



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)
In the given reaction sequence, the structure of Y would be:

The center of a disk of radius $ r $ and mass $ m $ is attached to a spring of spring constant $ k $, inside a ring of radius $ R>r $ as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following Hooke’s law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as $ T = \frac{2\pi}{\omega} $. The correct expression for $ \omega $ is ( $ g $ is the acceleration due to gravity): 
Let $ a_0, a_1, ..., a_{23} $ be real numbers such that $$ \left(1 + \frac{2}{5}x \right)^{23} = \sum_{i=0}^{23} a_i x^i $$ for every real number $ x $. Let $ a_r $ be the largest among the numbers $ a_j $ for $ 0 \leq j \leq 23 $. Then the value of $ r $ is ________.