform reaction). The molecule has two such groups, so 6 mol

I X_{2}

per mole. Total

= 0.25 \times 6 = 1.50 mol

; the organic products are 2 mol

CHI X_{3}

per mole and disodium butanedioate.