22 public questions tagged with this topic.
Sodium borohydride (NaBH₄) is a mild reducing agent that selectively reduces aldehydes to primary alcohols. This follows from NCERT principle where relation explains outcome clearly for students.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
Reduction of propanal (CH₃CH₂CHO) with NaBH₄ yields propan-1-ol (CH₃CH₂CH₂OH).
Ref: NCERT Class 12 Chemistry > Chapter 12: Aldehydes, Ketones and Carboxylic Acids > Topic: Physical Properties and Nucleophilic Addition Reactions
Reduction of glucose (an aldohexose) with hydrogen converts the aldehyde group to an alcohol, forming sorbitol.
Ref: NCERT Class 12 Chemistry > Chapter 14: Biomolecules > Topic: Carbohydrates - Monosaccharides - Structure and Properties
Butan-2-one (CH₃-CO-CH₂-CH₃) is reduced to butan-2-ol (CH₃-CH(OH)-CH₂-CH₃) using a reducing agent like NaBH₄, which selectively reduces ketones to secondary alcohols.
Ref: NCERT Class 12 Chemistry > Chapter 7: Alcohols, Phenols and Ethers > Topic: Preparation of Phenols and Ethers - Williamson and Other Methods
Acetophenone (C₆H₅COCH₃) is reduced to 1-phenylethanol (C₆H₅CH(OH)CH₃) using NaBH₄, a selective reducing agent for ketones.
Ref: NCERT Class 12 Chemistry > Chapter 7: Alcohols, Phenols and Ethers > Topic: Preparation of Alcohols - from Alkenes Haloalkanes Carbonyls Grignard
Phenol (C₆H₅OH) is reduced to benzene (C₆H₆) when heated with zinc dust, removing the -OH group as ZnO and H₂.
Ref: NCERT Class 12 Chemistry > Chapter 7: Alcohols, Phenols and Ethers > Topic: Chemical Reactions of Phenols and Ethers and Uses
KMnO 4 (Mn in +7 state) readily accepts electrons, making it a strong oxidant.
Ref: NCERT Class 12 Chemistry > Chapter 4: d and f Block Elements > Topic: Preparation and Properties of Potassium Dichromate and Permanganate
Cu²⁺ + e⁻ → Cu⁺ . 1 mole requires 1F.
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Corrosion and Applications of Electrochemistry
Reaction: 2H₂ + O₂ → 2H₂O . O₂ is reduced, so it’s the oxidizing agent.
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Batteries - Primary Secondary and Fuel Cells
Reaction: 2H₂ + O₂ → 2H₂O . O₂ is reduced, so it’s the oxidizing agent.
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Batteries - Primary Secondary and Fuel Cells
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O . 6 electrons per mole.
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Batteries - Primary Secondary and Fuel Cells
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O . 6 electrons per mole.
Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Batteries - Primary Secondary and Fuel Cells
