Interpretation:
The radical mechanism for the given process is to be written.
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.
Radical is an atom or molecule that has an unpaired valence electron. These unpaired electrons make radical highly chemically reactive.
Substitution reaction: A reaction in which one of the hydrogen atoms of a hydrocarbon or a functional group is substituted by any other functional group is called substitution reaction.
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.
Reduction is a process in which hydrogen atoms are added to a compound.
Homolytic fission is that fission in which each atom in the bond has an electron which results in species called free radical.
In heterolytic fission, when covalent bond is broken, the shared pair of electron is taken by one of the atoms.
A type of halogenation in which
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Organic Chemistry
- The chlorination of ethane produces many by-products. One by-product that forms in small amounts is 2-chlorobutane. Suggest a mechanism for the formation of 2-chlorobutane from the radical chlorination of ethane. (NOTES: intermediates can only have one radical, 2-chlorobutane should be the product of your termination step, any intermediate radical formed during the propagation phase can be used in the termination step)arrow_forwardDo not give handwriting solution.arrow_forwardAlcohols are acidic in nature. Therefore, a strong base can abstract the acidic hydrogen atom of the alcohol in a process known as deprotonation. The alcohol forms an alkoxide ion by losing the proton attached to the oxygen atom of the hydroxyl ( -OH) group. The alkoxide formed can act as a base or a nucleophile depending on the substrate and reaction conditions. However, not all bases can abstract the acidic proton of alcohols and not all alcohols easily lose the proton. Deprotonation depends on the strength of the base and the acidity of the alcohol. Strong bases, such as NaNH2, can easily abstract a proton from almost all alcohols. Likewise, more acidic alcohols lose a proton more easily. Determine which of the following reactions would undergo deprotonation based on the strength of the base and the acidity of the alcohol. Check all that apply. ► View Available Hint(s) CH3CH,OH + NH3 →CH,CH,O-NH CH3 CH3 H3C-C-H+NH3 → H3 C-C-H OH O-NH CH3CH2OH + NaNH, → CH3CH,O-Na* + NH3 CHC12 Cl₂…arrow_forward
- There is great excess of H2O in this reaction.arrow_forwardmethanol + CH3OH Suppose you were told that the above reaction was a substitution reaction but you were not told the mechanism. Evaluate the following categories to determine the reaction mechanism and then draw the structure of the major organic product. Type of alkyl halide: Type of nucleophile: Solvent: Is the product racemic?arrow_forwardCH3 CH3 Br- Br2 .CH3 CH2Cl2 CH3 H3C H3C Br Electrophilic addition of bromine, Brɔ, to alkenes yields a 1,2-dibromoalkane. The reaction proceeds through a cyclic intermediate known as a bromonium ion. The reaction occurs in an anhydrous solvent such as CH,Cl,. In the second step of the reaction, bromide is the nucleophile and attacks at one of the carbons of the bromonium ion to yield the product. Due to steric clashes, the bromide ion always attacks the carbon from the opposite face of the bromonium ion so that a product with anti stereochemistry is formed. Draw curved arrows to show the movement of electrons in this step of the mechanism. Arrow-pushing Instructions CH3 CH3 CH3 CH3 H3C H3C :Br: :Br:arrow_forward
- Alkyl halides can be reduced to alkanes by a radical reaction with tributyltin hydride, (C4H9)3SnH, in the presence of light (hv). Propose a radical chain mechanism by which the reaction might occur. The initiation step is the light-induced homolytic cleavage of the Sn-H bond to yield a tributyltin radical.arrow_forwardd) During the reaction between 3, 3-dimethylbutene (W) and HBr, intermediate species and Y are formed. CH3 CH;-C-CH=CH2 ČH3 HBr Y step 1 step 2 step 3 i) Write a mechanism for the formation of intermediate X? ii) Write the structure of intermediate Y? iii) Explain in your own words what happens in step 2 of the reaction? iv) Write a mechanism for the formation of Z from Y?arrow_forwardBr Brz CH3 CH3 H3C CH2CI2 H3C Br Electrophilic addition of bromine, Br2; to alkenes yields a 1,2-dibromoalkane. The reaction proceeds through a cyclic intermediate known as a bromonium ion. The reaction occurs in an anhydrous solvent such as CH,Cl). In the second step of the reaction, bromide is the nucleophile and attacks at one of the carbons of the bromonium ion to yield the product. Due to steric clashes, the bromide ion always attacks the carbon from the opposite face of the bromonium ion so that a product with anti stereochemistry is formed. Draw curved arrows to show the movement of electrons in this step of the mechanism. Arrow-pushing Instructions Br: :Br: .CH3 H3C H3C CH3 Br:arrow_forward
- Write a mechanism that accounts for the formation of ethyl isopropyl ether as one of the products in the following reaction. CI OEt HCI EtOH Write the mechanism for step one of this reaction. Show lone pairs and formal charges. Only the acidic hydrogen should be drawn out with a covalent bond. Write the mechanism for step two of this reaction (where the product of step one reacts with the solvent, ethanol). Show lone pairs and formal charges. Only the acidic hydrogen should be drawn out with a covalent bond. Write the mechanism for the last step of this reaction (formation of ethyl isopropyl ether). Show lone pairs and formal charges. Only the acidic hydrogen should be drawn out with a covalent bond. CI will act as the base in this reaction.arrow_forwardWhen the alkyl bromides (listed here) were subjected to hydrolysis in a mixture of ethanol and water (80% EtOH/20% H2O) at 55 °C, the rates of the reaction showed the following order: (CH3)3CBR > CH3Br > CH3CH2Br > (CH3)2CHBR Provide an explanation for this order of reactivity.arrow_forwardAlkynes do not react directly with aqueous acid as do alkenes, but will do so in the presence of mercury(II) sulfate as a Lewis acid catalyst. The reaction occurs with Markovnikov regiochemistry, so the OH group adds to the more highly substituted carbon and the H adds to the less highly substituted carbon. The initial product of the reaction is a vinyl alcohol, also called an enol. The enol immediately rearranges to a more stable ketone via tautomerization. Draw curved arrows to show the movement of electrons in this step of the mechanism. Arrow-pushing Instructions -X티 Hö: H-O -CH3 -CH3 H30*arrow_forward
- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning