Chemistry 220/221
Midterm Examination
November 13, 2003
Name:______________________________ ID#:________________ Lab:__________
Good morning, and welcome to the second organic chemistry picnic, tea and sale. Do all
the questions on this paper. You may use pencil or pen, as long as it’s clear, but you may
not use red pen. You may not use any notes, calculators, or other aids, but you may use
molecular models. If you change any answer, make sure I can tell what you want me to
mark, or I’ll mark the first thing I come to. Good luck!
1. Name or give structure as appropriate. Don’t forget your stereochemistry! (5 points
each):
(a)
(Z)-3-cyclopropyl-2-pentene (many people forgot the (E))
(b) (Z)-3-chloro-4-methyl-3-hexene
Cl
Many people got the stereochemistry backward. Check on the rules for E/Z!
(c)
(c) (S)-trans-4-methyl-2-hexene
H
Most people got the double bond stereochemistry OK (by accident?) but omitted the
chirality around carbon 4. Note that you need to show that hydrogen.
(d)
2,3,4-trimethylhexene. It’s not necessary to indicate the position of the double bond
(though not wrong to do it), because it’s at carbon 1. It has no stereochemistry, and none
is shown for the other potentially stereogenic carbons (3 and 4).
(e) trans-3,4-dimethyl-3-hexene
Most people got this one.
2. Define the following terms in the space provided. You may use paragraph or point
form, but be very clear! (5 points each)
(a) electrophile
An electrophile is an electron pair donor or Lewis base. It is not necessarily an anion,
though many are. It is not necessary for a species to be involved in substitution in order
to be an electrophile. For example, a proton in an acid-base reaction is an electrophile,
and it is an electrophile in the addition of H-X to a double bond as well.
(b) Sn1 reaction
This is a unimolecular substitution reaction in which a leaving group is replaced by a
nucleophile in a 2 step process. The first step is ionization to form the carbocation, then
the nucleophile forms a new bond to it. Sn1 reactions are favoured by polar reaction
conditions and are prone to rearrangements. The leaving group is not necessarily a
halide ion.
(c) Leaving group
A leaving group is simply an atom, ion or molecule which departs during a substitution
or elimination reaction. It again is not necessarily a halogen.
(d) Saytzeff’s rule
3. The rule states that, in elimination reactions in which there is a choice, the more
substituted double bond is formed in preference to the less. Note that the
substitution refers to the double bond, not the molecule as a whole. Many people
left out the double bond.
Give the reaction equation for the Sn1 substitution of 3-chloro-2,2dimethylbutane by hydroxide ion. Include all likely organic products (5 points).
Show the mechanism of this reaction. The mechanism should show how all the
organic products form. (20 points)
The reaction equation show the reactants, products and conditions, so for this reaction,
it’s:
OH
OH-
+
OH
Cl
Many people left all or part of this out. The mechanism shows the individual reaction
steps:
Cl
+
+
OH
OH-
AND ALSO
+
+
+
OH-
OH
Many people omitted the rearrangement step, which would be pretty easy. The
rearranged product would likely be the major on for this reaction. Elimination products
are superfluous here, since you were asked for the substitution reaction. Stereochemistry
is likewise unnecessary, though not incorrect.
4. In each of the reaction equations shown below, a reactant, product or condition
has been replaced by a question mark. In the space under each reaction, give the
missing component. Some parts may have more than one correct answer. (6
points each part)
Br
HBr
(a)
Br
This is an electrophilic addition reaction. The two products are equally likely because
both ends of the double bond are equally substituted.
base
(b)
n
Br
Elimination can occur only on one side because there are no hydrogens on the other
side for the base to pick off. If the double bond were placed there, there would be 5
bonds to carbon. Rerrangement could also occur under E1 conditions (polar solvent,
relatively weak base), leading to the product in brackets.
(c)
Strong acid
OH
Ideally, you would use a strong acid with a non-nuclephilic conjugate base, such as
sulphuric or phosphoric acid. The use of halogen acids like HCl or HBr might lead
to some substitution
Cl
HCl
(d)
This is the best answer, since Markovnikov's rule favours the desired product. The use
of 2-pentene would yield about a 50:50 mixture of 2- and 3-chloropentane, since there
is no reason to choose one over the other.
Br
(e)
O
O
This is a substitution reaction. Any good leaving group would do, I've just specified
bromine. Alkenes are not electrophilic enough to be attacked by alkoxide ions
under these conditions (but more on that in the second term).