Question 31 a the question stem gives us all of the

Question 31: A
The question stem gives us all of the information we will need to find a correct answer. The
reaction described is in fact a β­elimination: the CH
3
O ­ anion will remove a hydrogen from the
reactant, and its previously bonded electron will form a pi bond elsewhere on the molecule,
causing the elimination of I as a leaving group.
Execute (Predict): Since Zaitsev’s rule says that the most highly substituted alkene will be
favored in such a reaction, we expect compound 1, with three substituents to the double bond, to
be more abundant than compound 2, with only 2. (A) is correct.

Question 32: C
Bromination of a double bond is a common reaction in which a cyclic bromonium ion
intermediate is formed. The second bromine ion then attacks the more substituted carbon from
the backside of the bromonium, such that the bromine substituents end up in an anti
configuration, as shown below.
Execute (Predict): Because bromination of alkenes is an example of anti addition, we expect
the correct answer to exhibit this pattern. The only answer choice that does is (C) , which is
therefore correct.

Passage VII (Questions 33­39) (pages 416­419)
Sample Passage Outline
P1. Difficulties of preparing a standard solution for titration
P2. Primary standards: Na
2
CO
3
or KHP; criteria
P3. Prep of NaOH solution using KHP standard
F1. Reaction of NaOH with KHP
P4. Continuation of prep
Answer Key
33. A
34. A
35. D
36. A
37. D
38. C
39. A
Explanations
Question 33: A
The passage provides a list of criteria that make a good primary standard. Answer choices that
violate any of these criteria will NOT make good primary standards and should be eliminated.
Execute (Eliminate):
(A): NaCl, is stable, soluble in water, nontoxic, and inexpensive. Let’s keep this one and keep
going.
(B): The passage states explicitly that HCl is a poor primary standard because its solubility and
volatility is affected by temperature. Eliminate.

(C): KOH is similar to NaOH, which was also mentioned in the passage as being reactive with
air. This would make it a poor primary standard, so we should eliminate this.
(D): Hg
2
Cl
2
violates two of the cited criteria; it is both toxic and insoluble in water. This choice is
out too.
(A) is the only answer that does not violate a cited criterion, and is therefore correct.
Question 34: A
Le Châtelier's principle states that a system at equilibrium will shift to counteract any changes
placed on the system. The equilibrium of a solution of Ca(OH)
2
would appear as follows:
Ca 2+ + 2OH ­ ⇌ Ca(OH)
2
The answer choices all name compounds that contain hydroxide and that, according to the
question stem, are “fully soluble.”
Execute (Predict): If we add OH ­ ions to the solution, the equilibrium will shift to the right,
forming more Ca(OH)
2
and removing Ca 2+ ions from the solution. Thus, in order to change the
concentration the LEAST, we’ll want to add the FEWEST equivalents of OH ­ as possible. We are
given that the concentration of each of the solutions is the same, so (A) , which only adds one
hydroxide ion per molecule, is correct.