Concept explainers
(a)
Interpretation:
The respective reactant (methyl substituted piperidine) and respective product (diene) should be given.
Concept introduction:
Generally
These quaternary ammonium salt under goes an elimination reaction easily.
Hofmann elimination:
Quaternary ammonium ion undergoes elimination when using strong base like hydroxide ion this reaction is called as Hofmann elimination.
In the Hofmann elimination, abstraction of proton form β- carbon atom which is having more number of hydrogen to result the elimination product.
(b)
Interpretation:
The respective reactant (methyl substituted piperidine) and respective product (diene) should be given.
Concept introduction:
Hofmann elimination:
Quaternary ammonium ion undergoes elimination when using strong base like hydroxide ion this reaction is called as Hofmann elimination.
In the Hofmann elimination, abstraction of proton form β- carbon atom which is having more number of hydrogen to result the elimination product.
(c)
Interpretation:
The respective reactant (methyl substituted piperidine) and respective product (diene) should be given.
Concept introduction:
Hofmann elimination:
Quaternary ammonium ion undergoes elimination when using strong base like hydroxide ion this reaction is called as Hofmann elimination. Proton abstraction is takes place in β- carbon atom which is having more number of hydrogen.
(d)
Interpretation:
The respective reactant (methyl substituted piperidine) and respective product (diene) should be given.
Concept introduction:
Generally amines can’t undergo elimination reaction, therefore amines can be converted in to quaternary ammonium halide by treating with ethyl iodide in basic solution of potassium carbonate, and this ammonium halide is treating with silver oxide to give ammonium hydroxide salt.
These quaternary ammonium salt under goes an elimination reaction easily.
Hofmann elimination:
Quaternary ammonium ion undergoes elimination when using strong base like hydroxide ion this reaction is called as Hofmann elimination.
In the Hofmann elimination, abstraction of proton form β- carbon atom which is having more number of hydrogen to result the elimination product.
Want to see the full answer?
Check out a sample textbook solutionChapter 10 Solutions
EBK ORGANIC CHEMISTRY
- Draw the structures for penta-1,4-diene and cyclopenta-1,3-diene.arrow_forwardDraw the structure for these molecules. (a) 3-phenylmethylhex-4-en-1-yne; (b) 7-phenylcyclohepta-1,3,5-triene;(c) 3,5,6-trinitro-4-phenylmethylhepta-1,3,5-triene; (d) 5,5-dichloro-6-ethenyl-7-phenylcyclooct-3-en-1-ynearrow_forwardThe heat of combustion of decahydronaphthalene (C10H18) is -6286 kJ/mol. The heat of combustion of naphthalene 1C10H82 is -5157 kJ/mol. (In both cases CO2(g) and H2O(l) are the products). Calculate the heat of hydrogenation and the resonance energy of naphthalene.arrow_forward
- Draw the structures for each of the following molecules. (a) fluorobenzene; (b) 1-chloro-2-fluorobenzene;(c) 1-iodo-4-nitrobenzene; (d) 1,3-dibromobenzene; (e) 2,3-dimethyl-1-cyclopentylbenzene; (f) 4-ethoxy-1,2-dinitrobenzenearrow_forwardThe heat of combustion of decahydronaphthalene(C10H18) is -6286 kJ/mol. The heat of combustion ofnaphthalene (C10H8) is -5157 kJ/mol. (In both casesCO2(g) and H2O(l) are the products.) Calculate the heat of hydrogenationand the resonance energy of naphthalene.arrow_forward2. Please draw out structure of the following compounds. 1) 2,3-Methyl-1,4-pentadiene 2) (2E,4E)-2,4-heptadiene 3) (2Z,4Z)-3-Methyl-2,4-hexadienearrow_forward
- Based on the hydrogenation and the bromination reaction information, how many different alkene structures can you draw that could be Compound X? (If enantiomers are possible, count each pair of enantiomers as one structure.)arrow_forwardAlkenes can be hydrated to form alcohols by (1) hydroboration followed by oxidation with alkaline hydrogen peroxide and (2) acid-catalyzed hydration. Compare the product formed from each alkene by sequence (1) with those formed from (2). Q.)1-Methylcyclohexenearrow_forwardDraw the structure of the following compounds (E)-3,7-Dimethyloct-3-ene and Neopentylcyclohexanearrow_forward
- (a) Draw the two isomeric dienes formed when CH2 = CHCH2CH(Cl)CH(CH3)2 is treated with an alkoxide base, (b) Explain why the major product formed in this reaction does not contain the more highly substituted alkene.arrow_forward1,2,3,4,5,6-Hexachlorocyclohexane shows cis,trans isomerism. At one time, a crude mixture of these isomers was sold as an insecticide. The insecticidal properties of the mixture arise from one isomer, known as lindane, which is cis-1,2,4,5-trans- 3,6-hexachlorocyclohexane. Q.) Which of the alternative chair conformations of lindane is more stable? Explain.arrow_forward1,2,3,4,5,6-Hexachlorocyclohexane shows cis,trans isomerism. At one time, a crude mixture of these isomers was sold as an insecticide. The insecticidal properties of the mixture arise from one isomer, known as lindane, which is cis-1,2,4,5-trans- 3,6-hexachlorocyclohexane. Q.) Using a planar hexagon representation for the cyclohexane ring, draw a structural formula for lindane.arrow_forward
- Chemistry for Today: General, Organic, and Bioche...ChemistryISBN:9781305960060Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. HansenPublisher:Cengage LearningOrganic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning
- Organic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage Learning