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-EBOOK>I<
- 10. The most important reason why Br- is a better nucleophile than Cl-is ___. A. polarizability; B. size; C. solvation; D. basicity; E. polarity. Please include all steps. Thanks!arrow_forwardPredicting the qualitative acid-base properties of salts Consider the following data on some weak acids and weak bases: base acid Ка K₁₁ name formula name formula nitrous acid HNO2 4.5×10 4 pyridine CHEN 1.7 × 10 9 4 hydrofluoric acid HF 6.8 × 10 methylamine CH3NH2 | 4.4 × 10¯ Use this data to rank the following solutions in order of increasing pH. In other words, select a '1' next to the solution that will have the lowest pH, a '2' next to the solution that will have the next lowest pH, and so on. solution 0.1 M NaNO2 0.1 M KF pH choose one v choose one v 0.1 M C5H5NHBr 0.1 M CH3NH3CI choose one v ✓ choose one 1 (lowest) 2 ☑ 3 4 (highest) 000 18 Ararrow_forward4. The major product from treatment of 2-propanol with the Jonesreagent is ___.A. acetone; B. none of the other answers is correct C. propene; D.propanoic acid; E carbon dioxide. Please include all steps! Thank you!arrow_forward
- 7. All of the following compounds that are at the same oxidation levelare ___.u. methyl epoxide, v. propyne, w. propanal, x. propene,y. 2,2-dihydroxypropane, z. isopropanol?A. u,v,w,y; B. u,v,w; C. v,w,y,z; D. v, z; E. x,y,z Please include all steps. Thank you!arrow_forward9. Which one of the following substituents is the worst leaving group inan SN2 reaction? A. -NH2; B. -OH; C. –F; D. NH3; E. H2O Please include all steps. Thanks!arrow_forwardUsing the general properties of equilibrium constants At a certain temperature, the equilibrium constant K for the following reaction is 2.5 × 105: CO(g) + H2O(g) CO2(g) + H2(g) Use this information to complete the following table. Suppose a 7.0 L reaction vessel is filled with 1.7 mol of CO and 1.7 mol of H2O. What can you say about the composition of the mixture in the vessel at equilibrium? What is the equilibrium constant for the following reaction? Be sure your answer has the correct number of significant digits. CO2(9)+H2(g) CO(g)+H₂O(g) What is the equilibrium constant for the following reaction? Be sure your answer has the correct number of significant digits. 3 CO(g)+3H2O(g) = 3 CO2(g)+3H2(g) There will be very little CO and H2O. x10 There will be very little CO2 and H2. 000 Neither of the above is true. K = ☐ K = ☐ 18 Ararrow_forward
- 8. When ethane thiol is treated with hydrogen peroxide the product is___.A. ethane disulfide; B. diethyl sulfide; C. ethane sulfoxide; D. ethanesulfate; E. ethyl mercaptan. Please include all steps. Thanks!arrow_forward5. The major product of the three step reaction that takes place when 1-propanol is treated with strong acid is?A. dipropyl ether; B. propene; C. propanal; D. isopropyl propyl ether;E. 1-hexanol Please include all steps. Thank you!arrow_forward6. The formula of the product of the addition of HCN to benzaldehydeis ___.A. C8H7NO; B. C8H6NO; C. C14H11NO; D. C9H9NO; E. C9H8NO Please include all steps. Thank you!arrow_forward
- Predicting the qualitative acid-base properties of salts Consider the following data on some weak acids and weak bases: base acid K K a name formula name formula nitrous acid HNO2 4.5×10 hydroxylamine HONH2 1.1 × 10 8 hypochlorous acid HCIO 8 3.0 × 10 methylamine CH3NH2 | 4.4 × 10¯ 4 Use this data to rank the following solutions in order of increasing pH. In other words, select a '1' next to the solution that will have the lowest pH, a '2' next to the solution that will have the next lowest pH, and so on. 0.1 M KCIO solution PH choose one 0.1 M NaNO2 0.1 M CH3NH3Br 0.1 M NaBr choose one ✓ choose one v ✓ choose one 1 (lowest) ☑ 2 3 4 (highest)arrow_forwardFor this Orgo problem, don't worry about question 3 below it. Please explain your thought process, all your steps, and also include how you would tackle a similar problem. Thank you!arrow_forwardUsing the general properties of equilibrium constants At a certain temperature, the equilibrium constant K for the following reaction is 0.84: H2(g) + 2(g) 2 HI(g) = Use this information to complete the following table. Suppose a 34. L reaction vessel is filled with 0.79 mol of HI. What can you say about the composition of the mixture in the vessel at equilibrium? There will be very little H2 and 12. ☐ x10 There will be very little HI. Neither of the above is true. What is the equilibrium constant for the following reaction? Be sure your answer has the correct number of significant digits. 2 HI(g) H₂(9)+12(9) K = What is the equilibrium constant for the following reaction? Be sure your answer has the correct number of significant digits. 2 H2(g)+212(9) 4 HI(g) K = ☐ ☑arrow_forward
Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning