
Why is sulphuric acid not used during the reaction of alcohols with KI?
Answer
534k+ views
Hint: Sulphuric acid is a very strong oxidizing agent. So, its impact on any reaction will be very different from a weak oxidizing agent. The strong oxidizing nature of sulphuric acid forms the basis of many reactions which are not shown by its dilute counterpart.
Complete step by step solution:
${H_2}{SO_4}$ being a strong oxidising agent oxidises HI and produces $I_2$ . The reaction between alcohol (OH) and HI results in production of alkyl iodide. The motive of the reaction is to form alkyl iodides. When HI gets converted to $I_2$ , the motive of the reaction remains unaccomplished. This is the reason why sulphuric acid (${H_2}{SO_4}$) is not used during the reaction of alcohol (OH) with potassium iodide (KI).
In the presence of a dilute acid, KI would produce HI.
$ 2KI + {H_2}{SO_4}\to 2KHSO_4 + HI $
If the acid used is sulphuric acid, the HI gets used up to produce I2 gas.
$ 2HI + {H_2}{SO_4}\to I_2 + SO_2 + H_2O $
As a result, the action of alcohol on acid to produce alkyl iodide cannot occur. Therefore, sulphuric acid is not used for this reaction. Instead, a weak oxidizing agent like ${H_3}{PO_4}$ is preferred for this reaction.
Note:
Whenever there is a reaction involving a strong oxidizing agent, always consider the difference between reaction involving dilute acid and that involving concentrated acid. Concentrated acids will result in a different product than their dilute counterparts. Also, a strong oxidizing acid may cause destruction at higher temperatures.Also consider the reaction between HI and weak oxidizing agents.
Complete step by step solution:
${H_2}{SO_4}$ being a strong oxidising agent oxidises HI and produces $I_2$ . The reaction between alcohol (OH) and HI results in production of alkyl iodide. The motive of the reaction is to form alkyl iodides. When HI gets converted to $I_2$ , the motive of the reaction remains unaccomplished. This is the reason why sulphuric acid (${H_2}{SO_4}$) is not used during the reaction of alcohol (OH) with potassium iodide (KI).
In the presence of a dilute acid, KI would produce HI.
$ 2KI + {H_2}{SO_4}\to 2KHSO_4 + HI $
If the acid used is sulphuric acid, the HI gets used up to produce I2 gas.
$ 2HI + {H_2}{SO_4}\to I_2 + SO_2 + H_2O $
As a result, the action of alcohol on acid to produce alkyl iodide cannot occur. Therefore, sulphuric acid is not used for this reaction. Instead, a weak oxidizing agent like ${H_3}{PO_4}$ is preferred for this reaction.
Note:
Whenever there is a reaction involving a strong oxidizing agent, always consider the difference between reaction involving dilute acid and that involving concentrated acid. Concentrated acids will result in a different product than their dilute counterparts. Also, a strong oxidizing acid may cause destruction at higher temperatures.Also consider the reaction between HI and weak oxidizing agents.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Class 12 Question and Answer - Your Ultimate Solutions Guide

Complete reduction of benzene diazonium chloride with class 12 chemistry CBSE

How can you identify optical isomers class 12 chemistry CBSE

Trending doubts
Which are the Top 10 Largest Countries of the World?

What are the major means of transport Explain each class 12 social science CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

Differentiate between insitu conservation and exsitu class 12 biology CBSE

Draw a neat and well labeled diagram of TS of ovary class 12 biology CBSE

RNA and DNA are chiral molecules their chirality is class 12 chemistry CBSE

