Identify the major product of the following reaction.
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Wolff--Kishner reduction:
\[
R_2C=O
\rightarrow
R_2CH_2
\]
uses hydrazine and strong base under heating conditions. It reduces aldehydes and ketones to hydrocarbons while leaving nitro groups generally unaffected.
Step 1: Identify the reagents and name the reaction. The reagents are $N_2H_4$ (hydrazine), $NaOH$ (strong base), ethylene glycol as solvent, and heat. These are the Wolff-Kishner reduction conditions. This reaction selectively reduces a carbonyl group ($C{=}O$) to a methylene group ($-CH_2-$), leaving $-NO_2$ and $-OH$ groups intact. Step 2: Identify the carbonyl group in the substrate. The substrate is a benzene ring bearing three substituents: a hydroxyl group ($-OH$), a nitro group ($-NO_2$), and an acetyl group ($-COCH_3$). The only carbonyl group present is the ketone carbonyl of the acetyl substituent. Step 3: Write the Wolff-Kishner mechanism in outline. Step A: Hydrazine reacts with the ketone to form a hydrazone: $Ar{-}CO{-}CH_3 + N_2H_4 \rightarrow Ar{-}C(=N{-}NH_2){-}CH_3 + H_2O$. Step B: Under basic conditions and heat, the hydrazone decomposes with loss of $N_2$ and incorporation of two hydrogen atoms, giving the fully reduced product. Step 4: Apply the reaction to the acetyl group. \[ Ar{-}CO{-}CH_3 \xrightarrow{\text{Wolff-Kishner}} Ar{-}CH_2{-}CH_3 \] The $-COCH_3$ group is converted into $-CH_2CH_3$ (ethyl group). Step 5: Confirm which groups are unaffected. The phenolic $-OH$ and the nitro $-NO_2$ groups remain unchanged under Wolff-Kishner conditions, which are basic and mild enough not to affect these groups. Step 6: Name the final product. The product is a phenol bearing an ethyl group and a nitro group. Based on the relative positions in the starting material, the product is 4-ethyl-2-nitrophenol. Step 7: State the final answer. \[ \boxed{\text{4-Ethyl-2-nitrophenol}} \]
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