a)
Interpretation:
The products formed and the mechanism by which they are formed when 3-methoxypentane is treated with HBr is to be given.
Concept introduction:
Ethers are cleaved by strong acids. The cleavage takes place either by SN1 or SN2 mechanisms, depending upon the structure of the substrate. Ethers with only primary and secondary alkyl groups react by SN2 mechanism. The Br- or I- attacks the protonated ether at the less hindered side to yield a single alcohol and a single
To give:
The products formed and the mechanism by which it/they is/are formed when 3-methoxypentane is treated with HBr.
Answer to Problem 23MP
The products formed when 3-methoxypentane is treated with HBr are 3-pentanol and methyl bromide.
The mechanism of their formation is given below.
Explanation of Solution
The ether given has a secondary carbon and a primary carbon attached to the oxygen. The acid cleavage of the ether can take place through SN2 mechanism. The Br- attacks the protonated ether at the less hindered side to yield 3-pentanol and methyl bromide.
The products formed when 3-methoxypentane is treated with HBr are 3-pentanol and methyl bromide.
The mechanism of their formation is given below.
b)
Interpretation:
The products formed and the mechanism by which they are formed when ethyl isopropyl ether is treated with HI are to be given.
Concept introduction:
Ethers are cleaved by strong acids. The cleavage takes place either by SN1 or SN2 mechanisms, depending upon the structure of the substrate. Ethers with only primary and secondary alkyl groups react by SN2 mechanism. The Br- or I- attacks the protonated ether at the less hindered side to yield a single alcohol and a single alkyl halide. Ethers with a tertiary, benzylic or an allylic group cleave either by SN1 or E1 mechanism because these can produce a stable carbocations yielding alkenes and alcohols.
To give:
The products formed and the mechanism by which they are formed when ethyl isopropyl ether is treated with HI.
Answer to Problem 23MP
The products formed when ethyl isopropyl ether is treated with HBr are 2-propanol and methyl iodide.
The mechanism of their formation is given below.
Explanation of Solution
The ether given has a secondary carbon and a primary carbon attached to the oxygen. The acid cleavage of the ether can take place through SN2 mechanism. The I- ion attacks the protonated ether at the less hindered side to yield 2-propanol and methyl iodide.
The products formed when ethyl isopropyl ether is treated with HBr are 2-propanol and methyl iodide.
The mechanism of their formation is given below.
c)
Interpretation:
The products formed and the mechanism by which they are formed when ethyl phenyl ether is treated with HBr is to be given.
Concept introduction:
Ethers are cleaved by strong acids. The cleavage takes place either by SN1 or SN2 mechanisms, depending upon the structure of the substrate. Ethers with only primary and secondary alkyl groups react by SN2 mechanism. The Br- or I- attacks the protonated ether at the less hindered side to yield a single alcohol and a single alkyl halide. Ethers with a tertiary, benzylic or an allylic group cleave either by SN1 or E1 mechanism because these can produce a stable carbocations yielding alkenes and alcohols.
To give:
The products formed and the mechanism by which they are formed when ethyl phenyl ether is treated with HBr.
Answer to Problem 23MP
The products formed when ethyl phenyl ether is treated with HBr are phenol and ethyl iodide.
The mechanism of their formation is given below.
Explanation of Solution
The ether given has a benzene ring and a methyl group attached to the oxygen. The acid cleavage of the ether can take place through SN2 mechanism. The Br- ion attacks the protonated ether at the less hindered side to yield phenol and ethyl bromide.
The products formed when ethyl phenyl ether is treated with HBr are phenol and ethyl iodide.
The mechanism of their formation is given below.
d)
Interpretation:
The products formed and the mechanism by which they are formed when cyclopentyl propyl ether is treated with HI are to be given.
