Class Summary and Notes from 9/16/00 - CHEM3312

Topic: Preparation of Alcohols

Related Reading:

Summary:

Classification of reactions:

• Functional Group Interconversion
  Non-redox (addition/substitution) Redox

• Carbon-carbon bond forming reactions

• from alkyl halides (by substitution)
• from alkenes (by addition)
• from aldehydes/ketones with NaBH₄ (reduction)
• from acids/esters with LiAlH₄ (reduction)
• from aldehydes/ketones with Grignard (carbon-carbon bond forming) reactions

Problems Discussed In Class

How could you form 2-methyl-2-propanol from an alkyl halide?

(Starting questions:  what is the structure of the starting material, what mechanism do you want to use, what other reagents do you need?)

comments raised in class:  

1. Starting material could have bromide, chloride, fluoride, etc.  
2. Starting material selection should consider leaving group ability in this case as the tertiary halide will react only through an SN1 mechanism - a very good leaving group is needed in order to generate the carbocation intermediate.
3. Other reagent must have a nucleophilic oxygen to react with the carbocation intermediate - hydroxide was suggested first.  This will cause problems as it also acts as a strong base and will promote elimination by the E2 mechanism (no carbocation intermediate).  A less basic and less nucleophilic source of an OH group should be used.
4. Water can serve as the other reagent, it has lone pairs and a partial negative charge on the oxygen, and will wait for the carbocation to form before reacting.

How could you form 2-methyl-2-propanol from an alkene?

(same starting questions)

comments raised in class:

1. Starting material will be 2-methylpropene.
2. Regioselectivity of the reaction needs to be Markovnikov.
3. Possible other reagents:  H2SO4/H2O - problem with this choice is that this sets up an equilibrium between the alcohol and the alkene (same conditions used for elimination and addition)
4. Possible other reagents:  step 1:  Hg(OAc)2 in H2O/THF, step 2:  NaBH4  - this set of conditions has fewer reversibility problems
5. If the anti-markovnikov product had been desired:  step 1:  BH3 in THF, step 2:  peroxide in base

NaBH4 and LiAlH4 are both reducing agents that act by delivering hydride (H-).  One of these reduces only aldehydes and ketones, the other reduces all carbonyl groups.  Use the periodic table to try to determine which one is which.

comments raised in class:

1. Since hydride is delivered, the reducing agent that is less able to bear a negative charge should be the more reactive (and less discriminating) one.
2. Boron is above aluminum on the periodic table, and is more electronegative - it should be better able to have a negative charge and will be less reactive.

Grignard reagents have nucleophilic carbon atoms due to the bond between carbon and a less electronegative element (Mg), they have the general from RMgBr.  The alkyl group in a Grignard reagent acts like R:-, and is both nucleophilic and strongly basic.  Suggest products for the reaction between methyl magnesium bromide and 1. 2-propanone, 2. 2-propanol, 3. ethanoic acid, 4. methyl ethanoate.

comments raised in class:

1. 2-propanone has only one electrophilic carbon (the carbonyl carbon) and no acidic hydrogens - the product will be 2-methyl-2-propanol (after addition of water at the end of the reaction).
2. 2-propanol has an electrophilic carbon (the one with oxygen attached) and one acidic hydrogen (the one on the oxygen - acidity here is relative, the oxygen is better able to live with a negative charge than a carbon is), the product will be methane and the 2-propanol will be returned unchanged after the addition of water at the end of the reaction.
3. ethanoic acid is more acidic than 2-propanol, the Grignard will again act as a base.
4. methyl ethanoate has an electrophilic carbon and no acidic hydrogens, the methyl anion will attack the carbonyl.  The intermediate formed can then lose methoxide (not as unstable as the original methyl anion, so acceptable as a leaving group in this reaction) and a second molecule of the Grignard can add.  The product is 2-methyl-2-propanol.

Special Notes

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