Question

On prolonged heating with HI, glucose gives a compound 'C', which can be obtained by Wurtz reaction using sodium metal and compound 'D'. Identify 'D'

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

1. Prolonged heating of any carbohydrate (glucose, fructose, sucrose) with HI always results in n-hexane. This is a standard test for the straight-chain structure of hexoses. 2. The Wurtz reaction (2R-X + 2Na \(\rightarrow\) R-R) is best for synthesizing symmetrical alkanes (where R-R has an even number of carbons) from primary alkyl halides.

Answer 1

The Correct Option is C

Step 1: Identify Compound 'C': Glucose (\( \text{C}_6\text{H}_{12}\text{O}_6 \)) on prolonged heating with HI undergoes reduction to form n-Hexane. So, 'C' is n-Hexane (\( \text{C}_6\text{H}_{14} \)).
Step 2: Identify 'D' via Wurtz Reaction: Wurtz reaction involves the coupling of two alkyl halides (\( \text{R-X} \)) in the presence of Sodium/Dry ether to form an alkane with double the number of carbons (\( \text{R-R} \)). Reaction: \( 2\text{R-X} + 2\text{Na} \to \text{R-R} + 2\text{NaX} \). Here, product 'C' is n-Hexane (6 Carbons). Since n-Hexane is symmetric (\( \text{CH}_3\text{CH}_2\text{CH}_2-\text{CH}_2\text{CH}_2\text{CH}_3 \)), it is formed from a primary halide with 3 carbons. \( \text{R} \) must be n-Propyl group (\( \text{CH}_3\text{CH}_2\text{CH}_2- \)). So, 'D' must be 1-Chloropropane (n-Propyl chloride).
Step 3: Analyze Options (Visuals): Option 1: \( \text{Cl}-\text{CH}_2-\text{CH}_2-\text{CH}_3 \) (1-Chloropropane). Matches. Option 2: 2-Chloropropane? (Structure looks branched). Wurtz coupling of this gives 2,3-dimethylbutane (an isomer of hexane, but not n-hexane). Option 3: 1-Chlorobutane? (4 carbons). Would give Octane. Option 4: ? (3 carbons, attached at middle). Isopropyl chloride. Gives 2,3-dimethylbutane. Since glucose gives n-hexane, 'D' must be 1-Chloropropane.