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Chapter 4 Practice Chem 12A
- The major monobrominated product which results when ethylcyclohexane is subjected to free radical
bromination is ________.
A) a primary bromide
B) a secondary bromide
C) a tertiary bromide
D) a quaternary bromide
E) bromomethane
- Consider the bond dissociation energies listed below in kcal/mol.
CH
CH
CH
Br 68
(CH 3 ) 2 CH-Br 68
(CH 3 ) 3 C-Br 65
These data show that the carbon-bromine bond is weakest when bromine is bound to a ________.
A) methyl carbon
B) primary carbon
C) secondary carbon
D) tertiary carbon
E) quaternary carbon
- The bond dissociation energy is the amount of energy required to break a bond ________.
A) homolytically
B) heterolytically
C) so as to produce the more stable pair of ions
D) via hydrogenation
E) none of the above
- Rank the free radicals (I-III) shown below in order of decreasing stability (i.e., from most stable to least stable).
A) I > III > II B) II > III > I C) I > II > III D) II > I > III E) III > II > I
- Consider the reaction of A being converted into B at 25°C. If the ΔG° of this reaction is (^) +1.0 kcal/mol, the K eq
is
________ and the % conversion is ________.
A) 0.18; 15% B) 0.43; 30% C) 1.0; 50% D) 2.3; 70% E) 5.4; 84%
- Which of the following correctly expresses the standard Gibbs free energy change of a reaction in terms of the
changes in enthalpy and entropy?
A) ΔG° = ΔH° - TΔS°
B) ΔG° = ΔH° + TΔS°
C) ΔG° = ΔS° - TΔH°
D) ΔG° = ΔS° + TΔH°
E) none of the above
- Which of the following species is not formed through a termination reaction in the chlorination of methane?
A) CH
Cl B) HCl C) H 2
D) CH
CH
- In the first propagation step of the free radical chlorination of methane, which of the following occurs?
A) Cl 2
dissociates.
B) A chlorine radical abstracts a hydrogen.
C) A carbon radical reacts with Cl 2
D) A carbon radical reacts with a chlorine radical.
E) Two chlorine radicals combine.
- In the reaction of Cl 2
with ethane and UV light, which of the following reactions would be a chain termination
event(s)?
I) Cl∙ (^) + CH 3
- CH
→ CH
- CH
II) Cl∙ (^) + CH 3
- CH
→ CH
- H
C∙ (^) + HCl
III) Cl∙ (^) + CH 3
- H
C∙ → CH
- CH
IV) Cl 2
+ CH
- H
C∙ → CH
- CH
V) Cl 2 + UV light
→ C l∙ (^) + Cl∙
A) reaction V
B) reactions I and IV
C) reactions III and IV
D) reactions I and II
E) reaction III
- If the equilibrium constant (K eq
) of a reaction is 0.5 then which of the following that must be true?
A) The reaction will have an early transition state. B) Reaction equilibrium will favor the products.
C) Gibbs free energy (G) is positive. D) Gibbs free energy (G) is negative.
- Given the pK a
values for the two acids below, what would you expect the K eq
for this reaction to be?
A) K
eq
> 1 B) K
eq
C) K
eq
< 1 D) You can't tell from the information given.
- Species with unpaired electrons are called ________.
- What is the hybridization of the positively charged carbon in (CH 3
C
- When compound I, C 7
H
, was treated with chlorine and light it yielded 3 monochlorination products that
could be separated by chromatography. Two of the products were primary alkyl halides and the other was a
secondary alkyl halide. Provide a possible structure for compound I.
Remove an H+ from the following structure to create the most reactive (least stable) carbanion.
Provide the structure of the transition state in the first propagation step of the free radical chlorination of ethane.
What C 5 H 12 isomer will give only a single monochlorination product?
Consider the transformation of A to B (i.e., A →^ B). If at equilibrium at 25°C the concentration of A is 20% of the
initial concentration of A, determine the value of ΔG° (in kcal/mol) for this reaction. R (^) = 1.987 cal/mol∙K.
- When acetaldehyde (CH 3 CHO) is deprotonated, the resulting anion is stabilized by resonance. Draw the major
resonance contributing forms of this anion.
Use the Hammond Postulate to explain why free radical brominations are more selective than free radical chlorinations.
Write the structures of all of the monobromination products of 1,1,3,3-tetramethylcyclobutane.
Does one expect ΔS° in a propagation step of the free-radical chlorination of methane to be greater than zero, less
than zero, or approximately equal to zero? Briefly explain your choice.
- Consider the one-step conversion of F to G. Given that the reaction is endothermic by 5 kcal/mol and that the
energy difference between G and the transition state for the process is 15 kcal/mol, sketch a reaction-energy
diagram for this reaction. Make sure to show how the given energy differences are consistent with your sketch.
- When Br radical reacts with 1-butene (CH 3 CH 2 CH=CH 2 ), the hydrogen atom which is preferentially
abstracted is the one which produces a resonance stabilized radical. Draw the major resonance contributing
forms of this radical.
Answer Key
Testname: UNTITLED
1) C
2) D
3) A
4) A
5) A
6) A
7) C
8) B
9) E
10) C
11) C
radicals or free radicals
the activation energy or Ea
In an exothermic reaction, weaker bonds are broken and stronger bonds are formed.
Kinetics
2.0 kJ/mol
A is more stable because it has an additional resonance structure that can be drawn for it (resonance stabilization).
rate = k[(CH 3
CHCl]
- sp
(CH3)4C or neopentane or 2,2-dimethylpropane
- 3,400J/mol OR - 0.82 kcal/mol
Answer Key
Testname: UNTITLED
- The first propagation step in free radical bromination is endothermic while the analogous step in free radical
chlorination is exothermic. From the Hammond Postulate, this means that the transition state for the bromination is
product-like (ie, radical-like) while the transition state for the chlorination is reactant-like. The product-like transition
state for bromination has the C-H bond nearly broken and a great deal of radical character on the carbon atom. The
energy of this transition state reflects most of the energy difference of the radical products. This is not true in the
chlorination case where the transition state possesses little radical character.
- The propagation steps of the free-radical chlorination of methane are shown
below.
CH
∙ (^) + HCl
CH
∙ (^) + ClCl → (^) CH 3
Cl + Cl∙
In each of the steps, two reactant molecules generate two product molecules. This similarity in the number of molecular species
means that the disorder in the reaction is neither greatly increased nor diminished. Therefore, one expects ΔS° in either
propagation step to be approximately equal to zero.
32) CH
∙
C HCHCH
↔ CH
CHCH
∙
C H
- We consider chlorine radicals to be more reactive and because there are nine primary hydrogens as compared to one
tertiary hydrogen, the major product would come from abstraction of the primary hydrogen. However, bromination is
more selective, and the tertiary abstraction is 1600 time more likely than the primary, leading to
2 - bromo- 2 - methylpropane.
