Click the "draw structure" button to launch the drawing utility. under certain reaction conditions, 2,3−dibromobutane reacts with two equivalents of base to give three products, each of which contains two new π bonds. product a has two sp hybridized carbon atoms, product b has one sp hybridized carbon atom, and product c has none. what is the structure of b?

Explanation:

According to this. Let's analize the possible products a, b and c.

First, the problem states that we have 2 eq. of base, For this case, let's assume it's KOH. Now, As we are doing a reaction with base, means that this reaction can only take places under conditions of SN2 and E2, a fast reaction that is taking place in only 1 step.

With this in mind, let's analyze product a. This states that it has two sp hybridized carbon, in other words, a triple bond between two carbons. So the product is with no doubt, an alkyne.

Product b has only one sp hybridized carbon, which means that this carbon should cannot be an alkyne because we need two carbon atoms. The only way to have one atom of C sp hybridized, is with two double bonds, so product b would have to a alkene with two double bonds.

Product c do not have sp hybridized carbon, therefore, it only has two double bonds in two different Carbon atoms, so it's another alkene with two double bonds, but in two different atoms of carbon.

Picture attached show the product a, b and c. Hope this can help

um i think it might be a car?

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.