Substituição e Eliminação de Haletos

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Solutions for Alkyl Halides: Nucleophilic Substitution and Elimination
CH3(CH2)4CH2 OHCH3(CH2)4CH2 CNCH3(CH2)4CH2 OCH2CH3
11 
(a) Substitution—Br– is the leaving group; CH3O– is the nucleophile.
(b) Elimination—when OH is protonated, H2O is the leaving group.
(c) Elimination—both Br atoms are lost; iodide ion is a nucleophile that reacts at Br.
12 (a) (b) (c)
new rate
change in
NaOCH3
change in
C4H9Br
original rate
= 0.02 mole/L per secondx
( 1.0 M )
( 2.0 M )
( 0.5 M )
( 0.1 M )
( 0.05 mole/L per second ) x
13 (a) The rate law is first order in both 1-bromobutane, C4H9Br, and methoxide ion. If the concentration 
of C4H9Br is lowered to one-fifth the original value, the rate must decrease to one-fifth; if the concentration 
of methoxide is doubled, the rate must also double. Thus, the rate must decrease to two-fifths of the original 
rate, 0.02 mole/L per second:
rate =
A completely different way to answer this problem is to solve for the rate constant k, then put in new values 
for the concentrations.
rate = k [C4H9Br] [NaOCH3] 0.05 mole L–1 sec –1 = k (0.5 mol L–1) (1.0 mol L–1)
rate constant k = 0.1 L mol–1 sec–1
rate = k [C4H9Br] [NaOCH3] = (0.1 L mol–1 sec–1) (0.1 mol L–1) (2.0 mol L–1) = 0.02 mole L–1 sec –1
(b)
Br
H
H
N
H
H
H
nucleophile
N
H
H
H
C Br
H H
δ+ δ–
electrophile transition state
N
H
H
H
C
H
H
initial product
Br
leaving group
NH3
base
H2N C
H
H
+
+ NH4
final product
Br
(c) NaOCH2CH2CH2CH3 + CH3Br
CH3CH2CH2CH2O CH3 + NaBr
sodium butoxide bromomethane
1-methoxybutane
Br(CH3)2CHCH2 NH2
F
I
NH3Br
CH3CH2CH2 CN
(CH3)2CHCH2 NH3
HC C CH2CH2CH2CH3
(CH3)3C O CH2CH3
+ NH4
+
14 Organic and inorganic products are shown here for completeness.
(18-Crown-6 is the catalyst and does not change; CH3CN is 
the solvent.)
+ KBr
+ NaCl
(c)
(d)
+ NaCl(e)
(f) + KCl
(a)
(b)
+ NaI
The first NH3 replaces the Br. The second NH3 
removes H+ from the N, leaving R-NH2.
116