The condition for iodoform test is presence of CH3 group
and for fehling test is being aldehyde
Ethanal (CH3CHO) satisifes both the conditions.
option A is correct

The iodoform test and Fehling's test are both chemical tests used to identify the presence of certain functional groups in organic compounds. Let's analyze each option and determine which one answers both tests.

a) Ethanal:
- Ethanal, also known as acetaldehyde, has the chemical formula CH3CHO. It contains a carbonyl group (C=O) which is the functional group responsible for undergoing the iodoform test and Fehling's test.
- In the iodoform test, ethanal reacts with iodine in the presence of a base (such as sodium hydroxide) to form a yellow precipitate of iodoform (CHI3). This reaction confirms the presence of a methyl ketone functional group (R-CO-CH3), which ethanal possesses.
- In Fehling's test, ethanal can undergo oxidation by Cu2+ ions present in Fehling's reagent (a solution containing copper sulfate and sodium hydroxide). This results in the formation of a red precipitate of cuprous oxide (Cu2O), indicating the presence of an aldehyde functional group (R-CHO).

b) Propanone:
- Propanone, also known as acetone, has the chemical formula (CH3)2CO. It also contains a carbonyl group but differs from ethanal in the number of carbon atoms.
- Propanone can undergo the iodoform test since it contains a methyl ketone functional group. However, it does not react with Fehling's reagent because it lacks an aldehyde functional group.

c) Ethanol:
- Ethanol, with the chemical formula C2H5OH, does not contain a carbonyl group. Therefore, it does not undergo either the iodoform test or Fehling's test.

d) Methanol:
- Methanol, with the chemical formula CH3OH, also lacks a carbonyl group. Similar to ethanol, it does not react with either the iodoform test or Fehling's test.

From the analysis above, it is clear that the organic compound which answers both the iodoform test and Fehling's test is ethanal (option A).