CHM 221: Organic Chemistry II Review for Exam 1 Answers Electrophilic Addition Reactions

1. Draw the structure of the carbocation intermediate and the major reaction product for each of the following reactions. Be sure to include and pertinent stereochemistry.

Cl HCl Cl (a)

Br Br HBr (b)

I I HI (c)

Cl Cl Cl HCl HCl (d)

CH3 CH3 HBr CH3 1,2 H-shift CH3 (e) Br

20 carbocation 30 carbocation

2. Propose a mechanism to account for the following reactions. Show the structure(s) of all intermediates and use curved arrows to indicate the electron flow in each step.

Cl Cl HCl 1,2-H-shift (a)

Carbocations only rearrange to make if they become more stable. There is less ring strain by placing the carbocation center at a outside of the ring.

HCl 1,2-alkyl (b) shift Cl ring expansion Cl Expanding from a five-membered ring to a six-membered ring reduces overall ring strain. 3. Draw a reaction energy diagram for the addition of HBr to 1-pentene. Let one curve on the diagram show the formation of 1-bromopentane and another curve on the same diagram show the formation of 2- bromopentane. Label the positions for all reactants, intermediates and products. State which alkyl bromide would be the major product and why. [Notice! which curve has the higher energy carbocation intermediate and higher energy first transition state. Since these two possible reactions would be competing with each other, think about why there would be a much greater amount of one product relative to the other.]

10 carbocation intermediate

Red Line = formation of 1-bromopentane Blue Line = formation of 2-bromopentane Energy Pentene

20 carbocation

1-Bromopentane

2-Bromopentane

Reaction Coordinate

2-Bromopentane would be the major product because it has a lower activation energy and would be a faster reaction. There would essentially be no 1-bromopentane formed because once a molecule had enough energy to reach the transition state for the formation of the 20 carbocation, it would follow that path instead of continuing to a higher energy transition state which leads to a 10 carbocation.

4. The following reaction takes place in high yield. Even though you have never seen this reaction before, use your general knowledge of chemistry to propose a mechanism for the transformation. The reaction conditions contain no water.

H3CO 2C CO 2CH3 CO 2CH3 H CO 2CH3

Hg(OAc) 2

AcOHg AcOHg AcOHg

CO 2CH3

AcOHg

5. When 4-penten-1-ol is treated with aqueous bromine, a cyclic bromo ether is formed, rather than the expected bromohydrin (bromine and hydroxy groups on adjacent ). Propose a mechanism. Show the intermediate! H Br O Br O Br HO Br Br HO + HBr Br 6. Predict the products of the following reactions (also show the structure of intermediates A and B). Show stereochemistry!

Br2

HO HO Cholesterol Br Br

dilute Br 2

H2O

HO HO Br OH

HBr

HO HO Br

1. Hg(OAc) 2/H2O

2. NaBH 4 HO major product but also some HO product with OH pointed OH down

HO (Intermediate A) HO HgOAc

1. BH3

2. H2O2 major product but also some HO HO product with OH and H pointed up H OH

(Intermediate B) HO H BH2 7. When is treated with in ethanol, one of the reaction products is ethyl isopropyl ether [CH3CH2-O-CH(CH3)2]. Write a plausible mechanism that accounts for the formation of this product.

Cl H HCl HOCH2CH 3 O O + HCl

8. You know the mechanism of HBr addition to . Use this knowledge to predict which of the following two alkenes reacts faster with HBr. Explain your answer by drawing resonance structures of the carbocation intermediates.

CH3O O2N A B

Alkene A will react faster than B with HBr. Alkene A is able to make a more stable carbocation intermediate because the lone pair of electrons on the oxygen can be pushed through the π system to delocalize the positive charge. Alkene B contains an electron withdrawing group which would destabilize the benzylic carbocation.

O O

O O N N Unstable O O

9. Benzyl bromide is converted into benzaldehyde by heating in dimethyl sulfoxide. Propose a structure for the intermediate (A) and show the mechanisms of the two steps in the reaction. (Old reactions will not disappear)

Br O H S H SN Br 2 S O O + HBr + (CH3)2S

10. Suggest a plausible synthesis for each of the following molecules starting from 1-bromo-1- methylcyclohexane. You can use any reagents that we have encountered over the past two terms.

Br NaOCH 3 OH 1. BH3

HOCH 3 2. H2O2 (E2 Rxn)

O Br Cl OH O NaOCH 3 O Racemic Mixture HOCH 3 Br OCH 3 NaOCH 3 dilute Br 2 Br

HOCH 3 CH3OH

Br NaOCH 3 H2

HOCH 3 Pd-C

Br OH NaOCH 3 1. Hg(OAc) 2/ H2O

HOCH 3 2. NaBH4

Br OH NaOCH 3 OsO4 OH

HOCH 3

O Br OH Cl OH O NaOCH 3 O + OH H2O/H

HOCH 3

Br O 1. O3 NaOCH 3 H 2. Zn, HOAc HOCH 3 O

11. Briefly explain why oxymercuration of a terminal followed by treatment with aqueous acid produces a methyl ketone instead of an aldehyde.

Once the mercuric diacetate reacts with the π-bond to make the bridged mercuronium ion intermediate, the water will attack the carbon with the greatest amount of positive charge. This will be the more substituted carbon of the bridged intermediate. The resulting enol then tautomerizes into the methyl ketone. OAc Hg + Hg(OAc) 2 HgOAc H O H3C CCH H O H3C H 2 HO HO enol OH2

12. Briefly explain why the mercury does not have to be removed using NaBH4 when you react an alkyne with mercuric acetate but does when you react with an alkene.

The initial product from addition of mercuric diacetate and water to an alkyne is the vinyl mercury compound (see above mechanism) which reacts with acid to form the most stable carbocation. The most stable carbocation will be the resonance delocalized carbocation on the carbon bonded to the oxygen. The mercury then leaves in an elimination reaction to form the alkene which happens to be an enol. The enol tautomerizes to the carbonyl compound. When reacting with alkenes, the mercury has to be displaced in an SN2 type reaction using NaBH4.

13. Give the reaction conditions which will complete the following transformations.

CH3 Na/NH 3 Reduction to trans alkene

CH3 H2 Reduction to the cis alkene Pd/CaCO 3/Pb Lindlar's catalyst

CH3 Cl 1 eq Cl 2

Cl

H 1. R2BH H

2. H2O2/NaOH O