Problem 52 - A Complete Resource Book in Chemistry for JEE Main

Question

Which of the following is not correct? (a) ozone is used to identify the presence of unsaturation in alkenes (b) sodium thiosulphate reacts with iodine to form sodium suphate (c) iodine oxidizes sodium thiosulphate to sodium tetrathionate (d) sodium thiosulphate is soluble in water

Step-by-Step Solution

Answer

Option (b) is not correct.

Step 1: Analyze Option (a)

Ozone is used in the process of ozonolysis to break double bonds in alkenes, identifying unsaturation. Hence, option (a) is correct.

Step 2: Analyze Option (b)

Sodium thiosulphate reacts with iodine to produce sodium tetrathionate, not sodium sulphate. This means option (b) is incorrect.

Step 3: Analyze Option (c)

Iodine indeed oxidizes sodium thiosulphate to form sodium tetrathionate. Therefore, option (c) is correct.

Step 4: Analyze Option (d)

Sodium thiosulphate is a water-soluble salt, making option (d) true.

Key Concepts

Ozonolysis

Ozonolysis is a fascinating chemical reaction that involves the breakdown of alkenes using ozone. This reaction is crucial for identifying unsaturation within organic compounds. When ozone, a molecule consisting of three oxygen atoms, comes into contact with an alkene, it reacts to cleave the carbon-carbon double bonds within the molecule. This process converts alkenes into smaller compounds, typically carbonyl compounds such as aldehydes or ketones. The steps of ozonolysis are fairly straightforward:

• The first stage involves the formation of a molozonide intermediate through the addition of ozone to the alkene.
• This unstable intermediate rapidly rearranges into a more stable ozonide.
• Finally, in the presence of a reducing agent, the ozonide decomposes to yield the carbonyl compounds.

The importance of this reaction lies in its efficiency and its ability to help organic chemists determine the positions and presence of unsaturation in an alkene molecule. This powerful diagnostic tool is widely used in organic synthesis to modify and analyze complex organic structures.

Sodium Thiosulphate

Sodium thiosulphate is a versatile and water-soluble compound with the formula Na extsubscript{2}S extsubscript{2}O extsubscript{3}. It is widely used in various chemical applications, and its solubility makes it particularly useful in solution-based reactions. When sodium thiosulphate reacts with iodine, it performs as an excellent reducing agent.A remarkable feature of sodium thiosulphate is its ability to undergo redox reactions, such as with iodine. Despite common misconceptions, when it reacts with iodine, it does not form sodium sulphate but instead produces sodium tetrathionate and sodium iodide, following this equation:\[2 \, ext{Na}_2 ext{S}_2 ext{O}_3 + ext{I}_2 \rightarrow ext{Na}_2 ext{S}_4 ext{O}_6 + 2 \, ext{NaI}\]In addition to its application in chemical laboratories, sodium thiosulphate is also found in industries such as photography for fixing photographic papers. Its ability to neutralize halogens like iodine makes it an invaluable agent in various analytical methods, underlining its diverse utility in chemical processes.

Iodine Oxidation

Iodine oxidation is an intriguing process in which iodine acts as an oxidizing agent, particularly in reactions with other compounds. In the context of its interaction with sodium thiosulphate, iodine oxidizes this compound to form sodium tetrathionate.This redox reaction showcases iodine's ability to accept electrons. During the process:

• Iodine molecules are reduced, gaining electrons to form iodide ions.
• Simultaneously, sodium thiosulphate molecules are oxidized, resulting in the formation of sodium tetrathionate.

The balanced chemical equation demonstrating this transformation is:\[2 \, ext{Na}_2 ext{S}_2 ext{O}_3 + ext{I}_2 \rightarrow ext{Na}_2 ext{S}_4 ext{O}_6 + 2 \, ext{NaI}\]The ability of iodine to act as an oxidizing agent showcases its reactivity and its role in various chemical applications, not just limited to the production of sodium tetrathionate but also in broader contexts of redox reactions. This characteristic of iodine underlines its importance in both industrial and educational settings, serving as a critical component in studies related to oxidation and reduction processes.