Chloropropane is derived from propane by substituting Cl
for H on...

Question

Chloropropane is derived from propane by substituting Cl
for H on one of the carbon atoms. (a) Draw the structural
formulas for the two isomers of chloropropane.

Accepted Answer

welcome back everyone in this example we have the reaction of the butane with roaming in sunlight. It can form to mono substituted products and were told to draw the structures of the two summers. So our first step is to recall that in our prefix tells us the shape of our structure where the ending. Butane tells us that we have four carbons. So because we have the prefix, so we should have three methyl groups, so three lines. So we have 12 and then our third metal group here to represent our isso butane structure and we can count our carbons to make sure we have four carbons. Because recall that butte tells us that we have four carbons. So we have one carbon here, another carbon here, another carbon here and another carbon here and for each of these carbons to be stabled, we should recall that there's implied hydrogen in our structure. So we have for each end point of our structure, these three carbons bonded to three or having three bonds bonded to hydrogen. So that would be three atoms. They're bonded to where the central atom would be the fourth atom that they're bonded to to have their full octet. And then of course our central atom is bonded to these three carbon atoms and then we have an implied hydrogen bond here that I'll just draw in. So according to the prompt, we have bromine, we have abomination reaction here where we have bromine which we recall is represented by atomically as br two reacting in sunlight. And we should recall that to represent sunlight. We use the symbol H. V. Where we recall that H refers to Plank's constant which has a direct relationship with frequency. So this V symbol represents frequency and we should recall that this is describing a photon of sunlight. So that is why we represent H. B H. V. By sunlight here. So I'll just instead of drawing that arrow, we'll just say that this explains sunlight. So we want to recall that for a brahma nation reaction. Our first rate limiting step for bromine is endo thermic. And so we would recall that bro mean is highly selective and that means that it's only going to want to bond mainly at the most stable point or position in our molecule here. I so butane. So we want to recall our stability of our carbon atoms in our molecule where our central carbon here because it's bonded to recall that carbon stability is determined by the number of carbon atoms that are selected carbon atom is bonded to. So for this carbon atom we see that it's bonded to 1 to 3 other carbon atoms. And so we would mark it as a tertiary carbon And we want to recall that because it's tertiary this means very stable, Very stable carbon specifically. And then looking at our other carbons at the end. They're all just bonded to this one central carbon. And so these are all going to be one degree. So one degree would be low stability for carbon. So showing the products that we would form our bro. Main is going to select between bonding to one of the hydrogen is at the third degree or tertiary carbon position, or bonding to one of the hydrogen at the primary carbon position. Where for our tertiary choice, so I'll just do the primary in a different color actually for a tertiary choice, we're going to form the following product where we have bro mean now bonded in the location where our hydrogen was at, where it's not just going to be that mean, and that would be our first product. This is going to be the most stable choice for a position for bromine to bond to for our second product and I'll draw that in red here. We have our I still butane shape and then our roman is now just going to choose to bond to any of our primary carbon positions. So we'll just go with this one here and it's going to take the place of a hydrogen bonded to that carbon. So we'll just actually draw this in blue so that it's clear. So the second product would occur again if the bromine replaces one of the hydrogen on our three methyl groups here because they are all equal since they're all primary. And so for our final answers, these two ice climbers would result as our products. So what's boxed in yellow here is our final answer to complete this example. So I hope that everything that I reviewed was clear. If you have any questions, please leave them down below, and I will see everyone in the next practice video.

Chloropropane is derived from propane by substituting Cl for H on one of the carbon atoms. (a) Draw the structural formulas for the two isomers of chloropropane.

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