Neet Solved Paper 2013 Question 34
Question: Among the following ethers, which one will produce methyl alcohol on treatment with hot concentrated HI?
Options:
A) $ CH _3-CH _2-CH _2-CH _2-O-CH _3 $
B) $ CH _3-CH _2-\underset{\begin{smallmatrix} | \\ CH _3 \end{smallmatrix}}{\mathop{CH}},-O-CH _3 $
C) $ C{H_3}\text{-}\underset{\begin{smallmatrix} | \\ C{H_3} \end{smallmatrix}}{\overset{\begin{smallmatrix} C{H_3} \\ | \end{smallmatrix}}{\mathop{C}}},\text{-O-C}{H_3} $
D) $ C{H_3}\text{-}\underset{\begin{smallmatrix} | \\ C{H_3} \end{smallmatrix}}{\mathop{CH}},\text{-C}{H_2}\text{-O-C}{H_3} $
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Answer:
Correct Answer: C
Solution:
The ether, which gives more stable carbonation,
gives $ CH _3OH $ as one of the product with hot concentrated HI.
The order of stability of carbonation is $ 3{}^\circ >2{}^\circ >1{}^\circ $
Thus, $ C{H_3}-\underset{\begin{smallmatrix} | \\ C{H_3} \end{smallmatrix}}{\overset{\begin{smallmatrix} C{H_3} \\ | \end{smallmatrix}}{\mathop{C}}},-OC{H_3} $ gives $ C{H_3}OH $ as one of the reaction.
The reaction proceeds as $ {H_3}C-\underset{\begin{smallmatrix} | \\ C{H_3} \end{smallmatrix}}{\overset{\begin{smallmatrix} C{H_3} \\ | \end{smallmatrix}}{\mathop{C}}},-O-C{H_3}\text{+}{H^{+}} $
BETA
