Draw a Lewis structure for each of the following compounds:
(b)

Question

Draw a Lewis structure for each of the following compounds:

(b) Condensed structural formula of a brominated organic compound with a branched carbon chain.

Accepted Answer

Welcome back, everyone show the Lewis structure of the molecule below. Whenever we want to draw a loose structure, it is really important to understand that we want to show all of the bonding and non bonding electron pairs. So the easiest way to begin this problem is to first of all, find the longest continuous carbon chain, we have one to we or five carbon atoms in the longest continuous chain. So we want to show all of these carbon carbon bonds, 12, we four and five carbon atoms. Now from here, we can also add the remaining bonds, right? Let's ignore hydros for a moment and let's simply add that carbon which is bonded to carbon number two, right. So we are just going to draw that additional carbon carbon bond. And then as we can see, carbon number four is bonded to fluorine and carbon number five is bonded to another fluorine. Now, from here, this is where we can take care of hydrogens. The first carbon has three hydrogens, we have to recall that every carbon must have four bonds. So this makes sense, right? We now have a total of or bonds around the first carbon, the second carbon has a one hydrogen. So now we have a total of four bonds. Carbon. Number three, we need two hydros well done. Carbon. Number four, it has one hydrogen. Carbon, number 52, high jumps. Now let's go to C three. It is a muscle group. So as we can see, there are three hydrogens bonded to that specific carbon. Well done, we have added all of the hydrogen atoms. We can see that every carbon has four bonds. We simply want to take care of the long pairs of electrons. Now, we want to recall that every carbon will have four bonds and no lo pairs based on the octet rule based on the due at rule, hydrogen will have one bond and no alone pairs. So we only want to analyze the halogen atoms in this case, fluorine and fluorine. Well, it is a halogen, it is in group seven A and essentially it means that fluorine forms one bond and it contains three alone pairs of electrons, right, we have 1234567 if we count the bond as an additional electron. So essentially the octet rule will be satisfied for fluorine if we had one bond in three lone pairs. So this is applicable to every halogen and we have two fluorine atoms. So we simply want to add three lone pairs for each fluorine atom in the structure. And this gives us the Lewis structure. Let's label it as the final answer. Thank you for watching.

Draw a Lewis structure for each of the following compounds:
(b)

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Key Concepts

Lewis Structure

Valence Electrons

Octet Rule