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When ethyl bromide and $mathrm{n}$-propyl bromide is allowed to react with sodium, in ether, they form(a) mixture of four alkanes(b) mixture of three alkanes(c) mix...

Question

When ethyl bromide and $mathrm{n}$-propyl bromide is allowed to react with sodium, in ether, they form(a) mixture of four alkanes(b) mixture of three alkanes(c) mixture of two alkanes(d) single alkane

When ethyl bromide and $mathrm{n}$-propyl bromide is allowed to react with sodium, in ether, they form (a) mixture of four alkanes (b) mixture of three alkanes (c) mixture of two alkanes (d) single alkane



Answers

Based on the molecular formula, determine whether each compound is an alkane, alkene, or alkyne. (Assume that the hydrocarbons are noncyclic and there is no more than one multiple bond.)
a. C5H12
b. C3H6
c. C7H12
d. C11H22

Let's determine whether each of the following are alkaline Al Qaeda al kinds assuming non cyclical and no multiple bonds. So for an Al Kane, the general formula is C n each two n plus two. For an AL Keen, the general formula is C N H to n and for an al kind. The general formula is see N H two and minus two. So for a we have C five h 12, which fits the general definition of an Al Caine for B C three h six, which fits the definition of an AL keen for C C seven, age 12 fits the general definition of and al kind. And for D. C 11 h 22 it's the general definition of an AL Keen.

So now we'LL work on problem thirty four from Chapter twenty one. This problem here asks us to determined whether each compound is an Al Kane. All keen are all kind based on the molecular formula. And so we can assume that the hydrocarbons are not cyclic. And there's on ly one multiple bond, if at all, so we can go ahead and work on part A here Now The formula were given for part A S C eight h sixteen. Now here we need to remember the generic formulas. And if we do, we can say that it's in Al Keen. And that's because if I was the formula C and H to end so there's twice the number of hydrogen xas there's carbons. So this one is an alky, sir, where N is equal to eight sore. Next part Part B asks us about C for age six. Now we can see that this is hydrogen deficient. And if we look at our formula, we can say that it's an out kind because it follows the formula C N H to end minus two where n equals four heart See, it gives us see seven age six sixteen Ah, great deal of passion in here. We can say that it's an out Caine when we look at the formula si n age to n plus two hm. And in this case and is equal to seven in the last part is C to H too. And so this molecule isn't all kind because it follows once again the sea and H two n minus two. An end in this case is equal to two.

So now we'LL work on problem forty eight from chapter twenty one. This question here and gives us for reactions and it asks us to list all the possible products for the Al came substitution reactions and we assume mano substitution. So the first reactions that were given our ch for and seal too. And the question is, what does this produce? So one thing that we see here is that we have in this Ah, methane, all the hydrogen sze are equivalent So we're only going to get one product because no hydrogen is different from one another. And so if we substitute the chlorine for one of the hydrogen Sze, we get ch three CEO and h c e o So only one possibility for this reaction. So if we move to part B here, we're given Ah ethyl bromide ch three ch too pr. And then we add br too. Question is, what do we get? So in this case, we have two sets of hired guns. The hydrogen on the ch three group are the same in the hydrogen is on the ch two group of the same, so we could have addition to either the left or the right, and we would get different products. So going to go ahead and go down to the second line here and I'm gonna write the possible products we could add to the left carbon. And so we could substitute one of these hydrogen sze for a bro Ming. And then we still have the initial C H two b r hand. The second product possible product is if we add to the bro mean to the carbon, which already possesses a bro Me. One of these Hodgins is eliminated and that we now have to bro means here. So those two park products are possible, and we also form of course H p. R. So, for part, see, we're given we're given butane ch three ch too ch to ch three plus seal, too. So in this case, we could have addition to other the exterior carbons or the interior carbons. But since we're doing mono substitution, we only need to write two of them, because if it adds to the left side, and that would be equivalent to adding to the right side and rotating the molecule so two products are possible. So we're going to go ahead and write are possible React possible products here so we could add to the ending. Well, go ahead and add to the right side. So this is the same. And then we have a CH two on the end CEO. And then the second possibility is is is if we end to one of the middle carbons C is three ch to see h c l and then r ch three and then also will produce HCL. So if we had added to this carbon here, we would have produced this molecule here on DH. Then if we added to this carbon, we would have produced this molecule so we only have two possible products here. The last reaction that were given for party is ah, we have a ch three CHP are too. And then we add brew me b r too. Question is, what products do we get? So we'd already have to bro means on our right carbon. However, we still have a hydrogen which can be substituted for So we still have two possible products addition to the right or the left carbon so we can do the addition to the left carbon C h two b r and we still have our ch b r, too. And then the second possibility is that we add to the right carbon the sweets have r ch three c b r. Three. And of course, we also produce H B R.

No Friday in order to prove out from Elko hard, I'd reacted Alcohol highlight. It must eliminate the halogen on one of its hydrogen to create a double on. Therefore, we have a elimination reaction. Now for part B. The best way to produce an alcohol from an L team is with the addition of water. So we have a tradition reaction for a part. See, um, the formation of a nester from the reaction of carbo flipped out after it and alcohol is called condensation. And for party to produce a alco die highlight from Al peen, we need to add a college, and therefore we have again in addition reactions.


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