The Reduction of Benzil to Hydrobenzoin
Introduction
Figure 1 shows the reduction of a carbonyl group found in an aldehyde or
ketone to an alcohol. The reaction requires two steps. In Step 1, Na+ and H- add to
the carbonyl to make an intermediate salt. In the workup of Step 2, the salt is
converted into an alcohol by protonation of the negative oxygen. The proton of
Step 2 normally comes from a mineral acid such as HCl.
O
C
NaBH4
H+
O- Na+
C
H
O
C
H
H
H2O
Step2
Reduction of an Aldehyde or Ketone to an Alcohol by NaBH4
Step1
Figure 1. Reduction of a carbonyl group.
Reduction involves the loss of oxygen and/or the addition of hydrogen
atoms. Three common reducing agents are lithium hydride, LiH, lithium aluminum
hydride, LiAlH4, and sodium borohydride, NaBH4. Notice that all three contain
hydrogen in a negative or hydride form. A hydride ion, H-, is a nucleophile that
readily adds to a carbonyl carbon, which is positive relative to the oxygen of the
double bond. These three reducing agents all contain a metal; therefore, the
intermediate product of the reaction is an ionic salt, which must be acidified to
obtain the final organic covalent product. Figure 2 shows a simplified mechanism
for the reduction of a carbonyl group by a hydride ion, H-, which comes from
NaBH4.
O
C
H-
OC
H+
O
C
H
H
H2O
Step2
Mechanism of the Reduction of an Aldehyde or Ketone to an Alcohol
Step1
H
Figure 2. Mechanism of Reduction.
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The overall reduction of a carbonyl group to a hydroxyl group involves the
addition of two H atoms. The first H atom comes from a hydride, H-, of NaBH4.
The second comes from the workup of the reaction, which is normally conducted
in aqueous acid.
Sodium borohydride, NaBH4, is the mildest of the three hydride reagents and
is easy to use in the lab, because it is soluble in water, methanol and ethanol and
does not react with these solvents. Therefore, NaBH4 is the reagent of choice for
reducing aldehydes and ketones. Lithium aluminum hydride, LiAlH4, is a stronger
reducing agent than NaBH4, and LiAlH4 is used to reduce carboxylic acids,
epoxides, esters, lactones, nitro groups, nitriles, azides, amides and acid chlorides.
Lithium hydride, LiH, is a strong base as well as a reduction agent. For example,
LiH deprotonates an alcohol to make a lithium alkoxide and hydrogen gas. Figure
3 shows reduction reactions that require LiAlH4; each requires an acidification step
to convert the initially formed salt into a covalent organic compound. When two
steps are shown on one reaction arrow, they are shown in parenthesis [i.e., (1) and
(2)].
O
CH3CH2CH2COCH2CH3
(1) LiAlH4
(2) H3O+
ethyl butanoate
CH3CH2CH2CH2OH + HOCH2CH3
butanol
ethanol
two alcohols
an ester
OH
O
O
(1) LiAlH4
OH
+
(2) H3O
-valerolactone
a lactone
1,5-pentanediol
two alcohols
Figure 3. Reduction of an ester and a lactone.
Exceptionally strong reducing agents convert aldehydes and ketones into
alkanes. Three reactions are particularly useful. They are the Clemmensen
reduction, the Wolff-Kishner reduction, and the Raney nickel reduction reactions.
These three reactions may be conducted in acidic, basic and neutral solutions,
respectively. The Clemmensen reagents are zinc amalgam [Zn(Hg)] in
hydrochloric acid. The Wolff-Kishner reagents are hydrazine (H2NNH2) in base.
The use of Raney nickel involves first converting the carbonyl group into a cyclic
thioacetal and then reducing the sulfur compound with Raney nickel, a form of
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nickel that contains elemental hydrogen. Examples of reductions with these
reducing agents are shown in Figure 4.
O
CCH2CH3
Zn(Hg)
CH2CH2CH3
HCl
reflux
Clemmensen Reduction
O
CH
H2NNH2
CH3
KOH
DMSO (solvent)
Wolff-Kishner Reduction
H
O
S
SH
HS
S
H
Ni(R)
(H2)
H+
Raney Nickel Reduction
Figure 4. Reduction reactions of carbonyl groups to alkanes.
The Experiment
In this experiment, the diketone known by the common name benzil is the
organic substrate. Sodium borohydride reduces both carbonyl groups in benzil, and
acidification produces a diol or two secondary alcohols. See Figure 5.
