Vollhardt  Capítulo 6 (Haloalcanos)
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Vollhardt Capítulo 6 (Haloalcanos)

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to our original expectation!
Increasing basicity
F2 , Cl2 , Br2 , I2
Increasing nucleophilicity in aprotic solvents
Increasing polarizability improves nucleophilic power
The solvation effects just described should be very pronounced only for charged nucleophiles. 
Nevertheless, the degree of nucleophilicity increases down the periodic table, even for uncharged 
nucleophiles, for which solvent effects should be much less strong, for example, H2Se . H2S . 
H2O, and PH3 . NH3. Therefore, there must be an additional factor that comes into play.
This factor is the polarizability of the nucleophile (Section 6-1). Larger elements have 
larger, more diffuse, and more polarizable electron clouds. These electron clouds allow for 
more effective orbital overlap in the SN2 transition state (Figure 6-7). The result is a lower 
transition-state energy and faster nucleophilic substitution.
Table 6-6
 Relative rate
Formula Name Classifi cation (krel)
CH3OH Methanol Protic 1
HCONH2 Formamide Protic 12.5
HCONHCH3 N-Methylformamide Protic 45.3
HCON(CH3)2 N,N-Dimethylformamide Aprotic 1,200,000
Relative Rates of SN2 Reactions of lodomethane with Chloride 
Ion in Various Solvents
CH3I 1 Cl
2 uvy CH3Cl 1 I
Large 5p orbital, polarized toward
electrophilic carbon center
sp3 hybrid back lobe
Small 2p orbital,
relatively nonpolarized
\u3b4+ \u3b4\u2212\u3b4\u2212\u3b4+
Figure 6-7 Comparison of I2 and F2 in the SN2 reaction. (A) In protic solvents the larger iodide is a 
better nucleophile, in part because its polarizable 5p orbital is distorted toward the electrophilic carbon 
atom. (B) The tight, less polarizable 2p orbital on fl uoride does not interact as effectively with the 
 electrophilic carbon at a point along the reaction coordinate comparable to the one for (A).
6 - 8 S t r u c t u r e a n d S N 2 R e a c t i v i t y : T h e N u c l e o p h i l e
Nucleophilic substitution (SN2)
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238 C h a p t e r 6 P r o p e r t i e s a n d R e a c t i o n s o f H a l o a l k a n e s
Sterically hindered nucleophiles are poorer reagents
We have seen that the bulk of the surrounding solvent may adversely affect the power of 
a nucleophile, another example of steric hindrance (Section 2-8). Such hindrance may also 
be built into the nucleophile itself in the form of bulky substituents. The effect on the rate 
of reaction can be seen in Experiment 7.
Experiment 7
CH3I CH3OCH3\ufffd \ufffd
CH3CO \ufffd CH3OCCH3
CH3O \ufffd
CH3I \ufffd
\ufffd I \ufffd
\ufffd\ufffd I \ufffd
Conclusion. Sterically bulky nucleophiles react more slowly.
Exercise 6-24
Which of the two nucleophiles in the following pairs will react more rapidly with bromomethane?
CH3CHS\ufffdor (b) 
Exercise 6-25
Working with the Concepts: Suggesting a Reaction Product by 
Mechanistic Reasoning
Treatment of 4-chloro-1-butanol, :Cl
. . H, with NaOH in DMF solvent leads to 
rapid formation of a compound with the molecular formula C4H8O. Propose a structure for this 
product and suggest a mechanism for its formation.
Rather than immediately trying to fi gure out the structure of the product of the reaction, it is often 
more productive to \u201cthink mechanistically\u201d and consider the reaction pathways available. If the 
fi rst pathway doesn\u2019t work, try to refi ne the problem \u2014 what is the change that takes place in the 
molecule, and how could this change come about?
