hello everybody my name is Iman welcome back to my YouTube channel today we're going to do practice problems associated with nomenclature for MCAT organic chemistry let's go ahead and get started the first problem says which of the following lists the correct names for ethanol methanol and ethanol respectively all right so the common name for ethanol well it has this ethane or the beginnings of ethane here which references two carbons and it has an Al ending which refers to an aldehyde so ethanol is an aldehyde with two carbons and it looks something like this now we talked about this in lecture but the common name of ethanel is SC all right the next thing we're given is methanol methane meth here is for one carbon and this Al ending again indicates the presence of an aldehyde so if we go ahead and draw it one carbon all right and we draw an aldehyde this is what it would look like here now the common name of methanol like we like we covered in lecture is formaldehyde formel dehyde all right fantastic and the last thing we were given is ethanol again ethane here is for two carbons this ol ending references a alcohol group and so our molecule looks something like this two carbons and an alcohol group now for this molecule the common name of ethanol is ethyl alcohol all right ethyl alcohol so we're looking for an answer Choice all right which of the following lists the correct common names for ethanel methanol and ethanol we're looking for an answer choice that says acetyl aldehyde all right formaldehyde and then ethyl alcohol in the same order all right in the same order as it's referenced in the question all right and so the correct answer here for one is going to be answer Choice a fantastic two says which of the following are considered terminal functional groups aldehydes ketones carboxylic acids those are our three choices now from the lecture we remember that aldehydes and carboxylic acids are characterized by their positions at the end of carbon backbones and are thus considered terminal groups now as a result the carbons to which they are attached are usually designated carbon number one now ketones are not considered terminal functions ketones are internal by definition because there must be a carbon on either side of the carbonyl all right and so with that being said only aldehydes and carboxylic acids are referred and characterized all right as terminal groups so one and three is correct so two the answer Truth for two is C beautiful three says if all prefixes were dropped what would be the name of the parent root of this molecule all right so looking at this molecules just quickly write some stuff down here we see an aldehyde group right here aldehyde all right we see a carboxylic acid all right carboxylic acid now the highest priority functional group in this molecule is the carboxylic acid so this gives us a big hint all right this gives us a big hint that com that that this will be a component of the backbone and it's going to provide the suffix of the molecule now what we notice here is this molecule all right has this carboxylic acid and there is a total of one two three carbons in the main chain all right so it's asking us to forgive if all prefixes were chopped what would be the name of the parent chain or the parent root of this molecule it would be propane but replaced with what we know about carboxylic acid since that's the highest priority group for carboxylic acids we convert the E to OIC acid so this becomes propanoic acid all right the name of the parent root of this molecule is propanoic acid this is answer choice C beautiful 4 says what is the highest priority functional group in this molecule all right this molecule if we look at it looks like an anhydride the only other group are pretty much hydrocarbon chains which will provide part of the name of the parent root now keep in mind that when a carbonyl group is present with a leaving group the larger functional group takes priority over the carbonyl group alone all right just a reminder nevertheless this molecule is anhydride and just for fun this is an anhydride molecule all right if we divide it through the middle both sides are equal and if we're trying to name it we're going to look at how long the carbon chain is three carbons all right three carbons this is propane all right except this is an anhydride so we take the AE all right we convert it to propanoic all right and then instead of saying acid we say and hydride so just for fun that's the name of this molecule that's not what this question is asking at all just what the highest priority functional group is which is anhydride and then just for additional practice we went ahead and named it it's propanoic and hydride beautiful C says the IUPAC name for the following structure ends in what suffix all right now looking at this molecule you can go ahead and start to identify groups this is an alcohol group all right this is a halogen this is a double bond so an alkene all right this is a triple bond so an alkyne all right here we have an ether all right here we have a carboxylic among the functional groups presented here carboxylic acid has the highest priority all right it has the highest priority so we've determined that now now that we know that carboxylic acid is the highest priority here and the question it asks the name for the following structure ends in what suffix now we just reference what that was for carboxylic acids for carboxylic acids the compounds end with OIC acid that is going to be answer Choice C all right so 5 is C 6 says which of the two possibilities below correctly numbers the carbon backbone of this molecule we have numbering scheme one and numbering scheme two all right so looking at these molecules all right here is the