Chapter 10 Review Questions:
1. Classify the following alkyl halides:
I
Cl
Br
2º
Br
2º
2º
1º allylic
Cl
2º benzylic
I
3º
2. All about radicals:
a. What is the charge on a radical? Its neutral, no charge
b. Why are radicals reactive? The radical atom lacks an octet, has an odd number
of electrons
c. What is the hybridization of a radical? Sp2
d. What two factors stabilize radicals? Inductive effect and hyperconjugation
e. Radicals are similar to what other kind of intermediate? carbocations
f. What is the order of stability of radicals? 3º > 2º > 1º
g. List the three basic steps in a radical reaction. Initiation, propagation,
termination
3. Identify the type of radical step for each of the following:
a. Propagation
b. Initiation
c. Propagation
d. Termination
4. What problem is associated with working with radicals? Why are they not commonly
used in organic synthesis?
Radicals are difficult to control under most reaction conditions, leading to multiple
products.
5. How many types of radicals would form in each of the following molecules? (Could
have worded that as “How many monohalogenated products might form from each of
the following molecules?”)
3 types (aromatic
ring carbons don't
react)
6 types (methyls on
left are symmetrical)
9 types (no symmetry)
6 types (methyls on left
symmetrical, ethyls on right
also symmetrical)
6 types (3 in ring, 3
attached)
2 types (ethyl groups
are symmetrical)
6. Draw all the monochlorinated products for each:
Cl
Cl
Cl
Cl
Cl
Cl
Cl
3
Cl
7. If one sp C-H bond has a bond energy of 96 kcal/mol and another sp3C-H bond has a
bond energy of 93 kcal/mol, which bond is weaker, easier to break and why?
The bond with the lower energy value (93 kcal/mol) requires less energy to break. It
would require less energy to break because the carbon radical that forms is more stable.
8. Which of the following anions is the more reactive, stronger nucleophile and why?
The anion on the right does not participate in resonance and will be the stronger
nucleophile. Its charge is centered on one specific atom. The anion on the left does
participate in resonance and by doing so, delocalizes and spreads the anion over two
different carbons in the molecule, effectively splitting the anion’s charge over those
two carbons, weakening the effect of the anion on the left molecule.
9. The atoms involved in resonance must be of what hybridization? Why? How many pi
electrons are involved in the left molecule for Question 8?
Atoms involved in resonance must be sp2 hybridized – a perpendicular p orbital is
required on all atoms that are participating in resonance. It is inside the perpendicular
p orbitals that the pi electrons shift.
There are two pi electrons in the double bond itself and there are two electrons in the
lone pair giving a total of 4 electrons in the pi system for the molecule, on the left, in
Question 8.
10. Draw the resonance form for the following intermediates and note if they are
equivalent resonance forms, or not.
equivalent
non-equivalent
non-equivalent
equivalent
11. While NBS works well with benzylic systems, its less useful synthetically for allylic
systems. How many products will form in the following reaction (be careful!)?
Four different products from four different radicals (two initially formed in the
original allylic positions and then also the products formed from the resonance forms of
each:
Br
NBS, light
Br
Br
Br
12. Fill in the apppropriate reagent or product:
a. PBr3 (1º alcohol)
b. HCl (3º alcohol)
c. PBr3 (2º alcohol)
d.
HBr
OH
Br
Br
or
Br
or
and
Br
e.
H2O
Mg
Br
H
MgBr
or
f.
I
Mg
MgI
CH3OH
g. Acidic proton in alcohol will destroy Grignard as it forms:
Br
Mg
OH
OH
h. (CH3)2CuLi
i. (CH3CH2)2CuLi
j. The starting material for this reaction is optically active but the product is not
optically active. Why is it not?
Br
Ph2CuLi
chiral center isn't chiral now!