Concept introduction:
Ethers are cleaved by strong acids. The cleavage takes place either by SN1 or SN2 mechanisms, depending upon the structure of the substrate. Ethers with only primary and secondary alkyl groups react by SN2mechanism. The Br- or I- attacks the protonated ether at the less hindered side to yield a single alcohol and a single alkyl halide. Ethers with a tertiary, benzylic or an allylic group cleave either by SN1 or E1 mechanism because these can produce a stable carbocations yielding alkenes and alcohols.
To give:
The products formed and the mechanism by which they are formed when cyclopentyl propyl ether is treated with HI.
Answer to Problem 23MP
The products formed when cyclopentyl propyl ether is treated with HI are cyclopentanol and 1-iodopropane.
The mechanism of their formation is given below.
Explanation of Solution
The ether given has a secondary carbon and a primary carbon attached to the oxygen. The acid cleavage of the ether can take place through SN2 mechanism. The I- ion attacks the protonated ether at the less hindered side to yield cyclopentanol and 1-iodopropane.
The products formed when cyclopentyl propyl ether is treated with HI are cyclopentanol and 1-bromopropane.
The mechanism of their formation is given below.
All the reactions, (a), (b), (c) and (d) take place following SN2 mechanism and the attack of the halide ion on the protonated
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Chapter 18 Solutions
ORGANIC CHEMISTRY W/OWL
- Determine if the following salt is neutral, acidic or basic. If acidic or basic, write the appropriate equilibrium equation for the acid or base that exists when the salt is dissolved in aqueous solution. If neutral, simply write only NR. Be sure to include the proper phases for all species within the reaction LiNO3arrow_forwardAn unknown weak acid with a concentration of 0.410 M has a pH of 5.600. What is the Ka of the weak acid?arrow_forward(racemic) 19.84 Using your reaction roadmaps as a guide, show how to convert 2-oxepanone and ethanol into 1-cyclopentenecarbaldehyde. You must use 2-oxepanone as the source of all carbon atoms in the target molecule. Show all reagents and all molecules synthesized along the way. & + EtOH H 2-Oxepanone 1-Cyclopentenecarbaldehydearrow_forward
- R₂ R₁ R₁ a R Rg Nu R₂ Rg R₁ R R₁₂ R3 R R Nu enolate forming R₁ R B-Alkylated carbonyl species or amines Cyclic B-Ketoester R₁₁ HOB R R₁B R R₁₂ B-Hydroxy carbonyl R diester R2 R3 R₁ RB OR R₂ 0 aB-Unsaturated carbonyl NaOR Aldol HOR reaction 1) LDA 2) R-X 3) H₂O/H₂O ketone, aldehyde 1) 2°-amine 2) acid chloride 3) H₂O'/H₂O 0 O R₁ R₁ R R₁ R₁₂ Alkylated a-carbon R₁ H.C R₁ H.C Alkylated methyl ketone acetoacetic ester B-Ketoester ester R₁ HO R₂ R B-Dicarbonyl HO Alkylated carboxylic acid malonic ester Write the reagents required to bring about each reaction next to the arrows shown. Next, record any regiochemistry or stereochemistry considerations relevant to the reaction. You should also record any key aspects of the mechanism, such as forma- tion of an important intermediate, as a helpful reminder. You may want to keep track of all reactions that make carbon-carbon bonds, because these help you build large molecules from smaller fragments. This especially applies to the reactions in…arrow_forwardProvide the reasonable steps to achieve the following synthesis.arrow_forwardIdentify which compound is more acidic. Justify your choice.arrow_forward
- Provide the reasonable steps to achieve the following synthesis.arrow_forwardWhen anisole is treated with excess bromine, the reaction gives a product which shows two singlets in 1H NMR. Draw the product.arrow_forward(ii) Draw a reasonable mechanism for the following reaction: CI NaOH heat OH (hint: SNAr Reaction) :arrow_forward
Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning
EBK A SMALL SCALE APPROACH TO ORGANIC LChemistryISBN:9781305446021Author:LampmanPublisher:CENGAGE LEARNING - CONSIGNMENT