Cyclohexene adds to bromine to yield 1,2-dibromocyclohexane. Anti-addition occurs to give actually the racemic mixture of trans-1,2-dibromocyclohexane.
Bromine adds to cyclohexene through the steps as follows:
(Contd.)
In the first step, a bromine molecule transfers a bromine atom to cyclohexene to form a bromonium ion. In the second step, a bromide ion attacks the back side of C-2 (an SN2 process) in form I or II of the bromonium ion to produce the diaxial conformer of trans -1,2- dibromocyclohexane. This is rapidly converted into the more stable diequatorial form and when equilibrium is reached, the diequatorial form predominates. Nucleophilic attack at C-1 in I or II may also result in the formation of the diequatorial form, but the reaction does not proceed through this energetically unfavourable pathway involving a high energy twist boat conformation. The stereochemical outcome of the reaction is racemization because the two enantiomers of trans-1,2-dibromocyclohexane are formed in equimolar amounts.
1. What is called a stereospecifi c reaction? Give an example.
Solution A stereospecifi c reaction is one in which the stereochemically different starting materials lead to stereochemically different products. For example, cis-2-butene undergoes chlorination to give only (±)-2,3-dichlorobutane while trans-2-butene undergoes chlorination to give only meso-2,3-dichlorobutane. Since the two diastereoisomeric 2-butenes produce two different stereoisomeric products, the reaction is a stereospecifi c one.
2. Starting from acetylene outline a stereospecifi c synthesis of meso-3,4- dibromohexane.
Solution
3. Predict the product and the stereochemical outcome of the following reaction:
Cyclopentene+Br2 ổổặ
Solution Cyclopentene reacts with Br2 to give trans-1,2-dibromocyclopentane (a racemic mixture) as follows:
4. The addition of the elements Br and OR to a double bond is a common method for constructing ring containing oxygen atoms. The reaction is called bromoetherification. Draw a stepwise mechanism for the following intramolecular bromoetherification reaction:
Solution The reaction proceeds through the steps as follows:
In step 1, a bromonium ion is formed. In step 2, the bromonium ion undergoes intramolecular nucleophilic attack by –OH at the more substituted carbon (because it is more positive and hence, more electrophilic) to yield the conjugate acid of a cyclic ether. In step 3, loss of proton occurs to form the ether.
5. Predict the product and suggest a mechanism for each of the following reactions:
(a) (CH ) C3 2 ==CH2 +I—N== ==C O ổổặ (b)
CH3
+ ICl
Solution
(a) In this case, iodoisocyanate is formed. The reaction occurs as follows:
(CH ) C==CH3 2 2 (CH ) C— CH3 2 2 + :N==C==O ặ (CH ) C—CH I3 2 2 N==C==O
—
:I:+ :I—N==C==O
d+ d–
–
(b) Anti-addition of I Cl occurs to give the corresponding product as follows:
Since chlorine is more electronegative than iodine, therefore, iodine is the electrophilic end of the I —Cl
d+ d-
molecule. As a consequence, an iodonium ion is formed in the first step of the reaction. In the second step, Cl① ion attacks the more substituted and more positive (electrophilic) ring carbon to give the product.
6. Give the mechanism and the stereochemical outcome of the following reaction:
Solution The starting bridgehead alkene is chiral and so, trans-addition of Cl2 involving a chloronium ion intermediate leads to the formation of two optically active, i.e., chiral diastereoisomer I and II. Attack by Cl2 from the opposite face of the double bond does not take place because of steric crowding caused by the boat-axial or endo hydrogens at C-5 and C-6.
7. Give the mechanism of the following reaction and explain why it is slower than as compared to alkenes:
Solution The mechanism of the reaction is as follows:
The reason for the low reactivity of an alkyne is that the intermediate bromonium ion has a double bond and the ring is highly strained as compared to the bromonium ion obtained form an alkene.
8. How can it be proved that the following reaction involves a bromonium ion intermediate instead of an open carbocation?
