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Chapter 12: Problem 43

\- What products would you expect from the reaction of 1 -bromopropane with each of the following? (a) \(\mathrm{NaNH}_{2}\) (b) \(\mathrm{KOC}\left(\mathrm{CH}_{3}\right)_{3}\) (c) NaI (d) \(\mathrm{NaCN}\) (e) \(\mathrm{Mg}\), then \(\mathrm{H}_{2} \mathrm{O}\)

Short Answer

Expert verified
(a) Propene; (b) Propene; (c) 1-Iodopropane; (d) Butanenitrile; (e) Propane.

Step by step solution

01

Reaction with NaNH2

1-bromopropane reacts with sodium amide (\(\mathrm{NaNH}_2\)) to form propene and ammonium bromide. The strong base \(\mathrm{NH}_2^-\) abstracts a proton from the 尾-carbon, resulting in the elimination reaction that forms an alkene. Thus, the expected product is propene \(\mathrm{C_3H_6}\).
02

Reaction with KOC(CH3)3

1-bromopropane reacts with potassium tert-butoxide \(\mathrm{KOC}\left(\mathrm{CH}_3\right)_3\) via an E2 mechanism due to steric hindrance, leading to the elimination of \(\mathrm{HBr}\) and formation of an alkene. The major product is again propene \(\mathrm{C_3H_6}\).
03

Reaction with NaI

Sodium iodide \(\mathrm{NaI}\) in acetone facilitates a nucleophilic substitution (SN2) reaction because iodide is a good nucleophile. The expected product is 1-iodopropane \(\mathrm{C_3H_7I}\) owing to the displacement of bromide with iodide.
04

Reaction with NaCN

1-bromopropane reacts with sodium cyanide (\(\mathrm{NaCN}\)) via an SN2 mechanism, leading to the substitution of the bromine atom with a cyano group. The product is butanenitrile \(\mathrm{C_3H_7CN}\).
05

Reaction with Mg and H2O

1-bromopropane treated with magnesium forms a Grignard reagent, \(\mathrm{C_3H_7MgBr}\). The subsequent reaction with water causes the alkyl group to replace the magnesium halide with a hydrogen, forming propane \(\mathrm{C_3H_8}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Alkene Formation
When discussing organic chemistry, alkene formation is a crucial concept. It primarily involves the elimination reaction, where atoms or groups are removed from a molecule, resulting in the formation of a double bond. In the case of 1-bromopropane reacting with sodium amide (\(\mathrm{NaNH}_2\)) and potassium tert-butoxide (\(\mathrm{KOC}\left(\mathrm{CH}_3\right)_3\)), we observe typical elimination reactions known as the E2 mechanism.
  • With \(\mathrm{NaNH}_2\), the reaction is driven by the strong base \(\mathrm{NH}_2^-\) that abstracts a proton from the 尾-carbon, resulting in the removal of hydrogen bromide (HBr).
  • The potassium tert-butoxide is a bulky base, which favors an E2 elimination leading to the same product.
  • Both reactions result in the formation of an alkene, specifically propene (\(\mathrm{C_3H_6}\)).
Alkene formation is a foundational mechanism, explaining how organic compounds can transition to contain double bonds, which significantly alters their chemical behavior.
Substitution Reactions
Substitution reactions are another core element of organic chemistry, where one atom or group in a molecule is replaced by another. These reactions can occur via different mechanisms, but the most common ones are the \(\text{S}_\text{N}2\) and \(\text{S}_\text{N}1\) reactions.
  • In the reaction of 1-bromopropane with sodium iodide (\(\mathrm{NaI}\)), iodide acts as a nucleophile replacing bromide in an \(\text{S}_\text{N}2\) fashion, resulting in 1-iodopropane (\(\mathrm{C_3H_7I}\)).
  • Similarly, when 1-bromopropane reacts with sodium cyanide (\(\mathrm{NaCN}\)), the cyanide ion (\(\mathrm{CN}^-\)) substitutes bromide in another \(\text{S}_\text{N}2\) mechanism, producing butanenitrile (\(\mathrm{C_3H_7CN}\)).
The choice of nucleophile and the reaction conditions play a critical role in determining the type of substitution reaction and product formed, highlighting the versatility and significance of substitution reactions in synthetic chemistry.
Grignard Reagents
Grignard reagents are a pivotal tool in organic synthesis, formed by the reaction of an alkyl halide with magnesium in an ether solution. These organometallic compounds open up a vast array of synthetic possibilities. In the case of 1-bromopropane, when combined with magnesium, a Grignard reagent forms: \(\mathrm{C_3H_7MgBr}\).
  • This reagent is extremely versatile, particularly in forming carbon-carbon bonds, making it invaluable in complex molecule synthesis.
  • Upon subsequent reaction with water, the Grignard reagent is quenched, replacing the magnesium component with hydrogen, effectively converting the original halide into an alkane, such as propane \(\mathrm{C_3H_8}\).
Understanding Grignard reagents not only enriches comprehension of organic reactions but also equips chemists with fundamental techniques for creating new carbon frameworks, essential in pharmaceutical and materials chemistry.

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Most popular questions from this chapter

Organic solvents such as ether and chloroform are neither protic nor strongly polar. What effect would you expect these solvents to have on the reactivity of a nucleophile in \(\mathrm{S}_{\mathrm{N}} 2\) reactions?

Draw three resonance forms for the cyclohexadienyl radical.

The \(\mathrm{S}_{\mathrm{N}} 2\) reaction can occur intramolecularly (within the same molecule). What product would you expect from treatment of 4 -bromobutan-1-ol with base?

\- Which compound in each of the following pairs will react faster in an \(\mathrm{S}_{\mathrm{N}} 2\) reaction with \(\mathrm{OH}^{-}\) ? (a) \(\mathrm{CH}_{3} \mathrm{Br}\) or \(\mathrm{CH}_{3} \mathrm{I}\) (b) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{I}\) in ethanol or in dimethyl sulfoxide (c) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCl}\) or \(\mathrm{CH}_{3} \mathrm{Cl}\) (d) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CHBr}\) or \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCH}_{2} \mathrm{Br}\)

3 -Bromobut-1-ene and 1-bromobut-2-ene undergo \(\mathrm{S}_{\mathrm{N}} 1\) reaction at nearly the same rate even though one is a secondary halide and the other is primary. Explain.
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