Which of the following is not a good method to make $2, 2 - d i m e t h y l p r o p a n o i c$ acid?

(C H_{3})_{3} C C H_{2} O H K_{2} C r_{2} O_{7}, H_{2} S O_{4} ⟶ H_{2} O, H e a t

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(C H_{3})_{3} C B r M g ⟶ D i e t h y l e t h e r (i) C O_{2} ⟶ (i i) H_{2} O, H^{+}

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(C H_{3})_{3} C B r N a C N ⟶ H_{3} + O ⟶ H e a t

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(C H_{3})_{3} C C = C H_{3} O H_{2} ⟶

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Solution

The correct option is C $(C H_{3})_{3} C B r N a C N ⟶ H_{3} + O ⟶ H e a t$
Due to $+ I$ effect of 3 methyl groups, $3^{o}$ halide is very less reactive in $S_{N} 2$ reaction and this has very less tendency to form acid
$(C H_{3})_{3} C B r N a C N ⟶ H_{3} + O ⟶ H e a t$

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Similar questions

(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} O

α, β

C H_{3} C O O H + N a O H ⟶ C H_{3} C O O N a + H_{2} O . . . . Δ H = - 13.2 K c a l

H^{\oplus} + ⊖ O H ⟶ H_{2} O; . . . . Δ H = - 13.7 K c a l

N H_{4} O H

H C l + N H_{4} O H ⟶ N H_{4} C l + H_{2} O; Δ H = - 12.27 K c a l

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