Borane Reactions - Video Tutorials & Practice Problems
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concept
Borane Reactions
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Now remember, borane itself represents a boron atom connected to 3 hydrogens. And when it comes to Lewis dot structures, we know that a lot of the time elements are trying to fulfill the octet rule. Born itself though, only has 3 valence electrons. It can only make 3 bonds normally. As a result of this, we're gonna say, boron reactions, these reactions are driven by boranes high electron deficiency. They haven't fulfilled the octet rule. And they're also driven by diboranes high reactivity. Remember, we're using bridging hydrogens to create diboranes. These structures are very unusual for us, and they are highly reactive, not super stable. Now within this section we're gonna cover 2 types of reactions, and they're gonna have us reacting with water, and Lewis acid base reactions. One of which we know from past discussions, if you've seen my discussions on different types of acids and bases. Alright. So just remember, when we're talking about boron reactions it's only possible because boron itself is highly electron deficient, and if we ever covered diboron reactions they'd be reactive as well.
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Reaction with H2O
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Now when it comes to the reaction with water, we're gonna say, as a result of their high reactivity, diboranes readily react with liquid water. Now, gaseous diborane reacts with water to produce boric acid, and hydrogen gas. So this is the only reaction that's of importance when it comes to dibborane. So here we have dibborane as a gas reacting with liquid water, It's gonna form boric acid, which is h three b o three, and we're gonna have hydrogen gas being formed as well. Once we balance this out, we'd have 1, 6, 2, and 6. This will represent our reaction between our diborane molecule and liquid water.
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example
Borane Reactions Example
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Here in this example question it says, if the 466 kilojoules of energy is released for every mole of Diborne reacting with water, How much energy would be released when 150 grams of Diborane is submerged in excess water? Alright. So remember our balance equation is 1 mole of diborane gas reacts with 6 moles of water as a liquid to produce, 2 moles of boric acid plus 6 moles of hydrogen gas. And we're just told here that the amount of energy, which we're gonna say is Delta h, released means negative for 66 kilojoules. Right? So this becomes a thermo chemical question thermo chemical question because we're dealing with stoichiometry mixed with the enthalpy of the reaction. So the way we set this up is we have a 150 grams of Diborane. What we have to do first is we have to convert these grams into moles. 1 mole of diborane on top, grams of diborane on the bottom so they can cancel out. When you calculate the 2 Borons and the 6 Hydrogens involved, you get an overall molar mass of 27.668 grams. Grams cancel out for diborane, now we have moles of diborane. According to my balance equation, for every one mole of diborane, this is how much energy is released. Okay? So here we're not doing a mole to mole comparison, we're doing a Delta H to mole comparison. Everything cancels out, so what we have left at the end is negative 2526.38 kilojoules. This has 3 sig figs, this has 4 sig figs, so we'd have 3 sig figs at the end, so negative 25 20, or actually 2530 kilojoules at the end. This is how much heat would be released from this many grams of diborane.
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concept
Lewis Acid-Base Reaction
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Now, recall that a Lewis acid is an electron pair acceptor, and a Lewis base, we'll say this is the acceptor, and a Lewis base is an electron pair donor. Here we're gonna say because of their high electron deficiency, boranes are considered to be Lewis acids. So if we take a look here, we have ammonia molecule, nitrogen has a lone pair readily available, It could share that so it represents our Lewis base. Borane here, Boron is making 3 bonds because it has 3 valence electrons, but it hasn't fulfilled its octet rule, so there's room for it to accept an electron pair making it the acid, Lewis acid. What can happen here is nitrogen uses this lone pair and it shares it with the boron, creating this new bond here. As a result of this, we're adding them together, so we call this our adduct. Remember the adduct is just the product of our Lewis base and acid reaction. If we were to determine the formal charges of the nitrogen and boron, nitrogen is sharing its lone pair, so it's plus 1 for its formal charge. Boron is gaining new electrons, so it's negative one. Overall, the molecule will be neutral because positive and negative together, but here we're showing the formal charges on each of these atoms to be more accurate. Right? So alboborane is a new idea. Lewis acid based reactions have been talked in other sections dealing with acid based chemistry. So keep that in mind. A borane, because of its electron deficiency, can act as a Lewis acid and accept a lone pair from a Lewis base.
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example
Borane Reactions Example
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Here it says to draw the adduct product formed from the reaction between borane and a hydroxide ion. So remember, borane is b h 3. Hydroxide is o h minus 1, Oxygen itself has 3 lone pairs. We can use one of these to connect to the boron atom. So here goes our boron, still connected to its 3 hydrogens, and now it's making a new connection with the hydroxide ion. Oxygen still has its other 2 lone pairs. When oxygen makes its ideal number of bonds which is 2, its formal charge is 0. Boron here is gaining additional electrons and sharing them with the oxygen, so its formal charge becomes minus 1. This represents the newly created adduct product between the reaction of our borane molecule and our hydroxide ion.