What are the products of the following reactions? Show all ste...

Question

What are the products of the following reactions? Show all stereoisomers that are formed.

c.

d.

Accepted Answer

Welcome back, everyone. We need to draw the products of the reaction shown below. Considering stereo chemistry. Our first reactant is four methyl, three hex aone which reacts with an amine. We know we have an amine since we have nitrogen bond of hydrogen surrounded by two R groups where our two R groups represent carbon chains, meaning our nitrogen is surrounded by two carbon atoms. So this is a secondary amine and we have that lone pair on nitrogen that is implied. We also have an acidic work up where we have the presence of our third reactant being hydrochloric acid, which is present in a catalytic amount for this reaction. And due to the presence of this acid secondary amine and the reaction with our form me three hexen reactant, which we should recognize has a Carboni group surrounded by two alkyl chains which are asymmetrical. So we have ultimately an asymmetrical key tone due to the full reaction between these reactants and the presence of this catalytic acid, we will have an Inamine product recall that for our Inamine product, our major product is going to be the least substituted or the non zits product. So that would be our major product. And that will ensure that we have the least amount of olic strain in our final product for a more stable product. Recall that in our reaction, we have our first step which is to label our alpha carbon surrounding our carbonyl carbon, which are a part of the alkyl chains in form methyl three hexen. And in labeling these alpha carbons recall that they have 1/4 bond to an implicit hydrogen, which we are not given in our draw structure. However, we can draw those implicit hydrogens within our structure. Now, our dissociated strong acid will result in chloride ions being present in solution which will come and pluck off our proton or hydrogen from our asymmetrical ketone structure, which will allow a shift of the free rotating hydrogen bond here to shift and form a pi bond, which would be a carbon carbon pi bond. And that would cause our carbonyl pi bond to shift as electrons onto oxygen. Now, in our resultant structure for our first product, we draw our al chain and then we have our bond to our original carbonyl carbon, which now is no longer bonded to oxygen, but is now bonded to our Inamine where we have nitrogen now with three bonds. So it still has those two bonds originally to the methyl groups. And then it has its lone pair. We also now have that carbon carbon pi bond. And the key thing to note from this mechanism is that in the final steps due to the free rotation of that hydrogen bond that we drew in on our alpha carbon, the free rotation of that hydrogen bond allows for our, our group to be on either side of the pie bond. So we're going to begin our first product by drawing our resultant first product with its R group in an upward direction or on the same side of our pie bond. And then we would have a second product where we begin our alkyl chain, we have our carbonyl carbon. Now again bonded to our enemy nitrogen with it's three or two original bonds to methyl, as well as its bond to carbon. It's lone pair on nitrogen. And then we have our carbon carbon pi bond where then we can draw our methyl group or alkyl group in the opposite side or opposite direction of our pie bond. And again, this is key because in the final step of our mechanism, we have a Amini caton where nitrogen has a plus charge. And that ami cat become stabilized from the freely rotating alpha hydrogen from our alpha carbon of the carbonyl in our original structure where that alpha hydrogen is plucked off by a chloride anion and its remaining bond forms our carbon carbon pi bond in our structure. That free rotation of that alpha hydrogen allows for our remaining our group in our product structure to be on either side of the double bond and this is what influences the stereo chemistry of our products. Now take note that we need to label which stereo chemistry we have here. So we're going to recall that for estereo chemistry, our highest priority groups will be on opposite sides of the pie bond. Let's begin with our first product that we drew by focusing on our groups that are attached to the two carbons surrounding our carbon carbon pi bond in our product structure. So notice that we have our, I mean group here with the nitrogen and its three bonds and its lone pair, we have a methyl and we also have bonded to one of our pie bon carbons, a sec beutel group. So we have suck bey methyl and and I mean, so beginning with our first carbon in our carbon carbon pie bond and looking at the groups that we outlined that are bonded to that carbon, we have our suck beetle group versus our I mean group. So within our, I mean group, the bond that or the atom that is directly bonded to our carbon carbon pi bond is a nitrogen atom from our I mean group. So ultimately, we are comparing for this first structure, nitrogen versus the carbon that is a part of our sec bey group and directly bonded to the carbon in our carbon carbon pi bond. Recall that when assigning priority for stereo chemistry, we compare atomic numbers. Recall that nitrogen has an atomic number on the periodic table of seven. So we'll say Z for atomic number is equal to seven and carbon has a atomic number on the periodic table, equal to six or Z equal to six. Because nitrogen has a greater atomic number than carbon in our sex beetle group, which is bonded again to that carbon in the pi bond, we would observe that nitrogen takes priority number one. So we would label priority number one to nitrogen in our I mean group and priority number two to our carbon in our suck beetle group. Now, we're going to compare from the second carbon in our pi bond in our product structure. The group's so here we have carbon in our methyl group which is attached to our carbon within our pie bond of our product. And also note that we have an implicit hydrogen as the fourth bond for this carbon in our pi bond. And so ultimately attached also to this group is our implicit hydrogen. So in this case, we are comparing for this second carbon in our pi bond carbon versus hydrogen. Recall that carbon as we stated has an atomic number equal to six, whereas hydrogen has an atomic number equal to one. And so therefore, carbon takes priority number one, whereas hydrogen takes priority number two. So we can label our methyl group as priority number one as well. And notice that our two highest priority groups which weren't priority number one. Were our, I mean group and our methyl group that we outlined surrounding our pie bond in our product. And these groups are on the same side of our pie bond. Thus, we can label any groups that are highest priority. And on the same side of the pie bond as having Z stereo chemistry because we're called that for Z stereo chemistry, our highest priority groups are on the same side of the pie bond. And thus, that makes our second product that we drew as having estereo chemistry. Due to the fact that we can observe that our highest priority groups being our methyl and our nitrogen or I mean group are on opposite sides of the pie bond. So our final answer is going to be our two resultant structures with Z and E stereo chemistry as our final products of the reaction between four methyl, three hexen, a secondary amine and hydrochloric acid. I hope that this made sense and let us know if you have any questions.

What are the products of the following reactions? Show all stereoisomers that are formed.
c.
d.

Key Concepts

Imine Formation

Stereoisomerism

Catalysis in Organic Reactions

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