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Video Solution: Using the method outlined in Section 2.13, give an IUPAC nam...

Problem

Using the method outlined in Section 2.13, give an IUPAC name for each of the following alkyl groups, and classify each one as primary, secondary, or tertiary:

Video Transcript

Hey everyone, so it's our job in this problem to identify this alkyl group and give it an IUPAC name as well as identifying if it's primary, secondary and so forth, so let's do the first part. Notice that we have just something here that just ends, right? It's just a straight line, so what this represents is is going to be an alkyl group, right? So, in order to identify if it is going to be primary, secondary or so forth but what we need to do is look at this carbon that's the one that's bearing our the alkyl group, is this first one coming off, how many carbons is that one attached to? we're gonna say it's only attached to this one carbon right here, so that makes it primary, okay? when we attach to that carbon. Now, we know there's a primary alkyl substituent but what do we need to do now? we need to name it. So, say we had that line again, right? And now this is going to be the line that represents this one right here, okay? And I'm coming off that we can draw this, we can say we have CH2, CH2, CH and now I'm going to draw that CH2, CH3 group and then we have CH2, CH2, CH3, so this is going to be represented by this group right here and then our six carbon chain is going to be represented as followed, we have 1, 2, 3, 4, 5, 6 that corresponds to 6, 5, 4, 3, 2 and 1, okay? that should be at that carbon, this carbon one. So, whichever way these are you can translate it into this form and then maybe name the substituent that way and notice that I label the six carbons, that's our longest chain. So, yo should expect hexane but as an alkyl group, it gets the yl ending, so so far it should be called a hexyl but we also have a substituent, so how do we name that substituen? the same way we're used to, right? it's going to become 3-ethyl, hexyl. Okay, that's going to be the complete IUPAC name for our alkyl group, and already stated that its primary, so we answered all the questions for this problem, okay? So I hope this makes sense, let me know if you have any questions.

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