Which of the following secondary alcohols can be prepared by the ...

Question

Which of the following secondary alcohols can be prepared by the reaction of methyl formate with excess Grignard reagent?

A. CH3CH2CH(OH)CH3

B. CH3CH(OH)CH3

C. CH3CH(OH)CH2CH3

D. CH3CH2CH(OH)CH2CH3

Accepted Answer

Welcome back, everyone. We need to identify the secondary alcohol produced. When cyclopentolate reacts with an excess amount of a Grigor reagent. Choices A through D all give us secondary alcohols. Where in choice A, our point of attachment for our alcohol is adjacent to a methyl group as well as being adjacent to an isopropyl group. In choice B, our point of attachment for our alcohol is adjacent to a ethyl group as well as being adjacent to a cyclopia ring. Then in choice C, our point of attachment for our alcohol is adjacent between two identical alkyl chains which are both ethyl groups. And then in choice D our alcohol group is where its point of attachment is adjacent to a methyl group as well as another cyclo penal rink. So with all of our answer choices outlines, we're going to begin by writing out our reaction. We need to draw out our structure for cyclopentolate. So recall that our cyclopentolate is characterized by the presence of a carbon group where we have a double bond between carbon and oxygen that carbon will be bonded to a hydrogen. On the left. And on the right, we have an oxide where we have an oxygen bonded to our cyclo Penil ring. So we have a five carbon ring and sorry, this should be five carbons. So we have 1234 and then our fifth carbon here is at our carbon which is bonded to our oxygen. So this is our cycle or cyclo pencil format reactant. And we will show a reaction where above our reaction arrow, we have our Grigor reagent, which we would call a Grigor reagent is an organo metallic compound where we have an alch chain bonded to our metal, which is magnesium, which is bonded to a halogen. In this case, we'll use bromine. Now, we're going to need to go through our mechanism to understand which alcohol would be our final answer. So we're going to need to recognize that within our cyclopentolate structure, we have a carbonyl which is near an an alcoy group. So that would be our oxygen attached to our cyclops ring. And as a whole, this results in a group which we should recall as an Esther. Notice that within our group on our carbon portion, our carbon double bonded to that oxygen is less electron negative than oxygen and thus will have a partially positive charge where our oxygen has a partially negative charge. But we're only going to draw in our partially positive charge for that carbon. Because now looking at our grid nerd reagent notice that it's highly polar because we have our our group bonded to our metal magnesium in which our R group having a carbon is going to be more electron negative and carry a slightly negative charge where our magnesium will carry a slightly positive charge. So we can interpret that we have an reacting with our Griner re agent. And whenever we have an Esther reacting with a Grigor reagent, we want to recall that this will result in a addition reaction. So nucleic addition reaction where our grid reagent is going to be our strong nu file where specifically our alkyl group being our bond to our, our group in our grid reagent is going to attack being partially negative the partially positive carbon in our Carboni. Now recall that this carbon can only have four bonds. And so we're going to thus kick out one of our double bonds in our Carboni to oxygen so that it will fall as a lone pair on oxygen. And so as a result, we'll have the following structure where we have our hydrogen bonded to our carbon, which is now single bonded to oxygen. We recall that oxygen's bonding preference is to have two bonds and two lone pairs. So originally as a carbo, it carried two lone pairs on itself where now we have that third lone pair from moving our double bond in our Carboni as a set of lone electrons on oxygen. And now our carbon bonded to oxygen here is bonded to our alkyl group from our Grigor region, which is our carbon chain R. And then we still have that bond between our carbon to our al alco group where we have the oxygen bonded to our cyclops rink. We can go ahead and draw in those two lone pairs on this alco oxygen. Here. Notice that our oxygen in our original Carboni group here has now three lone pairs, which is a total of directly attached seven V electrons, which is one more than oxygen prefers to have being a group six a atom. And so thus, we have a negative one charge on our oxygen here. Now, whenever we have charges in our structures that is not going to be favorable. And so we're going to amend this by reestablishing our Carboni by using one of the lone pairs to form our double bond between our oxygen and carbon again, which is thus going to create five bonds on carbon, which we do not want. And so thus, we have to consider our leaving group, which is going to be our alcoy group due to the fact that it's going to