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Backside attack, and

Furthermore, 48 solvolyzed 350 times faster than its endo isomer 51. Similar high exo/endo rate ratios have been found in many other [2.2.1] systems. These two results—(1) that solvolysis of an optically active exo isomer gave only racemic exo isomers and (2) the high exo/endo rate ratio—were interpreted by Winstein and Trifan as indicating that the 1,6 bond assists in the departure of the leaving group and that a nonclassical intermediate (52) is involved. They reasoned that solvolysis of the endo isomer 51 is not assisted by the 1,6 bond because it is not in a favorable position for backside attack, and that consequently solvolysis of 51 takes... [Pg.414]

An elegant and clever experiment along these lines (Lieder and Brauman, 1974) confirms that the gas phase SN2 reaction proceeds by a predominant backside attack. Analysis of neutral products in the icr experiment shows that for reaction (27), backside attack, and consequently inversion at the reaction centre, occurs to the extent of 92 + 6%. For reaction (28) the same type of experiment indicates that inversion of configuration amounts to 87 + 7% of the reaction. [Pg.210]

Unsymmetrically substituted C=C double bonds are hydrated according to the same mechanism (Figure 3.49). The regioselectivity is high, and the explanation for this is that the mer-curinium ion intermediate is distorted in the same way as the bromonium ion in Figure 3.44. The H20 preferentially breaks the stretched and therefore weakened Csec—Hg bond by a backside attack and does not affect the shorter and therefore more stable C. —Hg bond. [Pg.149]

Figure 2.29 shows the mechanism of this reaction. A key intermediate is the alkylated phosphine oxide A, into which the carboxylate ion enters in a backside attack and displaces the leaving group 0=PPh3. [Pg.79]

Whenever you see retention of stereochemistry, you should think double inversion, and in fact double inversion occurs in this reaction. The Pd(0) complex acts as a nucleophile toward the allylic carbonate or acetate, displacing MeOCC>2 or AcO by backside attack and giving an allylpalladium(II) complex. The nucleophile then attacks the allylpalladium(II) complex, displacing Pd by backside attack to give the product and regenerate Pd(0). The regio-chemistry of attack (Sn2 or Sn2 ) is dependent on the structure of the substrate. [Pg.319]

Haloalkenes fail to undergo Sfjl reactions because the alkenyl carbocations produced through ionization of the carbon-halogen bond are too unstable. They fail to undergo Sn2 reactions because the planar geometry of the aikene does not allow backside attack, and the electron-rich nature of the aikene does not attract nucleophiles. [Pg.757]

The reactions presented thus far are examples of nucleophilic aliphatic substitution reactions, in which a nucleophile substitutes for a leaving group at an aliphatic carbon. Nucleophilic aliphatic substitution proceeds with backside attack and inversion of configuration in a bimolecular process. In other words, the nucleophile collides with the electrophilic carbon atom to initiate the reaction. A shorthand symbol is used to describe this reaction Sjfl, where S means substitution, N means nucleophile, and 2 means bimolecular, or nucleophilic bimolecular substitution. Once a reaction is identified as Sn2, back-side... [Pg.511]

When substitution occurs by an Sn2 mechanism, the nucleophile directly attacks the substrate, with the angle of approach being 180" to the C-L bond. This is called "backside attack," and the reaction proceeds with inversion of stereochemistry, the so-called "Walden inversion." The C-L bond is being broken concurrently with the formation of the C-Nu bond, so both the substrate, R-L, and the nucleophile are involved in the transition state of the rate-determining step. Reactions in which two reactants are involved in the transition state of the rate-determining step are termed bimolecular, and the rate of such processes depends on the concentration of the substrate and the nucleophile, as shown in Equation 14.5, where k2 is the second-order rate constant. [Pg.463]

Tertiary haloalkanes react through an Sjjl mechanism because the steric hindrance disfavors Sjj2 backside attack, and the attached alkyl groups stabilize a carbocation. [Pg.421]

The reaction shown in Eq. 12.67 gives complete inversion at the stereogenic center attached to Fe. This requires a backside attack, and the most logical way for this to occur is to start with an oxidative addition of the Br2, and then nucleophilic attack by bromide with Fe as the leaving group. The sequence of electrophilic addition of X2 to the metal followed by nucleophilic attack on the ligand is common for middle-to-late transition metals. Interestingly, when phenyl is in the 3-position, the reaction proceeds with retention. Retention is best explained by a double inversion, and the phenonium ion has been substantiated as the intermediate formed (Eq. 12.68). [Pg.736]

The first step is a reversible step in which the oxygen atom of the alcohol functional group is protonated. Because the a carbon is tertiary, the backside of the a carbon is not susceptible to backside attack and so substitution by S m2 does not occur. Instead, substitution occurs by S l. [Pg.1296]

Because the substrate is a 3° haloalkane (which disfavors backside attack) and the solvent is polar protic (and stabilizes a carbocation), the ether is formed by the S jl mechanism. [Pg.1422]


See other pages where Backside attack, and is mentioned: [Pg.768]    [Pg.474]    [Pg.320]    [Pg.578]    [Pg.210]    [Pg.9]    [Pg.140]    [Pg.461]    [Pg.763]    [Pg.62]    [Pg.651]    [Pg.344]    [Pg.888]   
See also in sourсe #XX -- [ Pg.2 , Pg.390 ]

See also in sourсe #XX -- [ Pg.2 , Pg.426 ]




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