Some chem questions. (1 Viewer)

[undalay]

A hydrocarbon X of molecular formula C6H14 was allowed to react with limited chlorine gas in the presence of light. The resulting mixture contained two products with different boiling points, as well as excess starting material. What is the systematic name of X?

A) 2,2-dimethylbutane
B) 2,3-dimethylbutane
C) 2-methylpentane
D) 3-methylpentane
E) hexane

The solubility product of Magnesium Hydroxide is 1 x 10^-11 mol^3 L^-3 at 25C. What is the solubility of magnesium hydroxide in a solution of pH 11.0?

A) 6g L^-1
B) 1g L^-1
C) 0.1 g L^-1
D) 0.0006 L^-1
E) 6x 10^-7 L^-1

 

[brenton1987]

B - 2,3-dimethylbutane
If you draw the molecule in full youll notice that there are only 2 unique types of carbon. Since chlorination is not a selective process the chlorine radical can attack either of them.

 

[xiao1985]

1) what a strange question. all 5 seems viable.

The reaction concerned here is halogen substitution (with prescence of light). And different boiling points refer to the different isomer formed after substitution. I guess i can only assume that only 1 hydrogen will be replaced by chlorine... in which case, there's B will be the answer, because there are only 2 possible products:

1-chloro-2,3-dimethylbutane
2-chloro-2,3-dimethylbutane

very vague question though...

2) hmmm supposingly in industrial chem? I'm assuming the equation you are after is:

Mg(OH)2 (s) ----> Mg2+ + 2 (OH) -


edit: breton: faster than me =p

but Cl may attack more than 1 carbon... that's why i'd say one need to assume only 1 carbon is attacked.
So the equilibrium constant = [Mg2+][OH-]^2 (hence the unit, mol^3 L^-3, MgOH2 is a solid, and the conc. maybe treated as constant)

at pH 11, [OH-] = 10^-3 mol/L
:. [mg2+] = 10^-5 mol/L

solubility = 10^-5 mol/L
mol mass of Mg (OH)2 ~60 g/mol
solubility ~ 6 E -4 g/L

ie D

 

[brenton1987]

but Cl may attack more than 1 carbon... that's why i'd say one need to assume only 1 carbon is attacked.

When there is limited halogen (excess hydrocarbon) only monochloro substitution occurs.

 

[undalay]

Sorry i still don't fully understand question 1. : S

 

[brenton1987]

2,3-dimethylbutane

Look at the above picture, would you agree that all of the red hydrogens are equivalent? Would you also agree that the 2 blue hydrogens are equivalent?
That means that the green carbons are equivalent as are the yellow carbons.

If you were to replace a red hydrogen with a chlorine atom, the molecule would be identical no matter which red hydrogen was replaced. Same as if you replaced a blue hydrogen, doesnt matter which.

Because there are only 2 unequilavent hydrogens, only 2 compounds can be formed when the parent compound is chlorinated.

If you follow the same logic through on all the other molecules you will find that only B satisfies the criteria of the question.

2,2-dimethylbutane (4 types of hydrogen)
2,3-dimethylbutane (2 types of hydrogen)
2-methylpentane (5 types of hydrogen)
3-methylpentane (4 types of hydrogen)
hexane (3 types of hydrogen)

Hope that makes a bit more sense.

 

[undalay]

Yeah it does, thank you. Isn't HCl a possible product too tho :S ?

 

[brenton1987]

Yeah it does, thank you. Isn't HCl a possible product too tho :S ?

Yes.
But since you are reacting liquid 2,3-dimethylbutane with gaseous chlorine, the HCl will boil as a gas at -85 degrees C because there is no water for it to form an aqueous solution.

 

[undalay]

oh right so the final products is exluding the chlrorine and HCl gases.

Thanks xiao1985 and brenton1987