Borane Reactions - Online Tutor, Practice Problems & Exam Prep

In chemistry, boron is unique due to its electron deficiency, as it does not fulfill the octet rule with only three valence electrons. This characteristic makes boron compounds, such as diborane (B₂H₆), highly reactive, especially in reactions with water and in Lewis acid-base interactions. Diborane reacts with water to form boric acid (H₃BO₃) and hydrogen gas, a notable reaction that highlights its reactivity. Furthermore, boranes are considered Lewis acids because they can accept an electron pair. For example, in a reaction with ammonia (a Lewis base), boron can form an adduct by accepting a lone pair from the nitrogen atom in ammonia, resulting in a new bond and formal charges on the nitrogen (+1) and boron (-1) atoms. This process exemplifies a Lewis acid-base reaction, where boron's ability to act as an electron pair acceptor is due to its high electron deficiency.

Borane Reactions

Now remember, boring itself represents a boron atom connected to three 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. Braun itself though only has three valence electrons and can only make three bonds. Normally.

As a result of this, we're going to say boring reactions. These reactions are driven by Borane's high electron deficiency. They haven't fulfilled the octet rule and they're also driven by Dybarans 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 going to cover two types of reactions and they're going to 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 basins, all right, so just remember, when we're talking about boring reactions, it's only possible because boring itself is highly electron deficient and if we ever covered dye borne reactions, they'd be reactive as well.

Reaction with H2O

Now when it comes to the reaction with water, we're going to say as a result of their high reactivity, Dybarans readily react with liquid water. Now, gaseous Dybborne reacts with water to produce boric acid and hydrogen gas South. This is the only reaction that's of importance when it comes to Dibori.

So here we have diboring as a gas reacting with liquid water. It's going to form boric acid, which is H₃BO₃, and we're going to have hydrogen gas being formed as well. Once we balance this out, we'd have:

1 / 6 , 2 : 00 , 6 : 00

This will represent our reaction between our dye boring molecule and liquid water.

Borane Reactions Example

Here in this example question it says if the 466 kilojoules of energy is released for every mole of Dybborne reacting with water, how much energy would be released when 150 grams of Dybborne is submerged in excess water? All right, so remember our balance equation is 1 mole of dye. Boring gas reacts with six moles of water as a liquid to produce 2 moles of boric acid +6 moles of hydrogen gas. And we're just told here that the amount of energy, which we're going to say is ΔH released means -466 kilojoules, right?

So this becomes a thermal chemical question, Thermochemical question, because we're dealing with stoichiometry mixed with the enthalpy of the reaction. So the way we set this up is we have 150 grams of Dybarrane. What we have to do first is we have to convert these grams into moles. 1 mole of Dybborne on top, grams of diborin on the bottom so they can cancel out. When you calculate the two borons and the six hydrogens involved, you get a overall molar mass of 27.668 grams. Cancel out for diborane.

Now we have moles of dye borne according to my balanced equation for everyone, mole of Dybborne. This is how much energy is released. OK, so here we're not doing a mole to mole comparison, we're doing a ΔH to mole comparison. Everything cancels out. So what we have left at the end is negative 2526.38 kilojoules. This has three sig figs. This has four sig figs. So we'd have three sig figs at the end. So -2520 or actually 2530 kilojoules at the end. This is how much heat would be released from this many grams of Dybborne.

Lewis Acid-Base Reaction

Now recall that a Lewis acid is an electron pair acceptor and a Lewis base. We'll see. This is the acceptor, and a Lewis base is an electron paired donor. Here we're going to 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 three bonds because it has three 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 one for its formal charge. Boron is gaining new electrons, so it's -1. Overall the molecule will be neutral, cause positive and negative together. But here we're showing the formal charges on each of these atoms to be more accurate, right?

So although borane is a new idea, Lewis acid base reactions have been talked in other sections dealing with acid base chemistry, so keep that in mind. Borane, because of its electron deficiency, can act as a Lewis acid and accept a lone pair from a Lewis base.

Borane Reactions Example

Here it says to draw the addict product form from the reaction between borane and a hydroxide ion. So remember, borane is BH3, Hydroxide is OH-1. Oxygen itself has three lone pairs. We can use one of these to connect to the boron atom.

So here goes. Our boron still connected to its three hydrogens, and now it's making a new connection with the hydroxide ion. Oxygen still has its other two 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 -1. This represents the newly created addict product between the reaction of our boring molecule and our hydroxide ion.