Dear
Dolphinsiu,
Please, tell me: Where have You dropped, -
in the meantime - , the
other “
6” (sex) “Hydrogens” of the second Reaction/Compound?
Is it not a “small” little toooo “radical” from you?
In secret, I hopped, that you have discovered, chem. Reactions have
something to do whit breaking bounds and building others instead.
But that means, that the whole thing depends on the Quality of the
bound to be broken, AND the Quality of the bound which has to be build.
In your case, it’s the bound between C and H, C-H.
So the Question in real is: How many bound of which Quality you have ??
(You may translate: Quality in this chem. sense as “
Strength”!!)
About “counting”: What You and I call counting is for Mother Nature,
HER liked “Likelihood” !
With simple words: As long as “
hv” is involved, AND you
replace
EVERY from You now
numbered “Hydrogens”
ONCE with a Chlorine, —
How many different products (mono-"Chlorides"), AND How much of each you should get?
But: Why can’t you find your “counted” ratio (= Likelihood for Mother N.)?
Are you now able to answer the Question of my fourth Hint?
Hope it may help
Good Luck!
ARGOS
++