xác định số đồng phân của c4h7cl, c4h8o, c5h12o

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1. a) \({C_4}{H_7}Cl\):

   \(C{H_2} = CH – C{H_2} – C{H_2}Cl\)

   \(C{H_2} = CH – CHCl – C{H_3}\)

   \(C{H_2} = CCl – C{H_2} – C{H_3}\)

   \(CHCl = CH – C{H_2} – C{H_3}\)

   \(CHCl – CH = CH – C{H_2}\) (đồng phân hình học)

   \(C{H_2} – CCl = CH – C{H_2}\) (đồng phân hình học)

   \(C{H_2} = C(C{H_3}) – C{H_2}Cl\)

   \(CHCl = C(C{H_3}) – C{H_3}\)

   b)\({C_4}{H_8}O\)

   Dạng ancol:

   \(C{H_2} = CH – C{H_2} – C{H_2}OH\)

   \(C{H_2} = CH – CHOH – C{H_3}\)

   \(C{H_2} = C(C{H_3}) – C{H_2}OH\)

   \(C{H_3} – CH = CH – C{H_2}OH\) (đồng phân hình học)

   Dạng ete:

   \(C{H_2} = CH – O – C{H_2} – C{H_3}\)

   \(C{H_2} = CH – C{H_2} – O – C{H_3}\)

   \(C{H_2} = C(C{H_3}) – O – C{H_3}\)

   \(C{H_3} – CH = CH – O – C{H_3}\) (đồng phân hình học)

   Dạng andehit:

   \(C{H_3} – C{H_2} – C{H_2} – CHO\)

   \(C{H_3} – CH(C{H_3}) – CHO\)

   Dạng xeton:

   \(C{H_3} – CO – C{H_2} – C{H_3}\)

   c) \({C_5}{H_{12}}O\)

   Dạng ancol:

   \(C{H_2}OH – C{H_2} – C{H_2} – C{H_2} – C{H_3}\)

   \(C{H_3} – CHOH – C{H_2} – C{H_2} – C{H_3}\)

   \(C{H_3} – C{H_2} – CHOH – C{H_2} – C{H_3}\)

   \(C{H_2}OH – CH(C{H_3}) – C{H_2} – C{H_3}\)

   \(C{H_3} – COH(C{H_3}) – C{H_2} – C{H_3}\)

   \(C{H_3} – CH(C{H_3}) – CHOH – C{H_3}\)

   \(C{H_3} – CH(C{H_3}) – C{H_2} – C{H_2}OH\)

   \(C{H_3} – C{(C{H_3})_2} – C{H_2}OH\)

   Dạng ete:

   \(C{H_3} – O – C{H_2} – C{H_2} – C{H_2} – C{H_3}\)

   \(C{H_3} – O – C{H_2} – CH(C{H_3}) – C{H_3}\)

   \(C{H_3} – O – CH(C{H_3}) – C{H_2} – C{H_3}\)

   \(C{H_3} – O – C{(C{H_3})_3}\)

   \(C{H_3} – C{H_2} – O – C{H_2} – C{H_2} – C{H_3}\)

   \(C{H_3} – C{H_2} – O – CH(C{H_3}) – C{H_3}\)

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