For doing reactions, it is necessary to memorize them in a general sense, but they will fall into categories such as "addition" and "elimination" and "substitution" for example. It's best to first draw all of the bonds to all of the atoms, even the hydrogens, when first writing the structure of the product of an organic reaction. Then you can be sure to get the correct number of bonds on the atoms. Four bonds on C, three bonds on N, two bonds on O, one bond on the halogens.
Cl Cl
| |
a) This one's OK. The structure is H–C––C–H
| |
H H
b) Count the number of bonds on the alkene carbons!
c) Again check the number of bonds on the carbons. There should now be a single bond between the carbons rather than a double bond.
General patterns to remember, alkyne + 2 AB
Adding the first "equivalent" of molecule AB gives
A B
\ /
H–C
C–H + A–B ––�?gt; C=C
/ \
H H
Adding the second equivalent of AB gives
A B A B
\ / | |
C=C + A–B ––�?gt; H–C––C–H
/ \ | |
H H A B
"A–B" can be H–H (H2), H–Cl, Cl–Cl (Cl2), H–OH (H2O), and H–OR (an alcohol) for example.
Remember, four bonds on carbon! 
d) Use the second general pattern above. Draw the structure of the starting alkene also, that helps. After a little practice you'll get used to the condensed structural formulas, but at first it's very easy to make mistakes with them since most of the bonds are not shown.
e) In this reaction (hydration, addition of water) "A–B" is "H–OH". Add H to one of the alkene carbons and the OH group to the other, bonded to the oxygen (same reaction pattern as above). This reaction obeys Markovnikov's rule which says, when adding "H–Y" to an alkene or alkyne, the H atom goes to the alkene or alkyne carbon that already has the greater number of hydrogens (this is the "less substituted carbon"). "Y" then bonds to the other alkene or alkyne carbon (the "more substituted" carbon). "Them that has, gets!" (the hydrogen).
The above reactions are all additions. Elimination is the opposite of addition. When you react chloroethane with a strong base like NaOH, the net result is removal or elimination of HCl although that's not literally how it happens. The actual reaction is
H Cl H H
| | \ /
H–C––C–H + NaOH ––�?gt; C=C + NaCl + H2O
| | / \
H H H H
Adding HCl back to the alkene product would give us back the original starting compound.
In the case of dehydration of alcohols, the net result is removal or elimination of water. This reaction is acid-catalyzed:
H OH H H
| | H2SO4 \ /
H–C––C–H –––––�?gt; C=C + H2O
| | heat / \
H H H H
Adding water back to the alkene would reform the original alcohol.
Not to complicate things too much, but there is also a rule regarding eliminations that's similar to Markovnikov's rule. It is called Zaitsev's rule and it simply says that when you can form more than one alkene product by an elimination reaction, the most stable alkene product is most favored to form, which is the "most highly substituted" alkene. Here's an example:
Cl
|
CH3–CH2–CH–CH3 + NaOH ––�?gt; CH3–CH=CH–CH3 + CH3–CH2–CH=CH2 ( + NaCl + H2O )
2-chlorobutane 2-butene 1-butene
major product minor product
(a disubstituted alkene) (a monosubstituted alkene)
Remember, "Four bonds on carbon!" 
Steve
Free Forum Hosting
Important Announcement
Prev Message
