Dehydration of 1- or 2-Butanol

Preparation: Mayo, et al., p 153-154; McM p. 233 paragraph starting with "Dehydration is often..." and Fig 17.5.

Dehydration is an acid catalyzed elimination reaction involving the loss of H2O from an alcohol to produce an alkene. The mechanism of the dehydration reaction is given for 1-methylcyclohexanol in McM Figure 17.5. The first step in the reaction is the protonation of the alcohol, an acid-base step:

  HC-C-OH   +   H2SO4   ®   HC-C-O+H2   +   HSO4-     (in McM Fig 17.5 the acid is H3O+)

From this point on the reaction is an E-1 elimination involving carbocation intermediate. In the second step, H2O is separated from the substrate to form a carbocation:

  H-C-C-O+H2   ®   H-C-C+   +   H2O

Characteristic of any carbocation, this intermediate may rearrange (see McM p.220) if there is a possibility of forming a more stable carbocation)

The final step in the elimination process is the loss of a proton from the carbon adjacent to the C+. This is another acid-base reaction. The proton is returned to restore the catalyst (or passed to another alcohol molecule to begin a new dehydration cycle). Note in Fig 17.5 that the electron pair that bonded this proton (blue) to carbon becomes the p bond of the alkene:

  HSO4-   +   H-C-C+   ®   H2SO4   +   C=C

In the laboratory, you will be dehydrating either 1-butanol or 2-butanol and measuring the relative amounts of 1-butene, cis-2-butene and trans-2-butene formed. In addition, you will interpret the data to determine the ratio of rates (and the difference in the activation energies) of two alternative reactions of the 1-butyl carbocation

Procedure:. Start warming the sand bath to about 80 oC (sandbaths can be shared by two students). To a 3 mL vial with a magnetic spin vane add 0.15 mL of the alcohol (Either 1-butanol or 2-butanol as assigned), and 2 drops (approx. 50 microliters) of concentrated H2SO4. Place the gas delivery tube, with an O ring and cap on the vial as in Fig 3.27.

To a 15x125 mm test tube add 2 mL of water and make a mark; then fill it and a gas collection tube (a simple glass tube with a rubber septum cap - obtain from stockroom) to their tops with water. Fill a 150 mL beaker with about 100 mL of water and use the thumb briefly to stopper both tubes while inverting them in the beaker. Both tubes should be upside down and filled, or almost filled with water. Set the beaker on the cold surface next to the hot part of the sand bath.

The vial is next placed in the sand bath (note the temperature) and the gas transfer tube is immersed in the beaker. Place the test tube on the transfer tube (clamps are not necessary), and collect gas to the 2 mL mark. Now exchange the test tube for the gas collection tube. Most of the gas will evolve when the temperature of the reaction vial is above 110 oC if the reactant is 2-butanol or 120 oC if the reactant is 1-butanol. Do not allow the temperature of the reaction vial to drop during this process.

Keep the open end of the collection tube under water. While you await the collection of the gas, it is a good idea to set up another run in the other 3 mL vial in case you need to repeat the experiment. When the collection tube is about half full of gas, it is ready for analysis by gas chromatography, but do not remove the apparatus until a chromatogram is completed. Then disassemble it and rinse carefully in the sink (watch that the spin vane isn't rinsed down the drain).

Analysis of the Data: For the GC analysis, use the DC-200 column at a temperature of about 10 oC. As usual, record all GC parameters on the chart and identify the alkene peaks. The peaks may not be well separated so it may be necessary to photocopy the chromatogram and then cut out and weigh the copy of each peak on a milligram balance to determine the percent distribution of the alkene products. If this is done, the peaks are stapled to a page in the notebook along with the original chromatogram. On the DC-200 column the order of elution follows the increasing boiling points of the alkenes: 1-butene, -6.3 oC; trans-2-butene, 0.9 oC; and cis-2-butene, 3.7 oC.


Interpretation of the Data: Data from both reactions are needed for the calculations, so information should be shared between designated student pairs.

Fill in the diagram below with your and your partner's data. Supply structures for the missing carbocation intermediates formed from each alcohol. Enter the product structures and the percent distributions of alkenes (rounded to the nearest percent)  you obtained from your alcohol (indicate which alcohol) and the percent distribution obtained by the person you are sharing data with (note who and the alcohol used by this person).

