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Chapter 13: Problem 286

Give the structures and names of the chief organic products expected from the reaction (if any) of n-butyl bromide with: (a) \(\mathrm{NaOH}(\mathrm{aq})\) (g) product (f) \(+\mathrm{D}_{2} \mathrm{O}\) (b) \(\mathrm{KOH}(\mathrm{alc})\) (h) dilute neutral \(\mathrm{KMnO}_{4}\) (c) cold conc. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (i) Nal in acetone (d) \(\mathrm{Zn}, \mathrm{H}^{+}\) (i) \(\mathrm{HC} \equiv \mathrm{C}^{-} \mathrm{Na}^{+}\) (e) Li, then Cul, ethyl bromide (f) \(\mathrm{Mg}\), ether (k) \(\mathrm{H}_{2} \mathrm{O}\) (1) \(\mathrm{NH}_{3}(\mathrm{aq})\)

Short Answer

Expert verified
(a) n-Butanol (b) n-Butanol (c) No reaction (d) n-Butane (e) n-Hexane (f) nC4H9MgBr (Grignard reagent) (g) n-Butyl deuterium oxide alcohol (h) No reaction (i) n-Butyl iodide (j) 1-Butyn-1-yl bromide (k) No significant reaction (l) n-Butylamine

Step by step solution

01

(a) Reaction of n-butyl bromide with aqueous NaOH

First, it's crucial to recognize that aqueous NaOH is a strong base. A primary alkyl halide, 1-bromobutane, will undergo nucleophilic substitution with a strong base, following the SN2 mechanism. The base will attack the electrophilic carbon and displace the bromide ion. The reaction is as follows: \(nC_{4}H_{9}Br + NaOH \rightarrow nC_{4}H_{9}OH + NaBr\). The structure of the main product will be that of n-butanol (primary alcohol) and the name of this organic product is n-butanol.
02

(b) Reaction of n-butyl bromide with alcoholic KOH

Alcoholic KOH, similar to aqueous NaOH, is a strong base. Again, a primary alkyl halide, 1-bromobutane, will undergo nucleophilic substitution with a strong base in an SN2 mechanism. The reaction is as follows: \(nC_{4}H_{9}Br + KOH \rightarrow nC_{4}H_{9}OH + KBr\). The structure of the main product will be that of n-butanol (primary alcohol) and the name of this organic product is n-butanol.
03

(c) Reaction of n-butyl bromide with cold concentrated H2SO4

No significant reaction would occur in this case. Cold concentrated H2SO4 is neither a strong nucleophile nor a strong base. It will not react with 1-bromobutane in any significant manner under these conditions.
04

(d) Reaction of n-butyl bromide with Zn and H+

In this reaction, zinc metal reduces the alkyl halide to an alkane, removing the halogen and replacing it with a hydrogen atom. The reaction is as follows: \(nC_{4}H_{9}Br + Zn + 2H^{+} \rightarrow nC_{4}H_{10} + Zn^{2+} + 2Br^{-}\). The structure of the organic product will be that of n-butane. The name of this organic product is n-butane.
05

(e) Reaction of n-butyl bromide with lithium metal, followed by copper(I) bromide and ethyl bromide

This reaction involves the formation of an organometallic compound, specifically an organolithium reagent, \(nC_{4}H_{9}Li\), when the alkyl halide reacts with lithium. The reaction is as follows: (1) \(nC_{4}H_{9}Br + 2Li \rightarrow nC_{4}H_{9}Li + LiBr\). The subsequent reaction with copper(I) bromide and ethyl bromide will lead to a coupling reaction forming a larger alkane: (2) \(nC_{4}H_{9}CuBr + nC_{2}H_{5}Br \rightarrow nC_{4}H_{9}C_{2}H_{5} + CuBr\). The structure of the final organic product will be that of n-hexane. The name of this organic product is n-hexane.
06

(f) Reaction of n-butyl bromide with magnesium and ether

This reaction results in the formation of a Grignard reagent: \(nC_{4}H_{9}MgBr\). The reaction is as follows: \(nC_{4}H_{9}Br + Mg \rightarrow nC_{4}H_{9}MgBr\). This Grignard reagent can be further utilized in various organic reactions depending on the conditions applied.
07

(g) Reaction of product (f) with D2O

The Grignard reagent formed in the previous reaction (f) can react with D2O to form an alcohol. The reaction is as follows: \(nC_{4}H_{9}MgBr + D_{2}O \rightarrow nC_{4}H_{9}OD + Mg^{2+} + 2Br^{-}\). The product's structure will be n-butanol where one hydrogen has been replaced by a deuterium (heavy hydrogen isotope). The name of this organic product is n-butyl deuterium oxide alcohol.
08

(h) Reaction of n-butyl bromide with dilute neutral KMnO4

No significant reaction would occur in this case. Dilute neutral KMnO4 is not likely to react with an alkyl halide like 1-bromobutane under these conditions.
09

(i) Reaction of n-butyl bromide with NaI in acetone

This reaction involves nucleophilic substitution via the SN2 mechanism. NaI in acetone serves as the nucleophile and replaces the bromide ion leading to a change in the halogen. The reaction is as follows: \(nC_{4}H_{9}Br + NaI \rightarrow nC_{4}H_{9}I + NaBr\). The main organic product yields n-butyl iodide. The name of this organic product is n-butyl iodide.
10

