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: Understanding the Concept:
This question involves two classic organic reactions:
1. The reduction of glucose with hydroiodic acid (HI), a powerful reducing agent that converts oxygen-containing groups to hydrogen.
2. The Wurtz reaction, which couples two alkyl halide molecules in presence of sodium metal in dry ether to form a larger alkane.
Step 2: Identifying Compound 'C':
Glucose (C\(_6\)H\(_{12}\)O\(_6\)) contains an aldehyde group and several hydroxyl groups. When heated for a long time with concentrated HI, all oxygen atoms are replaced by hydrogen atoms. This reduces glucose completely to a straight-chain hydrocarbon containing six carbon atoms — n-hexane.
\[ \text{C}_6\text{H}_{12}\text{O}_6 \xrightarrow{\text{HI, }\Delta} \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_3 \] This experiment demonstrates that all six carbon atoms in glucose are arranged in an unbranched chain. Therefore, ‘C’ = n-hexane.
Step 3: Formation of 'C' by Wurtz Reaction:
The Wurtz reaction is given by: \[ 2\text{R-X} + 2\text{Na} \rightarrow \text{R-R} + 2\text{NaX} \] To obtain n-hexane (C\(_6\)H\(_{14}\)), each alkyl halide must contain half the total number of carbon atoms, i.e., 3 carbons. Thus, the required alkyl halide is a propyl halide. When two molecules of 1-chloropropane react with sodium, they couple to form n-hexane: \[ 2\text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} + 2\text{Na} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2-\text{CH}_2\text{CH}_2\text{CH}_3 + 2\text{NaCl} \] Hence, the halide ‘D’ must be 1-chloropropane. Wurtz reaction gives the best yield with primary halides, making 1-chloropropane the most suitable choice.
Step 4: Evaluating the Options:
Let us analyze the given options:
(A) 1-Chloropropane: CH\(_3\)CH\(_2\)CH\(_2\)Cl — correct, forms n-hexane.
(B) 2-Chloropropane: (CH\(_3\))\(_2\)CHCl — gives 2,3-dimethylbutane (branched).
(C) 2-Chlorobutane: CH\(_3\)CH\(_2\)CH(Cl)CH\(_3\) — 4-carbon chain, incorrect.
(D) 1-Chloro-2-methylpropane: (CH\(_3\))\(_2\)CHCH\(_2\)Cl — branched 4-carbon halide, incorrect. Only option (A) can give a straight-chain C\(_6\)H\(_{14}\) via the Wurtz reaction.
Step 5: Conclusion:
The compound ‘C’ obtained from glucose is n-hexane, and to prepare it by the Wurtz reaction, the halide ‘D’ must be 1-chloropropane.
\[ \boxed{\text{C = n-Hexane, D = 1-Chloropropane (Option A)}} \]