For each molecular formula, draw all the isomeric alkynes, and...

Question

For each molecular formula, draw all the isomeric alkynes, and give their IUPAC names. Circle the acetylenic hydrogen of each terminal alkyne.

(b) C₆H₁₀ (seven isomers)

Accepted Answer

Hello everyone. Today we have the following problem. Draw all the possible alkaline isomers with the molecular formula of the following C seven H 12 give the systemic name of each isomer and highlight the aceto hydrogen if the isomer is a terminal alky. So in drawing the isomers, we know that we have seven carbons total. We wanna first start by drawing the seven carbon chain, followed by the six carbon chain and then the five carbon chain and so on with the substituents. So for the seven carbon chain, we can have our alpine view terminal wherein we have the seven carbons drawn out along with the associated hydrogens. But then because it is terminal, we have to make note of this a seal hydrogen. Now this name because it is composed of seven carbons and the alky is on carbon one or the first carbon. This gets the name of hept for seven carbons followed by one. I now moving on, we can move this alky on to the 2nd and 3rd carbon respectively. Such that if we were to move it to the second carbon, we would still have our seven carbons total. But our alky would be on carbon too instead. And this is an internal alky because it is on carbon two, we have the name Hept for seven carbons followed by two I and then as discussed, we can also move it to the third carbon, which is also an internal alpine. And because it is on carbon three, we have Hept 39. No, that is it for the seven carbon chains. For the six carbon chain, we can once again have it on an internal or on a terminal outline as the following. And then we have our six carbon chain thats actually draw the following. So we have our six carbon chain but to draw the additional carbon, we have to draw a methyl substituent, which we can draw on this particular carbon. And because this is a terminal outcome, we have to highlight the acetyl hydrogen as the following. And with naming this structure, we have the parent chain of the hex one I, but we also have a substituent. So if we number this al kine from left to right on carbon five, we have a methyl. So we have a five methyl, followed by the parent name of hex for six carbons one I. Now, as you do with the seven carbon structure, we can essentially move this alky or we can move the placement of the methyl group. So if we move the placement of the methyl group, we could move it from the carbon, it was originally on which was carbon five to a different carbon. So we can move it closer to the alpine or we could move it to the different carbon. But let's focus on this one first. As before, this will have a terminal al a terminal alky. So it will have that hydrogen. And since we moved the methyl group over one carbon, we have a new name which now it is on the carbon three. So we have three methyl followed by the parent name of hex one I because the alky is still in Carmen one. And then if we decide to move the alky into an internal alky, we would have the following wherein now we still have a substituent, the methyl group on carbon five. So we have a five methyl all by the parent name of Hex and then is on carbon two. So we have two I moving on, we can move that methyl group once again, such that we have the following. And this is keeping the alky internal, but now we have a new placement of that methyl group. So instead of carbons three or five, it is on carbon four. So we have four methyl and the al kine is once again still or this time it's on carbon two. So we have hex two I and then we can also move the alky such that it is further internal or on Carmen F three. And now we have two metal from left to right, followed by hex three I. And now that is it for the six carbon structures. For the five carbon structures, we can once again have our terminal Altin as the following we have in that terminal hydrogen. So we will highlight it as the following. And you may note that we have to draw our two substituents or two additional carbons as methyl groups to accommodate for the remaining two carbons. And so from left to right, we have to name this such that we name the substituents. First, we have two metals and carbon four. So we have 44 dimethyl and this will be followed by the apparent name of Pence for five carbons. And then one eye, we then can have a similar structure where we have a terminal alky were and we have a terminal alcy, but we just moved one of the metals over such that we have the following and we will highlight the hydrogen at the end. And so this one has had a 44 dime though it'll be 34 dimethyl hent one I and then we have just a few more that we can draw, we then we have our terminal alkin as before, but then we just move the two metals onto the carbon closest to the alkaline. So instead of 33 or so instead of 44 dimethyl, we can have 33 dimethyl, followed by the parent name of pint one, I And then if we wanted, we could draw our terminal alky with our two ethyl groups on that third carbon, such that we have the three ethyl pent one I. And then last, but not least, we can have an internal outline with five carbons such that we have the following. And the alka is on carbon two and we still have the 44 dimethyl. So we can have 44 dimethyl pint two I. And then of course, we have our aceto hydrogens highlighted as the following. And so with that, we have solved the problem overall, I hope this helps. And until next time.

For each molecular formula, draw all the isomeric alkynes, and give their IUPAC names. Circle the acetylenic hydrogen of each terminal alkyne.
(b) C₆H₁₀ (seven isomers)

Key Concepts

Isomerism

Alkynes

IUPAC Nomenclature

Watch next