Identify the weak diprotic acid. identify the weak diprotic acid. h2so4 hcooh
1 answer:
Among formic acid (HCOOH ) and sulfuric acid (H₂SO₄), formic acid is the weak acid. Acidic strength of any acid is the tendency of that acid to loose proton. Among these two acids formic acid has a pKa value of 3.74 greater than that of sulfuric acid i.e. -10. Remember! Greater the pKa value of acid weaker is that acid and vice versa. Below I have drawn the Ionization of both acids to corresponding conjugate bases and protons. The structures below with charges are drawn in order to explain the reason for strength. As it is seen in charged structure of formic acid, there is one positive charge on carbon next to oxygen carrying proton. The electron density is shifted toward carbon as it is electron deficient and demands more electron hence, attracting electron density from oxygen and making the oxygen hydrogen bond more polar. While, in case of sulfuric acid it is depicted that Sulfur attached to oxygen containing proton has 2+ charge, means more electron deficient as compared to carbon of formic acid, hence, more electron demanding and strongly attracting electrons from oxygen and making the oxygen hydrogen bond very polar and highly ionizable.
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Answer:
See explanation below
Explanation:
In this case we have reaction of addition. In this case a diene reacting with an acid as HBr. This reaction is known as Hydrohalogenation, and, as we have a diene, this kind of reaction can be done as 1,4 addition. Which means that the reaction will be undergoing with an adition in the carbon 1, and carbon 4.
At room temperature we can expect that this reaction can be done in thermodynamic conditions, Now, as the problem states that is forming 4 products, we can expect products of a 1,2 addition too. This product can be formed if the reaction is taking place in the most stable carbocation, and then, by resonance, we can expect the 1,4 product too.
Now, the HBr can be attacked by the double bond of the first position, giving two possible products or by the double bond of the third position giving the other two products. These products are all possible, obviously the most stable will be the major of all of them, but the other three are perfectly possible. One product is formed without doing much, and the other by resonance. Same happens with the other double bond.
In the picture below, you have the mechanism for all the 4 products.
Hope this helps
