What is the major product of the following reaction?

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What is the major product of the following
reaction?
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Predict the major product of the following
reaction2-methylbutane Br2/light energy ? A)
1-bromo-2-methylbutane B) 2-bromo-2-methylbutane
C) 2-bromo-3-methylbutane D) 1-bromo-3-methylbut
ane
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Which of the following alkanes will give more
than one monochlorination product upon treatment
with chlorine and light? A) 2,2-dimethylpropane
B) cyclopropane C) ethane D)
2,3-dimethylbutane
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Which of the following is the major product of
the chlorination of methane if a large excess of
methane is used? A) CH3Cl B) CH2Cl2 C) CH3CH2Cl
D) CCl4
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Consider the reaction of 2-methylpropane with a
halogen. With which halogen will the product be
almost exclusively 2-halo-2-methylpropane? A)
F2 B) Cl2 C) Br2 D) I2
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The alllyl carbon is the carbon adjacent to a
double bond.
The allylic radical is more stable than a 3?
radical.
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• Because allylic CH bonds are weaker than other
  sp3 hybridized CH bonds, the allylic carbon can
  be selectively halogenated using NBS in the
  presence of light or peroxides.

• NBS contains a weak NBr bond that is
  homolytically cleaved with light to generate a
  bromine radical, initiating an allylic
  halogenation reaction.
• Propagation then consists of the usual two steps
  of radical halogenation.

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• NBS also generates a low concentration of Br2
  needed in the second chain propagation step (Step
  3 of the mechanism).
• The HBr formed in Step 2 reacts with NBS to
  form Br2, which is then used for halogenation in
  Step 3 of the mechanism.

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Thus, an alkene with allylic CH bonds undergoes
two different reactions depending on the reaction
conditions.
How is the vicinal dibromide formed?
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• Halogens add to ? bonds because halogens are
  polarizable.
• The electron rich double bond induces a dipole in
  an approaching halogen molecule, making one
  halogen atom electron deficient and the other
  electron rich (X?X?).
• The electrophilic halogen atom is then attracted
  to the nucleophilic double bond, making addition
  possible.

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• Why does a low concentration of Br2 (from NBS)
  favor allylic substitution (over ionic addition
  to form the dibromide)?
• The key to getting substitution is to have a low
  concentration of bromine (Br2).
• The Br2 produced from NBS is present in very low
  concentrations.
• A low concentration of Br2 would first react with
  the double bond to 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 intermediatesthe
  bromonium ion and Br are low (as is the case
  here), the overall rate of addition is very slow,
  and the products of the very fast and facile
  radical chain reaction predominate.

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Predict the products.
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Halogenation at an allylic carbon often results
in a mixture of products. Consider the following
example

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

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Predict the products.
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Radical Additions to Double Bonds

• HBr adds to alkenes to form alkyl bromides in the
  presence of heat, light, or peroxides.
• The regioselectivity of the addition to
  unsymmetrical alkenes is different from that in
  addition of HBr in the absence of heat, light or
  peroxides.

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

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• HydrohalogenationElectrophilic Addition of HX
• The mechanism of electrophilic addition consists
  of two successive Lewis acid-base reactions. In
  step 1, the alkene is the Lewis base that donates
  an electron pair to HBr, the Lewis acid, while
  in step 2, Br is the Lewis base that donates an
  electron pair to the carbocation, the Lewis acid.

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• With an unsymmetrical alkene, HX can add to the
  double bond to give two constitutional isomers,
  but only one is actually formed

• This is a specific example of a general trend
  called Markovnikovs rule.
• Markovnikovs rule states that in the addition of
  HX to an unsymmetrical alkene, the H atom adds to
  the less substituted carbon atomthat is, the
  carbon that has the greater number of H atoms to
  begin with.

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• Note that in the first propagation step, the
  addition of Br to the double bond, there are two
  possible paths
• Path A forms the less stable 1 radical.
• Path B forms the more stable 2 radical.
• The more stable 2 radical forms faster, so Path
  B is preferred.

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

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• HBr adds to alkenes under radical conditions, but
  HCl and HI do not. This can be explained by
  considering the energetics of the reactions using
  bond dissociation energies.
• Both propagation steps for HBr addition are
  exothermic, so propagation is exothermic
  (energetically favorable) overall.
• For addition of HCl or HI, one of the chain
  propagating steps is quite endothermic, and thus
  too difficult to be part of a repeating chain
  mechanism.

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