Mechanisms

CH4

CHHHHCHHHHCHHHHHHHHC

SN2

Nu
LG
LG
HHH
Nu
LG
H
Nu
‡
δ+
δ–
δ–
Nu
Nu
LG
HHH
Nucleophile attacks an sp3 hybridized carbon.
‡
δ+
δ–
δ–
Nu–C bond forms as C–LG bond breaks through a single transition state.
Nu
LG
LG
HHH
‡
δ+
δ–
δ–
Nu–C bond forms as C–LG bond breaks through a single transition state.
Product forms with inverted stereochemistry. The leaving group departs with the bonding electrons.

Acetal Formation

OH
CH3O
HOHOHClHO+HOHCH3OHOHOHHHO+HHOHOHHOHO+HOHO+HCH3Cl−CH3OHO+HOHH
HCl
CH3OH
HOHOOH
CH3O
HCl
CH3OH
OHHOHClHOHO+Cl−HHOHOO+HCl−HHOO+HHOHCH3OHOHHOHCH3HO+HHO+HOHHO+HHO+HOHHHOHO+HO+HHHO+HOHCH3OHO+HCH3O+HOHO+HCH3Cl−CH3O+OHHCl−OHCH3
CH3O
OClH

OCN-

Step 1: Total the number of valence electrons
Step 1: Total the number of valence electrons
OCN–
Negative charge:
C:
N:
O:
Total ve:
4
5
6
1
16
Total ve:
16
Step 1: Total the number of valence electrons
Negative charge:
C:
N:
O:
4
5
6
1
OCN–
Step 2: Place least electronegative atom (not H)
Placed ve:
0
16
Remaining ve:
C
Step 2: Place least electronegative atom (not H)
Total ve:
16
Placed ve:
Placed ve:
0
0
16
16
Remaining ve:
Remaining ve:
C
Step 2: Place least electronegative atom (not H)
Step 3: Place remaining atoms around central atom
Total ve:
16
Placed ve:
0
16
Remaining ve:
CON
Step 3: Place remaining atoms around central atom
Total ve:
16
Placed ve:
0
16
Remaining ve:
CON
Total ve:
16
Placed ve:
0
16
Remaining ve:
Step 3: Place remaining atoms around central atom
Step 4: Draw a single bond between each terminal atom and the central atom.
CON
Total ve:
16
Placed ve:
Remaining ve:
Step 4: Draw a single bond between each terminal atom and the central atom.
0
16
4
12
2 e–
2 e–
CON
Total ve:
16
Placed ve:
4
12
Remaining ve:
2 e–
2 e–
Step 4: Draw a single bond between each terminal atom and the central atom.
Step 5: Distribute remaining electrons as lone pairs on terminal atoms. If any remain, place on central atom.
CON
Total ve:
16
Placed ve:
Remaining ve:
Step 5: Distribute remaining electrons as lone pairs on terminal atoms. If any remain, place on central atom.
4
12
16
0
6 e–
6 e–
CONCON
Total ve:
16
Placed ve:
16
0
Remaining ve:
Step 5: Distribute remaining electrons as lone pairs on terminal atoms. If any remain, place on central atom.
6 e–
6 e–
Step 6: If central atom has fewer than 8 electrons, convert one or more lone pairs from terminal atoms into bonding pairs to reach octet.
Option 1: Two lone pairs from oxygen.
CONCON
Step 6: If central atom has fewer than 8 electrons, convert one or more lone pairs from terminal atoms into bonding pairs to reach octet.
Option 1: Two lone pairs from oxygen.
CONCONCON
Step 6: If central atom has fewer than 8 electrons, convert one or more lone pairs from terminal atoms into bonding pairs to reach octet.
Option 1: Two lone pairs from oxygen.
Option 2: One lone pairs from oxygen and one from nitrogen.
CONCONCON
Step 6: If central atom has fewer than 8 electrons, convert one or more lone pairs from terminal atoms into bonding pairs to reach octet.
Option 2: One lone pairs from oxygen and one from nitrogen.
Option 2: One lone pairs from oxygen and one from nitrogen.
CONCONCONCON
Step 6: If central atom has fewer than 8 electrons, convert one or more lone pairs from terminal atoms into bonding pairs to reach octet.
Option 2: One lone pairs from oxygen and one from nitrogen.
Option 3: Two lone pairs from nitrogen.
CONCONCONCON
Step 6: If central atom has fewer than 8 electrons, convert one or more lone pairs from terminal atoms into bonding pairs to reach octet.
Option 3: Two lone pairs from nitrogen.
CONCONCONCON
Step 6: If central atom has fewer than 8 electrons, convert one or more lone pairs from terminal atoms into bonding pairs to reach octet.
Option 3: Two lone pairs from nitrogen.
Step 7: Since more than one valid arrangement exists, determine best structure by analyzing the formal charges on each atom.
CONCONCON
Step 7: Since more than one valid arrangement exists, determine best structure by analyzing the formal charges on each atom.
+1
0
–2
0
0
–1
–1
0
+1
CONCONCON
Step 7: Since more than one valid arrangement exists, determine best structure by analyzing the formal charges on each atom.
+1
+1
0
0
–2
–2
0
0
0
0
–1
–1
–1
–1
0
0
+1
+1
The best structure has the negative formal charge on the more electronegative atom and the positive formal charge on the less electronegative atom.
OCNCONCON
Step 7: Since more than one valid arrangement exists, determine best structure by analyzing the formal charges on each atom.
The best structure has the negative formal charge on the more electronegative atom and the positive formal charge on the less electronegative atom.
–1
0
+1
+1
0
–2
0
0
–1
CON
Step 7: Since more than one valid arrangement exists, determine best structure by analyzing the formal charges on each atom.
–1
0
+1
The best structure has the negative formal charge on the more electronegative atom and the positive formal charge on the less electronegative atom.
Step 8: Because the structure is ionic, encapsulate in square brackets and add the charge.
CON
Step 8: Because the structure is ionic, encapsulate in square brackets and add the charge.
−


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