CHEMISTRY 1AA3
TUTORIAL PROBLEM SET 8 Week of March 11, 2002
Answers
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1.
Using both sawhorse representations and Newman projections, draw the energetically
most favorable and least favorable conformers of 1,2-dichloroethane.
ANSWER
In the most favorable case, the CH2Cl groups are not only staggered, but also the
chlorines are in an anti orientation (at 180º to each other). In contrast, in the most
unfavorable situation, the two chlorines are mutually eclipsing.
most favorable
least favorable
staggered
eclipsed
H
H
Cl
H
H
H
Cl
Cl
Cl
H
H
H
sawhorse representations
Cl Cl
Cl
H
H
H
H
Cl
2.
H
H
H
H
Newman projections
Draw and name the most favorable conformers of:
(a) 1,4-dibromocyclohexane (b) 1,3-dibromocyclohexane.
ANSWER
H
H
H
Br H
H
H
H
H
H
H
Br
Note how the bulky bromines
are in equatorial positions; they
are also on opposite sides of the
six-membered ring, hence trans.
trans-1,4-dibromocyclohexane
H
H
Br
H
H
H
H
Br
H
H
H
H
cis-1,3-dibromocyclohexane
Note how the bulky bromines
again occupy equatorial positions;
but in this case they are on the
same face of the six-membered
ring, hence cis.
Identify all the sp2-hybridized atoms in:
3.
(a) 3-aminocyclobutanone
sp2
H 2N
O
sp2
(b) cis-3,4-dichlorocyclopentene
Cl
Cl
sp2
sp2
(c) E-1-bromo-3-chloro-1-fluoropropene
sp2
sp2
H
ClCH2
4.
Br
F
Note the priorities:
Br > F and CH2Cl > H.
The highest priority
substituents are on
opposite sides of the
double bond, hence E.
A molecule Q, CxHyOz, contains 44.12 % carbon and 8.82 % hydrogen, and is highly
symmetrical. Treatment with sodium liberates 2 moles of hydrogen per mole of Q.
Suggest a reasonable structure for Q. [Hint: recall the reaction of sodium with water.]
ANSWER
Let's determine the ratio of atoms in molecule Q:
For C: 44.12/12 = 3.67
For H: 8.82/1 = 8.82
For O: 47.06/16 = 2.94
To get the ratio of atoms, divide by the smallest of these numbers:
For C: 3.67/2.94 = 1.25
For H: 8.82/2.94 = 3
For O: 2.94/2.94 = 1
Since molecules only come with integer values of atoms, the smallest possible empirical formula
is C5H12O4 (C10H24O8 and higher multiples are also viable) and we are told that Q is
highly symmetrical. We also know that one mole of Q liberates two moles of H2 when
treated with sodium. This is reminiscent of the reaction of Na with water, and implies
that four hydroxyl groups are present, as in the molecule shown below.
CH2OH
C
HOCH2
CH2OH
CH2OH
5.
Assign hybridization states to each of the carbon, nitrogen and oxygen atoms in adenosine
(shown below).
Does this molecule contain any primary, secondary or tertiary alcohols (or amines) ?
HO
N
O
HO
N H2
N
N
N
OH
All the oxygens are sp3 hybridized.
The carbons and nitrogens indicated
by arrows are sp3 hybridized; the
others are sp2 hydridized.
10 alcohol
30 amine
HO
10 amine
N
O
HO
20 alcohol
N H2
N
OH
N
20 alcohol
N