carbon chain for this one one two three four five six and this is the carbon chain here one two three four five six all right now looking at numbering scheme one we have our first substituent at three and our second substituent at four all right so we have our first substituent at three all right and our second substituent at four beautiful now looking at numbering scheme 2 we have a substituent out position two and we have a substituent at position three all right now when naming alkanes you have to first locate the longest carbon chain all right you have to find the longest carbon chain this one has six carbons but this one has five carbons so what they've numbered here isn't even the longest carbon chain all right and then the numbering system should give the alkyl groups the lowest possible number all right so first thing you have to identify is have is have they chosen the correct longest carbon chain and the answer is no this one they're the same molecule this one I chose a carbon chain that has six carbons I'm terribly sorry about that all right this one is just identifying a five carbon chain all right and so with that it becomes easy to say that the correctly numbered scheme is going to be numbering scheme one all right so six is a beautiful seven says what is the proper structure for two three dihydroxy butane butane dioic acid or also commonly known as tartaric acid all right so we have two three dihydroxy butane dioic acid so butane that means four carbon chain all right two three dihyde hydroxy means we have two alcohol groups at position two and three all right and dioic acid means we have two carboxylic acids all right and obviously we have two carboxylic acids and we know that carboxylic acids or terminal groups then the carboxylic acids have to be at either end all right so we start by drawing a four carbon chain really quickly so here's our four carbon chain to each end of the chain we draw a carboxylic acid because it says dioic acid all right and again carboxylic acids are terminal group so we draw them at the last carbons on both sides of the four carbon chain now we number one two three four we can do it either way all right and at position two we put an alcohol group and at position three we have another alcohol group so this we have drawn now two three dihydroxy butane dioic acid looking at the answer choices the correct one that's exactly the same as the one we drew is B seven is B beautiful 8 says the common names for the aldehydes and carboxylic acids that contain only one carbon start with which prefix all right the answer to this is form form is a prefix shared by the common names of methanoic acid which is also known as formic acid and methanol also known as formaldehyde all right so those were things we mentioned that are common names that we had to memorize for the MCAT and what you notice for the one carbon aldehyde and carboxylic acid they start with the form prefix so eight is answer Choice B beautiful 9 says what is the IUPAC name for the following structure so the first task in naming a compound is identifying the longest carbon chain here I've highlighted the longest carbon chain you can try other combinations um by the way starting right here I'm going to highlight it in blue is completely equivalent to what we have already highlighted in yellow all right and so let's count now how many carbons we have in this longest carbon chain one two three four five six seven seven carbons all right so this gives us the alkane name heptane all right so heptane then we must make sure that the carbons are numbered so the substituent's position numbers are as small as possible this compound has two methyl groups minimizing their position numbers requires us to number the chain from right to left all right so we number it one two three four five six seven all right this way we have our first substituent at two and our second substituent at five at two we have a methyl group so two methyl and when at five we have a methyl group as well so 5-methyl all right so what we know now is we have a seven carbon chain we call that heptane there are no modifiers or functional groups all right there are two substituents we've numbered the chain and we have one substituent called two-methylene another one called five-methyl all we want to do now is put it together both are substituent are methyl groups so we can write dimethyl and right before dimethyl we could put the two numbers two and five all right and then at the end we cap it off with our chain name heptane all right two five dimethyl heptane all right that is going to be answer Choice a all right nine is a beautiful let's move on to 10 now 10 says what is the IUPAC name for the following structure all right again our first step is to identify the longest chain all right the longest chain is this one of course sometimes this will take some trial and error you might want to try this one two three four five you might want to try one two three four five six so on and so forth you just have to identify the longest carbon chain and that's this one and it has one two three four five six carbons so we have six carbons which means the parent name is hexane all right now second thing we notice is there's no modifiers there's no other functional groups this is just alkane all right so we don't have anything that's going to modify hexane because there's a functional group present the next step we do is we then number our carbon such that the lowest possible combination of numbers is given to the various substituents all right the way that we want to number this is so that the substituents get lower numbers so we're going to start from here one two our first substituent is that two second one is at three this one is four this is