What is the purpose of using a flame-dried apparatus and a drying tube during the hydroboration reaction?

lab 1 organic chem 2

Hydroboration/Oxidation of 1-Octene

Figure 1 Hydroboration/oxidation of an alkene
Hydroboration of alkenes followed by oxidation is an extremely useful sequence in organic synthesis. The net result of the two step process is addition of water (hydration) across a double bond. In contrast to acid-catalyzed hydration, the reaction is regioselective for the anti-Markovnikov product, meaning it favors the product where hydrogen attaches to the more substituted carbon and the hydroxy group attaches to the less substituted carbon (
Figure 2).

Figure 2 Regioselectivity of hydroboration/oxidation vs. acid-catalyzed hydration
This reversal in regioselectivity is due in part to electronic effects. Unlike addition of H-X or H 2O, where hydrogen is less electronegative than oxygen or a halogen, boron is actually less electronegative than hydrogen. As shown in
Figure 3, addition of the B-H bond across a C=C bond proceeds through a concerted mechanism. There isn’t an intermediate here, so we have to focus on the possible transition states. The resulting 4-centered transition state has some partial negative charge developing on boron and partial positive on the carbon that receives the new hydrogen. Of the two possibilities, the transition state where this partial positive is on a more substituted carbon is more stable. While the overall regiochemical outcome is the opposite of the acid-catalyzed hydration, both result from trends in carbocation stability.

Figure 3 Comparison of transition states leading to anti-Markovnikov and Markovnikov products Sterics also play a role in the regioselectivity, as the smaller hydrogen has better access to the more substituted carbon. This steric effect can be enhanced by starting with a bulky borane reagent such as disiamyl borane, [(CH 3) 2CH(CH 3)] 2B) or 9-BBN (9-borabicyclo[3.3.1]nonane or “banana borane”, see Figure 4).

Figure 4 9-borabicycylo[3.3.1]nonane, aka, banana borane]
Borane (BH 3), which is used in this experiment, is the smallest hydroboration reagent possible, and it would probably give poor regioselectivity if the reaction stopped after one addition. However, it has 3 B-H bonds that can add to the alkene. The regioselectivity improves after each addition, as the resulting mono- and dialkyl boranes are more sterically hindered (
Figure 5).

Figure 5 Each BH 3 can add up to 3 times to an alkene
In addition to being regioselective, hydroboration is a syn stereospecific reaction, where the boron and hydrogen are delivered to the same face of the alkene. The oxidation step preserves this syn stereochemistry. In the case of hydroboration/oxidation of the
monosubstituted 1-octene (this experiment), syn and anti addition lead to identical
products. An example where syn and anti products are not the same is shown in
Figure 6.

Figure 6 Hydroboration of 1-methylcyclopentene produces the trans-diastereomer.

(Note the -OH and -H added are on the same side of the ring).
Watch the four videos for the hydroboration of 1-octene – they’re all pretty short and
there’s more background on both steps of the reaction and new techniques involved. Take notes that include the quantities of the reagents used and the procedural steps of the synthetic sequence. There are tables below to assist with this process.

Hydroboration 1-octene (i.e. just the first step)

1. Equation for Synthesis Reaction Carried Out: (1 pt)

2. Results Tables (3pts)

Reactant
Volume used (mL) Literature
density (g/ml)
Mass used (g)
Molar mass

(g/mol) or
Molarity
(mol/L)

Moles Used

1-octene
g/mol
BH 3THF
n/a n/a M
Water
4 drops ~ 0.08 mL
limiting reactant =

theoretical yield of product (moles)=

3. Calculations – show your calculations below
Calculation of Masses of Reactants. Where applicable, use the volume and the
literature density of a liquid reactants to calculate the masses used – look up
the literature value for density, where needed. (2pts)

Calculation of Moles of Reactants and Product. Calculate moles for each reactant
and the product. Molar masses can be calculated from the molecular formula
and/or looked up. (3 pts)

Percent Yield Calculation. Based on moles of reactants and the stoichiometry of the
reaction, determined the limiting reactant, and theoretical yield in moles. Then determine the percent yield for the reaction. (3 pts)

Oxidation of the organoborane intermediate
1. Equation for Synthesis Reaction Carried Out: (1 pt)

2. Results Tables (3 pts)

Reactant
Volume used (mL) Concentration
(Molarity or
mass %)

Mass used (g)
Moles Used
Trialkylborane
n/a n/a n/a *
NaOH
n/a
H 2 O2

* For the trialkylborane, use the theoretical yield from the 1 st step.

limiting reactant =

theoretical yield of product (moles)=

actual yield of product (moles)= ___________________

percent yield =

Gas Chromatogram

Analysis of GC data (3 pts)
Component

Standard
Retention Time
(min)

Observed
Retention Time
(min)

Integration

(mV⋅min)

Percentage of
the mixture

THF

1-octene

2-octanol

1-octanol

Ratio of 1-octanol to 2-octanol =

3. Calculations – show your calculations below

Calculation of Reagent Masses. Where applicable, use the volume and the
literature density of a liquid reactants to calculate the masses used – look up
the literature value for density, where needed. (2pts)

Calculation of Moles of Reactants and Product. Calculate moles for each reactant
and the product. Molar masses can be calculated from the molecular formula
and/or looked up. (3 pts)

GC analysis. Use the integration values in the GC trace to calculate 1. The percentage of each component in the mixture and 2. the ratio of 1-octanol to 2-octanol (3 pts)

Percent Yield Calculation. Based on moles of reactants and the stoichiometry of the
reaction, determine the limiting reactant, and theoretical yield in moles. Use the mass of crude product and the GC trace to determine the actual yields and percent yields of 1- octanol and 2-octanol. (3 pts)

Your lab report outline:
Abstract: Provide a concise summary of the experiment and its results

Background: Discuss the regiochemical and stereochemical outcomes of the
hydroboration/oxidation reaction. Include a complete curved arrow mechanism for both the hydroboration step and the oxidation step.

Data: Complete the tables above and copy and paste it here
Analysis. Write the balanced chemical equation for the reaction. Identify the limiting
reagent in each step and determine the theoretical yield of the product. The crude mixture contains a mixture of compounds. Use the GC trace to determine the actual yield and percent yield of 1-octanol and 2-octanol.

Discussion: Reflect on how the reactions and workup/purification procedures were
performed. Considering the yield obtained for each, what improvements or modifications would you suggest? Support your ideas – why do you think this would improve the results?

Questions See next page

Questions:
1. What is the purpose of using a flame-dried apparatus and a drying tube during the
hydroboration reaction? What undesired reaction will occur if water is present?
(3pts)

2. Why is argon used to displace the air in the reaction apparatus for the hydroboration step? (2 pts)

3. Briefly describe how GC (gas chromatography) separates the compounds in the crude mixture. Based on this, why does 1-octanol have a higher retention time than 2-
octanol? (3 pts)

4. Show the first step in the mechanism for hydroboration E-3-methylpent-2-ene,
including the transition state. You don’t have to show the mechanism for the oxidation steps, but provide the alcohol (major product) produced after oxidation. (3 pts)

Show stereochemistry in the transition state and final product.

What is the purpose of using a flame-dried apparatus and a drying tube during the hydroboration reaction?
Scroll to top