Identify the major product of the following reaction
Show Hint
Clemmensen reduction \((Zn-Hg/HCl)\) converts aldehydes and ketones into hydrocarbons under strongly acidic conditions. Alcohol groups may undergo substitution reactions in concentrated \(HCl\).
Step 1: Identify the reagent from the text. The reaction uses zinc amalgam $Zn\text{-}Hg$ together with concentrated $HCl$, which is the classic combination for the Clemmensen reduction. Step 2: Recall what Clemmensen reduction does. Clemmensen reduction converts a carbonyl group $C=O$ of an aldehyde or ketone into a methylene group $CH_2$ under strongly acidic conditions: \[ R\text{-}CO\text{-}R' \xrightarrow[HCl]{Zn\text{-}Hg} R\text{-}CH_2\text{-}R'. \] Step 3: Examine the functional groups in the substrate. The given molecule contains a ketone carbonyl group along with an alcohol type group elsewhere in the structure. Step 4: Decide which group reacts. Clemmensen reduction is selective for the carbonyl group of aldehydes and ketones; it does not reduce an alcohol group. So only the $C=O$ is changed. Step 5: Apply the reduction. The ketone $C=O$ is converted into a $CH_2$ unit while the rest of the molecule, including the alcohol group, stays intact. Step 6: Conclude. The major product is the one in which only the carbonyl group has become a methylene group, which matches option 3. So the answer is \[ \boxed{\text{ketone } C{=}O \to CH_2 \text{ product (option 3)}} \]
Was this answer helpful?
0
Top Questions on Aldehydes, Ketones and Carboxylic Acids
