In the following reaction, we are likely to get:

$C H_{3} O N a + B r - t$ - $B u \to$

ether

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alkene

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alcohol

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none of the above

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Solution

The correct option is A alkene

Both the substitution product and the elimination product are formed by competing for SN1 and E1 mechanisms. This is for two reasons:

Sodium methoxide is a strong base and a strong nucleophile
The leaving group (Br−) leaves behind a tertiary carbocation, which is more stable than primary or secondary carbocations.
As a result, there are two major products: tert-butyl methyl ether (SN1) and 2-methylpropene (E1).

Hence, the correct option is $B$

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C H_{3} O N a

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