Напишите уравнения реакций, с помощью которых можно осуществить цепочки превращений. При необходимости укажите условия протекания взаимодействия.

**Ответ:** а) $Al_{4}C_{3} + 12H_{2}O \rightarrow 3CH_{4} + 4Al(OH)_{3} \downarrow$ $CH_{4} + Cl_{2} \xrightarrow{h\nu} CH_{3}Cl + HCl$ $2CH_{3}Cl + 2Na \rightarrow C_{2}H_{6} + 2NaCl$ $C_{2}H_{6} \xrightarrow{t, kat} C_{2}H_{4} + H_{2}$ $C_{2}H_{4} + H_{2}O \xrightarrow{H^{+}} C_{2}H_{5}OH$ $2C_{2}H_{5}OH + 2Na \rightarrow 2C_{2}H_{5}ONa + H_{2} \uparrow$ б) $CO + 2H_{2} \xrightarrow{p, t, kat} CH_{3}OH$ $CH_{3}OH + HBr \rightarrow CH_{3}Br + H_{2}O$ $2CH_{3}Br + 2Na \rightarrow C_{2}H_{6} + 2NaBr$ $C_{2}H_{6} + Cl_{2} \xrightarrow{h\nu} C_{2}H_{5}Cl + HCl$ $C_{2}H_{5}Cl + KOH_{(водн.)} \rightarrow C_{2}H_{5}OH + KCl$ в) $C_{2}H_{6} + Br_{2} \xrightarrow{h\nu} C_{2}H_{5}Br + HBr$ $C_{2}H_{5}Br + KOH_{(водн.)} \rightarrow C_{2}H_{5}OH + KBr$ $C_{2}H_{5}OH + CuO \xrightarrow{t} CH_{3}CHO + Cu + H_{2}O$ г) $CaC_{2} + 2H_{2}O \rightarrow C_{2}H_{2} + Ca(OH)_{2}$ $C_{2}H_{2} + H_{2}O \xrightarrow{Hg^{2+}, H^{+}} CH_{3}CHO$ $CH_{3}CHO + H_{2} \xrightarrow{Ni, t} C_{2}H_{5}OH$ $2C_{2}H_{5}OH \xrightarrow{H_{2}SO_{4}, t < 140^{\circ}C} C_{2}H_{5}-O-C_{2}H_{5} + H_{2}O$ д) $CH_{3}CH_{2}CH_{2}OH + HBr \rightarrow CH_{3}CH_{2}CH_{2}Br + H_{2}O$ $2CH_{3}CH_{2}CH_{2}Br + 2Na \rightarrow CH_{3}(CH_{2})_{4}CH_{3} + 2NaBr$ $CH_{3}(CH_{2})_{4}CH_{3} \xrightarrow{Pt, t} C_{6}H_{6} + 4H_{2}$ $C_{6}H_{6} + CH_{2}=CH-CH_{3} \xrightarrow{AlCl_{3}, H^{+}} C_{6}H_{5}-CH(CH_{3})_{2}$ е) $CH_{3}CH_{2}CH_{2}OH \xrightarrow{H_{2}SO_{4}, t} CH_{3}CH=CH_{2} + H_{2}O$ (X) $CH_{3}CH=CH_{2} + HBr \rightarrow CH_{3}CHBrCH_{3}$ (Y) $2CH_{3}CHBrCH_{3} + 2Na \rightarrow CH_{3}-CH(CH_{3})-CH(CH_{3})-CH_{3} + 2NaBr$ ж) $CH_{2}=CH-CH_{3} + HBr \rightarrow CH_{3}-CHBr-CH_{3}$ (X) $CH_{3}-CHBr-CH_{3} + KOH_{(водн.)} \rightarrow CH_{3}-CH(OH)-CH_{3} + KBr$ (Y) $CH_{3}-CH(OH)-CH_{3} \xrightarrow{H_{2}SO_{4}, t > 140^{\circ}C} CH_{2}=CH-CH_{3} + H_{2}O$