Question 1

What substrates give the haloform reaction?

Accepted Answer

The characteristic substrates are methyl ketones of the form CH3COR and secondary alcohols containing a CH3CH(OH)R group. Oxidation of the qualifying secondary alcohol gives the corresponding methyl ketone intermediate.

Question 2

Why are all three iodine atoms introduced at the same alpha-carbon?

Accepted Answer

In acetone, CH3COCH3, both methyl groups are equivalent. The lesson highlights one selected methyl carbon, which contains exactly three alpha-hydrogens. Successive deprotonation and iodination replace those three H atoms at that same site.

Question 3

What happens during hydroxide attack on the triiodomethyl ketone?

Accepted Answer

OH- attacks the carbonyl carbon, not the C-C bond directly. The carbonyl pi electrons move to oxygen, producing a tetrahedral intermediate containing a neutral OH group and an O- group. Collapse reforms the carbonyl and breaks the bond to the CI3 carbon, after proton transfer produces CHI3 and CH3COO-.

Question 4

Why is iodoform yellow while other haloforms are not treated the same way?

Accepted Answer

Iodoform, CHI3, is an insoluble yellow crystalline solid and gives a characteristic result in the iodoform test. The corresponding haloforms are CHX3 where X is chlorine, bromine, or iodine, but the reaction should not be represented as if every haloform is yellow.

Question 5

What is the balanced net reaction for acetone with iodine and sodium hydroxide?

Accepted Answer

The balanced net reaction is CH3COCH3 + 3I2 + 4NaOH producing CHI3 + CH3COONa + 3NaI + 3H2O. The organic products are iodoform and sodium acetate, while sodium iodide and water are the remaining products.

Haloform Reaction: Three Alpha Halogens and a Clean Break

About this lab

Practice