123.202: Organic and Biological Chemistry
Tutorial Answers for gjr’s Section – Sheet 1
Question 1.
Draw the Lewis structures (dot & cross diagrams) for the following FOUR
molecules:
SOCl2, POCl3, CH3(MeO)CCH2 & HNO3
SOCl2
O
O
x
x
Cl
Cl
x
x
Cl x S x Cl
S
Straight in with a nasty one! Both oxygen and sulfur are group 16 so 6
valence electrons & chlorine is group 17 so 7 valence electrons. For
oxygen to complete its octet it needs to form a double bond to the sulfur.
If we then form single bonds from sulfur to both the chlorine atoms we
complete the chlorine’s octet. At this point we still have 2 of sulfur’s
electrons left, so it must have a lone pair. But I hear you cry that makes it
have 10 electrons! Yup, it is in the third row so can expand the octet and
happily have more…
Alternatively you could have drawn the Lewis structure as:
O
x
x
O
x
x
Cl x S x Cl
Cl
S
Cl
This is equally acceptable, all atoms obey the octet rule but you MUST
have the formal charges! Oh yeah, notice that thionyl chloride is
tetrahedral and not trigonal planar. Probably a molecule that contains all
the tricks in the lecturer’s repertoire!
POCl3
O
O
x
Cl
P
Cl
Cl
Cl x P x Cl
Cl
Cl
x
Cl x P x Cl
Cl
O
O
x
x
x
x
P
Cl
Cl
Exactly the same except phosphorus is group 15 so 5 valence electrons
and hence it requires the extra chlorine atom. Just to make sure I continue
to confuse you, whilst an organic chemist will happily draw either of the
resonance forms of SOCl2 we never draw the single bond resonance
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structure of phosphine oxides. The reason for this is that phosphorus is
considered to be extremely oxaphilic and will form very strong bonds to
oxygen.
x
CH3(MeO)CCH2
H
H C H
x
x
x
x
C
xx
C
H
x
x
x
x
x
H xC
O
O
H
H
H
Hopefully that was an easy one….
HNO3
x
HxO
O
O
x
x Nx
x
O
O N
H
O
Question 2.
Draw the Lewis structures (dot & cross diagrams) for the following FOUR
molecules:
CH3CN, IF5, CH3NO2, [SO4]2–
CH3CN
x
xxx
x
H
Hx C xC N
H
IF5
The difficult one!
1. Count the number of valence electrons. I = 7 & F = 7. Total = 7 + (5
x 7) = 42
2. Draw fluorines around central iodine
2
F
F
F
I
F
F now add two electrons to bond each fluorine to the iodine
F
F
F
I
F
F
3. Now fill all the outer atoms (fluorine) so that they obey the octet rule
F
F
F
I
F
F
4. Have only used 40 of 42 electrons so have one lone pair. Place this
on the central atom
F
F
F
I
F
F
F
F
I
F
F
F
CH3NO2
x
O
H
xx
Hx C x N x
x
O
H
x
[SO4]2–
Another hard one! Remember to add 2 extra electrons for the 2– charge!
The formal charge on each oxygen is –1 & the formal charge on sulfur is
+2. So overall the compound’s charge is (4 x –1) + +2 = –2.
y
x
O
O x S xx O
O
y
x
x
Question 3.
3
Give the formal charge (if one exists) on each atom of the following:
H3C
O
S
H3C
CH3
O
H3C
S
H3C
CH3
O
O
H3C N
H3C
CH3
O
CH3
CH3
F
F
B
F
F
O
CH3
S
Sulfur: atomic number = 16 so it has 16 protons (16+). It has a full 1s
orbital (2-), a full 2s (2-), 3 full 2p orbitals (6-). Then we look at the
valence shell; 1 lone pair (2-) & 3 single bonds (so 1/2 of 6 electrons = 3-).
Add all the electrons (or negative charges) together (2+2+6+2+3)=15-.
The difference between 16+ & 15- is 1+ so sulfur has a formal charge of
+1.
Oxygen: atomic number 8 so 8+. Has full 1s (2-), then 3 lone pairs (6-) and
one bond (1/2 x 2 = 1-). Total negative charge is 9- and the difference
between 8+ & 9- is 1-. Therefore the oxygen has a formal charge of –1.
