Answer:
See explanation below
Explanation:
The drawing of the molecule and mechanism, you can see it in the attached pictures.
Now, answering the theorical questions:
The 1-butene is like this:
CH2 = CH - CH2 - CH3
If this molecule reacts with bromine (Br2) the reaction and product formed is:
CH2 = CH - CH2 - CH3 + Br2 -----------> Br-CH2 - CH(Br) - CH2 - CH3
The product formed is called 1,2 - dibromo - butane, and the reaction with halides like bromine is called halogenation. In this case, alkenes halogenation, so, we become a alkene like the 1-butene with a halide like bromine to form an alkane with halides. This reaction is taking place in conditions of Sn1, although this is an addition (Two steps, see picture below for mechanism).
The bromine, has a high electronegativity (2.9) this is even bigger than the iodine (2.7), so, when the bromine acts as a nucleophile in a SN2 or SN1 reaction (like this one), bromine atom becomes slightly more negative, and iodine atom becomes slightly more positive, so strictly speaking, the molecule is slightly polar. When the difference of the electronegativities is below of 0.4, we can say that the molecule is non-polar.
Because of the explanation above, the reaction is taking place with bromine, because it has a higher electronegativity, even more than the chlorine, so the molecule is more polar and can have a better reaction with the 1-butene than the chlorine. Has a better nucleophyle attack and also, is a great leaving group.
The picture below will show the mechanism:
