$R C H_{2} {C H}_{2} O H$ can be converted to $R C H_{2} {C H}_{2} C O O H$ by the following sequence of steps:

P {B r}_{3}, K C N, H_{3} O^{+}

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P {B r}_{3}, K C N, H_{2} / P

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K C N, H_{3} O^{+}

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H C N, P {B r}_{3}, H_{3} O^{+}

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Solution

The correct option is A $P {B r}_{3}, K C N, H_{3} O^{+}$
Option A represents the correct sequence of reagent for the reaction.

Option A is correct.

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Similar questions

R - C H_{2} - C H_{2} - O H

R - C H_{2} C H_{2} C O O H

(C H_{3})_{2} C H C O O H (i) L i A l H_{4} ⟶ (i i) H_{2} O P B r_{3} ⟶ K C N ⟶ D M S O H_{2} O, H^{+} ⟶ H e a t

H_{2} {P O}_{4}^{-}

H_{3} {P O}_{4} + H_{2} O \to H_{3} O^{+} + H_{2} {P O}_{4}^{-}

H_{2} {P O}_{4}^{-} + H_{2} O \to H {P O}_{4}^{2 -} + H_{3} O^{+}

H_{2} {P O}_{4}^{-} + {O H}^{-} \to H_{3} {P O}_{4} + O^{2 -}

H_{2} O + H_{3} {P O}_{4} ⇌ H_{3} O^{+} + H_{2} {P O}_{4}^{-}; p K_{1} = 2.15

H_{2} O + H_{2} {P O}_{4}^{-} ⇌ H_{3} O^{+} + H_{2} {P O}_{4}^{2 -}; p K_{2} = 7.20

0.01 M

N a H_{2} {P O}_{4}

ClC H_{2} OC H_{2} C H_{3}

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

S_{N} 2

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