Grignard reagent can be prepared by:

C H_{3} - C H_{2} - C l + M g d r y ⟶ e t h e r

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C H_{3} - C H ∣ C l - C H_{2} ∣ O H + M g d r y ⟶ e t h e r

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C H_{3} - C H_{3} ∣ C ∣ C H_{3} - O H + M g d r y ⟶ e t h e r

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C H_{3} - C H ∣ C l - C H_{2} ∣ C l + M g d r y ⟶ e t h e r

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Solution

The correct option is A $C H_{3} - C H_{2} - C l + M g d r y ⟶ e t h e r$
Grignard reagent can be prepared by,
$C H_{3} - C H_{2} - C l + M g d r y ⟶ e t h e r C H_{3} C H_{2} - M g - C l$

In this reaction, Mg atom inserts in C-Cl bond. It is necessary that the reaction should be carried out under dry conditions otherwise, Grignard reagent will get hydrolyzed.

So, option A is correct.

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Similar questions

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

C H_{3} - C H_{2} - C l + K C N Δ - \to

C H_{3} - C H = C H_{2} + H I \to

C H_{3} - C H_{2} - O H + N a B r + H_{2} S O_{4} \to

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

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

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