Câu 1: 

Dẫn xuất halogen: `CH_3Cl, CH_3CH_2Cl`

Alcohol: `C_6H_5CH_2OH, CH_3CH_2CH-2OH`

Aldehyde: `C_2H_5CHO, CH_3CHO`

Ketone: `CH_3COCH_3, C_2H_5COCH_3`

Câu 2:

Dẫn xuất halogen của hydrocarbon: `C Cl_4, CH_3CH_2CH(Br)CH_3, CH_2BrCl, CH_2=CHCl`

Câu 3:

$a) CH_3CH_2CH_2Br + NaOH \xrightarrow{t^o} CH_3CH_2CH_2OH + NaBr$
$b) CH_3CH(Cl)CH_3 + NaOH \xrightarrow{t^o} CH_3CH(OH)CH_3 + NaCl$

Câu 4: 

$a) CH_3CH(Cl)CH_2CH_3 + NaOH \xrightarrow{C_2H_5OH,t^o} \left[\begin{matrix} CH_2=CHCH_2CH_3 (spp) + NaCl + H_2O\\ CH_3CH=CHCH_3 (spc) + NaCl + H_2O \end{matrix}\right.$

$b) CH_3CH_2CH_2Br + KOH \xrightarrow{C_2H_5OH,t^o} CH_3CH=CH_2 + KBr + H_2O$

Câu 5:

`m_{\text{tinh bột}} = 1000.38\% = 380(kg)`

Ta có sơ đồ:
`(C_6H_{10}O_5)_n -> 2nC_2H_5OH`

`162n (g)-> 92n(g)`

`380(kg) ->x (kg)`

`=> x = (92n.380)/(162n) = (17480)/(81) (kg)`

`=> m_{C_2H_5OH} = (17480)/(81) .81\% = 174,8(kg)`

Câu 6:

`0,8g//mL = 0,8 kg//L`

`m_{C_2H_5OH} = 10 . 0,8 . (46)/(100) = 3,68 (kg)`

Ta có sơ đồ:

`(C_6H_{10}O_5)_n -> 2nC_2H_5OH`

`162n (g) -> 92n(g)`

`x (kg) -> 3,68(kg)`

`=> x = (3,68.162n)/(92n) = 6,48(kg)`

`=> m = (6,48)/(80\% . 75\%) = 10,8(kg)`