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Chapter 9 Test Practice
Organic
9.10 Methyl iodide (0.1 M) and hydriodic acid (HI, 0.1 M) are allowed to react in ethanol solution with 0.1
M sodium ethoxide. What products are observed?
9.16 The following hydroxide-catalyzed β-elimination takes place by a carbon–anion stepwise mechanism.
Show the carbon–anion intermediate and explain its stability. Think in terms of a polar effect (Sec.
3.6C). Recalling also that resonance structures imply heightened stability (Sec. 1.4), draw a resonance
structure for this anion as well.
9.21 Draw the structure of the starting material that would undergo anti-elimination to give the E isomer of
the alkene product in the E2 reaction of Eq. 9.40.
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9.22 What nucleophile or base and what type of solvent could be used for the conversion of isobutyl bromide
into each of the following compounds?
(a) (CH3)2CHCH2S+(CH3) 2 Br-
(b) (CH3) 2CHCH2SCH2CH3
(c) (CH3) 2C=CH2
9.35 Give the products expected when isopentyl bromide (1- bromo-3-methylbutane) or the other substances
indicated react with the following reagents.
(a) KI in aqueous acetone
(b) KOH in aqueous ethanol
(c) K+ (CH3) 3C-O— in (CH3) 3C-OH
(d) product of part (c) + HBr
(e) CsF in N,N-dimethylformamide (a polar aprotic solvent)
(h) Li in hexane, then ethanol
i) sodium methoxide in methanol
(j) Mg and anhydrous ether, then D2O
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9.52 (a) Two isomeric SN2 products are possible when sodium thiosulfate is allowed to react with one
equivalent of methyl iodide in methanol solution. Give the structures of the two products. (Thiosulfate
is an example of an ambident, or “twotoothed,” nucleophile.)
(b) In fact, only one of the two possible products is formed. Which one is formed, and why?
9.55 When methyl bromide is dissolved in methanol and an equimolar amount of sodium iodide is added, the
concentration of iodide ion quickly decreases, and then slowly returns to its original value. Explain.
9.56 Consider the following experiments with trityl chloride, Ph3C-Cl, a very reactive tertiary alkyl halide:
(1) In aqueous acetone, the reaction of trityl chloride follows a rate law that is first order in the alkyl
halide, and the product is trityl alcohol, Ph3C-OH.
(2) In another reaction, when one equivalent of sodium azide (Na+ N3 — ; see Table 9.3) is added to a
solution that is otherwise identical to that used in experiment (1), the reaction rate is the same as in (1);
however, the product isolated in good yield is trityl azide, Ph3C-N3.
(3) In a reaction mixture in which both sodium azide and sodium hydroxide are present in equal
concentrations, both trityl alcohol and trityl azide are formed, but the reaction rate is again unchanged.
Explain why the reaction rate is the same but the products are different in these three experiments.
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9.58 An optically active compound A has the formula C8H13Br. Compound A gives no reaction with Br2 in
CH2Cl2, but it reacts with K+(CH3)3C-O— to give a single new compound B in good yield. Compound B
decolorizes Br2 in CH2Cl2 and takes up hydrogen over a catalyst. When compound B is treated with
ozone followed by aqueous H2O2, dicarboxylic acid C is isolated in excellent yield; notice its cis
stereochemistry.
Identify compounds A and B, and account for all observations. (If you need a refresher on how to solve
this type of problem, see Study Guide Links 4.3 and 5.3.)
9.69 In 1975, a report was published in which the reaction given in Fig. P9.69 was observed. The -OBs
(brosylate) group is a leaving group conceptually like halide. (Think of this group as you would -Br.)
Notice that the reaction conditions favor an SN2 reaction.
(a) This result created quite a stir among chemists because it seemed to question a fundamental principle
of the SN2 reaction. Explain.
(b) Because the result was potentially very significant, the work was reinvestigated very soon after it
was published. In this reinvestigation, it was found that after about 10 hours’ reaction time, the product
consisted almost completely of trans-P. Only on standing under the reaction conditions did cis-P form
(and trans-P disappear) to give the product mixture shown in Fig. P9.69. Furthermore, when the trans
isomer of S was subjected to the same conditions, mostly cis-P was formed after 10 h, but, after 5 days,
the same 75:25 cis:trans product mixture was formed as in Fig. P9.69. Finally, subjecting pure cis-P or
pure trans-P to the reaction conditions gave, after five days, the same 75:25 mixture. Explain these
results.
(c) Why is cis-P favored in the product mixture?
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