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Part]; On treatment with McLculs acid, which ofthe following carbonyl compounds will be expected to have the greatest amount o hydrate at equilibrium? (# polnts)Ide...

Question

Part]; On treatment with McLculs acid, which ofthe following carbonyl compounds will be expected to have the greatest amount o hydrate at equilibrium? (# polnts)Identily ALL acetals that are NOT derived from an aldehyde (6 polnts)Several ofthe following arylbromides cannot he used generate Grignard reagents Identily ALL trylbromides that ARE suitable for Kenerating Grignard rexgent by treatinert with Mg in dry cther: (6 point)Identily the correct order of increasing pKa ofthe following carboxyli

Part]; On treatment with McLculs acid, which ofthe following carbonyl compounds will be expected to have the greatest amount o hydrate at equilibrium? (# polnts) Identily ALL acetals that are NOT derived from an aldehyde (6 polnts) Several ofthe following arylbromides cannot he used generate Grignard reagents Identily ALL trylbromides that ARE suitable for Kenerating Grignard rexgent by treatinert with Mg in dry cther: (6 point) Identily the correct order of increasing pKa ofthe following carboxylic acids (4 points) COOH COOH COOH COOH OzN H,co (Im) (I) < (II) < (III) < (IV) (I) < (I) < () < (IV) (I) < () = (IV) < (I) (IV) (II) (I) - (III) < (IV) Which ofthe following carbonyl compounds will have the highest enol content? (4 points) OEI



Answers

Which among the following compounds will give a secondary alcohol on reacting with Grignard reagent followed by acid hydrolysis? (I) HCHO (II) $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{CHO}$ (III) $\mathrm{CH}_{3} \mathrm{COCH}_{3}$ (IV) $\mathrm{HCOOC}_{2} \mathrm{H}_{5}$ Select the correct answer using the codes given below: (a) III only (b) II only (c) I and IV (d) II and IV

I guess you wouldn't do a mechanism for strong reaction. And we're reacting from Allah Hood with some acid and water. So the first step is at auction of the alcohol is going to, it's part of the acid and it's going to be positive charge because it's going to pair of electrons are negative and the water is going to attack the carbondale. So water would attack the carbon will directly address be pertinent first these waters not very strong base and the the door bonds to break with the addition of oh H two because of charge. And the last step is for because your base of HC boss. It's gonna be oh H two is going to deeper in the water. And I'll give us to our calls. So the formula for each step of the reaction, we have to hide regions carbon and auction. So that's going to be um, Coh two. They're reacting with each three plus in water. And it's gonna give us ceo each three for the species. And so it's going to be So we have 12345 hot regions and two auctions. Then this one is going to be CO two 123 fractions the following products.

Hi. In this video, we're going to be going over a carbon. You'll chemistry specifically how we convert a compound with a carbon steel group to a hydrate. So these creations are very common. But before I begin, let me first go over what a carbon steel group it really is. So a carbon heel is a carbon that is double bonded to an oxygen. The important thing we have to remember is because of the difference in electro negativity. Is oxygen being a lot more of electro negative than carbon? We're going to have a intrinsic polarity in which carbon is partially positive, whereas oxygen is partially negative because it holds onto these electrons tighter and it hogs them. And so, in light of this knowledge, let's go over this reaction. So I'm converting formaldehyde, which is the most simple out to hide to its hydrate in the presence of water as well as an acid catalyst. So this reaction occurs in three main steps. First, we're going to protein eight, the carbon Neil oxygen. Then we're going to form a Covalin bond between a nuclear file as well as an electro filic region within this formaldehyde and then we're going to D protein eight water at the end. And this will be more clear as I draw this mechanism. So first, let's draw formaldehyde again. So in the first step, right in the first step, we have the acid catalyst. We have the acid catalyst and this carbon. Neil, this, uh, Carbonell oxygen is going to It's going to attack this hydrogen. And then these two electrons are going to be kicked off and put on this oxygen, uh, of this acid right here forming water. So what we have formed is this intermediate in which oxygen has a formal charge of plus wind. I realize here, very importantly, is that this carbon right here of this Carbonell group is still partially positive, making it the electro electro Filic region, right? It wants electrons because it's partially positive. So what ends up happening is we have water it here, which has which is going to be our nuclear file. It's a weak nuclear file, but that's going to attack this electro Philip carbon kicking these two electrons off onto oxygen so that oxygen can routine. It's formal charge of zero. And then we have this lovely water group as well as the two hydrogen. So we just formed a Covalin bond right right here, right here. Because of the reaction between our nuclear file water and the Electra Filic carbon region right here. And so in our last step, all we have to do is de protein. Eight is to de Protein eight. This water group, such that, uh, such that oxygen can then have a formal charge of zero. How we do that is water is going to then the protein eight, this water group right here and then what we're left with is formal in as we saw earlier. And then we also remake our acid as well. So that is the mechanism for this reaction. And I hope that this made a made this reaction more clear, as well as give you a better understanding of Carbondale chemistry.

In this problem, increasing all the given alcohol in order of increasing is of dehydration ego, be high duration order, increasing Egypt dehydration order are as follow. So just look at it carefully the list it for at the CCH two oh edge after that, Cl three CCH 20 H. After that CS three C h 20 H and largely, but C685 CH 20H. So according to the option option C, each correct answer.

In this question, we will understand about the basic properties of Cowboy Did and polymers like So in this question, there are some compounds we have to identify the compounds. The question is an organic compound that gives violet color with neutral fell Fell slide like very closely dance was labour managed. This compound on catholic reduction gives me so we can identify that. Yes, funeral C two H 50 H. Is less acidic than final. Thank you. So the presence of electron donating group reduces facility while the electron withdrawing groups increases the I. C. D. D. Hence the order of facilities are C two H. Five words, then a medal that is paramount in final, then the compound A. Which is final and then another night of it all. Yeah, so this is the order of the for the facilities of the uh different compounds.


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