Assertion (A): On prolonged heating with HI, glucose forms n-hexane.
Reason (R): All the six carbon atoms of glucose are linked in a straight chain.
In the light of the above statements, choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Both (A) and (R) are true and (R) is the correct explanation of (A).
On prolonged heating, HI reduces glucose completely: every oxygen-containing group is removed and only the carbon skeleton with hydrogen atoms is left. The hydrocarbon obtained therefore shows how the carbon atoms of glucose are joined.
\( \displaystyle \ce{CHO-(CHOH)4-CH2OH} \xrightarrow[\Delta]{\ce{HI}} \ce{CH3-CH2-CH2-CH2-CH2-CH3} \)
\( \displaystyle \ce{C6H12O6 + 14HI -> C6H14 + 6H2O + 7I2} \)
Assertion (A): On prolonged heating with HI, glucose forms n-hexane. — True. The product is the unbranched hydrocarbon n-hexane.
Reason (R): All the six carbon atoms of glucose are linked in a straight chain. — True. Glucose is an aldohexose with an unbranched six carbon chain.
Why R explains A?
HI removes the oxygen but does not change the carbon chain.
Because the six carbon atoms of glucose are in a straight chain, the hydrocarbon left is n-hexane, not a branched hexane; this is how the result was used to show the straight chain.
NCERT (2026-27 edition):
10.1.2.1 Glucose – “2. On prolonged heating with HI, it forms n-hexane, suggesting that all the six carbon atoms are linked in a straight chain.”