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Dehydration of Cyclohexanol
- Dehydration is the elimination of water from an alcohol to form an alkene. In the course of the reaction, cyclohexene forms in the presence of the phosphoric acid catalyst (H 3 PO 4 ), H 2 O, and unreacted cyclohexanol. Drierite® is anhydrous calcium sulfate. It absorbs the water as it is formed, preventing it from contaminating the product cyclohexene. It also serves as column packing for a fractional distillation.
- Here is the equation for the dehydration of cyclohexanol:
H 3 PO 4 , 85% OH +^ H^2 O CaSO 4 (anhydrous)
The mechanism of this reaction is as follows. This is review from the CHEM 2210 lecture.
+ H 2 PO 4
OH
+ H 2 O
+ H 3 PO 4 OH 2 + H 2 PO 4
OH 2
+ H 3 PO 4
Here are some things to note about the kinds of mechanisms that express the ideas behind them best, and which therefore get full credit on exams. (1) The steps are written as separate, balanced equations. Nothing is left out and there is no shorthand. (2) The equations for the mechanism (three in this case), when added together, give the overall reaction. (3) The catalyst (H 3 PO 4 in this case) reacts in the first step and is made again in a later step. This is the way catalysts behave in mechanisms. (4) The protonated alcohol, H 2 PO 4 – , and the carbocation appear as reaction products and then react in later steps. This means that they are reactive intermediates.
Sample problem: Propose a mechanism for the following reaction. (The answer is given on the next page).
H 3 PO 4
+ H 2 O
OH
The mechanism of the reaction at the bottom of the previous page is:
+ H 3 PO 4 + H 2 PO 4
+ H 2 O
OH OH 2
OH 2
a rearrangement step
+ H 2 PO 4 + H 3 PO 4
Here are some more mechanism hints. If a carbocation can rearrange by way of a 1,2 shift to give a more stable carbocation, it will. More generally, if bond-line formulas still give you trouble, propose your mechanism using condensed or Lewis structures. Your instructor won't mind.
H 3 C
H 3 C CH^3
+ H 3 PO 4
+ H 2 O
a rearrangement step
+ H 2 PO 4
H 3 C OH
CH 3 —C—C—CH 3
H 3 C H
H 3 C OH 2
CH 3 —C—C—CH 3
H 3 C H
+ H 2 PO 4
H 3 C OH 2
CH 3 —C—C—CH 3
H 3 C H
H 3 C
CH 3 —C—C—CH 3
H 3 C H
H 3 C
CH 3 —C—C—CH 3
H 3 C H
H 3 C
CH 3 —C—C—CH 3
H 3 C H
H 3 C
CH 3 —C—C—CH 3
H 3 C H
C C
CH 3
+ H 3 PO 4
Instructors like mechanisms that contain arrows showing electron movement, as used in the mechanism version here.
special kit for cleaning up mercury spills. You will use a sand bath during this experiment. As you know by now, portions of this apparatus become very hot during normal use. Handle sand baths with care, and don't touch the hot parts. They also use electricity for heating; do not use water or other solvents carelessly around them. If the wires on any electrical apparatus are frayed, the shock and fire hazards are multiplied; do not use the apparatus.
Notes for topics that may be included in quizzes given after the experiment:
- To predict the product of a dehydration reaction, you will often need to know the mechanism of the reaction. This is because rearrangements will occur. The mechanism steps are as follows (see section 2 above for examples): 1. Protonate the alcohol using the acid catalyst. 2. Lose H2O to form a carbocation. 3. If the carbocation can rearrange by a 1,2 shift to a more stable carbocation, do so. 4. Remove a proton from a carbon atom adjacent to the one with the positive charge to form the alkene product. Sample problem: The alcohol (CH 3 ) 2 CHCHOHCH 3 reacts in the presence of H 2 SO 4 (catalyst) to give the alkene (CH 3 ) 2 C=CHCH 3 as the major product. What other alkene is formed?
Solution: The answer may seem obvious, but let’s go through the mechanism. Here are the first two steps:
+ H 2 SO 4
+ H 2 O
H 3 C OH
CH 3 —C—C—CH 3
H H
H 3 C OH 2
CH 3 —C—C—CH 3
H H
+ HSO 4
H 3 C OH 2
CH 3 —C—C—CH 3
H H
H 3 C
CH 3 —C—C—CH 3
H H
The carbocation formed in the second step above is 2°, and will rearrange to form a 3° carbocation in the next step: H 3 C | CH 3 —C—C—CH 3 | | H H
H 3 C H
CH 3 —C—C—CH 3
H
(The mechanism is continued on the next page)
The 3° carbocation can then lose a proton from either carbon adjacent to the positive charge. The carbon to the left of the carbon with the positive charge gives the answer; the carbon to the right of the positive charge gives the major product (structure shown in the statement of the problem).
H 3 C H
CH 3 —C—C—CH 3
H
H 3 C H
CH 2 —C—C—CH 3
H
H 3 C H
CH 3 —C—C—CH 3
answer
major product (more stable)
Note that only the alkene in the center of the last step above is the answer to the question as stated. The rest of the mechanism should be on scratch paper only.
