Using [TABLE 7-2]

Question

Using [TABLE 7-2]

as a guide, predict which member of each pair is more stable, as well as by about how many kJ/mol or kcal/mol. a. cis,cis-hexa-2,4-diene or trans,trans-hexa-2,4-diene b. 2-methylbut-1-ene or 3-methylbut-1-ene c. 2-methylbut-1-ene or 2-methylbut-2-ene d. cis-4-methylpent-2-ene or 2-methylpent-2-ene

Accepted Answer

Hey everyone and welcome back in today's video. We are looking at this problem which reads identify which member in each pair is more stable than the other. Also calculate their energy difference in college joules per mole. Using the table given below. Now in this problem we are seeing four pairs of Calkins and for each we will determine which of the two members is more stable. And we have to recall two main principles to successfully solve this problem. Principle number one is that stability increases with substitution. In other words, mono substituted Elkins, our least able, followed by die try and tetra where tetra substituted Elkins would be considered most stable. And rule number two trans eyes MERS are more stable dancers keeping these two factors in mind, we will consider the stability of each member. Now for a if we look at the two al kinds shown, we noticed that the first structure has a double bond with two hydrogen atoms. That these two positions. We have two hydrogen atoms that are implicit and here we have an additional hydrogen atom. So we only notice one al kill substitution. Over here. We may classify this as a mono substituted Elkin. However, if we look at the second structure, we may expand these two implicit hydrogen. So we may state that because we are seeing two alcohol substations bonded to the double bond, this is a di substituted Elkin and based on Rule number one, we noticed that stability increases with substitution. So a di substituted Elkin would be considered to be more stable than a mono substituted Elkin and that's sufficient for us to state that the second structure would be considered more stable for B. We have two structures and we may notice that they have different spatial arrangement of the metal substitutions across the double bond. The first structure is considered to be says because the two method groups are on the same side of the double bond and the second one is trans because the two method groups are pointing in opposite directions with respect to the double bond, they're equally substituted di substituted Calkins. And because of that, we may immediately move to Rule number two, which states that trans eyes murders are more stable than Cis. And we may also indicate that the trans one is more stable. So the second Elkin is more stable than the first for C. We noticed that we have to al kinds one of them is tri substituted because we see one hygiene which is implicit at this position and three L kill substitute wants the second one is di substituted because there are two implicit hydrogen atoms and the two alcohol groups that we are seeing. Now this is a triceps tooted and this is a di substituted al king between the two. A try substitute would be more stable based on rule number one, meaning we may indicate that the first structure is more stable than the second one and finally moving on to the we have a tetra substituted Elkin Because there are four methods substitue ints for the second structure we have a di substituted elkin which is in transposition but don't worry about it being trans factor number one is the primary factor in comparing stability as our major goal in getting the right answer. So when we don't, when we do not worry about the system trans configuration, we essentially first of all, once we valued whether they're equally substituted or not and in this case a debt was substituted as a more substituted alky while a di substituted one is less substituted. So we may say that the first structure is more stable than the second one because it's more substituted. Then the 2nd 1, we now have our more stable al canes labeled. We are ready to calculate the energy differences between them. To do that. We are going to use our table for part a. We first will notice that we have a mono substituted alk in and according to the table it has an a hydrogenation energy of 100 and 24 killer jewels per mole. The second structure is essentially germinal, di substituted alki in which has a hydrogenation energy of 117 kg per month. So if we subtract these two numbers, we essentially get an energy difference of -117 Killer Joules Per Mole. This gives us an energy difference of seven kg joules per mole. Now moving on to be The first structure that we are seeing is a cyst is more which is di substituted. A cyst di substituted I zimmer shows an energy the molar heat of hydrogenation of kelly jo's Brummel And the second structure is a trans di substituted elkin which has the molar heat of hydrogenation of 114. If we subtract these two numbers, which essentially noticed that the energy difference between them yes, Equal to -114 Killer Juice Per Moment. That would give us an energy difference of four kg per mall for C. We noticed that we have a tri substituted elkin which has the molar heat of hydrogenation equal to 111 kg per mole. Based on the table. In the second structure Is basically di substituted elkin and in particular that's a germinal di substituted Elkin which has the molar heat of hydrogenation according to the table equal to killer jobs per month. We calculate the energy difference. We may take a greater numbers. Our first number because we just want to find the difference is the absolute value. We already know which one of them is more stable. So we may take 117 minus kg joules per mole, Which gives us six killer jewels from all. And D. We have a tetra substituted elkin with the moller heats of hydrogenation equivalent to 100 and some. And the second one is a die substitute at al kin which also has a trans arrangement. The trans di substituted elkin shows an energy. The molar heat of hydrogenation equal to kg promote. Based on the table. If we find that difference is the absolute value, we get 114 -100. and 14 minus 100. And some killer joules per mole. That gives us four kg joules per mole. We may now add these numbers, our final answers for a. We have seven, Then we have four sex and four. Let's now conclude our findings or a. The second structure is more stable and the energy difference is seven kg per mole for B. The second structure is more stable. The energy difference is four kg per mole. For see the first structure is more stable and the energy differences six kg per mole. And for the the first structure is more stable and the energy differences for killing joules per mole. The correct answer to this multiple choice question is D And if you have any questions, please feel free to drop them down below in the comment section and thank you for watching

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