Reactions of Alkenes

Question 1

Markovnikov Addition of Nucleophilic Acid

Answer 1

1) π bond electrons attack the acidic hydrogen to protonate one double-bonded carbon.
2) Rearrangement occurs to place the positive charge on the most stable carbon.
3) Nucleophile attacks the carbocation to form the R—Nu bond at the most stable carbon.

The acidic hydrogen substitutes the least stable carbon of the double bond.

Question 2

Halogen Addition (w/o Reactive Solvent)

Answer 2

1) π bond electrons attack the partially positive halogen atom to form a cyclic halonium ion across the former double bond.
2) The negative halogen ion attacks the cyclic halonium ion (opposite of the ring plane) at the most substituted carbon to yield a di-halogenated alkane product.

Heterolytic cleavage of the halogen diatomic allows for π bond electrons to attack the positive halogen atom.

Question 3

Halogen Addition w/ Reactive Solvent

Answer 3

1) π bond electrons attack the partially positive halogen atom to form a cyclic halonium ion across the former double bond.
2) The reactive solvent ion attacks the cyclic halonium ion (opposite of the ring plane) at the most substituted carbon to yield a di-halogenated alkane product.

The absolute outnumbering of the solvent over the negative halogen atom result in the addition of the solvent (and not the halogen) to the alkane compound.

Question 4

Is Halogen Addition to an Alkene Anti-Addtion or Syn-Addition?

Answer 4

Anti-Addition

Halogen addition of an alkene in either a reactive solvent or a non-reactive solvent both result in anti-addition.

Question 5

Hydrogenation of Alkene (w/ Metal Catalyst)

Answer 5

1) Each hydrogen of the hydrogen diatomic bonds to the metal catalyst surface.
2) A hydride attacks one double-bonded carbon to create a protonated sp³ carbon while the π bond electrons attack the metal catalyst surface to form a C—Metal bond.
3) The second hydride attacks the carbon of the C—Metal bond to yield a second hydrogen-substituted sp³ carbon.

Question 6

Is Alkene Hydrogenation Syn-Addition or Anti-Addition?

Answer 6

Syn-Addition

Question 7

Specificity of Symmetrical Alkene Hydrogenation

Answer 7

Stereospecific

• Alkene hydrogenation of the Z isomer results in a meso compound.
• Alkene hydrogenation of the E isomer results in a pair of enantiomers.

Question 8

Alkene + Aqueous Non-Nucleophilic Acid

Answer 8

1) π bond electrons attack the acidic hydrogen to protonate one double-bonded carbon.
2) Rearrangement occurs to place the positive charge on the most stable carbon.
3) Water attacks the carbocation via SN1 to form the R—OH₂ bond at the most stable carbon.
4) A base deprotonates the OH₂ groups to yield an alcohol.

Question 9

Specificity of Symmetrical Alkene Halogen Addition

Answer 9

Stereospecificity

• The cis alkene will yield a racemic mixture (a pair of enantiomers).
• The trans alkene will yield a meso compound (if the solvent is non-reactive).

Question 10

Specificity of Unsymmetrical Alkene Halogen Addition

Answer 10

Regiospecificity

The X^– ion or the reactive solvent attacks the most substituted carbon of the double bond.

Question 11

3 Mechanisms for Synthesizing Alcohols from Alkenes

Answer 11

• Hydration of Alkene
• Oxymercuration-Demercuration Reaction
• Hydroboration Reaction

Question 12

Specificity of Oxymercuration-Demercuration Reaction

Answer 12

Regioselective

The nucleophile attacks the most substituted carbon of the double bond.

Question 13

Does Oxymercuration-Demercuration Cause Syn-Addtion or Anti-Addition?

Answer 13

Anti-Addition

Question 14

mCPBA Synthesis of Epoxide

Answer 14

1) π bond electrons attack the electrophilic peroxide oxygen while the O—H σ bond electrons attack one double-bonded alkene carbon.
2) The O—O σ bond electrons transfer to the O—C peroxide-adjacent bond (to form a double bond) while the mCPBA carbonyl π bond electrons attack the acidic peroxide hydrogen.
3) An epoxide (across the former π bond) and the benzoic carboxylic acid form.