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This question tests understanding of alcohol reactions with HI. Primary alcohols (like CH₃CH₂OH) undergo SN2 substitution to form alkyl iodides. Secondary/tertiary alcohols undergo SN1 substitution. With Red P + HI, alcohols are reduced to alkanes. The correct option shows a primary alcohol (CH₃CH₂OH) reacting with HI alone to form the iodide (CH₃CH₂I), which is accurate.
Final answer: CH3–CH2–OH CH3–CH2–I