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)
Let $ S $ denote the locus of the point of intersection of the pair of lines $$ 4x - 3y = 12\alpha,\quad 4\alpha x + 3\alpha y = 12, $$ where $ \alpha $ varies over the set of non-zero real numbers. Let $ T $ be the tangent to $ S $ passing through the points $ (p, 0) $ and $ (0, q) $, $ q > 0 $, and parallel to the line $ 4x - \frac{3}{\sqrt{2}} y = 0 $.
Then the value of $ pq $ is
Let $ y(x) $ be the solution of the differential equation $$ x^2 \frac{dy}{dx} + xy = x^2 + y^2, \quad x > \frac{1}{e}, $$ satisfying $ y(1) = 0 $. Then the value of $ 2 \cdot \frac{(y(e))^2}{y(e^2)} $ is ________.
Let $ \mathbb{R} $ denote the set of all real numbers. Then the area of the region $$ \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x > 0, y > \frac{1}{x},\ 5x - 4y - 1 > 0,\ 4x + 4y - 17 < 0 \right\} $$ is