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HO
O O
C C
benzil
*
H
HO
(1) NaBH4
(2) H2O (H+)
H
HO
*
Ph Ph
meso
*
H
HO
Ph
HO
Ph
*
H
1S,2S
*
Ph
HO
H
H
*
Ph
1R,2R
hydrobenzoin
Figure 5. The sodium borohydride reduction of benzil (Ph = phenyl).
The product of the reaction contains two chirality centers shown by asterisks
(*). The maximum number of stereoisomers for a compound with two chirality
centers is four (22). However, when the two centers contain the exact same four
groups, a meso form is possible, and the total number of stereoisomers is three.
Thus, the reduction of benzil produces two chirality centers but only three
stereoisomers. Figure 5 shows the structures of the three stereoisomers. Though all
three stereoisomers are produced, the meso form is the major product of this
reduction reaction.
Benzil is a yellow solid and hydrobenzoin is white; therefore, the reduction
reaction is evident by the color change. Reduction of an aldehyde or ketone
carbonyl group with NaBH4 is a general, two-step, reaction that is frequently seen
in reaction schemes on standardized examinations.
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Procedure
Reminder: Record all data directly in your notebook and turn in your data
sheet with your lab report!
1. Add approximately 0.5 g of benzil (yellow solid) to a 50-mL Erlenmeyer flask
and record the exact mass of the benzil in your notebook.
2. Add 5 mL of 95% ethanol to the Erlenmeyer flask that contains the benzil.
3. Cool the Erlenmeyer flask by rotating it under a stream of tap water.
4. Weigh 0.1-g NaBH4 on a waxed paper and transfer the white solid to the
Erlenmeyer flask that contains benzil.
5. Gently swirl the Erlenmeyer flask until the benzil dissolves and the yellow color
disappears. The exothermic reaction is over within three minutes.
6. After the yellow color is gone, warm the flask on a hot plate until the solution
begins to boil.
7. As soon as the solution boils, carefully move the hot Erlenmeyer flask from the
hot plate to the bench top, using hot hands or tongs to handle the hot glassware.
8. Allow the flask to cool on the bench top until crystals no longer appear to form.
Then, place the Erlenmeyer flask in an ice and water bath for an additional five
minutes.
9. Collect the crystals (mp 136-137 oC) on a Bűckner or Hirsh funnel, show the
crystals to the instructor, and proceed as directed by the instructor.
10. Clean all glassware and return it to the same location where you found it. Clean
your bench top and return all equipment to its storage location.
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Problems Reduction
Stu No___ Sec___Last name_____________________________, First ___________________
Multiple choice questions have one or more correct answers.
1. Calculate the oxidation number for the starred carbons in each of the following
structures. All must be correct for credit.
O
O
* O
* OH
* OH
2. Reduction reactions of aldehydes and ketones with NaBH4 and LiAlH4 require
an acidification step to:
__A. convert a salt into a covalent compound
__B. convert an alkoxide into an alcohol
__C. convert a hydride into an alcohol
__D. protonate an oxide
3. What is the name of the enantiomer of (1R,2R)-hydrobenzoin?
__A. (1S,2S)-hydrobenzoin
__B. (1S,2R)-hydrobenzoin
__C. (1R,2S)-hydrobenzoin
__D. (meso)-hydrobenzoin
4. Which of the following are diastereomers of (1R,2R)-hydrobenzoin?
__A. (1S,2S)-hydrobenzoin
__B. (1S,2R)-hydrobenzoin
__C. (1R,2S)-hydrobenzoin
__D. (meso)-hydrobenzoin
Show the major organic product(s) for each of the following equations.
5.
O
H2NNH2/KOH
(DMSO)
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6.
Zn(Hg)/HCl
O
OH
(reflux)
7.
O
(1) LiAlH4
O
(2) H+
8.
O
O
(1) xs NaBH4
(2) H+
9. The reduction of ethyl acetate by lithium aluminum hydride followed by an
acidification step produces what products?
__A. ethanol and acetic acid
__B. ethanoic acid and ethanol
__C. ethanoic acid and ethyl alcohol
__D. acetic acid and ethyl alcohol
10. Draw a complete structure for each of the following families. No credit is given
for partial (incomplete) structures.
___________________ ________________ _______________ ___________
aldehyde
ether
ester
lactone
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