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Question Number 92016 by peter frank last updated on 04/May/20

Answered by Rio Michael last updated on 04/May/20

$$\left(\mathrm{ii}\right){\mathrm{C}}_4{\mathrm{H}}_9\mathrm{Br}+\mathrm{NaOH}\to {\mathrm{C}}_4{\mathrm{H}}_9\mathrm{OH}+\mathrm{NaBr}$$$$\left(\mathrm{iii}\right){CH}_3{CH}_2{CH}_{2}Cl+NaOH\to {CH}_3{CH}_2{CH}_{2}OH+NaCl$$$$\left(\mathrm{iv}\right){CH}_3{CH}_2{CH}_{2}C\equiv N+NaOH+H_{2}O\to {CH}_3{CH}_2{CH}_{2}COONa+{NH}_3$$

Commented by peter frank last updated on 10/May/20

$$thankyou$$

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