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Question

Find the product of:


CH3CH2OCH2CH2OCH2C6H5+HI(excess)?

A
OHCH2CH2OH,C6H5CH2I,CH3CH2I
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B
C6H5CH2OH,CH3CH2I,ICH2CH2OH
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C
ICH2CH2I,C6H5CH2I,CH3CH2OH
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D
HOCH2CH2OH,C6H5CH2I,CH3CH2OH
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Solution

The correct option is B OHCH2CH2OH,C6H5CH2I,CH3CH2I
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 CO bond in protonated ether.

Since C6H5CH+2 is more stable due to resonance. (SN1 mechanism)

Thus the product for given equation are C6H5CH2I,CH3CH2I,HOCH2CH2OH

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