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Chemical Properties of Ethers

On the treatm...

Question

On the treatment of the following compound with a strong acid, the most susceptible site for bond cleavage is:

O_{2} - C_{3}

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O_{5} - C_{6}

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C_{4} - O_{5}

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C_{1} - O_{2}

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Solution

The correct option is B $O_{5} - C_{6}$
On the treatment of the given compound with a strong acid, the most susceptible site for bond cleavage is $O 5 - C 6$ .
The lone pair of electrons on $O 2$ is involved in resonance with $C = C$ . Hence, $O 2$ will not be protonated.
The lone pair of electrons on $O 5$ is not involved in resonance with $C = C$ . Hence, $O 5$ will be protonated. Chloride ion will then attack least substituted C atom $(C 6)$

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Similar questions

N_{2} O_{5} ⟶ 2 N O_{2} + 1 / 2 O_{2}

- \frac{d [N_{2} O_{5}]}{d t} = k_{1} [N_{2} O_{5}]

\frac{d [N O_{2}]}{d t} = k_{2} [N_{2} O_{5}]

\frac{d [O_{2}]}{d t} = k_{3} [N_{2} O_{5}]

k_{1}, k_{2}

k_{3}

N_{2} O_{5} \to 2 N O_{2} + \frac{1}{2} O_{2} .

G i v e n : - \frac{d [N_{2} O_{5}]}{d t} = K_{1} [N_{2} O_{5}]

\frac{d [N O_{2}]}{d t} = K_{2} [N_{2} O_{5}]

A n d \frac{d [O_{2}]}{d t} = K_{3} [N_{2} O_{5}]

The relation between $K_{1}, K_{2}$ and $K_{3}$ is

N_{2} O_{5} \to 2 N O_{2} + \frac{1}{2} O_{2},

\frac{- d}{d t} [N_{2} O_{5}] = k_{1} [N_{2} O_{5}], \frac{d}{d t} [N O_{2}] = k_{2} [N_{2} O_{5}] a n d

\frac{d}{d t} [O_{2}] = k_{3} [N_{2} O_{3}],

N_{2} O_{5} ⟶ 2 {N O}_{2} + \frac{1}{2} O_{2}

- \frac{d [N_{2} O_{5}]}{d t} = K_{1} [N_{2} O_{5}]

- \frac{d [{N O}_{2}]}{d t} = K_{2} [{N O}_{2}]

- \frac{d [O_{2}]}{d t} = K_{3} [O_{2}]

K_{1}, K_{2}, K_{3}

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