Let's analyze the problem step-by-step. --- # **Given:** - The structure is **ethanol (C₂H₅OH)**, with the molecular structure shown. - The reaction involves **acidic conditions** (H⁺). --- # **Objective:** Predict the products of the **elimination reaction** when ethanol is treated with an acid (H⁺). --- # **Step 1: Understand the structure** Ethanol (C₂H₅OH): ``` H H | | H — C — C — OH | | H H ``` The structure shows a hydroxyl group (-OH) attached to the second carbon. --- # **Step 2: Recognize the type of reaction** - Under acidic conditions, **elimination reactions** typically involve removal of a proton (H⁺) and a leaving group (like -OH), leading to the formation of an **alkene**. - In alcohols, especially under acid catalysis, the hydroxyl group can be protonated to form a better leaving group (water, H₂O). --- # **Step 3: Protonate the hydroxyl group** - The hydroxyl group gets protonated: \[ \text{Ethanol} \rightarrow \text{Protonated ethanol} \] - The protonated form looks like: \[ \mathrm{C_2H_5–OH_2^+} \] which is a good leaving group. --- # **Step 4: Loss of water** - The protonated hydroxyl group leaves as water: \[ \mathrm{C_2H_5–OH_2^+} \rightarrow \mathrm{C_2H_5^+} + \mathrm{H_2O} \] - The resulting carbocation is a **primary carbocation**: \[ \mathrm{CH_3–CH_2^+} \] --- # **Step 5: Elimination (Deprotonation)** - A proton (H⁺) from a neighboring carbon (preferably from the carbon adjacent to the carbocation) is removed, forming a double bond: \[ \mathrm{CH_3–CH_2^+} + \text{base} \rightarrow \text{alkene} \] - The most stable alkene is formed by removing a hydrogen from the carbon adjacent to the carbocation. --- # **Step 6: Final product prediction** - Removing a hydrogen from the methyl group (attached to the CH₂) yields **ethene (C₂H₄)**: \[ \boxed{ \text{Product: Ethene (C}_2\text{H}_4) } \] --- # **Final answer:** **The elimination of water from ethanol under acidic conditions produces ethene (C₂H₄).**