four and the third one is at five so we have if we number starting here we have substituents at two three and five if we go the other way one two three four five six we would have substituents at two four and five so obviously the two three five would be better because we want as the lowest numbers possible so then our numbering is like so all right we have a substituent at two all right it's two methyls so it's two two dimethyl for position two then we have an ethyl group at three all right so three ethyl and then at position five we have another methyl group now we want to put it all together all right we want to put it all together all right in alphabetical order don't forget alphabetical order so first we have three ethyl all right then what we can do is write two two five trimethyl because we have actually three total methyl substituents and this would be an easier way to write it so three ethyl two two five trimethyl and then we cap it off with the parent name hexane all right so the name for this is three ethyl 225 trimethyl hexane all right that makes the correct answer here all right answer Choice d alright so 10 is d beautiful eleven says the IUPAC name for the following structure starts with what prefix all right it starts with what prefix so we can identify a carbon chain one two three four all right that's a butane but we have an alcohol group so this becomes butane null all right and now we want to make sure that this butanol this alcohol group gets the lowest number so we number it one two three four all right and now it becomes two butanol all right but that's not it for the name we also have a substituent at position two and it's a methyl so we say 2-methyl then we can say two butanol all right methyl is that substituent so the name for this molecule is two methyl two butanol what is the prefix they're asking for the substituents here it's going to be two methyl so the correct answer for 11 here is answer Choice B beautiful 12 says nadh is a coenzyme that releases high energy electrons into the electron transport chain it is known as nicotinamide adenine dinucleotide or diphospho pyridine nucleotide what functional groups exist in this molecule phosphate amide or anhydride so in the name the suffix amide in Nico nicotinamide indicates that this compound contains in a mild functional group so it definitely has a functional group then in the other name that it gives us right there's two names diphospho indicates that there are two phosphate groups so this molecule also has two phosphate groups all right even if we you know we did not study the prefix phospho from this chapter but you can recognize that phospho here in the name is a phosphate group all right there is no indication though of an anhydride group in the naming all right and therefore we can only say that there is a phosphate a phosphate and in the mind and the correct answer here for 12 is going to be C beautiful 13 says pyruvic acid one of the end products of glycolysis is commonly called acetylformic acid based on its common name the structure of pyruvic acid must be blank all right so we can use the name acetyl formic acid to figure out what our functional groups are so the prefix a c refers to two carbon unit with one carbon in a carbonyl group so think of acetic acid or acetic anhydride or acetyl aldehyde all right then the carbonyl group then the carbonyl carbon is the point of attachment to another functional group formic acid is a single carbon carboxylic acid if you remember from our lecture therefore acetyl formic acid is an acetyl group directly attached to formic acid all right so that means all right two carbons with one having a carbonyl all right directly bonded to a carboxylic acid all right with three carbons all right and therefore the correct answer here is going to be answer Choice a beautiful 14 says consider the name two three diethylpentane based on the structure implied by the name the correct IUPAC name for this molecule is blank so we can go ahead and draw it out all right so it's telling us two three diethyl pentane so pentane is five carbons one two three four five one two three four five and we have um two ethyl groups one here at position two and one here at position three based on the structure applied by this name what is the correct IUPAC name so when we draw it here all right we're drawing pentane it's telling us that this chain right here is the parent chain all right now then it tells us what's the correct IUPAC name so obviously this structure that's not its name all right we have to actually observe it a little more if it's not the name that means they probably got something from the naming wrong and usually your first point to check is did they find the longest carbon chain so if they named it in the way that um you know as two three die ethyl pentane and we were able to draw it like so all right that's what they thought it was named and they thought this was the longest carbon chain right here this is four or five carbons but their naming is wrong which means they probably got the parent chain the longest carbon chain wrong so if we observe further what we notice is that this part right here is the longest carbon chain it's six carbons long all right if you have six carbons then you then your root name is hexane not pentane and your substituents are going to be a methyl group at position four and an ethyl group at position three and so you have three fl4 methyl hexane that is the correct name for this molecule and that is going to be answer Choice C all right fantastic alrighty that's all I have for you all right let me know if you have any questions comments concerns down below other than that good luck happy studying and have a beautiful beautiful day future doctors