In the following reaction sequence, the major product Q is
\(\text{L-Glucose}\xrightarrow[ii)Cr_2,775K,10-20 \text{atm}]{i)HI, \Delta}P\xrightarrow[UV]{Cl_2(excess)}Q\)
\(\text{L-Glucose}\xrightarrow[ii)Cr_2,775K,10-20 \text{atm}]{i)HI, \Delta}P\xrightarrow[UV]{Cl_2(excess)}Q\)
The Correct Option is D
Solution and Explanation
To solve the problem, we analyze the reaction sequence starting from L-glucose to find the major product Q.
1. Step i) Reaction with HI and heat (\(\Delta\)):
- The sugar undergoes cleavage of C-O bonds by HI.
- This reduces glucose to a mixture of iodoalkanes (polyiodo compounds).
- The carbon skeleton remains but oxygen atoms are replaced by iodine.
2. Step ii) Oxidation with Cr\(_2\)O\(_3\), high temperature and pressure:
- Iodoalkanes are oxidized to form fully chlorinated cyclic hydrocarbons (via halogen exchange).
- Oxygen atoms are removed, and the carbon framework becomes a chlorinated cyclic structure.
3. Step iii) Treatment with excess Cl\(_2\) under UV:
- Further chlorination occurs, leading to a hexachlorinated cyclohexane derivative.
4. Structure of the final product Q:
- Hexachlorocyclohexane (all positions chlorinated).
- This corresponds to structure (D).
Final Answer:
Option (D)
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