(j) Reaction of n-butyl bromide with sodium acetylide (HC≡CNa)

Sodium acetylide is a strong base and nucleophile. This reaction involves nucleophilic substitution through SN2 mechanism adding the acetylide ion to the electrophilic carbon. The reaction is as follows: \(nC_{4}H_{9}Br + HC \equiv CNa \rightarrow nC_{4}H_{9}C \equiv CH + NaBr\). The main organic product is 1-butyn-1-yl bromide. The name of this organic product is 1-butyn-1-yl bromide.
11

(k) Reaction of n-butyl bromide with water (H2O)

H2O can act as a weak nucleophile; however, this reaction would be very slow, and the expected yield would be extremely low. No significant product formation would occur in this case.
12

(l) Reaction of n-butyl bromide with aqueous NH3

Aqueous ammonia (NH3) can act as a weak nucleophile and base. In this case, it can react with 1-bromobutane via an SN2 mechanism, replacing the bromide ion with the amine group. The reaction is as follows: \(nC_{4}H_{9}Br + NH_{3} \rightarrow nC_{4}H_{9}NH_{2} + HBr\). The structure of the organic product will be that of n-butylamine. The name of this organic product is n-butylamine.

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Most popular questions from this chapter

Ethyl chloride \((0.1 \mathrm{M})\) reacts with potassium iodide \((0.1 \mathrm{M})\) in acetone solution at \(60^{\circ}\) to give ethyl iodide and potassium chloride at a rate of \(5.44 \times 10^{-7}\) mole/liter/sec (a) If the reaction proceeded by an \(\mathrm{S}_{\mathrm{N}} 2\) mechanism, what would the rate of the reaction be at \(0.01 \mathrm{M}\) concentrations of both reactants? Show your method of calculation. (b) Suppose the rate were proportional to the square of the potassium iodide concentration and the first power of the ethyl chloride \(\left(\mathrm{S}_{\mathrm{N}} 3\right)\). What would the rate be with \(0.01 \mathrm{M}\) reactants? (c) If one starts with solutions initially \(0.1 \mathrm{M}\) in both reactants, the rate of formation of ethyl iodide is initially \(5.44 \times 10^{-7}\) mole/liter/sec but falls as the reaction proceeds and the reactants are used up. Make plots of the rate of formation of ethyl iodide against the concentration of ethyl chloride as the reaction proceeds (remembering that one molecule of ethyl chloride consumes one molecule of potassium iodide) on the assumption that the rate of reaction is proportional to the first power of the ethyl chloride concentration; and to (1) the zeroth power, (2) the first power, and (3) the second power of the potassium iodide concentration. (d) What kind of experimental data would one need to tell whether the rate of the reaction of ethyl chloride with potassium iodide is first order in each reactant or first order in ethyl chloride and zero order in potassium iodide?

HCN has \(\mathrm{pK}_{\mathrm{a}}=9.21 ;\) acetic acid has \(\mathrm{pK}_{\mathrm{a}}=4.76 .\) (a) What is the difference in the standard free energies \(\left(\Delta \Delta \mathrm{G}^{\circ}\right)\) for these two acid-base equilibria? (b) What is the equilibrium constant and \(\Delta \mathrm{G}^{\circ}\) for the reaction \(\mathrm{HCN}+\mathrm{CH}_{3} \mathrm{CO}_{2}^{-} \rightarrow \mathrm{CN}^{-}+\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\)

Explain why \(\left(\mathrm{CH}_{3}\right)_{4} \mathrm{~N}^{+}\) is neither a nucleophile nor an electrophile.

(a) In the liquid form, tert-buty1 fluoride and isopropyl fluoride gave the following nmr spectra; tert-butyl fluoride: doublet, \(\delta 1.30, \mathrm{~J}=20 \mathrm{~Hz}\) isopropyl fluoride: two doublets, \(\delta 1,23,6 \mathrm{H}\), \(\mathrm{J}=23 \mathrm{~Hz}\) and \(4 \mathrm{~Hz}\) two multiplets, \(\delta 4.64,1 \mathrm{H}\) \(\mathrm{J}=48 \mathrm{~Hz}\) and \(4 \mathrm{~Hz}\) How do you account for each of these spectra? (b) When the alkyl fluorides were dissolved in liquid \(\mathrm{SBF}_{5}\), the following nmr spectra were obtained* tert-butyl fluoride: singlet, \(\delta 4.35\) isopropyl fluoride: doublet, \(\delta 5.06,6 \mathrm{H}, \mathrm{J}=4 \mathrm{~Hz}\) multiplet, \(\delta 13.5,1 \mathrm{H}, \mathrm{J}=4 \mathrm{~Hz}\) To what molecule is each of these spectra due? (Hint: What does the disappearance of just half the peaks observed in part (a) suggest?) Is the very large downfield shift what you might have expected for molecules like these?

(a) What product would be formed if the reaction of cis- 4-bromocyclohexanol with \(\mathrm{OH}^{-}\) proceeded with inversion? (b) Without inversion? (c) Is it always necessary to use optically active compounds to study the stereochemistry of substitution reactions?
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