Match each of the fatty acids with the appropriate melting poi...

Question

Match each of the fatty acids with the appropriate melting point:

Table showing melting points of fatty acids: 13°C, 45°C, -5°C, 76°C, 63°C, with room temperature at 25°C.

a) CH₃(CH₂)₁₈COOH ______

b) CH₃(CH₂)₁₄COOH ______

c) CH₃(CH₂)₁₀COOH ______

d) CH₃(CH₂)₇CH=CH(CH₂)₇COOH ______

e) CH₃(CH₂)₄CH=CH-CH₂-CH=CH(CH₂)₇COOH ______

Accepted Answer

Alright, so this practice problem wants us to match each of the fatty acids that we see down below right here with the appropriate melting point that we see over here. And notice that we're given the room temperature as 25 degrees Celsius. Now, the first thing that we need to notice here is that for all of these fatty acids, notice that they all have a carboxylic acid group that I'm highlighting in green, And really the rest of this that we see are all just hydrocarbon chains and so these are indeed fatty acids. The next thing that you guys should note is that these first three fatty acids do not have any double bonds being indicated, which means that these first three are going to be saturated fatty acids. And so what we can do is go ahead and label these first three as being saturated. And then what you should notice is that these down here actually do have the presence of double bonds and so because they have double bonds these 2 are going to be of course unsaturated. And then, of course, what we need to recall is that saturated fatty acids tend to be solids at room temperature and unsaturated fatty acids tend to be liquids at room temperature. And so, if you are going to be a solid at room temperature, that means that your temperature must be above the room temperature and so, notice that there are 3 temperatures that are above the room temperature of 25 degrees Celsius and so these three temperatures must correspond with these 3 fatty acids and then we have 2 temperatures that are below the room temperature and so these would correspond with being liquids at room temperature. And so these 2 must correspond with these two down below. These two temperatures must correspond with these, down below. And so, the next thing that we need to do is, for these saturated fatty acids is consider the total number of carbon atoms or the length of the chain. Now notice that for, for fatty acid a here, there's, 1 plus 18 plus 1, which is 20 carbon atoms in the first one. So we'll put 20 c here. For fatty acid b, we have 1+14+1 which is 16carbon atoms and then for fatty acid c down below we have, 1+10+1 which is 12carbon atoms. And so recall from our previous lesson videos that the greater the length of the chain, the higher the temperature is going to the melting temperature is going to be. And so, of these 3 here, that we've designated for these positions, the highest one is 76 degrees and that's gonna correspond with, the one that has the longest chain. So we can go ahead and cross that off. The next highest is 63 degrees and that's gonna correspond with the next highest chain. So, we can cross that off and then, the next highest is going to be, 45 degrees. So, that's gonna correspond with this chain right here and so we can cross that off. And then looking at, these down below here, what we need to notice is uh-uh count the number of carbon atoms as well. So, for fatty acid D here, we have 1, plus 7, which is 8. Here's 9, 10, plus another 7 is 17, and then plus 1 is 18. So this has 18 carbons. And then, counting the carbon atoms here in fatty acid e, we have 1 plus 4, which is 5, plus 1, which is 6, here's 7, 8, 9, 10, plus another 7 is 17, plus another one is, 18. And so notice that they both have 18 carbons, which means that their length, the length of the chain is exactly the same, and that's not what's going to differentiate these 2, melting points. And so instead, what we need to take a look at is the number of double bonds. Notice that fatty acid d only has one double bond, whereas fatty acid e actually has 2 double bonds. And recall that the greater the degree of saturation or the more double bonds there are, the lower the temperature is. And so because fatty acid e has more double bonds, it's gonna have the lower melting temperature of negative 5 degrees Celsius. And then of course, that leaves the 13 degrees Celsius for this last one and that concludes our practice problem here, and so I'll see you guys in our next video.

Difficoltà con Biochemistry?

Match each of the fatty acids with the appropriate melting point:

a) CH₃(CH₂)₁₈COOH
b) CH₃(CH₂)₁₄COOH
c) CH₃(CH₂)₁₀COOH
d) CH₃(CH₂)₇CH=CH(CH₂)₇COOH
e) CH₃(CH₂)₄CH=CH-CH₂-CH=CH(CH₂)₇COOH __

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Set di pratica su Fatty Acids

Difficoltà con Biochemistry?

Match each of the fatty acids with the appropriate melting point:a) CH3(CH2)18COOH ______b) CH3(CH2)14COOH ______c) CH3(CH2)10COOH ______d) CH3(CH2)7CH=CH(CH2)7COOH ______e) CH3(CH2)4CH=CH-CH2-CH=CH(CH2)7COOH ______

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Set di pratica su Fatty Acids

Match each of the fatty acids with the appropriate melting point:

a) CH₃(CH₂)₁₈COOH
b) CH₃(CH₂)₁₄COOH
c) CH₃(CH₂)₁₀COOH
d) CH₃(CH₂)₇CH=CH(CH₂)₇COOH
e) CH₃(CH₂)₄CH=CH-CH₂-CH=CH(CH₂)₇COOH __