\u2022 The most obvious mechanism to try is an SN2 trajectory for the substrate and hydroxide:
HO \ufffd HOCH2CH2CH2CH2OHClCH2HOCH2CH2CH2 \ufffd\ufffd OO Cl \ufffd
Unfortunately, the product in this equation cannot be correct, because its molecular formula is 
C4H10O2, not C4H8O.
\u2022 Let\u2019s consider another approach. The substrate has the molecular formula C4H9OCl. Therefore, 
the change in its conversion to C4H8O is loss of one hydrogen and the chlorine \u2014 that is, a mole-
cule of the strong acid HCl. How could we effect this change?
\u2022 Hydroxide is a base as well as a nucleophile; therefore, a reasonable alternative to the (incorrect) 
SN2 reaction above is an acid-base reaction with the most acidic hydrogen in the substrate:
HO \ufffd H \ufffd OCH2CH2CH2CH2ClCH2Cl \ufffd\ufffd O H2OOCH2CH2CH2
Exercise 6-23
Which species is more nucleophilic: (a) CH3SH or CH3SeH; (b) (CH3)2NH or (CH3)2PH?
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 C h a p t e r 6 239
Nucleophilic substitutions may be reversible
The halide ions Cl2, Br2, and I2 are both good nucleophiles and good leaving groups. 
Therefore, their SN2 reactions are reversible. For example, in acetone, the reactions between 
lithium chloride and primary bromo- and iodoalkanes form an equilibrium that lies on the 
side of the chloroalkane products:
CH3CH2CH2CH2I LiCl\ufffd CH3CH2CH2CH2Cl LiI\ufffd
This result correlates with the relative stabilities of the product and starting material, which 
favor the chloroalkane. However, this equilibrium may be driven in the reverse direction 
by a simple \u201ctrick\u201d: Whereas all of the lithium halides are soluble in acetone, solubility of 
the sodium halides decreases dramatically in the order NaI . NaBr . NaCl, the last being 
virtually insoluble in this solvent. Indeed, the reaction between NaI and a primary or sec-
ondary chloroalkane in acetone is completely driven to the side of the iodoalkane (the 
reverse of the reaction just shown) by the precipitation of NaCl:
CH3CH2CH2CH2Cl NaI\ufffd CH3CH2CH2CH2I NaClg
in acetone
The direction of the equilibrium in reaction 3 of Table 6-3 may be manipulated in 
exactly the same way. However, when the nucleophile in an SN2 reaction is a strong base 
(e.g., HO2 or CH3O
2; see Table 6-4), it will be incapable of acting as a leaving group. In 
such cases, Keq will be very large and displacement will essentially be an irreversible process 
(Table 6-3, reactions 1 and 2).
In Summary Nucleophilicity is controlled by a number of factors. Increased negative 
charge and progression from right to left and down (protic solvent) or up (aprotic solvent) 
the periodic table generally increase nucleophilic power. Table 6-7 compares the reactivity 
Exercise 6-26
The formula of the organic product of this transformation is C4H8OCl
2, only a chloride ion away 
from the correct product. How can we induce chloride ion to leave without adding any external 
species? By displacement using the nucleophilic negatively charged oxygen at the opposite end 
of the molecule, forming a ring:
 O ClCH2CH2CH2CH2 \ufffdOOO Cl \ufffd
Indeed, such intramolecular SN2 reactions are widely used for the synthesis of cyclic compounds.
 You may ask why this reaction proceeds as it does. There are two main reasons. First, Brønsted-
Lowry acid-base reactions \u2014 proton transfers from one basic atom to another \u2014 are generally faster 
than other processes. So removal of a proton from the hydroxy group of the substrate by hydrox-
ide ion (second equation) is faster than that same hydroxide displacing chloride in an SN2 process 
(fi rst equation). Second, reactions that form fi ve- and six-membered rings are generally favored, 
both kinetically and thermodynamically, over mechanistically analogous processes between two 
separate molecules. Thus, the internal displacement of chloride by alkoxide in the last equation is 
preferred over
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