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Predict the products of the following reaction
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- When HBr adds to (CH3)2C = CH2 under radical conditions, two radicals are possible products in the first step of chain propagation. Draw the structure of both radicals and indicate which one is formed. Then draw the preferred product from HBr addition under radical conditions.Compounds containing a phenol group may work as ANTIOXIDANTS to prevent free radical damage. This is accomplished when a free radical (or UV light) encounters a phenol group, turning the phenol group into a radical. However, contrary to typical radical behavior, the structure of the phenol radical can neutralize (or quench) the unpaired electron. Specifically, the phenol structure neutralizes (or quenches) the unpaired radical electron by doing the following: taking the electron and The correct name (or abbreviation) of an example compound (discussed in the lecture videos) containing a phenol group with antioxidant properties is:In free-radical substitution reaction of alkanes with halogens under uv light, the photolytic breaking of the halogen is the rate determining step. the formation of alkylradical is the rate determining step. the formation of halogen radical is the rate determining step. the abstraction of hydrogen from alkane by the halogen radical is the rate determining step
- In nucleophilic aromatic substitution reactions that proceed by the nucleophilic addition-elimination mechanism, the reaction rate increases as the electronegativity of the halogen leaving group increases: Ar-I < Ar-Br < Ar-CI < Ar-F. Which step does this suggest is the rate-determining step of the mechanism-the addition step or the elimination step? Explain.A benzene ring alters the reactivity of a neighboring group in the so-called “benzylic” position, similarly to how a double bond alters the reactivity of groups in the “allylic” position. Benzylic cations, anions, and radicals are all more stable than simple alkyl intermediates. a) Use resonance structures to show the delocalization of the positive charge, negative charge, and unpaired electron of the benzyl cation, anion, and radical.The SN 1 mechanism starts with the rate-determining step which is the dissociation of the alkyl halide into a carbocation and a halide ion. The next step is the rapid reaction of the carbocation intermediate with the nucleophile; this step completes the nucleophilic substitution stage. The step that follows the nucleophilic substitution is a fast acid-base reaction. The nucleophile now acts as a base to remove the proton from the oxonium ion from the previous step, to give the observed product. Draw a curved arrow mechanism for the reaction, adding steps as necessary. Be sure to include all nonzero formal charges. Cl: Add/Remove step G Click and drag to start drawing a structure.
- The SN 1 mechanism starts with the rate-determining step which is the dissociation of the alkyl halide into a carbocation and a halide ion. The next step is the rapid reaction of the carbocation intermediate with the nucleophile; this step completes the nucleophilic substitution stage. The step that follows the nucleophilic substitution is a fast acid-base reaction. The nucleophile now acts as a base to remove the proton from the oxonium ion from the previous step, to give the observed product. Draw a curved arrow mechanism for the reaction, adding steps as necessary. Be sure to include all nonzero formal charges. Br. Add/Remove step ك Click and drag to start drawing a structure.Fill both dashes in the paragraph Compounds containing a phenol group may work as ANTIOXIDANTS to prevent free radical damage. This is accomplished when a free radical (or UV light) encounters a phenol group, turning the phenol group into a radical. However, contrary to typical radical behavior, the structure of the phenol radical can neutralize (or quench) the unpaired electron. Specifically, the phenol structure neutralizes (or quenches) the unpaired radical electron by doing the following: taking the electron and ---------. The correct name (or abbreviation) of an example compound (discussed in the lecture videos) containing a phenol group with antioxidant properties is: ---------.CH3 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:
- The rate law for addition of Br2 to an alkenes is first order in Br2 and first order in the alkene. Does this information suggest that the mechanism of addition of Br2 to an alkene proceeds in the same matter as for addition of HBr? Explain.3. Please draw the most important (stable) resonance form of the carbonation that mediates the following reaction: OH H3O+ O 4. Draw the structure of the alkene isomer (with the molecular formula of C5H10) that is most reactive one in the addition reaction with Br₂.The rate law for addition of Br2Br2 to an alkene is first order in Br2Br2 and first order in the alkene. Does this information suggest that the mechanism of addition of Br2Br2 to an alkene proceeds in the same manner as for addition of HBr? Explain.