Tell how many different products you would obtain on hydrogenation of the triacylglycerol in Problem 23.49 if:

a. One double bond was converted to a single bond
b. Two double bonds were converted to single bonds
c. Three double bonds were converted to single bonds
d. All four double bonds were converted to single bonds

Question

Tell how many different products you would obtain on hydrogenation of the triacylglycerol in Problem 23.49 if:

a. One double bond was converted to a single bond

b. Two double bonds were converted to single bonds

c. Three double bonds were converted to single bonds

d. All four double bonds were converted to single bonds

Accepted Answer

Welcome back, everyone identify the number of different products obtained upon hydrogenation of the triazole glycerol shown below. Using the following numbers of double bonds that are converted into single bonds. And we're given four parts. The first one is one, the second one is two, the third one is four and the fifth one is five. So basically, we want to analyze the given structure and we want to focus on the carbon, carbon double bonds. We have 123 or, and five double bonds. So let's begin with the first part. How many products can we get if we only convert one double bond into a single bond? Well, essentially because we can convert any of them, we can have five products. So in other words, we reduce one double bond any of the five and we leave the remaining double bonds, meaning we have five possibilities. Now, for the second part, we are essentially focusing on two double bonds and this becomes more difficult because we can reduce the first two, we can reduce the first one and the third one, the first one and the fourth one, the first one and the fifth one maybe it's the second one and the third one. So the problem becomes complicated and we're going to use the formula four combinations, which essentially tells that the number of combinations would be the factorial of N divided by the factorial of K multiplied by the difference between N and K factorial. So in this case, the total number of double bonds is five. And that's our N. So we take five factorial and the number of double bonds that we are turning into single bonds is K, which is two. So we take two factorial and five minus two gives us weve, well, essentially fact, we can find that using a calculator, the factorial of five is simply five multiplied by four, multiplied by three, multiplied by two, multiplied by one. So all the natural numbers going down until we reach one and then two factorial. Well, that's two multiplied by one, multiplied by three factorial, which is three multiplied by two, multiplied by one. So we can essentially get rid of those ones, we can get rid of two. And while we can get rid of, we, we get 20 divided by two, which gives us 10. So there would be 10 products. And now for part number, we were going to use the same formula. Our N, in this case is the number of double bonds in total, we have five and we want to reduce four double bonds. So we take the formula and we get five factorial divided by four factorial multiplied by one factorial because five minus four gives us one and essentially five factorial is just five, multiplied by four factorial. And at the bottom, we have four factorials. So those cancel each other out and we end up with five. So there would be five products. And of course, or number four, if we have five products. Well, I'm sorry, if we reduce five double bonds, we reduce all of them. And we only get one single product. We can verify that using the same formula, right? If we take five for N and 54 K, we simply end up with five vector divided by five vector multiplied by zero vector. And this gives us five divided by five, multiplied by one, which is a one. So basically we have our final answers. Let's label them and conclude the video. Thank you for watching.

Verified Solution

Key Concepts

Triacylglycerol Structure

Hydrogenation Process

Isomer Formation