Questions dealing with elimination reactions

  1. When 3-bromo-2,2-dimethylbutane is heated with a dilute solution of ethanol or sodium ethoxide in ethanol, reaction follows first-order kinetics. Along with substitution, elimination occurs to yield 2,3-dimethyl-2-butene and 2,3-dimethyl-1-butene. Propose a mechanism for their formation.
  2. When 3,3-dimethyl-1-butene is treated with HCl there is obtained a mixture of 3-chloro-2,2-dimethylbutane and 2-chloro-2,3-dimethylbutane. Propose a mechanism for this reaction.
  3. Propose a mechanism for the formation of (Z)-2-bromo-2-butene from both (R,R) and (S,S)-2,3-dibromobutane via a dehydrobromination reaction.
  4.   Try it with (S,S).

  5. Base treatment of (2R,3S)-2-bromo-3-deuteriobutane gave, in addition to the deuterated 1-butene, only (Z)-2-deuterio-2-butene and (E)-2-butene. Explain.

  7. The cis and trans isomers of 1,2-dibromocyclohexane are treated separately with NaOH in aqueous ethanol, leading predominantly to elimination products. The cis isomer gives only 1-bromocyclohexene while the trans isomer is converted to 3-bromocyclohexene, which underwent further elimination to 1,3-cyclohexadiene. Explain.
  8. The cis isomer has one axial and one equatorial bromine. Only the axial bromine has the necessary anti arrangement for elimination.

    The trans does have the diaxial conformation, albeit unstable, available to it. Now both bromines are ani to a hydrogen and can eliminate. Try drawing the structure.

  9. trans-1-Bromo-4-t-butylcyclohexane, in the presence of sodium methoxide, undergoes an E2 elimination more slowly than the corresponding cis isomer. Explain.
  10. The cis isomer has the bromine in an axial position as the t-butyl group will be equatorial due to steric considerations. In order to get the bromine axal in the trans isomer, the t-butyl group must also occupy an axial position which is sterically hindered – hence the difference in rate

  11. Bromohydrins react with base to form epoxides according to the following equation. Propose a mechanism.
  12. In the laboratories of the firm "Halides ‘R Us" has been found a compound "A" in a vial labeled only "achiral alkyl halide C10H17Br."

    Two units of unsaturation

    The management feels that the compound might be useful as a pesticide, but they need to know its structure. You have been called in as a consultant at a handsome fee. Compound "A", when treated with KOH in ethanol, yields two compounds "B" and "C", each with molecular formula C10H16.

    Three units of unsaturation now present. A dehydrobromination has occurred to give a new double bond.

    Compound "A" rapidly reacts in aqueous ethanol to give an acidic solution which, in turn, gives a precipitate of AgBr when tested with AgNO3 solution.

    Suggests a tertiary bromide which forms a tertiary carbocation and bromide ion

    Ozonolysis of "A" followed by treatment with (CH3)2S gives (CH3)2C=O as one of the products plus an unidentified halogen-containing product.

    Original hydrocarbon contains (CH3)2C= group

    Hydrogenation of either "B" or "C" gives a mixture of both trans- and cis-1-isopropyl-4-methylcyclohexane.

    This is the carbon skeleton of "A"

    Compound "A" reacts with one equivalent of Br2 to give a mixture of two separable compounds "D" and "E", both of which can be shown to be achiral. Finally, ozonolysis of "B" gives (CH3)2C=O and diketone:

    Propose structures for compounds A through E.

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This page is maintained by Dr. Ed Blackburn (Ed.Blackburn@UAlberta.CA), course instructor.

Updated August 23, 2002