Question

Here's my synthesis:

HBr
NaEtO
OsO4/NMO
But the problem is that for step 2, there are 3 possible alkene locations, with all of them having 3 substituents, so I would have all 3 forming (although 2 of them would result in the same thing). So I checked my professor's solution, but he used HBr twice. However, wouldn't the first HBr result in a hydride shift, making the 2nd HBr unnecessary?

Either way, this solution also allows the formation of alkene in 3 sites, just like my solution. Is that allowed?

This is my prof's solution:

Answer 1

In this scenario, tertiary alcohols react with HBr using an SN1 mechanism forming stable carbocations, primary alcohols react through an SN2 mechanism, and hydride shifts occur for more stable carbocations. Elimination reactions compete with substitution reactions to form alkenes, which can add HBr to form the same product as substitution under certain conditions.

Step-by-step explanation:

The scenario involves a series of chemical reactions in organic chemistry where hydrobromic acid (HBr) is added to different alcohols to form alkyl bromides. The mechanism by which HBr adds to alcohols can result in either the formation of an alkyl halide via an SN1 mechanism or an SN2 mechanism, depending on the structure of the alcohol.

For tertiary alcohols, the reaction proceeds through an SN1 mechanism, where the alcohol first forms a carbocation intermediate which then reacts with bromide. For primary alcohols, the reaction proceeds via an SN2 mechanism, where the substitution happens in one step with the backside attack of bromide.

In some cases, hydride shifts can occur if it leads to a more stable carbocation during the SN1 mechanism. However, if the initial carbocation formed is already stable, as in the case of a tertiary alcohol, additional hydride shifts are not necessary.

When it comes to forming alkenes through elimination reactions, the most substituted alkene product is typically favored due to its stability, known as Zaitsev's rule. Alkene formation is possible when elimination competes with substitution, and in some conditions, the alkene formed can further react to form the same final product as the direct substitution.