What is the major product of the following reaction?

Predict the major product of the following reaction:2-methylbutane + Br2/light energy ?

A) 1-bromo-2-methylbutane

B) 2-bromo-2-methylbutane

C) 2-bromo-3-methylbutane

D) 1-bromo-3-methylbutane

H3CH2CCH3CH3Br

Which of the following alkanes will give more than onemonochlorination product upon treatment with chlorine and light?A) 2,2-dimethylpropane

B) cyclopropane

C) ethane

D) 2,3-dimethylbutane

H3CCH3CH3CH3Cl2hvH3CCH3CH3CH3Cl+H3CCH2ClCH3CH3

Which of the following is the major product of the chlorination ofmethane if a large excess of methane is used?

A) CH3Cl

B) CH2Cl2

C) CH3CH2Cl

D) CCl4

Consider the reaction of 2-methylpropane with a halogen. Withwhich halogen will the product be almost exclusively 2-halo-2-methylpropane?

A) F2

B) Cl2

C) Br2

D) I2H3CCH3CH3Br2hvH3CCH3CH3BrCl2hvH3CCH3CH3Cl+ClH2CCH3CH3

The alllyl carbon is the carbon adjacent to a double bond.

The allylic radical is more stable than a 3° radical.

•Because allylic C—H bonds are weaker than other sp3 hybridizedC—H bonds, the allylic carbon can be selectively halogenatedusing NBS in the presence of light or peroxides.

•NBS contains a weak N—Br bond that is homolytically cleavedwith light to generate a bromine radical, initiating an allylichalogenation reaction.

•Propagation then consists of the usual two steps of radicalhalogenation.

•NBS also generates a low concentration of Br2 neededin the second chain propagation step (Step [3] of themechanism).

•The HBr formed in Step [2] reacts with NBS to form Br2,which is then used for halogenation in Step [3] of themechanism.

Thus, an alkene with allylic C—H bonds undergoes twodifferent reactions depending on the reaction conditions.

How is the vicinal dibromide formed?

•Halogens add to π bonds because halogens are polarizable.

•The electron rich double bond induces a dipole in an approachinghalogen molecule, making one halogen atom electron deficient andthe other electron rich (Xδ+—Xδ–).•The electrophilic halogen atom is then attracted to the nucleophilicdouble bond, making addition possible.

Why does a low concentration of Br2 (from NBS) favor allylicsubstitution (over ionic addition to form the dibromide)?

•The key to getting substitution is to have a low concentration ofbromine (Br2).

•The Br2 produced from NBS is present in very lowconcentrations.

•A low concentration of Br2 would first react with the double bondto form a low concentration of the bridged bromonium ion.

•The bridged bromonium ion must then react with more bromine(in the form of Br¯) in a second step to form the dibromide.

•If concentrations of both intermediates—the bromonium ion andBr¯ are low (as is the case here), the overall rate of addition is veryslow, and the products of the very fast and facile radical chainreaction predominate.

NBShv

Predict the products.Br

H2CCHCH3Br2 H2CHCCH3BrBr

NBShv

H2CCHCH2Br

Halogenation at an allylic carbon often results in a mixture ofproducts. Consider the following example:

•A mixture results because the reaction proceeds by way of aresonance stabilized radical.

H3CCHCHCH3NBShvBrH2CCHCHCH3+

H2CCHHCCH3Br

H2CCH2CCH3H2CCH3NBShv

H2CCCHCH3H2CCH3Br+

BrH2CCHCCH3H2CCH3

Predict the products.

CH2NBShv CH2Br+ CH2Br

Radical Additions to Double Bonds

•HBr adds to alkenes to form alkyl bromides in the presence ofheat, light, or peroxides.

•The regioselectivity of the addition to unsymmetrical alkenes isdifferent from that in addition of HBr in the absence of heat, lightor peroxides.

•The addition of HBr to alkenes in the presence of heat, light orperoxides proceeds via a radical mechanism.

Hydrohalogenation—Electrophilic Addition of HX•The mechanism of electrophilic addition consists of two successiveLewis acid-base reactions. In step 1, the alkene is the Lewis basethat donates an electron pair to H—Br, the Lewis acid, while in step2, Br¯ is the Lewis base that donates an electron pair to thecarbocation, the Lewis acid.

•With an unsymmetrical alkene, HX can add to the doublebond to give two constitutional isomers, but only one isactually formed:

•This is a specific example of a general trend calledMarkovnikov’s rule.

•Markovnikov’s rule states that in the addition of HX to anunsymmetrical alkene, the H atom adds to the lesssubstituted carbon atom—that is, the carbon that has thegreater number of H atoms to begin with.

•Note that in the first propagation step, the addition of Br•to the double bond, there are two possible paths:

1.Path [A] forms the less stable 1° radical.

2.Path [B] forms the more stable 2° radical.

•The more stable 2° radical forms faster, so Path [B] ispreferred.

•The radical mechanism illustrates why the regio-selectivity of HBr addition is different depending on thereaction conditions.

•HBr adds to alkenes under radical conditions, but HCl and HI donot. This can be explained by considering the energetics of thereactions using bond dissociation energies.

•Both propagation steps for HBr addition are exothermic, sopropagation is exothermic (energetically favorable) overall.

•For addition of HCl or HI, one of the chain propagating steps isquite endothermic, and thus too difficult to be part of a repeatingchain mechanism.

H3CCHCH2HBr H3CHCCH3Br

H3CCHCHBrROORCH3CH3 H3CHCCHCH3CH3Br

HClhvCl