
What would be the final product of the following reaction?$C{H_3}C{H_2}COOH\xrightarrow[{\operatorname{Re} d\;P}]{{B{r_2}}}A\xrightarrow{{Aq.KOH}}B$
Answer
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Hint:As we are well aware of Hell Volhard Zelinsky reaction where halogenation of carboxylic acid with chlorine or bromine in the presence of Red Phosphorus giving alpha acids. And aqueous potassium hydroxide is involved in dissociation reaction to produce a hydroxide ion.
Complete step-by-step answer:As we know that Carboxylic acids having at least one alpha or more alpha hydrogen atoms reacts with chlorine or bromine in the presence of a small amount of red phosphorus resulting in the formation of alpha- carboxylic acids is a halogenation reaction given by Hell Volhard Zelinsky.
So we can show this reaction with the help of equation as shown below:
$C{H_3} - C{H_2} - COOH\xrightarrow[{\operatorname{Re} d\;P}]{{B{r_2}}}C{H_3} - CH(Br) - COOH$
Thus, the product A is an alpha-bromopropanoic acid or $2 - bromopropanoic\;acid$.
Now, comes the next reagent which is aqueous potassium hydroxide and we know that it is alkaline in nature and helps to dissociate the compound to produce a hydroxide ion which acts as strong nucleophile and replace the halogen atoms in an alkyl halide. Similarly, the hydroxide ion from aqueous potassium hydroxide will remove the bromine atom present in the compound A so formed and gets attached in its place to form the compound B.
We can show this reaction as below:
$C{H_3} - CH(Br) - COOH\xrightarrow{{Aq.KOH}}C{H_3} - CH(OH) - COOH$
Therefore, the compound B which is our final product is $2 - hydroxypropanoic\;acid$.
Note:Remember that $2 - hydroxypropanoic\;acid$ is commonly known as the lactic acid which is an organic acid widely found in living organisms. Lactic acid is a principal metabolic intermediate in most of the living organism and it is evenly produced in muscles when they respire in lack of oxygen. Also, the resultant formed by the reaction of aqueous solution is formed due to the nucleophilic substitution reaction.
Complete step-by-step answer:As we know that Carboxylic acids having at least one alpha or more alpha hydrogen atoms reacts with chlorine or bromine in the presence of a small amount of red phosphorus resulting in the formation of alpha- carboxylic acids is a halogenation reaction given by Hell Volhard Zelinsky.
So we can show this reaction with the help of equation as shown below:
$C{H_3} - C{H_2} - COOH\xrightarrow[{\operatorname{Re} d\;P}]{{B{r_2}}}C{H_3} - CH(Br) - COOH$
Thus, the product A is an alpha-bromopropanoic acid or $2 - bromopropanoic\;acid$.
Now, comes the next reagent which is aqueous potassium hydroxide and we know that it is alkaline in nature and helps to dissociate the compound to produce a hydroxide ion which acts as strong nucleophile and replace the halogen atoms in an alkyl halide. Similarly, the hydroxide ion from aqueous potassium hydroxide will remove the bromine atom present in the compound A so formed and gets attached in its place to form the compound B.
We can show this reaction as below:
$C{H_3} - CH(Br) - COOH\xrightarrow{{Aq.KOH}}C{H_3} - CH(OH) - COOH$
Therefore, the compound B which is our final product is $2 - hydroxypropanoic\;acid$.
Note:Remember that $2 - hydroxypropanoic\;acid$ is commonly known as the lactic acid which is an organic acid widely found in living organisms. Lactic acid is a principal metabolic intermediate in most of the living organism and it is evenly produced in muscles when they respire in lack of oxygen. Also, the resultant formed by the reaction of aqueous solution is formed due to the nucleophilic substitution reaction.
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