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Question: Which of the following gives the iodoform test? A) \(C{H_3}C{H_2}OH\) B) \({C_2}{H_5}CHO\) C) ...

Which of the following gives the iodoform test?
A) CH3CH2OHC{H_3}C{H_2}OH
B) C2H5CHO{C_2}{H_5}CHO
C) (CH2OH)2{\left( {C{H_2}OH} \right)_2}
D) None of these

Explanation

Solution

Basically iodoform test is being performed to check the presence of carbonyl compound with the structure having R−CO−CH3R - CO - C{H_3} or some alcohols which have a structural composition R−CH(OH)−CH3R - CH\left( {OH} \right) - C{H_3}. One can identify the structures which give a positive test by analyzing the presence of these structures.

Complete step by step answer:

  1. The iodoform test can be elaborated as the reaction of iodine and base with methyl ketone which gives yellow precipitate.
  2. This test also comes positive for some secondary alcohols which have at least one methyl group at the alpha position of the functional group.
  3. Iodoform test is also used for the identification of aldehyde and ketone groups. An aldehyde group having compound gives the positive test for iodoform reaction then it can be only acetaldehyde as it has the group CH3−C=OC{H_3} - C = Oin its structure.
  4. A general example of iodoform reaction is as follows:
    R−CHOH−CH3→NaOHI2R−CO−O−Na++CHI3R - CHOH - C{H_3}\xrightarrow[{NaOH}]{{{I_2}}}R - CO - {O^ - }N{a^ + } + CH{I_3}
    Here, in the above reaction R=H,CH3,C2H5,.....etcR = H,C{H_3},{C_2}{H_5},.....etc
  5. Hence, if we put HH in place RR in the above general reaction then it will show ethanol as a chemical compound which will give iodoform test positive.
    Therefore, CH3CH2OHC{H_3}C{H_2}OH will give a positive iodoform test which represents option A as a correct option.

Additional information: Iodoform test for the aldehyde or ketone group will be as follows:
R−CO−CH3→NaOHI2R−CO−O−Na++CHI3R - CO - C{H_3}\xrightarrow[{NaOH}]{{{I_2}}}R - CO - {O^ - }N{a^ + } + CH{I_3}
Here, in the above reaction R=H,CH3,C2H5,.....etcR = H,C{H_3},{C_2}{H_5},.....etc
In the iodoform reaction, the CHI3CH{I_3} occurs due to three iodine atoms replace with the HH atoms of CH3−C=ORC{H_3} - C = OR the group breaking the C−CC - C bond because of the electron-withdrawing effect of the three iodine atoms. Therefore CHI3CH{I_3} gets removed forming a R−CO−O−R - CO - {O^ - } group that later gets stabilized by the sodium ion.

Note:
For iodoform reaction to occur there must be a −CH3 - C{H_3} group present at the alpha position of the functional group. Tertiary alcohols cannot give a positive iodoform test as they cannot be oxidized.