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larisa86 [58]
4 years ago
14

Draw the major monobromination product formed by heating the following alkane with bromine.

Chemistry
1 answer:
yawa3891 [41]4 years ago
8 0
A monobromination reaction of an alkane involves an alkane and bromine. The position of the hydrogen atom that will be substituted by the bromine free radical will depend on the order of the alkane. The bromine will attach to the carbon that has the most substituents.
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Answer:

2RbNO₃ + BeF₂ → Be(NO₃)₂ + 2RbF, because Be keeps a 2+ charge throughout the reaction

Explanation:

2RbNO₃ + BeF₂ → Be(NO₃)₂ + 2RbF, because Be keeps a 2+ charge throughout the reaction

Rb is a +1 cation, NO3 is a -1 anion, Be is a +2 cation and F is a -1 anion.

In writing an ionic compound the charge of the cation becomes the subscript of the anion and the charge of the anion becomes the subscript of the cation.  

So the ionic compound formed between Be2+ and F- is BeF2. The ionic compound formed between Be2+ and NO3- is Be(NO₃)₂.  

As there are two NO₃ on the product side it is balanced by writing a 2 coefficient before RbNO₃ on the reactant side.  

And as there are two F on the reactant side it is balanced by writing a 2 coefficient before RbF on the product side.  

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