Concept explainers
Videos
Name the following
a)
Interpretation:
The alkyne shown is to be named and the products formed when it reacts with 1) H2 in the presence of Lindlar catalyst and 2) H3O+ in the presence of HgSO4 is to be predicted.
Concept introduction:
The longest carbon chain which contains the carbon-carbon triple bond is chosen. The chain is numbered from the end that gives the lowest number to the carbon in triple bond. Compounds with more than one triple bond are called diynes, triynes and so forth. The substituents present, if any are written in the alphabetical order.
When reduced with Hydrogen in the presence of Lindlar catalyst the reduction of alkynes stops in the alkene stage. When treated with H3O+ in the presence of HgSO4, alkynes undergo hydration following Markovnikov regiochemistry to give an enols which will tautomerize to yield aldehydes (terminal alkynes) or ketones (internal alkynes).
To give:
The name of the alkyne shown and to predict the products formed when it reacts with 1) H2 in the presence of Lindlar catalyst and 2) H3O+ in the presence of HgSO4.
Answer to Problem 14VC
The name of the alkyne shown is 4,4-dimethyl-1-hexyne.
The product formed when it reacts with H2 in the presence of Lindlar catalyst is 4,4-dimethyl-1-hexene.
The product formed when it reacts with H3O+ in the presence of HgSO4 is 4,4-dimethyl-2-hexanone.
Explanation of Solution
The compound has a six carbon straight chain with two methyl groups on C4 with a triple bond between C1 & C2. Hence it’s name is 4,4-dimethyl-1-hexyne.
When reduced with hydrogen in the presence of Lindlar catalyst, the triple bond becomes a double bond as each of the two carbons in the triple bond gets attached to a hydrogen and an alkene, 4,4-dimethyl-1-hexene,is thus produced.
When treated with H3O+ in the presence of HgSO4, the addition of water takes place in the triple bond following Markovnikov regiochemistry. The OH group adds to more highly substituted carbon and H adds to the less highly substituted carbon in triple bond resulting in the formation of an enol which undergoes tautomerization to yield the ketone, 4,4-dimethyl-2-hexanone.
The name of the alkyne shown is 4,4-dimethyl-1-hexyne.
The product formed when it reacts with H2 in the presence of Lindlar catalyst is 4,4-dimethyl-1-hexene.
The product formed when it reacts with H3O+ in the presence of HgSO4 is 4,4-dimethyl-2-hexanone.
b)
Interpretation:
The alkyne shown is to be named and the products formed when it reacts with 1) H2 in the presence of Lindlar catalyst and 2) H3O+ in the presence of HgSO4 is to be predicted.
Concept introduction:
The longest carbon chain which contains the carbon-carbon triple bond is chosen. The chain is numbered from the end that gives the lowest number to the carbon in triple bond. Compounds with more than one triple bond are called diynes, triynes and so forth. The substituents present, if any are written in the alphabetical order.
When reduced with Hydrogen in the presence of Lindlar catalyst the reduction of alkynes stops in the alkene stage. When treated with H3O+ in the presence of HgSO4, alkynes undergo hydration following Markovnikov regiochemistry to give an enols which will tautomerize to yield aldehydes (terminal alkynes) or ketones (internal alkynes).
To give:
The name of the alkyne shown and to predict the products formed when it reacts with 1) H2 in the presence of Lindlar catalyst and 2) H3O+ in the presence of HgSO4.
Answer to Problem 14VC
The name of the alkyne shown is 2,7-dimethyl-4-octyne.
The product formed when it reacts with H2 in the presence of Lindlar catalyst is cis-2,7-dimethyl-4-octene.
The product formed when it reacts with H3O+ in the presence of HgSO4 is 2,7-dimethyl-4-octanone.
Explanation of Solution
The compound has an eight carbon straight chain with two methyl groups on C2 & C7 with a triple bond between C4 & C5. Hence it’s name is 2,7-dimethyl-4-octyne.
When reduced with hydrogen in the presence of Lindlar catalyst, the triple bond becomes a double bond as each of the two carbons in the triple bond gets attached to a hydrogen and an alkene, cis- 2,7-dimethyl-4-octene.is thus produced.
When treated with H3O+ in the presence of HgSO4, the addition of water takes place in the triple bond. As the alkene is symmetrical the Markovnikov regiochemistry cannot be applied. The OH group adds to one carbon and H adds to the other carbon in triple bond resulting in the formation of an enol which undergoes tautomerization to yield the ketone, 2,7-dimethyl-4-octanone.
The name of the alkyne shown is 2,7-dimethyl-4-octyne.
The product formed when it reacts with H2 in the presence of Lindlar catalyst is cis-2,7-dimethyl-4-octene.
The product formed when it reacts with H3O+ in the presence of HgSO4 is 2,7-dimethyl-4-octanone.
Want to see more full solutions like this?
Chapter 9 Solutions
OWLv2 with Student Solutions Manual eBook, 4 terms (24 months) Printed Access Card for McMurry's Organic Chemistry, 9th
- Explain the importance of having a sampling plan with respect to food analysis. Explain the importance of having a sampling plan with respect to food analysis. Provide examples.arrow_forwardPlease predict the products for each of the following reactions. Clearly show the regiochemistry (Markovnikov vs anti-Markovnikov) and stereochemistry (syn- vs anti- or both). If a mixture of enantiomers is formed, please draw all the enantiomers. cold KMnO4, NaOH 2. DMS 1. 03 CH3OH Br2 1. 03 2. (CH3)2S H₂ Pd or Pt (catalyst) HBr 18 19 20 1 HBr ROOR (peroxide) H₂O H₂SO4 HCI HI 17 16 6 15 MCPBA 1. BH3 THF 2. H₂O2, NaOH 1. OsO4 2. H₂O₂ 110 CH3CO₂H (peroxyacid) 1. MCPBA 2. H₂O* Br2 H₂O BH3 THF B12 EtOH Pd or Ni (catalyst) D₂ (deuterium) Bra A B C D H OH H OH OH H OH α α α OH H OH OH фон d H "Harrow_forwardBriefly indicate the models that describe the structure of the interface: Helmholtz-Perrin, Gouy-Chapman, Stern and Grahame models.arrow_forward
- Using Benzene as starting materid show how each of the Following molecules Contel Ve syntheswed CHI 9. b -50311 с CHY 503H Ночто d. อ •NOV e 11-0-650 NO2arrow_forwardThe molecule PYRIDINE, 6th electrons and is therefore aromatre and is Assigned the Following structure contering Since aromatk moleculoy undergo electrophilic anomatic substitution, Pyridine shodd undergo The Following reaction + HNO3 12504 a. write all of the possible Mononitration Products that could Result From this reaction 18. Bared upon the reaction mechanison determime which of these producty would be the major Product of the hegetionarrow_forwarda. Explain Why electron withdrawing groups tend to be meta-Directors. Your answer Should lyclude all apropriate. Resonance contributing Structures fo. Explain why -ll is an outho -tura drccton even though chlorine has a very High Electronegativityarrow_forward
Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning