Predict the products of reaction of pent-1-yne with the follow...

Question

Predict the products of reaction of pent-1-yne with the following reagents.

g. cold, dilute KMnO₄

h. warm, concd. KMnO₄, NaOH

i. Na, liquid ammonia

Accepted Answer

Hey everyone and welcome back in today's video, we are going to solve the following problem determine the identity of products obtained in the reaction of butte 19. With each of the regions shown below. Number one, we have the loot cold solution of potassium permanganate. Number two, concentrated warm solution of potassium permanganate and sodium hydroxide than hydrochloric acid and number three sodium and liquid ammonia. We may start this problem by drawing beat 19. This is an ALC. In so we have a triple bond which starts at position number one. And because the parent is B time, there will be two more carbon atoms giving us a total of four. And we have three reactions. So we may draw three arrows representing each reaction separately. The first one, The 2nd 1 and then the 3rd 1. We will now list all of our regions for reaction number one, we're treating the AL kind with the loot cold solution of Cayman oh four that's in permanganate, K. Mn 04 which is dilute and cold solution. Reaction number two, we have two steps first of all we're using but as soon permanganate also label this as step number one, but as soon permanganate warm and also concentrate that with sodium hydroxide Step two, addition of hydrochloric acid. Now, for the third reaction we're treating the al Qaeda with sodium and liquid ammonia. Now we may start analyzing our reactions and in particular reactions number one and two, they're very similar because each of them involves but as in permanganate but at the same time, the first reaction involves they loot camera four while the second one has a warm concentrated solution. So number one is a weak oxidation because the solutions as dilute and it's a cold solution. You may wish to recall that whenever we are using weak oxidation, we're not really breaking the carbon carbon bond. Instead we are forming to carbonell groups. The triple bond will be gone and in particular we will be forming two carbonell groups. One over here, another one over here. That's everything that we need to know so far. So the triple bond will disappear. We will only have a single carbon carbon bond and we will just add two oxygen atoms double bonded to carbons. And that that will be our first intermediate actually. So let's draw our intermediate in this reaction. As we previously assad, we will be adding to carbonell groups. Nothing will change with respect to that hydrogen. So we may just indicate that carbon hydrogen bond and the first carbon atom will have a double bond with oxygen. So that forms the first carbonell group. Now the 2nd 1 also forms a carbon eel group and we still have two more carbon atoms left unaffected. This is an AL to hide or more specifically keto alga hide because it involves both in the alga hide group and a keto group. Now, if we continue our reaction, recall that Aldo heights can be oxidized further into carb oxalic acids because the oxidizing agent is strong. That will happen. And our final product will actually be a keto asset, keto asset. Well done. So that's our first product. We may just label this as our first product for this problem. And now we are ready to look at the second reaction. In the second reaction we are treating the AL kind with but as in permanganate that is warm and concentrated and sodium hydroxide, which implies that this is a strong oxidation. And now we will be breaking the carbon carbon bond in particular, if we are breaking the carbon carbon bond, one of the oxygen's will be added at this position on the right there will be a double bond. That oxygen to that carbon. So we will be forming oxygen. Carbon oxygen double bond. That's very important to understand. And the other carbon will be separated from the molecule because it's a terminal AL kind. It's very important to understand that we are forming C. 02. The carbon atom that is separated will have an ability to bond to T oxygen atoms, one from the right, another from the left, while this carbon on the right will have one oxygen bond, it's it. And after the separation of the triple bond, there will be an ability to add one more. So we will be adding a total of four carbon atoms. My apologies, four oxygen atoms, one double bonded, one single bonded, and similarly to this carbon with due to the presence of a terminal al Qaeda will be adding to double bonded oxygen because there's no presence of any carbons nearby, there's just a single carbon atom. So knowing all of this, we may now start drawing our structure for the product of this reaction. Number two, if we just start with the right part of the AL kind, we have a total of three carbon atoms. So that's what we can see. We may draw three carbon atoms. But we also have to keep in mind that the first carbon or this carbon in particular will have two oxygen's bonded to it now. So the first auction will be double bonded and the second one will involve a single carbon oxygen bond because that would make already four bonds total on that carbon. That's carbon number one. Number two. Number three, This is our fragment with a negative charge. And now, as we previously said, just the presence of a terminal AL kind, we'll be forming a molecule of carbon dioxide because there would be two oxygen's bonded to carbon via double bonds. And now, because the second step is the acidic work up, we are simply adding a proton to this an ion to make it an asset. Right? So we're going to state that the final product that we get from this one is the acid. We get a car book select asset plus C. O. T. So these two are our final products for reaction number two. Now let's move on to reaction number three. Reaction number three involves addition of sodium in liquid ammonia. And you may recall that addition of sodium and liquid ammonia to AL kinds forms al keane's in trans arrangement. A trans al keane would be produced. But is it really important notice that because this is a terminal AL kind, there's already a presence of one hydrogen. So we add one more. We will not have an ability to form either CIS or trans is a more is um So we we may simply say that the AL kind shown becomes an AL keen. We may start by drawing a double bond and then we would add two more carbon atoms because we have a total of four. So our final product is an L. Keen and we may label this as our product for reaction number three. We may now conclude that the correct answer to this multiple choice question is C. However, if you have any questions, don't hesitate, leave your comment down below in the comment section and thank you for watching

Predict the products of reaction of pent-1-yne with the following reagents.
g. cold, dilute KMnO₄
h. warm, concd. KMnO₄, NaOH
i. Na, liquid ammonia

Key Concepts

Alkyne Reactivity

Oxidation Reactions

Reduction Reactions

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