Overall the molecule is neutral.
oxygen
H 3C
O
H 3C
CH3
1s
= 21lp
= 23bonds = 1/2 x 6- = 3Overall
oxygen
O
H3C
O
nitrogen
CH3
CH3
O
H 3C
CH3
= 8+
O
H 3C
O
= 1= 7+
1s
= 20 lp
= 04bonds = 1/2 x 8- = 4Overall
H 3C
= 1+
1s
= 23lp
= 61bond = 1/2 x 2- = 1Overall
H3C N CH3
= 8+
CH3
H3C N CH3
CH3
= 1+
4
boron
1s
= 20 lp
= 04bonds = 1/2 x 8- = 4-
F
F
B
= 5+
F
F
Overall
F
F
B
F
F
= 1-
Question 4.
Draw the molecular orbitals, label the orbitals and give the bond angles for
the following:
CH3OH, CH3–, CH3+, (CH3)2C=NNH2
CH3OH
! (1s+sp3)
sp3 lone pair
H
H
sp3
C, 109.5˚
C
H
O
! (1s+sp3)
H
! (sp3+sp3)
sp3 O, 104.5˚
CH3–
! (1s+sp3)
H
H
sp3
C, 109.5˚
sp3 lone pair
C
H
CH3+
H
H
sp2 C, 120˚
C
empty p
orbital
! (1s+sp2)
H
(CH3)2C=NNH2
5
H
H
! (sp3+sp2)
H
sp3 C, 109.5˚
! (sp2(O)+sp2(N))
C
sp2 lone pair
sp2 C, 120˚
! (1s+sp3)
H
sp3 lone pair
N
C
C
N
H
H
H
" (pz+pz)
sp2 N, 120˚
! (1s+sp3)
H
sp3 N, 107.3˚
! (sp3+sp2)
Question 5.
Draw the molecular orbitals, label the orbitals and give the bond angles for
the following:
O
O
O
N
H
N
the molecular orbitals, label the orbitals and give the bond angles for the
following.
H
! (sp3+sp2)
H
sp2 C, 120˚
H
C
! (1s+sp3)
C, 109.5˚
!
(1s+sp3)
H
sp2 lone pairs
C
H
sp3
sp3 C, 109.5˚
! (sp2(C)+sp2(O))
C
O
O
O
O
sp2 O, 120˚
C
H
H
H
sp3 O
" (pz+pz)
! (sp3+sp3)
Nasty one! Will accept:
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! (sp2(C)+s(H))
! (sp3+sp3)
sp2
C, 120˚
C, 109.5˚
H
O
C
H
H
O
sp2 lone pairs
H
! (1s+sp3)
sp3
! (sp2(C)+sp2(O))
N
C
sp2 O, 120˚
H
sp3 N, 107.3˚
C
H
H
" (pz+pz)
H
But reality closer to:
!*
p lone pair
O
R
N sp2
N
R
R
1
Next two are also a little tricky…
p & " overlap
(parallel)
3
" (pz+pz)
sp C, 109.5˚
H
H
!
H
H
H
C
H
C
C
(sp3+sp2)
! (1s+sp2)
! (1s+sp3)
H
C
sp2 C, 120˚
H
C
sp3 C, 109.5˚
! (1s+sp2)
sp2 C, 120˚
H
empty p orbital
90˚ to bonds
! (sp2(C)+sp2(C))
and…
p & " overlap
(parallel)
" (pz+pz)
! (1s+sp3)
sp3
H
H
C, 109.5˚
H
C
C
sp2 C, 120˚
! (1s+sp2)
! (sp2(C)+sp2(C))
H
H
C
H
C
sp3 C, 109.5˚
H
H
! (sp3+sp2)
C
H
full p orbital 90˚
to bonds
7
! (1s+sp2)
sp2 C, 120˚
N
! (sp2(C)+sp2(C))
sp2 C, 120˚
H
H
C
H
C
C
H
" (pz(C)+pz(N))
C
H
! (sp2(C)+1s(H))
C
H
sp2 lone pair
N
H
C
N
C
C
H
H
! (sp2(C)+sp2(N))
N
" (pz(C)+pz(C))
Answers are available at:
http://www.massey.ac.nz/gjrowlan
8
C
C
H