Draw a constitutional isomer for C₄H₁₁N ...

Question

Draw a constitutional isomer for C₄H₁₁N containing a 1° amine at a

(a) 1° carbon,

(b) 2° carbon, and

(c) 3° carbon.

Accepted Answer

Hey everyone and welcome back in today's video. We are going to draw all the possible eyes mars with the molecular formula C four H nine cl and determine how many are primary alcohol. Hey lights, secondary alcohol. Hey lights and tertiary alcohol. Hey lights. Let's begin solving this problem by finding the hydrogen deficiency index which is calculated as two multiplied by the number of carbon atoms plus two plus the number of nitrogen atoms minus the number of hydrogen atoms minus the number of religions in our structure And everything is divided by two. So for this structure, The hydrogen deficiency index is equal to two multiplied by four plus to you Plus 0 -9 - divided by two. This gives us now in the numerator we have eight plus two. That's 10 minus nine minus one. That's zero, divided by two. That's zero. The hygiene deficiency index of zero implies that there will be no double bonds and no rings, no structure, there will be no triple bonds, the structure is fully saturated and it would have no rings. So let's start drawing our possible eyes MERS first of all, we are not going to worry about parts A B and C and we are just going to draw all the possible eyes murders and then we will assign them as primary, secondary and tertiary. Our strategy is to, first of all, start with the longest chain that we can join. This problem. If we have four carbon atoms, we are going to draw a straight chain of four carbon atoms, that's 1234. And we are going to assign chlorine at position one. Okay. So that's our first possible structure. Now, the question is, can we still keep our four member chain and add chlorine to it? The answer is yes, we can still keep our four member chain of carbon atoms. But now, instead of assigning chlorine to position one, we are going to assign it to position two. And those would be our only eyes numbers that correspond to a four member chain. Because if we had a chlorine here, that would correspond to the same structure, it has this line of symmetry which passes through our compound. And therefore, those are the only possibilities for us whenever we have our longest chain of carbon atoms. So now we are going to start shortening it and we are going to draw a three member chain. So if we make it a three member chain and supposing that we add chlorine at position one, we need to add one more carbon atom. And the only possibility for us to do so is to add it at position two, right? Because if we added at position three, we would get, we would essentially get our four member chain right? Once again. So this is our next structure when we have 123 carbon atoms in the parent. And now just as we did it before we can change the position of that glory in atom, right, we can still keep our longest chain with three carbon atoms. So let's do that 123, we are going to have a methyl group at position two. But now we're going to assign chlorine to position too as well, right? So this is still our three member chain, but now glory in and the metal group, their bonnets the same second carbon atom. So we got all of our possible structures note that the parent cannot be short and even more because if we have a two member parents, we cannot assign two carbon atoms such that we get a two member parents still. So those would be our four possibilities. And now let's understand what we want to define in this part, starting with part A, we want to understand which are primary alcohol. Hey, lights and primary means that the chloral substitution must be bonded to a carbon atom that only has one neighbor. So for example, the first structures primary because our chloral substitution has bonded to carbon, that only has one neighbor on the left side. Similarly, if we look at the other structure below it, specifically this this one below the previous one, the chloral substitution is bonded to a carbon atom that only has one neighbor to the left. So this one is also primary. Now if you look at the next one Chloral substitution is bonded to carbon that has two neighbors on the left and on the right. So that would be secondary and the final one close a position is bonded to carbon that has three neighbors. So that's a tertiary alcohol highlight. So that means for a, we have two answers, that would be the first structure and the third one that we have to go on. So we have a total of two. Now, for B we want to define our secondary ones and we only have one. So let's throw it out. There's one secondary al kill hey line. And finally, for C tertiary, we also have only won one of these. So that's basically the following structure. We can label that we have one and 1. And then we got our answers, we may essentially label them. We may conclude that we have two primary LKL. Hey lights. For A for B we have one secondary alcohol, hey light and for C we have one tertiary alcohol. Hey light. Now, the correct answer to the multiple choice question is a. However, if you have any questions don't have states, leave a comment down below and thank you for watching.

Draw a constitutional isomer for C₄H₁₁N containing a 1° amine at a
(a) 1° carbon,
(b) 2° carbon, and
(c) 3° carbon.

Key Concepts

Constitutional Isomers

Primary, Secondary, and Tertiary Amines

Carbon Chain Structure

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