Find the product for CH3CH2-O-CH2-CH2-O-CH2-C6H5+HI (excess)

Question

Find the product for CH3CH2-O-CH2-CH2-O-CH2-C6H5+HI (excess) (A) HO-CH2CH2OH, C6H5CH2-I, CH3CH2-I (B) C6H5CH2-OH, CH3CH2-I, I-CH2CH2-OH (C) I-CH2CH2-I, C6H5CH2-I, CH3CH2-OH (D) HO-CH2CH2-OH, C6H5CH2-I, CH3CH2-OH

Tardigrade Admin · Accepted Answer

Presence of excess of HI favours SN1 mechanism. So, formation of products is controlled by the stability of the carbocation resulting in the cleavage of C - O bond in protonated ether. Thus the product for given equation are C6H5CH2I, CH3CH2I, HOCH2 - CH2OH

Q. Find the product for
$C H_{3} C H_{2} - O - C H_{2} - C H_{2} - O - C H_{2} - C_{6} H_{5} + H I$
(excess)

$H O - C H_{2} C H_{2} O H, C_{6} H_{5} C H_{2} - I, C H_{3} C H_{2} - I$

$C_{6} H_{5} C H_{2} - O H, C H_{3} C H_{2} - I, I - C H_{2} C H_{2} - O H$

$I - C H_{2} C H_{2} - I, C_{6} H_{5} C H_{2} - I, C H_{3} C H_{2} - O H$

$H O - C H_{2} C H_{2} - O H, C_{6} H_{5} C H_{2} - I, C H_{3} C H_{2} - O H$

Q. Find the product for CH3​CH2​−O−CH2​−CH2​−O−CH2​−C6​H5​+HI (excess)

HO−CH2​CH2​OH,C6​H5​CH2​−I,CH3​CH2​−I

C6​H5​CH2​−OH,CH3​CH2​−I,I−CH2​CH2​−OH

I−CH2​CH2​−I,C6​H5​CH2​−I,CH3​CH2​−OH

HO−CH2​CH2​−OH,C6​H5​CH2​−I,CH3​CH2​−OH

Presence of excess of HI favours SN​1 mechanism. So, formation of products is controlled by the stability of the carbocation resulting in the cleavage of C−O bond in protonated ether. Thus the product for given equation are C6​H5​CH2​I,CH3​CH2​I,HOCH2​−CH2​OH

Q. Find the product for
$C H_{3} C H_{2} - O - C H_{2} - C H_{2} - O - C H_{2} - C_{6} H_{5} + H I$
(excess)

$H O - C H_{2} C H_{2} O H, C_{6} H_{5} C H_{2} - I, C H_{3} C H_{2} - I$

$C_{6} H_{5} C H_{2} - O H, C H_{3} C H_{2} - I, I - C H_{2} C H_{2} - O H$

$I - C H_{2} C H_{2} - I, C_{6} H_{5} C H_{2} - I, C H_{3} C H_{2} - O H$

$H O - C H_{2} C H_{2} - O H, C_{6} H_{5} C H_{2} - I, C H_{3} C H_{2} - O H$