Interpretation:
The reason for the failure of
Concept introduction:
Electrophiles are electron-deficient species, which has positive or partially positive charge. Lewis acids are electrophiles, which accept electron pair.
Nucleophiles are electron-rich species, which has negative or partially negative charge. Lewis bases are nucleophiles which donate electron pair.
Substitution reaction: A reaction in which one of the hydrogen atoms of a hydrocarbon or a
Elimination reaction: A reaction in which two substituent groups are detached and a double bond is formed is called elimination reaction.
Addition reaction: It is the reaction in which unsaturated bonds are converted to saturated molecules by the addition of molecules.
Dehydration of alcohols includes the first step as the protonation of alcohol and formation of carbocation.
The order of stability of carbocation is such that the tertiary carbocation is most stable, whereas the primary carbocation is least stable and secondary carbocation lies between primary and secondary carbocation.
For naming bicyclo-compounds:
Compounds containing two fixed or bridged rings are known as bicycloalkanes.
Then number of carbon atoms in each bridge is written in square brackets in decreasing order.
For numbering substituent groups, start numbering from the longest chain, then the moderate chain, and at last, the shortest chain.
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Organic Chemistry
- Provide an IUPAC name for each of the compounds shown. (Specify (E)/(Z) stereochemistry, if relevant, for straight chain alkenes only. Pay attention to commas, dashes, etc.)arrow_forwardBromine reacts with alkenes in methanol according to the equation (see image 1). When this reaction was carried out with 4-tert-butylcyclohexene, only one isomer was formed with the molecular formula C12H23BrO (80% yield) a) Which of the following is the structure more reasonable for this compound? (see image 2) b) Explain your reasoning through a corresponding mechanismarrow_forwardBi and ii I need help witharrow_forward
- A)I,ii and iii pleasearrow_forward1) Show how you could synthesize (Z)-5-octene-2-ol using precursors that contain fewer than five carbon atoms (<5). You may use starting materials containing more than five carbon atoms if no more than four of the carbon atoms from that precursor remain in the final product. Note: (Z)-allylic bromides and chlorides are not stable and spontaneously rearrange to the (E) isomers.arrow_forwardCompound X, C,4H12Br2, is optically inactive. On treatment with strong base, X gives hydrocarbon Y, C14H10: Compound Y absorbs 2 equivalents of hydrogen when reduced over a palladium catalyst to give z (C14H14) and reacts with ozone to give one product, benzoic acid (C,Hg02). Draw the structure of compound Z. • Use the wedge/hash bond tools to indicate stereochemistry where it exists. • Ignore alkene stereochemistry. • If more than one structure fits the description, draw them all. • Draw one structure per sketcher. Add additional sketchers using the drop-down menu in the bottom right corner. • Separate structures with + signs from the drop-down menu. ChemDoodlearrow_forward
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- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning