What is the major alkene formed in the following Hofmann elimination?

C H_{3} C H = C H_{2}

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C H_{3} - C | | C H_{2} - C H_{3}

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C H_{3} C H_{2} C H = C H_{2}

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C H_{3} C H | C H_{3} C H = C H_{2}

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C H_{3} C H_{2} C | C H_{3} = C H_{2}

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Solution

The correct option is A $C H_{3} C H = C H_{2}$
Hoffmann elimination provides less substituted alkene
$C H_{3}$ $C H_{3}$
$C H_{3} - C H_{2} - C H_{2} -^{\oplus} - | C | - C H_{3}$
$C H_{3}$ $C H_{3}$
The product will be less substituted alkene $C H_{3} C H = C H_{2}$ .

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Similar questions

An alkane with molecular formula $C_{6} H_{14}$ gives two monochloro alkanes on chlorination. What is the structure of alkane?

S_{N} 2

c o n c . H C l :

(a) C H_{3} - C H_{2} - O H A n h y Z n C l_{2} / △ - -------- \to C H_{3} - C H_{2} - C l

(b) C H_{3} - C H_{3} | C | C H_{3} - C H_{2} O H A n h y Z n C l_{2} / △ - -------- \to C H_{3} - C H_{1} | C | C H_{3} - C H_{2} - C H_{3}

(c) C H_{3} - O H | C | C H_{3} - C H_{3} A n h y Z n C l_{2} / △ - -------- \to C H_{3} - C l | C | C H_{3} - C H_{3}

(d) C H_{2} = C H - C H_{2} - O H A n h y Z n C l_{2} / △ - -------- \to C H_{2} = C H - C H_{2} - C l

i) C H_{3} - C H_{2} - C H_{3} | C H - C H_{2} - O = C H_{2} - C H_{3}

i i) C H_{3} - C H_{2} - C H_{2} - O - C H_{3} | C | C H_{3} - C H_{3}

C H_{3} - C H_{2} - O | | C - C H_{2} - C H_{3}

C H_{3} - C H_{2} - C H_{2} - C H_{2} - C ∥ C H_{2} - C H_{2} - C H_{2} - C H_{3}

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