Give two names for each of the following alkyl halides and ind...

Question

Give two names for each of the following alkyl halides and indicate whether each is primary, secondary, or tertiary:

c. Structural formula of an alkyl halide with a chlorine atom, indicating branching and carbon chain length.

d. Structural formula of an alkyl halide with a fluorine atom, indicating its branching and connectivity.

Accepted Answer

Welcome back, everyone, write the IOP name and indicate if the alky Hali this primary secondary or she, we're given two structures. The first one is an alky bromide and the second one is an alky chloride. We're going to start with the first one. And our first step is to identify the parent. Let's recall that al kill Hay lights will have halogens as substitutions while the parent will correspond to an ALCA. So our goal is to find the longest continuous carbon chain and that would be 12345 carbon atoms. We have to recall that this would be pense number two. We want to enumerate our parent properly and to enumerate this properly, we want to minimize the coefficients of our substations as well as the sum of those coefficients. So if we enumerate it from left to right, we have two substitutions on carbon number two. So that'd be 22. And then we have a substituent on carbon number four, 224. If we go from right to left, we would then have 244, right. So essentially, we noticed that the second coefficient becomes greater, right, it becomes four instead of two, meaning we are going to enumerate it from left to right. So we are going to delete our second pathway and we are just going to focus on the first one from left to right. OK. Our third step is to identify the substitutions on carbon. Number two, we have two metal groups, right? Because we have two solid lines which indicate the presence of two metal groups because we have two of them, we're going to use a prefix dye that's really important. And we are going to state the position of a carbon atom to which they are bonded. So that's carbon number two. And in addition to that, we have our bromo substituent bonded to carbon number four. Well done, we have all of the information we need. Now we want to start naming our structure by defining our substitutions. So based on our analysis, if we have two metal groups, we're going to say dimethyl, of course, we're going to state the carbon atom that they are bonded to you, that's carbon number two. And we have to repeat it two times because there are two metal groups, meaning our name will have 22 dimethyl as well as or bromo, right? Because the bromo substituent is bonded to carbon number four. And now to name it, we want to arrange them alphabetically, we're going to exclude the prefix from the comparison. So we have a comparison between M and B B gets prioritized alphabetically. So we're going to start naming it as for BROMO, followed by 22. And we are going to finish with our parent, which is well done. We got the name for the first structure. So we're going to label it as the first one. And of course, we can identify the Alky Halite as primary, secondary or tertiary for that purpose. We're going to locate our promo group and we understand that it is bonded to a carbon atom that has two neighbors or two neighboring carbon atoms to be specific. And that means it would be secondary. So let's say that we have a secondary, I kill Hali because the halogen is bonds, the carbon atom that has two neighboring carbon atoms well done. And now we are ready to move on to the second problem. Let's start with the parent. Our longest continuous carbon chain has 12345 carbon atoms. So once again, we have pense right D L D halogen does not get prioritized, right. We are always naming an can. So our second step is to enumerate it. Whether we go from left to right or right to left, we will get the one and only substituent bonded to carbon number three, because the structure is symmetrical. So when we want to enumerate that we can go from left to right or right's left, it doesn't matter. Our next step is to identify our substitutions. And as we said, we only have one Chloro substituent bonnet two carbon number three. So now we are ready to name it. Of course, we start with the position of the substituent. In this case, that would be three. The name of the substituent is Chloro, our parent is Pen. And now we want to understand if it's primary secondary or third sharing. So we identify our Chloro substituent which is bonded to carbon number three and the carbon number three has two neighboring carbons. So that means it is secondary as well. And that will be our answer for the second part. Well, then, so let's conclude our answers for the first one. The name is four bromo, 22 dimethyl Penan, which is a secondary alkyl halite. And for the second one, the name is three Chloro Penan, which is also secondary. Thank you for watching.

Give two names for each of the following alkyl halides and indicate whether each is primary, secondary, or tertiary:
c.
d.

Key Concepts

Alkyl Halides

Nomenclature

Classification of Carbon Atoms

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