3 3 2 2 3 3 2
(CH ) C—CH==CH ổổổBr ặ(CH ) C—CHBrCH Br
Solution If an open carbocation is involved, the reaction is expected to give 1,3-dibromo- 2,3-dimethylbutane along with 1,2-dibromo-3,3-dimethylbutane. However, the former compound is not at all obtained. This observation suggests that the reaction proceeds through a bromonium ion intermediate instead of an open carbocation.
The reaction actually occurs as follows to give only 1,2-dibromo-3,3-dimethylbutane.
9. Give the mechanism and the stereochemical course involved in the following reaction:
Solution Dihydrofuran undergoes electrophilic attack by Br2 to give initially a resonance- stabilized carbocation instead of a bromonium ion as intermediate. The bromide ion (Br①) then attacks the carbocation from its two faces to give cis- and trans-2,3-dibromotetra hydrofuran which are two diastereoisomers. Each diastereoisomer is obtained as a racemic mixture. The reaction proceeds as follows:
10. Suggest a mechanism for the following reaction:
Solution The mechanism of the reaction involves the steps as follows:
11. Explain the formation of the product:
Solution
12. Provide mechanisms that explain the formation of the following products:
Solution The mechanistic courses of these two reactions may be given as follows:
Unlike the fi rst case, the ring does not participate due to the presence of the electron- withdrawing –CF3 group.
1. When ethylene is passed through an aqueous solution of bromine containing NaCl, the products of the reaction are as follows:
Br Br ; Br OH ; Br Cl
Write mechanisms to show the formations of these products.
2. Arrange the following compounds in order of increasing rate of addition to ethylene and give your reasoning:
I— I, I— Br, I— Cl
3. Draw the structural isomers expected to be formed in each of the following reactions:
(a) ==CH2 Br2
H O2 (b)
4. Cis- and trans-2-pentene reacts with Br2 to yield two pairs of enantiomers.
Explain.
[Hint: The cis-isomer yields (R,R) and (S,S) enantiomers, while the trans-isomer yields (R, S) and (S, R) enantiomers. See also article no. 4.2.4.5.]
5. Account for the following observation:
[Hint: Formation of a bromonium ion followed by intramolecular SN2.]
6. Give mechanism of each of the following reactions:
(a)
(b)
[Hint:
(a) Formation of a trans-bromohydrin followed by an intramolecular SN2 involving alkoxy oxygen.
(b) Formation of a bromonium ion followed by an intramolecular SN2.]
7. Give the mechanism and the stereochemical outcome of the following reaction:
H C
H CH3 H C3
H C H2 5
Br2
8. Account for the following observations:
(a) Addition of bromine to vinyl chloride (CH2== CHCl) occurs at a faster rate in acetic acid medium than in CCl4 medium.
(b) When treated with Br2/H2O, allyl bromide gives chiefly the primary alcohol, BrCH2CHBrCH2OH, in contrast, propene under the same reaction conditions gives the secondary alcohol, CH3CHOHCH2Br.
9. Predict the major product and suggest a mechanism for each of the following reactions:
(a) CH3CH == CHCH3 + CH3S Cl ổổặ (b) + NOCl 10. Draw a stepwise mechanism for each of the following reactions:
(a) (b)
11. Draw the product(s) of each of the following reactions, including stereochemistry:
(a) HO C2 —C∫∫C—CO H2 ổổổổặ(1 mole)Br (b)
12. Identify the compounds A, B and C. Give mechanisms for these conversions.
13. Addition of bromine to acetylenes give only trans-dibromides. However, phenylacetylene gives both cis- and trans-dibromides. Explain this observation.
14. Predict the product(s) and suggest a mechanism for each of the following reactions:
(a)
—
H
COOH
Br2
(b) Br2
15. Bromine adds to 2-butene at a faster rate in the presence of the Lewis acid AlBr3, but at a slower rate in the presence of KBr. Account for these observations.
[Hint: AlBr3 polarizes Br2 molecule to a greater extent and as a result electrophilic addition takes place more easily. In the presence of KBr, Br2 combines with Br①to form the complex polyhalide ion Br3①. As a result, the tendency of Br2 molecule to act as an electrophile decreases. ]