be weakly basic since we have those lone electrons on oxygen. And so we would kick off the bond between our carbon and our alcoy group to fall as a lone pair on our alcoy oxygen here. And as a result in our next line, we would draw the resulting structure where we would transition from this ketone intermediate. Recall that our ketone is our carbon bonded to oxygen and then bonded to two our groups. So we'll label this as a ketone. So again, our resulting structure from this rearrangement is going to be the reformation of our Carboni where we have carbon double bonded to oxygen, we still have a bond to hydrogen and then we still have carbon bonded to our first R group from the al chain on our grind re agent. So now note that we are told in our prompt, we have excess grig reagent. So if we have access and we'll note that down, we have excess Griner reagent. That means we can take another molecule where for that slightly negative charge on our alkyl chain in our grig nerd reagent, which is our, our group here, it's going to be attracted to the partially positive charge on our carbon in our Carboni. So we would have a second nucleic addition reaction in which our alkyl group from our Gregor region will attack that partially positive carbon. Thus, kicking off our double bond between our carbon and oxygen and our Carboni group back on as a lone pair to oxygen. And thus, we have another ketone intermediate where we now have oxygen with three lone pairs on itself a single bond to our carbon and our carbon bonded to the hydrogen that it was originally bonded to and now bonded to two R groups. So we'll distinguish the second R group as our prime So now notice that in this al oxide structure, we have, after we have completed the second nucleic addition, we have our oxygen within our ketone group here again, because we have an oxygen bonded to a carbon bonded to our groups. This oxygen has three lone pairs. And thus, we have a set or a set net net negative charge of minus one here due to those three lone pairs on oxygen. So at this point, we're going to need to quench our reaction. And we should recall that for a grig nerd reaction, we quench with a aqueous acidic solution. And this is going to allow us to generate hydro num H 30 plus, which is important because if we generate hydro, notice that with that plus one charge, we would be able to prod donate our alco side here so that we can form an alcohol creating a stable product. And so in this case, if we add hydrochloric acid, recall that when it reacts with water, it's going to donate a proton to water being an acid. And thus we would form chloride as a spectator and then our hydro num as our second product. So ultimately, we can understand that quenching our reaction is just going to create a stable product so that we no longer have this reactive alcide intermediate with this negative charge on oxygen. So just so it's helpful to understand, let's draw hydro num out so that we have the actual structure shown where we have three hydrogens bonded to an oxygen and sorry, this should be hydrogen. Here, we have a lone pair on that oxygen. Recall that oxygen with one lone pair in itself and three bonds will have a plus one charge. And so our negative charge from our ketone and our Aloxi intermediate is going to be attracted to that ne that positive charge on our hydro and correction. It's going to be attracted to our proton within our hydro num because again, this is a protonation step. And so our oxide intermediate is going to be weakly basic, since we have that negative charge and the abundance of electrons on that oxygen. So it's going to actually being weakly basic be an electron pair donor. So forming a bond with one of our protons and hydro and thus that bond between our proton and our oxygen and hydro is going to fall as a lone pair on oxygen. And so as a result, we would have now stabilized our oxide intermediate and now formed an alcohol product in which we have our carbon in the center now bonded to a alcohol group or hydroxide. From that protonation step, we still have our bonds to our two art groups. And actually let's draw it as we did before. So we have a bond to our two, our groups and then we have our bond to hydrogen. So we're going to need to look at our answer choices and choose our alcohol product that is bonded to two, our groups that are identical most likely. So looking at our answer choices, choice C would make the most sense since we have our alcohol, our secondary alcohol bonded to two identical R groups, which we identified as eyl groups. And so choice C would be our final answer as the secondary alcohol that is produced when cyclopentolate reacts with an excess amount of Agner re agent. I hope this made sense and let us know if you have any questions.

Which of the following secondary alcohols can be prepared by the reaction of methyl formate with excess Grignard reagent? A. CH3CH2CH(OH)CH3 B. CH3CH(OH)CH3 C. CH3CH(OH)CH2CH3 D. CH3CH2CH(OH)CH2CH3

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