Question

\[\alpha \]-Hydroxypropanoic acid can be prepared from ethanal by following the steps given in the sequence.
A. Treat with $\text{HCN}$ followed by acidic hydrolysis
B. Treat with $\text{NaHS}{{\text{O}}_{4}}$ followed by reaction with $\text{N}{{\text{a}}_{2}}\text{C}{{\text{O}}_{3}}$
C. Treat with ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$ followed by hydrolysis
D. Treat with ${{\text{K}}_{2}}\text{C}{{\text{r}}_{2}}{{\text{O}}_{7}}$ in presence of sulphuric acid

Answer 1

Hint: For this problem, we have to write the complete reaction of the formation of \[\alpha \]-Hydroxypropanoic acid from ethanal which is a two-carbon containing compound. After which we can easily choose the correct option.

Complete Step-by-step answer:
- In the given question, we have to explain the formation of \[\alpha \]-Hydroxypropanoic acid from ethanal step by step.
- As we know that ethanal consists of two carbons and one aldehyde group attached to the first carbon.
- So, we have to attach the cyanide group on the first carbon of ethanal by breaking a double bond.
- For which when the double bond breaks, the positive charge comes on carbon and negative charge on oxygen.
\[\text{C}{{\text{H}}_{3}}\text{ - CHO + HCN }\to \text{ C}{{\text{H}}_{3}}-\text{H}{{\text{C}}^{+}}\text{(}{{\text{O}}^{-}}\text{)}\]
- Now, the hydrogen will attach to oxygen and the cyanide group will attach with a carbon atom.
- The product formed is known as cyanohydrin and the chemical reaction is shown below:
\[\text{C}{{\text{H}}_{3}}\text{ - CHO + HCN }\to \text{ C}{{\text{H}}_{3}}-\text{CH(OH)CN}\]
- Now, cyanohydrin will undergo acidic hydrolysis due to which the oxidation of the cyanide group attached with carbon takes place.
- Oxidation is the process of the addition of the oxygen or removal of hydrogen and also loss of the electron takes place.
- Due to the oxidation, the cyanide group will convert carboxylic acid as the oxygen atom attaches.
- The chemical reaction is:
\[\text{C}{{\text{H}}_{3}}-\text{CH(OH)CN }\xrightarrow{\text{Hydrolysis}}\text{ C}{{\text{H}}_{3}}-\text{CH(OH)COOH}\]

Therefore, option A is the correct answer.

Note: Here, in the compound alpha is used to indicate the position of the hydroxyl group. As we know that alpha carbon is adjacent to the carbon with the functional group so the group attached with the alpha carbon is indicated as alpha.