Chapter 18.SE, Problem 23MP

Interpretation Introduction

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 alkyl halide. Ethers with a tertiary, benzylic or an allylic group cleave the ethers 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 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.

Conclusion

The products formed when 3-methoxypentane is treated with HBr are 3-pentanol and methyl bromide.

The mechanism of their formation is given below.

Interpretation Introduction

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.

Conclusion

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.

Interpretation Introduction

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.

Conclusion

The products formed when ethyl phenyl ether is treated with HBr are phenol and ethyl iodide.

The mechanism of their formation is given below.

Interpretation Introduction

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.

Conclusion

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 epoxide takes place from the less hindered side.