The most suitable reagent for the conversion of primary alcohol into aldehyde with the same number of carbon is:
(A) Acidified $ K_2Cr_2O_7 $
(B) Acidified $ KMnO_4 $
(C) Alkaline $ KMnO_4 $
(D) Pyridinium chlorochromate (PCC)
(E) $ CrCO_3 $
Answer
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Hint :Primary alcohol is an organic compound with only one carbon atom. For it to be converted into an aldehyde, the primary alcohol needs to be oxidized in the presence of a suitable reagent. Also, the alcohol should not be too oxidized otherwise it may go over the aldehyde phase and form another, more oxidised, compound.
Complete Step By Step Answer:
The oxidation of primary alcohols in the presence of reagents such as acidified potassium dichromate, acidified potassium permanganate, pyridinium chlorochromate and chromic anhydride.
The reaction of primary alcohol to give aldehyde is:
$ RCH_2\ OH+(O)\Rightarrow RCHO+H_2\ O $
Here, the (O) is provided from the
Depending on the strength of the oxidising agent, the reaction could also go forward and convert the aldehyde into carboxylic acid.
Alkaline $ KMnO_4 $ is the least efficient form of oxidising agent as strongly alkaline mediums do not allow for the proper transfer of electrons. Therefore, option (C) can be ruled out.
Acidified mediums are best for oxidisation. However, in the above question, primary alcohol needs to be oxidised until it forms aldehyde only and not further than that.
Acidified $ KMnO_4 $ and acidified $ K_2 Cr_2 O_7 $ are both too strong oxidising agents and will convert the formed aldehyde into carboxylic acid as well. In this case, an extra distillation method will be required to separate the aldehyde, which is hard to achieve.
Therefore, options (A) and (B) can also be ruled out.
Chromic acid ( $ CrCO_3) $ is also a stronger oxidising agent and thus oxidises the aldehyde to further form carboxylic acid. Therefore, option (E) can also be ruled out.
Option (D) pyridinium chlorochromate (PCC) is the most suitable choice for the conversion of primary alcohol into aldehyde.
Therefore, the answer to this question is option (D).
Note :
Such questions are very specific and fact-based. Make sure you learn your chapters thoroughly and take note of such details. Also, by having knowledge of the reactions of organic compounds and the reagents used in those reactions will help answer these questions.
Complete Step By Step Answer:
The oxidation of primary alcohols in the presence of reagents such as acidified potassium dichromate, acidified potassium permanganate, pyridinium chlorochromate and chromic anhydride.
The reaction of primary alcohol to give aldehyde is:
$ RCH_2\ OH+(O)\Rightarrow RCHO+H_2\ O $
Here, the (O) is provided from the
Depending on the strength of the oxidising agent, the reaction could also go forward and convert the aldehyde into carboxylic acid.
Alkaline $ KMnO_4 $ is the least efficient form of oxidising agent as strongly alkaline mediums do not allow for the proper transfer of electrons. Therefore, option (C) can be ruled out.
Acidified mediums are best for oxidisation. However, in the above question, primary alcohol needs to be oxidised until it forms aldehyde only and not further than that.
Acidified $ KMnO_4 $ and acidified $ K_2 Cr_2 O_7 $ are both too strong oxidising agents and will convert the formed aldehyde into carboxylic acid as well. In this case, an extra distillation method will be required to separate the aldehyde, which is hard to achieve.
Therefore, options (A) and (B) can also be ruled out.
Chromic acid ( $ CrCO_3) $ is also a stronger oxidising agent and thus oxidises the aldehyde to further form carboxylic acid. Therefore, option (E) can also be ruled out.
Option (D) pyridinium chlorochromate (PCC) is the most suitable choice for the conversion of primary alcohol into aldehyde.
Therefore, the answer to this question is option (D).
Note :
Such questions are very specific and fact-based. Make sure you learn your chapters thoroughly and take note of such details. Also, by having knowledge of the reactions of organic compounds and the reagents used in those reactions will help answer these questions.
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