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Question: Conversion of \( C{H_3}C{H_2}OH \) to \( C{H_3}CHO \) in the presence of: (A) \({\text{N}}{{\text{...

Conversion of CH3CH2OHC{H_3}C{H_2}OH to CH3CHOC{H_3}CHO in the presence of:
(A) Na2Cr2O7 and NaOH{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}{\text{ and NaOH}}
(B) Na2Cr2O7 and aq H2SO4{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}{\text{ and aq }}{{\text{H}}_2}S{O_4}
(C) NaOH{\text{NaOH}}
(D) Fe in presence of NaOH{\text{Fe in presence of NaOH}}

Explanation

Solution

We need to convert alcohol to aldehyde. We need to identify the reagent here. This question requires the concepts of organic chemistry. The reaction requires an oxidising agent and also dehydration, i.e. elimination of water molecules.

Complete step by step solution:
We already know that,
Alcohols are oxidised to aldehydes by mixture of Na2Cr2O7 and aq H2SO4{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}{\text{ and aq }}{{\text{H}}_2}S{O_4} ​. This mixture produces nascent oxygen.
Thus, the reaction is as follows:
CH3CH2OH→Na2Cr2O7 and aq H2SO4CH3CHO+H2OC{H_3}C{H_2}OH\xrightarrow{{{\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{r}}_{\text{2}}}{{\text{O}}_{\text{7}}}{\text{ and aq }}{{\text{H}}_2}S{O_4}}}C{H_3}CHO + {H_2}O
So, we need to select the correct option.
The correct option is B.

Note:
Sodium Dichromate is an orange to red coloured, crystalline, inorganic compound that emits toxic chromium fumes upon heating. Sodium dichromate is highly corrosive and is a strong oxidizing agent. Atoms, ions, and molecules that have an unusually large affinity for electrons tend to be good oxidizing agents. Elemental fluorine, for example, is the strongest common oxidizing agent. It is such a good oxidizing agent that metals, quartz, asbestos, and even water burst into flame in its presence. Sulfuric Acid is a strong mineral acid that is colourless when pure. It is a strong acid which reacts by donating an H+{H^ + } ion. Strong acids are proton donors and electron acceptors. So, it will accept an electron during reaction.by accepting an electron it is undergoing reduction so it is an oxidizing agent. it is always an oxidizing agent. In sulphuric acid, sulphur is in +6+ 6 oxidation state. To act as a reducing agent, one should lose electrons and the atom should go from lower positive oxidation state to higher positive oxidation state.
Thus, we can convert alcohol to aldehyde using the following reagents.