Draw structural formulas of at least two alkynes of each molec...

Question

Draw structural formulas of at least two alkynes of each molecular formula.

1. C₆H₁₀

2. C₈H₁₂

3. C₇H₈

Accepted Answer

Hey everyone. Let's do this problem. It says give structural formulas of two different AL kind molecules that correspond to each of the following molecular formulas C four H six, C five H six and C nine H 12. Well, I am a fan of process of elimination. So the easiest way to eliminate some answer choices is to double check and make sure they have the molecular formula that is specified in the problem. So I like to just count the number of carbons. So we can count starting with C four H 612341234 carbons 1234512345. Okay, so this one doesn't follow that molecular formula 1234512345. Okay, neither does C 12341234. All right, C five H 61234512345 five and five and C nine H 123456789123456789, nine and 9. So just by counting carbons, we've already eliminated 50% of the answer choices. Okay, now down to the real stuff. So the problem says that we are given structural formulas for AL kind molecules but we're just given the molecular formula. So how do we know how many triple bonds there are or if there are double bonds or rings? Well, we can calculate the index of hydrogen deficiency to tell us what types of structural characteristics there are that lead to unsaturated in And we can use the formula two N. Plus two minus H divided by two. Where N. Is the number of carbons and H. Is one for every hydrogen or halogen. Sorry, zero for every oxygen and negative one for every nitrogen. Okay, but here we only have carbons and hydrogen. So we'll simply just count the hydrogen and once we calculate that, how will that translate into our structure? Well, for the index of hydrogen deficiency, every value of one can equal a double bond or a ring and every two values can equal a triple bond. Okay, so let's see what the index of hydrogen deficiency is for each of these molecular formulas. And then we can see which structures we have. So 1st C four H 6. We have two times four carbons plus two -6 hydrogen that gives us two. So our value is two that could translate into a two double bonds or a double bond or a ring. But we know that these are all al kind molecules so that too has to go for one triple bond. Okay, so that means our structure could look like this. If we have four carbons we could place one triple bond on the end carbons or we could place it in the middle. Right. Those are our only two options just adding the hydrogen is here. So with that said let's see if both of our answer choices have that structure. So we have our triple bond on the end triple bond in the middle. Triple bond at the end triple bond in the middle. So both of these satisfy C. Four H. Six. Okay let's go to C. Five H. Six. Index of hydrogen efficiency equals two times five plus two minus six divided by two. And that will equal three. So we can have one triple bond. We know it has to have a triple bond which would account for two of these values. So that means we can either have one double bond or a ring. Well looking at the answer choices, we know there are no rings. So let's give it a double bond. So we can have one 45. And we can also do four carbons with a branched carbon to total five. So we can add a triple bond at the end in a double bond in the middle or in this case where we have our four carbons and then our branch carbon we can have a double bond on that branch and a triple bond on this end carbon. And there are other possibilities but we only have to draw one or sorry to. So this satisfies both A. And D. Alright so it's gonna come down to C. Nine H. 12. So I. H. D. Equals two times nine plus two minus divided by two is going to be four. Okay so we know we have to have one triple bond and that means we can either have one more triple bond or two double bonds or two rings Or one double bond in one ring. But based on our answer choices, it looks like it's either a triple bond or double bond. So let's see what we can formulate here.  aN:aN:000NaN 123456789. We could put two triple bonds at the ends of these chains, oops. Or 23456789. We could do one triple bond at the end and one towards the middle. Alright, so let's look at our answer choices. Well, answer choice A looks like the one we just described. And answer choice D. Shows a triple bond and a double bond. Well, that would only count for two plus one. Is three index of hydrogen deficiency. So this does not satisfy the index of hydrogen deficiency calculation. So that leaves a to be the correct answer for this problem. That's it for this one. I hope this video is helpful. And thanks for watching

Draw structural formulas of at least two alkynes of each molecular formula.
1. C₆H₁₀
2. C₈H₁₂
3. C₇H₈

Key Concepts

Alkyne Structure

Molecular Formula and Degree of Unsaturation

Isomerism in Alkynes

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