Identify the stereochemically correct product from the following:

The correct answer is option 1.

The reactant in the question is ester.

Esters are close relatives of aldehydes and ketones: they consist of a carbonyl group directly attached to an OR group. As you might expect, they react with Grignards in a similar fashion to aldehydes and ketones: with formation of a new \(C-C\) bond and breakage of a \(C-O\) (pi bond).

The Mechanism

This reaction incorporates the second most important mechanism of carbonyls (next to “addition”), namely, “elimination“. In fact, “elimination” is the exact reverse of “addition” There are 4 steps:

In the first step, the Grignard performs an addition reaction on the ester, forming \(C-C\) and breaking \(C-O\) (pi), giving us an intermediate with a negatively charged oxygen. We’ve seen this type of reaction before in the addition of Grignards to aldehydes and ketones.

Now comes the new step: elimination (sometimes, “1,2 elimination”). This intermediate has a reasonably good leaving group (\(OCH_2CH_3\) in the case below). What happens next is reformation of the \(C-O\) pi bond with expulsion of the leaving group (\(CH_3CH_2O^–\) in the case below). In other words, we form \(C–O\) π and break a \(C–O\) single bond. The new product is a ketone.

Elimination does not occur in addition to aldehydes and ketones because the leaving group would have to be the extremely strong bases H(-) or R(-). It is reasonably favorable for esters because the leaving group RO(-) is of comparable basicity to the negatively charged oxygen of the tetrahedral intermediate.

After Step 2, we have a new ketone. As we’ve seen before, Grignards will react quickly with ketones in yet another addition reaction [Step 3]. Here, as in Step 1, we form \(C–C\) and break \(C–O\) (pi). The result is a tertiary alkoxide (the conjugate base of a tertiary alcohol).