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Đáp án:
Giải thích các bước giải:
1. C5H12 → C3H6 + C2H6 (cracking, t,xt)
C3H6 + Cl2 → CH2Cl-CH2Cl-CH3 ( CCl4)
C3H6 + H2 → C3H8(Ni, t)
C3H8 → C3H6 + H2(xt, t)
nCH2=CH-CH3 → (CH2-CH-CH3)n ( xt,t)
C5H12 → C2H4 + C3H8 (xt,t)
nCH2=CH2 → (CH2-Ch2)n (xt,t)
2. C6H14 → CH3-CH=CH-CH3 + C2H6(xt,t)
CH3-CH=CH-CH3 + Br2 → CH3-CH(Br)-CH(Br)-CH3 (CCl4)
C6H14 → C2H6 + C4H8 (xt,t)
C2H6 → CH2=CH2 + H2 (xt,t)
CH2=CH2 + HCl → CH3-CH2Cl (xt,t)
CH3-CH2Cl + H2O → CH3-CH2-OH + HCl
CH3-CH2-OH → CH2=CH2 + H2O (HgSO4,170)
nCH2=CH2 → (CH2-CH2)n (xt,t)
3. CaCO3 → CaO + CO2 (t)
CaO + 3C → CaC2 + CO (t)
CaC2 + H2O → Ca(OH)2 + CH≡CH
2CH≡CH → CH≡C-CH=CH2 (NH4Cl, CuCl,t)
CH≡C-CH=CH2 + H2 → CH2=CH-CH=CH2 (Pd/PbCO3, t)
nCH2=CH-CH=CH2 → (CH2-CH=CH-CH2)n (trùng hợp, t,xt)
Những cái trong ngoặc là điều kiện phản ứng nha!
Hãy giúp mọi người biết câu trả lời này thế nào?

Các phản ứng xảy ra:
1)
\({C_5}{H_{12}}\xrightarrow{{{t^o},xt}}{C_3}{H_8} + {C_2}{H_4}\)
\(n{C_2}{H_4}\xrightarrow{{{t^o},xt}}{( - C{H_2} - C{H_2} - )_n}\)
\({C_5}{H_{12}}\xrightarrow{{{t^o},xt}}{C_3}{H_6} + {C_2}{H_6}\)
\(C{H_2} = CHC{H_3} + C{l_2}\xrightarrow{{}}C{H_2}ClCHClC{H_3}\)
\({C_3}{H_6} + {H_2}\xrightarrow{{Ni,{t^o}}}{C_3}{H_8}\)
\({C_3}{H_8}\xrightarrow{{{t^o},xt}}{C_3}{H_6} + {H_2}\)
\(nC{H_2} = CHC{H_3}\xrightarrow{{{t^o},xt}}{( - C{H_2} - CH(C{H_3}) - )_n}\)
2)
\({C_6}{H_{12}}\xrightarrow{{{t^o},xt}}{C_2}{H_6} + C{H_3}CH = CHC{H_3}\)
\(C{H_3}CH = CHC{H_3} + B{{\text{r}}_2}\xrightarrow{{}}C{H_3}CHB{\text{r}}CHB{\text{r}}C{H_3}\)
\({C_2}{H_6}\xrightarrow{{{t^o},xt}}{C_2}{H_4} + {H_2}\)
\({C_2}{H_4} + HCl\xrightarrow{{}}{C_2}{H_5}Cl\)
\({C_2}{H_5}Cl + NaOH\xrightarrow{{}}{C_2}{H_5}OH + NaCl\)
\({C_2}{H_5}OH\xrightarrow{{{H_2}S{O_4},{t^o}}}{C_2}{H_4} + {H_2}O\)
\(n{C_2}{H_4}\xrightarrow{{{t^o},xt}}{( - C{H_2} - C{H_2} - )_n}\)
3)
\(CaC{{\text{O}}_3}\xrightarrow{{{t^o}}}CaO + C{O_2}\)
\(CaO + 3C\xrightarrow{{{t^o}}}Ca{C_2} + CO\)
\(Ca{C_2} + 2{H_2}O\xrightarrow{{}}Ca{(OH)_2} + {C_2}{H_2}\)
\(2{C_2}{H_2}\xrightarrow{{{t^o},xt}}CH \equiv C - CH = C{H_2}\)
\(CH \equiv C - CH = C{H_2} + {H_2}\xrightarrow{{P{\text{d}}/PbC{O_3},{t^o}}}C{H_2} = CH - CH = C{H_2}\)
\(nC{H_2} = CH - CH = C{H_2}\xrightarrow{{{t^o},xt}}{( - C{H_2} - CH = CH - C{H_2} - )_n}\)
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