Does the given molecule show an iodoform test?
 $ {C_2}{H_5} - O - {C_2}{H_5} $

Hint :Iodoform test is a type of haloform reaction in which methyl ketones i.e., compounds having structures of type $ R - COC{H_3} $ undergo exhaustive halogenation in the presence of a dilute base i.e., sodium hydroxide in order to form iodoform $ (CH{I_3}) $ along with the removal of respective sodium salts.

Complete Step By Step Answer:
As per the question, the given molecule is diethyl ether and structurally it is represented as follows:

For a compound to show iodoform test, it is mandatory to have a carbonyl group with a methyl group present in the compound at the position adjacent to carbonyl group i.e., compound must be of type $ R - COC{H_3} $ to show positive haloform test. As there is no carbonyl group present in the given compound, the given compound will not give an iodoform test.
Therefore, $ {C_2}{H_5} - O - {C_2}{H_5} $ will not give an iodoform test.

Additional Information:
When iodine reacts with methyl ketones in the presence of aqueous sodium hydroxide, then they give yellow precipitate with a characteristic odour. It is a specific test for methyl ketones and for alcohols which can be oxidized to form a compound with a structure of type $ R - COC{H_3} $ and this test is specifically known as an iodoform test. Except fluoroform $ (CH{F_3}) $ , all other halogens can be prepared in a haloform reaction because it would require the presence of highly unstable hypofluorite ions.

Note :
It is important to note that all methyl ketones and secondary alcohols which are oxidizable to methyl ketones can show positive iodoform tests. In the case of alcohols and aldehydes, only ethanal and ethanol molecules have a tendency to undergo haloform reaction and show positive iodoform tests.