Question

Which of the following will not give iodoform test?
A.

B.

C.

D.

Answer 1

Hint: Iodoform test is given by all alcohols, acetaldehyde and all methyl ketones. The test is declared positive or negative by looking at the colour obtained by the precipitate after the completion of the reaction.

Complete answer:
-There are many functional groups that can attach themselves with the carbon chains to form compounds which show different characteristics and behavior compared to normal compounds.
-The functional groups are differentiated with one another through different reagents. They form different products on oxidation, reaction with 1 mole of grignard’s reagent, reaction with Tollen’s reagent, reaction of Benedict’s solution and many more reactions.
-Pure carbonyl compounds are aldehydes and ketones. Aldehyde gives black silver mirror with Tollen’s reagent $\left( AgN{{O}_{3}}+N{{H}_{4}}OH \right)$ and red precipitate with Fehling’s solution while ketone does not give this test. So they can be distinguished through these tests.
-Iodoform test is a test which can be given by alcohols also along with the methyl ketones and aldehyde. All alcohols give this test and form yellow precipitate after the completion of the reaction which tests for the presence of the functional groups.
-In the question, we are given alcohol, methyl ketone and ethyl ketone. All alcohols give positive iodoform test giving yellow precipitate but the iodoform test for ketones is selective. Only methyl ketones can give this test.
-As the test is positive only for the methyl ketones and not any other ketones, it is not given by ethyl ketones also. The ketones with methyl groups attached to any side of the carbon ring are methyl ketones and others are not methyl ketones.
-The reaction for alcohols and methyl ketones can be shown as

Therefore the correct option is B.

Note:
The iodoform test reaction occurs such that the product formed contains 1 less carbon atom as that carbon atom is the one which links to the iodine group leading to the formation of $CH{{I}_{3}}$ which verifies the iodoform test by giving the yellow precipitate.