Lab 6: Determination of the Ratio of Product Alcohols in the Luche Reduction of (-)-Menthone

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Last updated 8:35 PM on 4/30/26
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30 Terms

1
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what is the Luche reduction used for

for the regioselective reduction of alpha enones

2
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what is the term to describe the catalyst needed (what does it do)

Lanthanide catalyst (it activates the alcohol solvent)

3
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in what formation does Luche reduction occur in?

1, 2 fashion

4
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what is the purpose of this experiment?

The Luche reduction can increase the stereoselectivity of chiral ketones

5
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what does the stereoselectivity of chiral ketone depend on?

the concentration of catalyst

6
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reagents used

L-Menthone (structure, MW, BP, density)
Cerium chloride heptahydrate (structure, MW, MP)
Sodium borohydride (structure, MW, MP)
Ethanol (structure, MW, BP, density)
Diethyl ether (structure, MW, BP, density)
Magnesium sulfate (structure, MW, MP)
(-)-Menthol (structure, MW, BP, density, α°)
(+)-Neomenthol (structure, MW, BP, density, α°)

7
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broadly describe the protocol

  1. dissolve menthone into ethanol into small beaker

  2. place in water bath then add NaBH4 and let the H2 bubbles go away. stir for about 5-10 mins

  3. dilute in DI water and extract with diethyl ether (aka add diethyl ether to aqueous)

  4. combine the organic layers and wash with (1) water then (2) brine (add these to organic)

  5. dry with magnesium sulfate and remove solid with vacuum filtration (and any other remaining solvent with rotovap)

  6. take IR

  7. dissolve samples in 15 mL 95% ethanol and transfer to polarimetry tube. get the degree of rotation

8
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what is the difference between luche and standard reduction reaction

Luche adds cerium catalyst to a standard sodium borohydride reaction. the purpose is the allow for a more regio and stereoselective reduction reaction. also, standard adds in 1,4 addition while luche does 1,2 addition

9
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what kind of addition to Luche reaction favor

1,2 addition

10
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normal sodium borohydride reactions will produce a racemic mixture. what will luche do?

it favors one an will produce enantiomeric excess of one

11
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what position does Luche favor the most?

equatorial alcohols when reducing substituted, cyclic ketonesh

12
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how do you find enantiomeric excess ratio?

  1. find aº of mixture

  2. then plug into second equation to find x (make sure you have the aº for methanol and ethnaol)

  3. once you have x, (aka it will be x:1) ad one then find % methanol by taking the x value of methanol (numerator) over the total

13
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what does x represent

methanol: neomethanol

14
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polarimetry procedure

see photo

<p>see photo</p>
15
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what dextrorotatory mean

when the darker half circle is to the right

16
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what does levorotatory mean

darker half circle is on the left

<p>darker half circle is on the left</p><p></p>
17
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what instrument will you be using

SR-6 polarimeter

18
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what are the alcohols formed in this experiment

menthol (wedge) and neomenthol (dash)

<p>menthol (wedge) and neomenthol (dash)</p>
19
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what is the reactant

L-menthone

<p>L-menthone</p>
20
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When both the organic and aqueous layers have been transferred into the separatory funnel in the Luche reduction experiment, which layer will be on the bottom?

aqueous

21
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How will the residual water remaining in the organic layer be removed prior to distillation in the Luche reduction experiment?

magnesium sulfate

22
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What solvent will your sample be dissolved in for determining the optical rotation?

95% ethanol

23
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general balanced equation

see photo

<p>see photo</p>
24
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general reaction mechanism

  1. you have to form your B(OEt)3H (aka the first three steps get the OEt group to connection with B

  2. the bottom is the O-Ce, nuc attack by B(OEt)3H to push electrons from bond to O. then HOEt will protonate to form hydroxyl

<ol><li><p>you have to form your B(OEt)3H (aka the first three steps get the OEt group to connection with B</p></li><li><p>the bottom is the O-Ce, nuc attack by B(OEt)3H to push electrons from bond to O. then HOEt will protonate to form hydroxyl</p></li></ol><p></p>
25
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limiting reagent

L-menthone

26
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did i use IR to find final product? or the ratio? what about to tell if reaction went to completion

no you cannot find the ratio because they have the same functional groups

you will see the disapperance of the ketone group and the formation of the broad intense peak for OH

27
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<p>predict major/minor reactions </p>

predict major/minor reactions

1, 2 addition is the major

<p>1, 2 addition is the major</p>
28
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The Luche reduction can also be stereoselective, as you saw in this lab which gave you a mixture of diastereomers. Predict the major and minor reduction products considering only the stereochemistry (for this question, you may disregard regioisomers).

see image

<p>see image</p>
29
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<p>Why is only one product observed from this reaction? Explain your answer in complete sentences and include an image of the chair structure that adequately shows any relevant steric or electronic interactions that impact this reactivity.</p>

Why is only one product observed from this reaction? Explain your answer in complete sentences and include an image of the chair structure that adequately shows any relevant steric or electronic interactions that impact this reactivity.

Here, the tert-butyl group is bulky, so it needs to be in the equatorial position to minimize steric hindrance. Putting it in the axial positions results in overcrowding and steric hindrance, even if it is just two adjacent hydrogen groups. Having it in the equatorial position will result in a more stable, lower-energy chair configuration. Additionally, because the front side of the carbonyl is crowded by the two wedge methyl groups, the hydride shift must occur on the backside—the least sterically hindered face. The hydride will attack from the back, and ultimately, after protonation, the OH will end up on the opposite side, ending up on the front side. This results in the OH being in the equatorial position in the chair configuration.

<p><span>Here, the tert-butyl group is bulky, so it needs to be in the equatorial position to minimize steric hindrance. Putting it in the axial positions results in overcrowding and steric hindrance, even if it is just two adjacent hydrogen groups. Having it in the equatorial position will result in a more stable, lower-energy chair configuration. Additionally, because the front side of the carbonyl is crowded by the two wedge methyl groups, the hydride shift must occur on the backside—the least sterically hindered face. The hydride will attack from the back, and ultimately, after protonation, the OH will end up on the opposite side, ending up on the front side. This results in the OH being in the equatorial position in the chair configuration.</span></p><p></p>
30
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<p>draw the non chair configuration with this</p>

draw the non chair configuration with this

see photo

<p>see photo</p>