
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
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- Experiment 27 hates & Mechanisms of Reations Method I visual Clock Reaction A. Concentration effects on reaction Rates Iodine Run [I] mol/L [S₂082] | Time mo/L (SCC) 0.04 54.7 Log 1/ Time Temp Log [ ] 13,20] (time) / [I] 199 20.06 23.0 30.04 0.04 0.04 80.0 22.8 45 40.02 0.04 79.0 21.6 50.08 0.03 51.0 22.4 60-080-02 95.0 23.4 7 0.08 0-01 1970 23.4 8 0.08 0.04 16.1 22.6arrow_forward(15 pts) Consider the molecule B2H6. Generate a molecular orbital diagram but this time using a different approach that draws on your knowledge and ability to put concepts together. First use VSEPR or some other method to make sure you know the ground state structure of the molecule. Next, generate an MO diagram for BH2. Sketch the highest occupied and lowest unoccupied MOs of the BH2 fragment. These are called frontier orbitals. Now use these frontier orbitals as your basis set for producing LGO's for B2H6. Since the BH2 frontier orbitals become the LGOS, you will have to think about what is in the middle of the molecule and treat its basis as well. Do you arrive at the same qualitative MO diagram as is discussed in the book? Sketch the new highest occupied and lowest unoccupied MOs for the molecule (B2H6).arrow_forwardQ8: Propose an efficient synthesis of cyclopentene from cyclopentane.arrow_forward
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- Organic ChemistryChemistryISBN:9781305580350Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. FootePublisher:Cengage LearningEBK A SMALL SCALE APPROACH TO ORGANIC LChemistryISBN:9781305446021Author:LampmanPublisher:CENGAGE LEARNING - CONSIGNMENT


