a)
Interpretation:
The products formed and the mechanism by which they are formed when tert-butyl ethyl 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
To give:
The products formed and the mechanism by which they are formed when tert-butyl ethyl ether is treated with HBr is to be given.
Answer to Problem 24MP
The products formed when tert-butyl ethyl ether is treated with HBr are ethanol and tert-butylbromide.
The mechanism by which they are formed is given below.
Explanation of Solution
The reaction occurs following SN1 mechanism. The acid protonates the ether initially and the protonated ether eliminates ethanol to produce a stable tert-butyl carbocation. In the next step the bromide ion attacks the carbocation to yield tert-butyl bromide as the product.
The products formed when tert-butyl ethyl ether is treated with HBr are ethanol and tert-butylbromide.
The mechanism by which they are formed is given below.
b)
Interpretation:
The products formed and the mechanism by which they are formed when tert-butyl 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 tert-butyl phenyl ether is treated with HBr.
Answer to Problem 24MP
The products formed when tert-butyl phenyl ether is treated with HBr are phenol and tert-butyl bromide.
The mechanism by which they are formed is given below.
Explanation of Solution
The reaction occurs following SN1 mechanism. The acid protonates the ether initially and the protonated ether eliminates phenol to produce a stable tert-butyl carbocation. In the next step the bromide ion attacks the carbocation to yield tert-butyl bromide.
The products formed when tert-butyl phenyl ether is treated with HBr are phenol and tert-butyl bromide.
The mechanism by which they are formed is given below.
c)
Interpretation:
The products formed and the mechanism by which they are formed when tert-butyl isopropyl ether is treated with HI 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 tert-butyl isopropyl ether is treated with HI.
Answer to Problem 24MP
The products formed when tert-butyl isopropyl ether is treated with HI are tert-butyl bromide and 2-propanol.
The mechanism by which they are formed is given below.
Explanation of Solution
The reaction occurs following SN1 mechanism. The acid protonates the ether initially and the protonated ether eliminates 2-propanol to produce a stable tert-butyl carbocation. In the next step the iodide ion attacks the carbocation to yield tert-butyl iodide.
The products formed when tert-butyl isopropyl ether is treated with HI are tert-butyl bromide and 2-propanol.
The mechanism by which they are formed is given below.
d)
Interpretation:
The products formed and the mechanism by which they are formed when ethyl 1-methylcyclohexyl ether is treated with HI 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 1-methylcyclohexyl ether is treated with HI.
Answer to Problem 24MP
The products formed when ethyl 1-methylcyclohexyl ether is treated with HI are 1-iodo-1-methylcyclohexane and ethanol.
The mechanism by which they are formed is given below.
Explanation of Solution
The reaction occurs following SN1 mechanism. The acid protonates the ether initially and the protonated ether eliminates ethanol to produce a stable tert-butyl carbocation. In the next step the iodide ion attacks the carbocation to yield tert-butyl iodide.
The products formed when ethyl 1-methylcyclohexyl ether is treated with HI are 1-iodo-1-methylcyclohexane and ethanol.
The mechanism by which they are formed is given below.
All the reactions (a), (b), (c) and (d) occur through SN1 mechanism.
All the reactions, (a), (b), (c) and (d) take place following SN1 mechanism. The protonated
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Chapter 18 Solutions
Organic Chemistry
- Predict the major products of the following organic reaction: + A ? Some important notes: • Draw the major product, or products, of the reaction in the drawing area below. • If there aren't any products, because no reaction will take place, check the box below the drawing area instead. • Be sure to use wedge and dash bonds when necessary, for example to distinguish between major products that are enantiomers. Explanation Check Click and drag to start drawing a structure. C © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Centearrow_forwardPolar solutes are most likely to dissolve into _____, and _____ are most likely to dissolve into nonpolar solvents. A. nonpolar solutes; polar solvents B. nonpolar solvents; polar solvents C. polar solvents; nonpolar solutes D. polar solutes; nonpolar solventsarrow_forwardDeducing the Peactants Can the molecule on the right-hand side of this organic reaction be made in good yield from no more than two reactants, in one step, by moderately heating the reactants? ? Δ If your answer is yes, then draw the reactant or reactants in the drawing area below. You can draw the reactants in any arrangement you like. If your answer is no, check the box under the drawing area instead. Explanation Check Click and drag to start drawing a structure. © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center Xarrow_forward
- Draw all 8 stereoisomers, circling each pair of enantiomer(s)/ mirror image compound(s)arrow_forwardBookmarks Profiles Tab Window Help Chemical Formula - Aktiv Che X + → C 11 a app.aktiv.com Google Chrome isn't your default browser Set as default Question 12 of 16 Q Fri Feb 2 Verify it's you New Chrome availabl- Write the balanced molecular chemical equation for the reaction in aqueous solution for mercury(I) nitrate and chromium(VI) sulfate. If no reaction occurs, simply write only NR. Be sure to include the proper phases for all species within the reaction. 3 Hg(NO3)2(aq) + Cг2(SO4)3(aq) → 3 Hg₂SO (s) + 2 Cr(NO3), (aq) ean Ui mate co ence an climate bility inc ulnerabili women, main critic CLIMATE-INI ernational + 10 O 2 W FEB 1 + 4- 3- 2- 2 2 ( 3 4 NS 28 2 ty 56 + 2+ 3+ 4+ 7 8 9 0 5 (s) (1) Ch O 8 9 (g) (aq) Hg NR CI Cr x H₂O A 80 Q A DII A F2 F3 FA F5 F6 F7 F8 F9 #3 EA $ do 50 % 6 CO & 7 E R T Y U 8 ( 9 0 F10 34 F11 川 F12 Subr + delete 0 { P }arrow_forwardDeducing the reactants of a Diels-Alder reaction n the molecule on the right-hand side of this organic reaction be made in good yield from no more than two reactants, in one step, by moderately heating the reactants? ? Δ • If your answer is yes, then draw the reactant or reactants in the drawing area below. You can draw the reactants in any arrangement you like. • If your answer is no, check the box under the drawing area instead. Explanation Check Click and drag to start drawing a structure. >arrow_forward
- Predict the major products of the following organic reaction: + Some important notes: A ? • Draw the major product, or products, of the reaction in the drawing area below. • If there aren't any products, because no reaction will take place, check the box below the drawing area instead. • Be sure to use wedge and dash bonds when necessary, for example to distinguish between major products that are enantiomers. Explanation Check Click and drag to start drawing a structure.arrow_forwardif the answer is no reaction than state that and please hand draw!arrow_forward"I have written solutions in text form, but I need experts to rewrite them in handwriting from A to Z, exactly as I have written, without any changes."arrow_forward
Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage Learning