CH 3 CHCH CH
H: O OCH O
The positive (δ) part of the addendum is an electrophile (E) which forms CH3CHCH2E rather than CH3CHECH2. The Nu: part then forms a bond with the carbocation.
The Eis in a box; the Nu:is encircled.
Halogens (Cl2, Br2 only) Hydrohalic acids Hypohalous acids Sulfuric acid (cold) Water (dil. H3O+) Borane
Peroxyformic acid
Mercuric acetate, H2O
Ethylene dihalide Ethyl halide Ethylene halohydrin Ethyl bisulfate Ethyl alcohol Ethyl borane Ethylene glycol
Ethanol
CH2XCH2X CH3CH2X CH3XCH2OH CH3CH2OSO3H CH3CH2OH
[CH3CH2BH2]→ (CH3CH2)3B X:X
H:Xδ+ δ−
δ+ δ−
X:OH
δ+ δ−
H:OSO3OH
δ+ δ−
H2B:H H:OHδ+ δ−
δ+ δ−
H:O OCH O
H2C HO Hg
CH2 NaBH4
NaOH
HO H O
COCH3
H2C CH2 [CH2OHCH2OCH] HOCH2CH2OH
O
δ+ δ−
Hg(O2CCH3)2 H2O
REAGENT
NAME STRUCTURE NAME STRUCTURE
PRODUCT TABLE6.1
Problem 6.20 Account for the anti-Markovnikov orientation in Problem 6.19(f).
The electron-deficient B of BH3, as an electrophilic site, reacts with the π electrons of CC, as the nucle-ophilic site. In typical fashion, the bond is formed with the C having the greater number of H’s—in this case, the terminal C. As this bond forms, one of the H’s of BH3begins to break away from the B as it forms a bond to the other doubly bonded C atom, giving a four-center transition state shown in the equation. The product from this step, CH3CH2CH2BH2(n-propyl borane), reacts stepwise in a similar fashion with two more molecules of propene, eventually giving (CH3CH2CH2)3B.
ΔH ΔH
Enthalpy
Reaction Progress adds to C
1 adds to C
2
(CH3)2CHCH2 (1o)
(CH3)2C2 = C1H2 + H (CH3)3C+ (3o)
+
+
‡ ‡
Figure 6.4
This reaction is a stereoselective and regioselective syn addition.
Problem 6.21 (a) What principle is used to relate the mechanisms for dehydration of alcohols and hydration of alkenes? (b) What conditions favor dehydration rather than hydration reactions?
(a) The principle of microscopic reversibility states that every reaction is reversible, even if only to a micro-scopic extent. Furthermore, the reverse process proceeds through the same intermediates and transition states, but in the opposite order:
(b) Low H2O concentration and high temperature favor alkene formation by dehydration, because the volatile alkene distills out of the reaction mixture and shifts the equilibrium. Hydration of alkenes occurs at low tem-perature and with dilute acid, which provides a high concentration of H2O as reactant.
Problem 6.22 Why are dry gaseous hydrogen halides (HX) acids and not their aqueous solutions used to prepare alkyl halides from alkenes?
Dry hydrogen halides are stronger acids and better electrophiles than the H3Oformed in their water solu-tions. Furthermore, H2O is a nucleophile that can react with Rto give an alcohol.
Problem 6.23 Arrange the following alkenes in order of increasing reactivity upon addition of hydrohalogen acids: (a) H2CCH2, (b) (CH3)2CCH2, (c) CH3CHCHCH3.
The relative reactivities are directly related to the stabilities of the intermediate R’s. Isobutylene, (b), is most reactive because it forms the 3° (CH3)2CCH3. The next-most reactive compound is 2-butene, (c), which forms the 2° CH3CHCH2CH3. Ethylene forms the 1° CH3CH2and is least reactive. The order of increasing reactivity is (a) (c) (b).
Problem 6.24 The addition of HBr to some alkenes gives a mixture of the expected alkyl bromide and an isomer formed by rearrangement. Outline the mechanism of formation and structures of products from the reaction of HBr with (a) 3-methyl-1-butene, (b) 3,3-dimethyl-1-butene.
No matter how formed, an Rcan undergo H: or :CH3(or other alkyl) shifts to form a more stable R′.
Problem 6.25 Compare and explain the relative rates of addition to alkenes (reactivities) of HCl, HBr, and HI.
The relative reactivity depends on the ability of HX to donate an H(acidity) to form an Rin the rate-con-trolling first step. The acidity and reactivity order is HI HBr HCl.
Problem 6.26 (a) What does each of the following observations tell you about the mechanism of the addi-tion of Br2 to an alkene? (i) In the presence of a Clsalt, in addition to the vic-dibromide, some vic-bro-mochloroalkane is isolated but no dichloride is obtained. (ii) With cis-2-butene, only rac-2,3-dibromobutane is formed. (iii) With trans-2-butene, only meso–2,3-dibromobutane is produced. (b) Give a mechanism compati-ble with these observations.
(a) (i) Br2adds in two steps. If Br2added in one step, no bromochloroalkane would be formed. Furthermore, the first step must be the addition of an electrophile (the Brpart of Br2) followed by addition of a nucle-ophile, which could now be Bror Cl. This explains why the products must contain at least one Br. (ii) One Br adds from above the plane of the double bond; the second Br adds from below. This is an anti (trans) addition. Since a Brcan add from above to either C, a racemic form results.
(iii) This substantiates the anti addition.
The reaction is also stereospecific because different stereoisomers give stereochemically different prod-ucts—for instance, cis→ racemic and trans → meso. Because of this stereospecificity, the intermediate cannot be the free carbocation CH3CHBrCHCH3. The same carbocation would arise from either cis- or trans-2-butene, and the product distribution from both reactants would be identical.
(b) The open carbocation is replaced by a cyclic bridged ion having Brpartially bonded to each C (bromonium ion). In this way, the stereochemical differences of the starting materials are retained in the intermediate. In the second step, the nucleophile attacks the side opposite the bridging group to yield the anti addition product.
Br2does not break up into Brand Br. More likely, the π electrons attack one of the Br’s, displacing the other as an anion (Fig. 6.5).
Figure 6.5
Problem 6.27 Alkenes react with aqueous Cl2or Br2to yield vic-halohydrins,⎯CXCOH. Give a mechanism for this reaction that also explains how Br2and (CH3)2CCH2give (CH3)2C (OH)CH2Br.
The reaction proceeds through a bromonium ion [Problem 6.26(b)], which reacts with the nucleophilic H2O to give
This protonated halohydrin then loses Hto the solvent, giving the halohydrin. The partial bonds between the C’s and Br engender δ charges on the C’s. Since the bromonium ion of 2-methylpropene has more partial positive charge on the 3° carbon than on the 1° carbon, H2O binds to the 3°C to give the observed product. In general, X appears on the C with the greater number of H’s. The addition, like that of Br2, is anti because H2O binds to the C from the side away from the side where the Br is positioned.
Problem 6.28 (a) Describe the stereochemistry of glycol formation with peroxyformic acid (HCO3H) if cis-2-butene gives a racemic glycol and trans-2-butene gives the meso form. (b) Give a mechanism for cis.
(a) The reaction is a stereospecific anti addition similar to that of addition of Br2. (b)
Dimerization and Polymerization
Under proper conditions, a carbocation (R), formed by adding an electrophile such as Hor BF3to an alkene, may add to the CC bond of another alkene molecule to give a new dimeric R′; here, Racts as an elec-trophile and the π bond of CC acts as a nucleophilic site. R′may then lose an Hto give an alkene dimer.
Problem 6.29 (a) Suggest a mechanism for the dimerization of isobutylene, (CH3)2CCH2. (b) Why does (CH3)3Cadd to the “tail” carbon rather than to the “head” carbon? (c) Why are the Brönsted acids H2SO4and HF typically used as catalysts, rather than HCl, HBr, or HI?
(b) Step 2 is a Markovnikov addition. Attachment at the “tail” gives the 3° R′; attach at the “head” would give the much less stable, 1° carbocation CH2C(CH3)2CMe3.
(c) The catalytic acid must have a weakly nucleophilic conjugate base to avoid addition of HX to the CC.
The conjugate bases of HCl, HBr, and HI (Cl, Br, and I) are good nucleophiles that bind to R. The newly formed R′may also add to another alkene molecule to give a trimer. The process whereby simple molecules, or monomers, are merged can continue, eventually giving high-molecular-weight molecules called polymers. This reaction of alkenes is called chain-growth (addition) polymerization. The repeating unit in the polymer is called the mer. If a mixture of at least two different monomers polymerizes, a copolymer is obtained.
Problem 6.30 Write the structural formula for (a) the major trimeric alkene formed from (CH3)2CCH2, labeling the mer; (b) the dimeric alkene from CH3CHCH2. [Indicate the dimeric R.]
Stereochemistry of Polymerization
The polymerization of propylene gives stereochemically different polypropylenes having different physical properties:
The C’s of the mers are chiral, giving millions of stereoisomers, which are grouped into three classes depending on the arrangement of the branching Me(R) groups relative to the long “backbone” chain of the polymer (Fig. 6.6).
CH3H
Isotactic (Me or R all on same side)
Syndiotactic (Me or R allternate side)
Atactic (Me or R randomly distributed) CH3H CH3H CH3H CH3H CH3H
CH3H CH3H HCH3CH3H HCH3 HCH3