What is an Iodoform test?

Hint :In order to answer this question, first we will explain the iodoform test for the presence of carbonyl compounds in brief. And then we will explain and discuss the whole mechanism of iodoform test with the reaction process.

Complete Step By Step Answer:
The iodoform test is a test for the presence of carbonyl compounds with the structure $ RCOC{H_3} $ and alcohols with the structure $ RCH\left( {OH} \right)C{H_3} $ .
A solution of $ {I_2} $ is added to a small amount of your unknown, followed by just enough $ NaOH $ to remove the colour.
The formation of a pale yellow precipitate of iodoform (with a characteristic "antiseptic" smell) is a positive result.
Now, we will explain the whole mechanism:-
(1.) $ O{H^ - } $ removes an acidic $ {\mathbf{\alpha }} - {\mathbf{hydrogen}} $ .
 $ RCOC{H_3} + O{H^ - } \rightleftharpoons RCOC{H^{2 - }} + {H_2}O $
(2) The enolate ion displaces an $ {I^ - } $ from $ {I_2} $
 $ RCOC{H_2}^ - + I - I \rightleftharpoons RCOC{H_2}I + {I^ - } $
The process repeats twice more.
3. $ RCOC{H_2}I + O{H^ - } \rightleftharpoons RCOCH{I^ - } + {H_2}O $
4. $ RCOCH{I^ - } + I - I \to RCOCH{I_2} + {I^ - } $
5. $ RCOCH{I_2} + O{H^ - } \rightleftharpoons RCOC{I_2}^ - + {H_2}O $
6. $ RCOC{I_2}^ - + I - I \to RCOC{I_3} + {I^ - } $
7. A hydroxide ion adds to the base of the carbonyl.
 $ RCOC{I_3} + O{H^ - } \rightleftharpoons RC - O\left( {OH} \right)C{I_3} $
8. The carbonyl group re-forms and eliminates a $ C{I_3}^ - $ ion.
 $ RC - O\left( {OH} \right)C{I_3} \rightleftharpoons RCOOH + C{I_3}^ - $
9. The acidic $ RCOOH $ and the strongly basic $ C{I_3}^ - $ ion are neutralized.
 $ RCOOH\left( {aq} \right) + C{I_3}^ - (aq) \to RCOO - \left( {aq} \right) + HC{I_3}\left( s \right) $
The overall equation is:-
 $ RCOC{H_3} + 3{I_2} + 4O{H^ - } \to RCO{O^ - } + 3{I^ - } + 3{H_2}O + HC{I_3} $

Note :
The iodoform test reveals the presence of an aldehyde or ketone in which a methyl group is one of the groups immediately connected to the carbonyl carbon. Such a ketone is referred to as methyl ketone. The unknown is allowed to react with a mixture of excess iodine and excess hydroxide in the iodoform assay.