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What will be the product of the reaction...

What will be the product of the reaction ?
`H_(3)C-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-Br+Na-O-CH_(3) rarr`

A

`CH_(3)-underset(CH_(3))underset(|)(C)=CH_(2)`

B

`CH_(3)-O-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-CH_(3)`

C

`CH_(3)-CH_(2)-CH_(2)-CH_(3)`

D

`CH_(3)-underset(CH_(3))underset(|)(CH)-CH_(3)`

Text Solution

Verified by Experts

The correct Answer is:
A

When teritary alkyl halide is treated with sodium alkoxide than elimination reaction competes over substitution reaction because alkoxides are not only nucleophiles but strong base as well. Therefore, alkenes are formed instead of ethers.
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Knowledge Check

  • CH_(3)-Br+NaO-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-CH_(3)to ?

    A
    `CH_(3)-O-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-CH_(3)`
    B
    `CH_(3)-underset(CH_(3))underset(|)overset(CH_(2))overset(||)(C)`
    C
    `CH_(2)=CH_(2)`
    D
    `CH_(3)-O-underset(CH_(3))underset(|)(CH)-CH_(3)`
  • The products formed during the following reaction are : CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-O-CH_(3)+HIrarr?

    A
    `CH_(3)OH+CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-I`
    B
    `CH_(4)+CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-OI`
    C
    `CH_(3)I+CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-OH`
    D
    `CH_(3)OI+CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-H`
  • The major product formed in the reaction CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C )-CH_(2)underset(..)(ddotC)l:overset(Ag^(+))underset(H_(2)O)(rarr) is

    A
    B
    `CH_(3)-overset(CH_(3))overset(|)underset(OH)underset(|)(C )-CH_(2)CH_(3)`
    C
    `CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C )-CH_(2)OH`
    D
    `HO-CH_(2)overset(CH_(3))overset(|)(C )HCH_(2)CH_(3)`
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