MediumChemistryHigh School (11–12)Organic Compounds with Functional Groups › Aldehydes, Ketones and Carboxylic AcidsNumerical

2.90 g of propanone (58.08 g/mol) is warmed with iodine and sodium hydroxide…

Contributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

$2.90\ \text{g}$ of propanone ( $58.08\ \text{g/mol}$ ) is warmed with iodine and sodium hydroxide. The equation is $\text{CH}_3\text{COCH}_3 + 3\,\text{I}_2 + 4\,\text{NaOH} \to \text{CHI}_3 + \text{CH}_3\text{COONa} + 3\,\text{NaI} + 3\,\text{H}_2\text{O}$ . The yield of iodoform ( $393.73\ \text{g/mol}$ ) is $60\ \%$ . Find the amount of propanone, the theoretical mass of iodoform and the mass obtained. Then find the amount of iodine ( $\text{I}_2$ ) used up.

Need a reminder? The key ideaShow

NaBH $_4$ , LiAlH $_4$ and H $_2$ with a metal reduce C=O to an alcohol. Zn-Hg with conc. HCl (Clemmensen) or NH $_2$ NH $_2$ then KOH (Wolff-Kishner) reduce it to CH $_2$ . Tollens' reagent gives 2 Ag for each aldehyde group, and Fehling's solution gives 1 Cu $_2$ O. The iodoform test is given by CH $_3$ CO- groups and CH $_3$ CH(OH)- groups.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

← All Problems

← 2.20 g of ethanal (44.05 g/mol) is used in two separate experiments…Six compounds are given: ethanal, propanal, benzaldehyde, pentan-2-one, pentan-3-one and butan-2-ol… →

More onOrganic Compounds with Functional Groups › Aldehydes, Ketones and Carboxylic Acids

Discussion (0)

Loading…