1. Consider the radical halogenation reaction shown below to answer parts a-c. Note that "Ph" refers to a benzene ring. BDE (kJ/mol) a. Draw the major monobrominated organic product for the reaction. Initiation Br₂, heat Propagation Step 1. Major Organic + HBr Product Step 2. b. Draw an arrow-pushing mechanism that shows the reaction pathway leading to the formation of the major organic product. If multiple stereoisomers are formed, only draw the mechanism for one of them. Bond CH3CH₂-H (CH3)3C-H PhCH₂-H CH3CH₂-Br (H3C)3C-Br PhCH₂-Br Termination (Draw one termination step.) H-Br Br-Br 423 404 378 303 304 264 368 190
Basics in Organic Reactions Mechanisms
In organic chemistry, the mechanism of an organic reaction is defined as a complete step-by-step explanation of how a reaction of organic compounds happens. A completely detailed mechanism would relate the first structure of the reactants with the last structure of the products and would represent changes in structure and energy all through the reaction step.
Heterolytic Bond Breaking
Heterolytic bond breaking is also known as heterolysis or heterolytic fission or ionic fission. It is defined as breaking of a covalent bond between two different atoms in which one atom gains both of the shared pair of electrons. The atom that gains both electrons is more electronegative than the other atom in covalent bond. The energy needed for heterolytic fission is called as heterolytic bond dissociation energy.
Polar Aprotic Solvent
Solvents that are chemically polar in nature and are not capable of hydrogen bonding (implying that a hydrogen atom directly linked with an electronegative atom is not found) are referred to as polar aprotic solvents. Some commonly used polar aprotic solvents are acetone, DMF, acetonitrile, DMSO, etc.
Oxygen Nucleophiles
Oxygen being an electron rich species with a lone pair electron, can act as a good nucleophile. Typically, oxygen nucleophiles can be found in these compounds- water, hydroxides and alcohols.
Carbon Nucleophiles
We are aware that carbon belongs to group IV and hence does not possess any lone pair of electrons. Implying that neutral carbon is not a nucleophile then how is carbon going to be nucleophilic? The answer to this is that when a carbon atom is attached to a metal (can be seen in the case of organometallic compounds), the metal atom develops a partial positive charge and carbon develops a partial negative charge, hence making carbon nucleophilic.
Need help with step g
![1. Consider the radical halogenation reaction shown below to answer parts a-c. Note that "Ph"
refers to a benzene ring.
BDE (kJ/mol)
a.
Draw the major monobrominated organic product for
the reaction.
Initiation
Br₂, heat
Propagation
Step 1.
Major
Organic + HBr
Product
Step 2.
Bond
CH3CH₂-H
(CH3)3C-H
PhCH₂-H
b. Draw an arrow-pushing mechanism that shows the reaction pathway leading to the
formation of the major organic product. If multiple stereoisomers are formed, only draw
the mechanism for one of them.
Termination (Draw one termination step.)
CH3CH₂-Br
(H3C)3C-Br
PhCH₂-Br
H-Br
Br-Br
423
404
378
303
304
264
368
190](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5cd73367-8561-4729-9c21-fa26176abb9c%2F2b9ba6a9-253b-4009-b60a-95bde90b5d85%2F256vunn_processed.jpeg&w=3840&q=75)
![c. Calculate AH° for the first propagation step for this reaction. Show your work and
include units.
d. Calculate AH° for the second propagation step for this reaction. Show your work and
include units.
e. Draw a reaction coordinate diagram for the two-step propagation cycle for the reaction.
Include the two AH values that you calculated previously in the diagram. Label the rate-
determining step. Label each step as exothermic or endothermic.
f. Draw the transition state for the rate-determining step. Does the transition state look most
like the alkane starting material, the radical intermediate, or halogenated product?
Explain your answer. (Hint: Hammond Postulate.)
g. Bromination is 80× more selective for the secondary position than the primary position.
Using this information, complete the chart and determine what percentage of the
experimental product yield will have substitution at each position. Only consider
monobromination products.
Primary
Secondary
Statistical
Experimental Yield
(%)
Selectivity](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5cd73367-8561-4729-9c21-fa26176abb9c%2F2b9ba6a9-253b-4009-b60a-95bde90b5d85%2F2kuy6q_processed.jpeg&w=3840&q=75)
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