            ®       ®   C-C+-C-C    
        2-butanol       protonated
                      / | \  
                    a   b   c
                ___%   ___%   ___%
                    1-butene   trans-2-butene   cis-2-butene


  ___%   ___%
                    / \   |   /
                  "elim" a'   b'   c'  
                /     \ |   /
    ®       ®   C+-C-C-C   ®
1-butanol       protonated

A. Calculate the ratio of cis- to trans-2-butene formed from each alcohol. 

    From 2-butanol:  ___________                        From 1-butanol : ___________.

This ratio should not depend on which butanol was reacted; it reflects the preference of the 2-butyl carbocation to form the cis- over the trans-2-butene (or is it the other way?). Compare your cis/trans ratios to the ratio obtained by others, particularly those reacting the "other" alcohol. Are the ratios substantially different?


B. The 1-butyl carbocation has two fates: rearrangement to form the 2-butyl carbocation or direct elimination to form 1-butene. We can calculate the relative rates of these rates, Raterearr/Rateelim in the following way:

1. Let us assume that the 2-butyl carbocation formed during the reaction of 1-butanol undergoes the same fate as the 2-butyl carbocation formed from 2-butanol. In particular, assume that the ratio of 1-butene to combined 2-butene products is consistent no matter which alcohol was used. That is: a'/(b'+c') = a/(b+c).

Using your team's values obtained above, calculate the value for a' by filling the blanks.

a' = a(b' + c')
(b + c)
= ___________
= _________ = the % 1-butene from 2-butyl carbocation from 1-butanol

2. Now the % 1-butene formed from the 1-butyl carbocation can be calculated by subtraction. This is the amount of 1-butene formed by direct elimination from the 1-butyl carbocation, "elim". 

all the 1-butene from 1-butanol - a'   = ________% - ________% = __________% = "elim"

3. The 1-butyl carbocation either directly forms 1-butene by "elim" or it rearranges, "rearr", to form the 2-butyl carbocation. The percent of the 1-butyl carbocation that rearranges to form the 2-butyl carbocation can also be calculated by subtraction:

  100 %   -   "elim" = 100% - ________% ___________% "rearr"

The relative rates of rearrangement to elimination is the same as the relative percentages calculated above:

= %rearr
= ______%
 = _________ = % 1-butly carbocation -> 2 butyl carbocation
     % 1-butyl carbocation -> 1-butene

C. The activation energy, deltaG± (McM p. 175), is related to the rate and the temperature, T, of a reaction by

    rate   =   PZe-deltaG±/RT

where P and Z are factors relating to probability factors such as steric effects and concentrations, etc. For our calculations below, these factors will be cancelled because we are working with relative rates of optional reactions of a single intermediate, the 1-butyl carbocation.

We will calculate the difference in deltaG± for rearrangement and deltaG± for elimination. For each rate,

  Raterearr   =   PZe-deltaG± for Rrgt./RT       and       Rateelim   =   PZe-deltaG± for Elim./RT

And the ratio of rates:

    Raterearr/Rateelim   =   PZe-deltaG± for rearr./RT
PZe-deltaG± for elim./RT
  =   e+deltaG± for elim/RT
e+deltaG± for rearr/RT

Taking the natural logarithm of both sides

    Ln (Raterearr/Rateelim )   =   deltaG± for elim. /RT - deltaG± for rearr./RT
  or   RT·Ln (Raterearr/Rateelim )   =   deltaG± for elim.- deltaG± for rearr.

If the reaction is carried out at about 120 oC

(8.31 joules/mole oK)·(393 oK)·Ln(Raterearr/Rateelim ) = deltaG± for elim. - deltaG± for rearr. 

Thus the difference in the activation energies for the two reaction paths from the 1-butyl carbocation can be calculated. Fill in the rate ratio you determined in part B above and the actual temperature the reaction was carried out:

deltaG± for elim - deltaG± for rearr = 

=(8.31 joules/mole oK)·(393 oK)·Ln( the rate ratio you calculated above)

= ___________ · __________ · Ln ( __________) = ________________ joules/mole

= _________kJ/mole


1.Calculate, using the ideal gas law, the number of mL of gas theoretically possible from this reaction; show your work.

2. Sketch an energy drawing starting with the 1-butyl carbocation that shows the two alternative steps, either to 2-butyl carbocation or to 1-butene. Mark and label the distances for their energies of activation. Finally, insert the kilojoules/mole value calculated above.

3. What are the major sources of error in your measurements?


"Mayo et al.": Mayo, D.W., Pike, R.M., Butcher, S.S. and Trumper, P.K. Microscale Techniques for the Organic Laboratory; Wiley: New York, 1991

"McM": McMurry, J. Organic Chemistry, 5th ed., Brooks/Cole Publishing Company, Pacific Grove, CA. 1999

Rev. January 2001