Dehydrohalogenation of Alkyl Halides Organic Chem LAB Please HELP! Will rate...

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Dehydrohalogenation of Alkyl Halides

Organic Chem LAB

Please HELP! Will rate...

see here for REACTIONS: http://imgur.com/gallery/AHBHd

5. What is the solid material that precipitates as the eliminations proceed?

6. If all the elimination reactions in the experimental section had proceeded by the E1 mechanism, would the results have been different from those actually obtained? Why?

8. In the miniscale procedures, what purpose is served by packing the Hempel column with Raschig rings or other packing materials?

From book...

The electronegative halogen atom of an alkyl halide polarizes the carbon-halogen
bond so the carbon atom bears a partial positive charge, +, and the halogen atom a
partial negative charge, . This polarization may be transmitted through the
-bond network, a phenomenon referred to as an inductive effect, to enhance the
acidity of hydrogen atoms on the -carbon atom. Removing a proton from this
atom, and simultaneous departure of the halide ion, which is a leaving group, from
the -carbon atom forms the carbon-carbon -bond of the alkene in a concerted
reaction (Eq. 10.1). An important characteristic of a good leaving group is that it
should be a weak base; in contrast, the base B: should be strong. The transition
state 4 for the reaction is shown in Equation 10.2, in which the curved arrows symbolize
the flow of electrons. As shown in 4, the preferred dihedral angle between
the -hydrogen being removed and the leaving group X is 180º, an angular relationship
called anti-periplanar. This transformation of alkyl halides to form alkenes
is called dehydrohalogenation.

The concerted reaction depicted in Equation 10.2 is classified as an E2 process,
where E stands for elimination and 2 refers to the molecularity of the ratedetermining
step of the reaction. For E2 processes, the rate of the reaction depends
upon the concentrations of the organic substrate and the base (Eq. 10.3), so both
reactants are involved in the transition state of the rate-determining step. This
bimolecularity is illustrated in 4 (Eq. 10.2).

Rate = k2[alkyl halide][B:-]

Because it bears a partial positive charge, the -carbon atom of an alkyl halide
is electrophilic and thus also subject to attack by nucleophiles, which, as Lewis
bases, are electron-rich and frequently anionic species. This process produces
substitution rather than elimination products (Eq. 10.4), so a possible competition between the two types of reactions must be considered. Examining transition states
4 and 5 for the two reactions (Eqs. 10.2 and 10.4) suggests that steric factors may
play a critical role in the competition. In support of this hypothesis, it is observed
experimentally that elimination is favored as the degree of substitution on the
-carbon atom increases. The resulting steric hindrance inhibits direct attack of the
base, or nucleophile, on the -carbon atom, and removal of the -proton and formation
of elimination products becomes favored. Thus, when the -carbon atom is
tertiary, elimination to form alkenes is generally the only observable reaction.

Dehydrohalogenation may give a mixture of products if the halogen is
unsymmetrically located on the carbon skeleton. For example, 2-bromo-
2-methylbutane (6), the substrate you will use in this experiment, yields both
2-methyl-2-butene (7) and 2-methyl-1-butene (8) on reaction with strong base
(Eq. 10.5). Because such elimination reactions are normally irreversible under these
experimental conditions, the alkenes 7 and 8 do not undergo equilibration subsequent
to their production. Consequently, the ratio of 7 and 8 obtained is defined
by the relative rates of their formation. These rates, in turn, are determined by the
relative free energies of the two transition states, 9 and 10, respectively, rather
than by the relative free energies of the alkenes 7 and 8 themselves.

In the absence of complicating factors, the predominant product in an E2 elimination
is the more highly substituted alkene, which is 7 in the present example; this
observation is the source of Zaitsev’s rule. The trend is observed because an
increase in the number of alkyl substituents on the double bond almost always
increases the stability—that is, lowers the free energy—of an alkene. Those factors
that stabilize the product alkenes also play a role in stabilizing the respective transition
states in which partial double-bond character is developing between the two
carbon atoms. Thus, the enthalpy of activation, [   H‡], for forming the more stable
alkene 7 is less than that for the less stable alkene 8.

The relative free energies of transition states of competing elimination
reactions may be influenced by steric factors that increase the energies of some
transition states relative to others. For instance, the less stable alkene may become
the major product if steric factors raise the energy of the transition state leading to
the more highly substituted alkene more than that for forming the less highly
substituted. This may occur when sterically demanding substituents on the carbon
atom from which the proton is being removed offer hindrance to the approaching
base. Removal of a proton from a less substituted carbon to give a less substituted
alkene then becomes more favorable. As more sterically demanding, bulky bases
are used, abstraction of the more accessible proton will also be favored, thereby
leading to the formation of even greater amounts of the less substituted alkene.
Transition states 9 and 10 serve as an illustrative example of these principles. In
9, which is the transition state leading to the formation of 7, there can be an unfavorable
steric interaction between the methyl group on the -carbon atom and the base.
There is no comparable steric interaction in 10, which produces the thermodynamically
less stable 8. Consequently, as the steric bulk of the base is increased, 8 will be
formed in increased amounts.

In the two experiments that follow, you will explore these principles by
performing base-promoted elimination reactions of 2-bromo-2-methylbutane (6).
Using this tertiary alkyl bromide eliminates problems that might be associated with
forming by-products from substitution reactions and allows a study of effects on
product ratios when bases having different steric requirements are used. The
dehydrohalogenation of 6 may be effected by using various base/solvent combinations.
For example, some members of your class may use potassium hydroxide as
the base and 1-propanol as the solvent for the dehydrohalogenation of 6, whereas
others may use potassium tert-butoxide as the base and tert-butyl alcohol as the
solvent. By comparing the ratios of the alkenes produced by the two different
methods, you can explore how varying the reagents and solvents can influence the
course of a chemical reaction.

Explanation / Answer

5) base which we are using in elimination reactions can be precipitated as salt after the reaction.

6) The results will be different because of an intermediated existed between the reactions. This carbonium ion intermediate may results both substitution and elimination. So we get the corresponding products.

8) In minscale procedures,Hempel packing provides a contact between a continupusly moving liquid and vapor systems and at the same time allowing for good throigh-put of descending liquid.

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