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
Want to see more full solutions like this?
Chapter 18 Solutions
Student Value Bundle: Organic Chemistry, + OWLv2 with Student Solutions Manual eBook, 4 terms (24 months) Printed Access Card (NEW!!)
- Correct each molecule in the drawing area below so that it has the skeletal ("line") structure it would have if it were dissolved in a 0.1 M aqueous solution of HCI. If there are no changes to be made, check the No changes box under the drawing area. No changes. HO Explanation Check NH, 2 W O :□ G ©2025 M unter Accessibilityarrow_forwardAn expression for the root mean square velocity, vrms, of a gas was derived. Using Maxwell’s velocity distribution, one can also calculate the mean velocity and the most probable velocity (mp) of a collection of molecules. The equations used for these two quantities are vmean=(8RT/πM)1/2 and vmp=(2RT/M)1/2 These values have a fixed relationship to each other.(a) Arrange these three quantities in order of increasing magnitude.(b) Show that the relative magnitudes are independent of the molar mass of the gas.(c) Use the smallest velocity as a reference for establishing the order of magnitude and determine the relationship between the larger and smaller values.arrow_forwardThe reaction of solid dimethylhydrazine, (CH3)2N2H2, and liquefied dinitrogen tetroxide, N2O4, has been investigated for use as rocket fuel. The reaction produces the gases carbon dioxide (CO2), nitrogen (N2), and water vapor (H2O), which are ejected in the exhaust gases. In a controlled experiment, solid dimethylhydrazine was reacted with excess dinitrogen tetroxide, and the gases were collected in a closed balloon until a pressure of 2.50 atm and a temperature of 400.0 K were reached.(a) What are the partial pressures of CO2, N2, and H2O?(b) When the CO2 is removed by chemical reaction, what are the partial pressures of the remaining gases?arrow_forward
- One liter of chlorine gas at 1 atm and 298 K reacts completely with 1.00 L of nitrogen gas and 2.00 L of oxygen gas at the same temperature and pressure. A single gaseous product is formed, which fills a 2.00 L flask at 1.00 atm and 298 K. Use this information to determine the following characteristics of the product:(a) its empirical formula;(b) its molecular formula;(c) the most favorable Lewis formula based on formal charge arguments (the central atom is N);(d) the shape of the molecule.arrow_forwardHow does the square root mean square velocity of gas molecules vary with temperature? Illustrate this relationship by plotting the square root mean square velocity of N2 molecules as a function of temperature from T=100 K to T=300 K.arrow_forwardDraw product B, indicating what type of reaction occurs. F3C CF3 NH2 Me O .N. + B OMearrow_forward
- Benzimidazole E. State its formula. sState the differences in the formula with other benzimidazoles.arrow_forwardDraw product A, indicating what type of reaction occurs. F3C CN CF3 K2CO3, DMSO, H₂O2 Aarrow_forward19) Which metal is most commonly used in galvanization to protect steel structures from oxidation? Lead a. b. Tin C. Nickel d. Zinc 20) The following molecule is an example of a: R₁ R2- -N-R3 a. Secondary amine b. Secondary amide c. Tertiary amine d. Tertiary amidearrow_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