13 public questions tagged with this topic.
2,3-Dimethylbut-2-ene ((CH₃)₂C\=C(CH₃)₂) cleaves symmetrically to form two molecules of acetone (CH₃COCH₃), a ketone.
Ref: NCERT Class 12 Chemistry > Chapter 12: Aldehydes, Ketones and Carboxylic Acids > Topic: Reactions due to α-Hydrogen - Aldol, Cannizzaro and Acidity of Carboxylic Acids
4-Methylpent-1-ene (CH₂\=CHCH₂CH(CH₃)₂) cleaves at the double bond, forming methanal (HCHO) and 3-methylbutanal ((CH₃)₂CHCH₂CHO).
Ref: NCERT Class 12 Chemistry > Chapter 12: Aldehydes, Ketones and Carboxylic Acids > Topic: Reactions due to α-Hydrogen - Aldol, Cannizzaro and Acidity of Carboxylic Acids
The compound has a 4-carbon chain with an aldehyde group (butanal) and a double bond between C-3 and C-4, named but-3-enal.
Ref: NCERT Class 12 Chemistry > Chapter 12: Aldehydes, Ketones and Carboxylic Acids > Topic: Chemical Reactions - Reduction, Oxidation and Tests - Tollens, Fehling, Iodoform
The compound has a 3-carbon chain with an aldehyde group (prop-2-enal) and a phenyl group on C-3: 3-phenylprop-2-enal.
Ref: NCERT Class 12 Chemistry > Chapter 12: Aldehydes, Ketones and Carboxylic Acids > Topic: Chemical Reactions - Reduction, Oxidation and Tests - Tollens, Fehling, Iodoform
2-Methylbutan-2-ol (CH₃C(OH)(CH₃)CH₂CH₃) dehydrates to 2-methylbut-2-ene (CH₃C(CH₃)=CHCH₃) as the major product, following Saytzeff’s rule.
Ref: NCERT Class 12 Chemistry > Chapter 7: Alcohols, Phenols and Ethers > Topic: Chemical Reactions of Alcohols - Dehydration Oxidation Esterification
Dehydration of butan-2-ol gives a mixture of alkenes, with but-2-ene as the major product according to Saytzeff’s rule.
Ref: NCERT Class 12 Chemistry > Chapter 7: Alcohols, Phenols and Ethers > Topic: Chemical Reactions of Alcohols - Dehydration Oxidation Esterification
Dehydration of pentan-2-ol gives a mixture of alkenes, with pent-2-ene as the major product according to Saytzeff’s rule.
Ref: NCERT Class 12 Chemistry > Chapter 7: Alcohols, Phenols and Ethers > Topic: Chemical Reactions of Alcohols - Dehydration Oxidation Esterification
Dehydrohalogenation of 2-bromopropane with alcoholic KOH follows Zaitsev's rule, forming the more substituted alkene, CH₃CH=CH₂ (propene).
Ref: NCERT Class 12 Chemistry > Chapter 6: Haloalkanes and Haloarenes > Topic: Polyhalogen Compounds - Chloroform Iodoform DDT Freons
Elimination (E₂) with alcoholic KOH favors tertiary halides due to stable alkene formation. (CH₃)3CBr reacts fastest.
Ref: NCERT Class 12 Chemistry > Chapter 6: Haloalkanes and Haloarenes > Topic: Polyhalogen Compounds - Chloroform Iodoform DDT Freons
A vinylic halide has the halogen on a carbon of a C=C bond. CH₃CH=CHBr (1-bromopropene) is vinylic.
Ref: NCERT Class 12 Chemistry > Chapter 6: Haloalkanes and Haloarenes > Topic: Nucleophilic Substitution Reactions - SN1 and SN2 Mechanisms
With peroxide, HBr adds to 2-butene anti-Markovnikov, forming CH₃CH₂CH₂CH₂Br (1-bromobutane).
Ref: NCERT Class 12 Chemistry > Chapter 6: Haloalkanes and Haloarenes > Topic: Nucleophilic Substitution Reactions - SN1 and SN2 Mechanisms
Alkenes react with hydrogen halides (e.g., HBr ) via electrophilic addition to form alkyl halides. CH₃CH₂CH₂Br can be made from CH₃CH=CH₂ with HBr (peroxide conditions).
Ref: NCERT Class 12 Chemistry > Chapter 6: Haloalkanes and Haloarenes > Topic: Classification and Nomenclature and Isomerism of Haloalkanes
