The final product A formed in the following multistep reaction sequence is
A reaction sequence starting with styrene undergoing Markovnikov hydration, followed by oxidation with CrO3, and finally reduction with hydrazine and KOH.
A.
A reaction sequence starting with styrene undergoing Markovnikov hydration, followed by oxidation with CrO3, and finally reduction with hydrazine and KOH.
B.
A reaction sequence starting with styrene undergoing Markovnikov hydration, followed by oxidation with CrO3, and finally reduction with hydrazine and KOH.
C.
A reaction sequence starting with styrene undergoing Markovnikov hydration, followed by oxidation with CrO3, and finally reduction with hydrazine and KOH.
D.
A reaction sequence starting with styrene undergoing Markovnikov hydration, followed by oxidation with CrO3, and finally reduction with hydrazine and KOH.
Solution & Explanation
Core Logic
The reaction sequence proceeds in three distinct steps from the starting material, styrene (Ph-CH=CH₂$Ph-CH=CH_2$).
Step 1: Acid-catalyzed Hydration
Styrene reacts with H₂O, H^+$H_2O, H^+$ to undergo electrophilic addition. Protonation yields the more stable secondary benzylic carbocation. Attack by water followed by deprotonation gives 1-phenylethanol (Ph-CH(OH)-CH₃$Ph-CH(OH)-CH_3$).
Step 2: Oxidation
1-phenylethanol is a secondary alcohol. Treatment with chromium trioxide (CrO₃$CrO_3$, Jones reagent condition) oxidizes the secondary alcohol to a ketone. This yields acetophenone (Ph-CO-CH₃$Ph-CO-CH_3$).
Step 3: Wolff-Kishner Reduction
Acetophenone is treated with hydrazine (NH₂-NH₂$NH_2-NH_2$) and a strong base (KOH$KOH$) under heating. This is the classic Wolff-Kishner reduction, which completely reduces the carbonyl group (C=O$C=O$) to a methylene group (-CH₂-$-CH_2-$). The final product is ethylbenzene (Ph-CH₂-CH₃$Ph-CH_2-CH_3$).
Step 1: Overall Reaction Pathway
A reaction sequence starting with styrene undergoing Markovnikov hydration, followed by oxidation with CrO3, and finally reduction with hydrazine and KOH.
The final product A is ethylbenzene.
Chapter Mix
Class 11 Chemistry: Hydrocarbons
Class 12 Chemistry: Alcohols Phenols and Ethers
Class 12 Chemistry: Aldehydes Ketones and Carboxylic Acids
Keywords:#styrene#hydration#CrO3#Wolff-Kishner
More Aldehydes Ketones and Carboxylic Acids Previous-Year Questions — Page 9
Qjee_main_2024_30_jan_morningPreparation of Aldehydes
This reduction reaction is known as:
The image shows the catalytic hydrogenation of benzoyl chloride to benzaldehyde in the presence of Pd-BaSO4.
The reaction depicts the partial reduction of an acid chloride (benzoyl chloride) to an aldehyde (benzaldehyde) using hydrogen gas in the presence of a poisoned palladium catalyst (Pd supported on BaSO₄$BaSO_4$).
The image shows the catalytic hydrogenation of benzoyl chloride to benzaldehyde in the presence of Pd-BaSO4.
This specific reaction is known as the Rosenmund reduction.
Pattern Recognition
Acid Chloride + H₂, Pd/BaSO₄$H_2, Pd/BaSO_4$arrow$\rightarrow$ Aldehyde is strictly the Rosenmund reduction. The BaSO₄$BaSO_4$ poisons the catalyst to prevent over-reduction to an alcohol.
Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Pathway 1 (Upper): Toluene is treated with Reagent 'A' and acetic anhydride (CH₃CO)₂O$(CH_3CO)_2O$ to form an intermediate (benzylidene diacetate), which on hydrolysis gives benzaldehyde. The reagent used here is Chromic oxide (CrO₃$CrO_3$). Thus, A is CrO₃$CrO_3$.
Pathway 2 (Lower): Toluene is treated with Reagent 'B' in CS₂$CS_2$ to form a chromium complex intermediate, which on hydrolysis yields benzaldehyde. This is the Etard reaction, and the reagent used is Chromyl chloride (CrO₂Cl₂$CrO_2Cl_2$). Thus, B is CrO₂Cl₂$CrO_2Cl_2$.
The image shows the oxidation of toluene to benzaldehyde via two different pathways requiring reagents A and B.
Step 1: Selection
Therefore, A = CrO₃$CrO_3$ and B = CrO₂Cl₂$CrO_2Cl_2$.
Pattern Recognition
Etard reaction always uses Chromyl chloride (CrO₂Cl₂$CrO_2Cl_2$). Oxidation of toluene with acetic anhydride uses Chromic acid (CrO₃$CrO_3$). Both stop the oxidation at the aldehyde stage via intermediate formation.
Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Ethanal (CH₃CHO$CH_3CHO$) reacts with semicarbazide (H₂N-NH-CO-NH₂$H_2N-NH-CO-NH_2$) via nucleophilic addition followed by elimination of water to form a semicarbazone.
Step 1: Product Analysis
The product is Ethanal semicarbazone: CH₃-CH=N-NH-CO-NH₂$CH_3-CH=N-NH-CO-NH_2$.
Counting the nitrogen atoms in this structure:
The imine nitrogen (=N-$=N-$)
The amine nitrogen (-NH-$-NH-$)
The amide nitrogen (-NH₂$-NH_2$)
Total = 3 Nitrogen atoms.
Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Qjee_main_2024_31_jan_eveningPreparation of Aldehydes and Ketones
Identify the name reaction.
The image shows the conversion of benzene to benzaldehyde using CO, HCl, and Anhydrous AlCl3/CuCl.
The reaction of benzene with carbon monoxide (CO$CO$) and hydrogen chloride (HCl$HCl$) in the presence of anhydrous aluminium chloride (AlCl₃$AlCl_3$) and cuprous chloride (CuCl$CuCl$) to give benzaldehyde is known as the Gatterman-Koch reaction.
The image shows the conversion of benzene to benzaldehyde using CO, HCl, and Anhydrous AlCl3/CuCl.
Pattern Recognition
CO + HCl arrow$CO + HCl \rightarrow$ Formyl chloride intermediate (in situ) with Lewis acid arrow$\rightarrow$ formylation of benzene. This is definitively the Gatterman-Koch formylation.
Chapter Mix
Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
