Give a systematic (IUPAC) name for each alcohol. Classify each...

Question

Give a systematic (IUPAC) name for each alcohol. Classify each as primary, secondary, or tertiary.

(f) Structural formula of an alcohol showing an -OH group, indicating its classification as primary, secondary, or tertiary.

(g) Chemical structure of cyclopentanol with the formula CH2OH, illustrating alcohol nomenclature and classification.

Accepted Answer

All right. Hello everyone. So this question is asking us to determine the Iupac name of the following alcohols and classify them as either primary secondary or tertiary alcohol. Here, we have two different skeletal structures labeled one and two respectively. Now recall first and foremost that in a skeletal structure, carbon is not explicitly shown and neither are any hydrogen atoms connecting two carbon atoms. Instead, carbon carbon bonds are showcased as lines with every corner or open end representing carbon atoms with enough hydrogens to complete carbon's for bond requirement. Now hetero atoms and hydrogen atoms bound to hetero atoms are shown in skeletal structures. But with this in mind, let's recall a few rules when it comes to determining an Iupac Nate. First, the easiest first step is to locate the parent chain and name it accordingly. The parent chain is the longest continuous chain of carbon atoms that contains our primary functional group. So based on that primary functional group, the suffix of the parrot name must be change accordingly. However, once the parrot name has been identified and any modifiers depending on functional groups have been added, we can proceed to step two, which is all about locating substituents and naming them accordingly. Recall that substituents are listed in alphabetical order before the name of the parent chain and each substituent must have a number indicating where on the parent chain it's located. Now, before we talk about the actual Iupac name for both of these compounds. Let's start by determining the degree of each alcohol, whether it's primary, secondary or tertiary. Recall that the terms primary, secondary and tertiary, describe how many other carbon atoms are attached to one particular carbon atom. So in this case, were going to focus on the Corbin Adams that are connecting directly to the hydroxy group in both alcohols here, depending on how many other carbons this hydroxy carbon is connecting to that determines its priority. For example, looking at compound number one here, the carbon atom that contains the hydroxy group is attached to only one other carbon atom. This makes it a primary carbon and therefore a primary alcohol. If there were two carbons attached, it would be a secondary alcohol. And if there were three carbons attached, it would be a tertiary alcohol. But with this in mind, we can see that compound number two is also a primary alcohol. So both are examples of primary alcohols. And now let's begin with the Iupac name of compound one as mentioned before, our first step should always be to locate the parent chain. Now, it just so happens that we have two different functional groups in compound one, we have our alcohol, of course, and an alkene as well. However, because the alcohol functional group has a hetero atom, it is of a higher priority than the all key. So the parrot chain should make sure that the hydroxy group has a smaller position number. And the parent name is going to have a suffix based on the presence of the alcohol as the higher priority functional group. So here if I count the longest possible parrot chain that comes out to seven carbons. Now, even though the alcohol is the primary functional group, there is going to be a modifier based on both the all keen and the alcohol. However, the alcohol should come less since it is higher priority. Now because the power chain is seven carbons that would be hept and normally an alcohol with seven carbons in its parent cane would be referred to as Hep tool with being the suffix for the presence of the alcohol functional group. However, because we also have an alkene in this compound, I would drop the A in A N as in hepton, no and replace it with an E het eno. But we do have to be a bit more specific because we should always mention where these functional groups are located in the parent chain. So the double bond is in between carbons four and five. So I would describe that as for E and the hydroxy group is on carbon one. So I would describe that as one all. So with the modifications to the parent name that comes out to HEPT for in one all notice how I'm using dashes to separate numbers from letters. But now let's talk about substituents that are present in the parent chain here, disregarding the hydroxy group and the alkene, we can see that there are a total of four substituents, all of which happened to be the same, we have four methyl groups present in this compound. Now recall that when there are multiple copies of the same substituent present in a molecule. Another prefix is necessary before the substituents name to specify how many there are. So that would be D or excuse me, die for two, try for three and tetra for four. So in my Iupac name instead of writing simply metal, I would write tetra methyl since there are four methyl groups in the same compound. But even though it's the same substituent, each one of them should have a position number. So I have one on carbon four, another on carbon five and two more on carbon six. So that comes out to four comma five, comma six, comma six, you should always make sure to have one position number for each substituent present even if one number is repeated. So after placing my position numbers in front of the substituent name, this comes out to 4566 tetra methyl helped for in one all, notice how here, I'm using commas to separate numbers from each other. But there's one more thing missing. There's one more step to complete the Iupac name for part one here because recall that some alkenes can have E and Z isomers. So we have to be specific and list which one is present in the given molecule. So this means that we have to consider the higher priority group attached to each carbon of the starting double W. When considering carbons four and five, we can see that there is a larger substituent connecting to each carbon compared to a methyl group. So I'm going to circle both of these because they would be the higher priority group on both sides or on each side because the higher priority groups are on opposite sides of the double bond. That would be the E configuration. So in front of my preliminary Iupac name, I would write E in parentheses and then separate numbers from letters with a dash. So the true Iupac name is E 4566 tetra methyl hep for in 10. And now lets talk about part two or compound two. Now, this particular case is a little bit more complex because we still have an alkene and an alcohol. But the all keen is inside of a ring and the alcohol is part of a straight chain. However, because the alcohol is considered the higher priority functional group, the straight chain of carbon that it is connected to would be considered the parent chain with the ring considered a substituent. So if I make sure that my hydroxy group has the smallest possible position number, that would be carbon number one of a three carbon parent chain. Now here a three carbon chain containing an alcohol functional group would be known as propanol. Once again, OL is the suffix for the alcohol functional group. And this time, the A N in propanol would stay the same because there are only single bonds in between the carbons of the parent chain. However, because we should always strive to be specific since the hydroxy group is on carbon one, I would say propane one all instead, oops. So now let's talk about how to name our substituent. Now, the substituent is relatively complex on its own. So it's almost as if we have an Iupac name within an Iupac name because the substituent itself is a five carbon ring with a double bond in between carbons one and two of the rink. So that modifies the name from cyclops to cyclops. Now, specifically because the double bond is in between carbons one and two, that would be described, let me actually make some space here, but that would be described as cyclops one E. However, because this is a substituent, I cannot leave my name here as is for my substituent because carbon one of the ring is connecting to the parrot chain, I would drop the very last E clop one E and replace that with one hill. So that's cyclops one E one L. So now I enclose the name of my substituent in parentheses to avoid confusion and add the number three and a dash in front of it because my substituent is attached to carbon three of the parrot chain. So here the complete iupac name is three cyclen one in one prop 10 for part two and both of these are primary alcohols and there you have it. So if you watch this video all the way through, thank you so very much for doing so. And I hope you found this helpful.

Give a systematic (IUPAC) name for each alcohol. Classify each as primary, secondary, or tertiary.
(f)
(g)

Key Concepts

IUPAC Nomenclature

Classification of Alcohols

Functional Groups

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