Alkohol Fenol 5b2535

No

Procedure

1

Solubility Test tube 1 : 0,5 mL ethanol Test tube 2 : 0,5 mL n-buthyl alcohol Test tube3 : 0,5 mL siclohexanol Test tube 4 : 0,5 mL ethylen-glycol Test tube 5 : 0,5 mL phenol 

Put in test tube 1-5  Added with 2mL aquadest in each test tube  Shaked  Observed Result

Result Before: ethanol is colourless n-buthyl alcohol is colourless siclohexanol is colourless ethylen-glycol is colourless phenol is orange after:

hypothesis

 tube 1 : CH3CH2OH(aq)

Conclution

The solubility priority: Ethylen Glycol> ethanol + H2O(l)  > n-Buthyl alcohol > CH3CH2OH(aq) syclohexanol > fenol  tube 2 : C4H9OH(aq) + ethylen glycol more soluble in water because H2O(l)  C4H9OH(aq)  ethanol is dissolved in water with  tube 3 : C6H11OH(aq) + it has 2 hydroxyls and has short carbonil chain

solubility (+++) and the solution H O(l)  C H OH(aq) 2 6 11 colour is colourless  n-buthyl alcohol is dissolved in  tube 4 : HOC2H4OH(aq) water with solubility (++) and the + H2O(l)  colour of solution is colourless HOC2H4OH(aq)  siclohexanol is dissolved in water  tube 5 : C H OH(aq) + 6 5 with solubility is (+) and the H2O(l)  C6H5OH(aq) solution colour is colourless  ethylen-glycol is colourless and dissolved in water with solubility (++++)  phenol is in soluble solution among other those solution tested and the colour is white turbid and there is white precipitate

N o 2

Procedure Test tube 1 : 0,5 mL n-buthyl alcohol Test tube 2 : 0,5 mL syclohexlnol Test tube 3 : 0,5 mL fenol Tabung 4 : ± 0,5 g naftol  Added with 5 mL NaOH 10%  Shaked  Observed

Result

Result

Hypothesis

Conclution

The priority of alcohol  Test tube 1 : and fenol to react with  NaOH is colourlessnol C4H9OH(aq) + NaOH(aq) ≠ alkaly are:  n-Buthyl alcohol is  Test tube 2 : 1. Fenol colourless C6H11OH(aq) NaOH(aq)  2. Naftol  fenol is orange C6H11ONa(aq) + H2O 3. N-buthyl  syclohexanol is colourless  Test tube 3 : alcohol  naftol is colourless C6H5OH(aq) + NaOH(aq)  4. Siclohexanol After: C6H5ONa(aq) +H2O(aq) The acidic of alcohol is  n-Buthyl alcohol+ NaOH:  Test tube 4 : lower than fenol so not react but form two layer the C10H7OH(aq) + NaOH(aq)  alcohol cannot react with upper layer is turbid and the lower NaOH strong base C10H7ONa(S) + H2O(l) Before:

layer is colourless, time needed is 17.88 s

 Syclohexanol + NaOH:

Not react and form two layer the upper layer is turbid and the lower layer is colourless time needed to react is 18,85 S

 Fenol +NaOH : Greenish brown solution and time needed to react is 3,02 s  Naftol +NaOH: Brown solution wi th time to react is 4,63 s

N o 3

Procedure Test tube 1 : 2 mL ethanol Test tube 2 : 2 mL 1-propanol Test tube 3 : 2 mL 2-propanol

 Put sodium metal litle ammount  Observed  Added 2 drops of indicator PP  Observed

Result

Result

Hypothesis

Conclution

Before: Na(s) = silver metal PP indicator is colourless

 Tube 1 : C2H5OH(aq)+Na(s)C2H5ONa(a q) + H2(g)

All alcohol group thet react with Na(s) produce gas H2 The fastest solution that react with Alkaly metal is primer alcohol

1. Ethanol= colourless 2. 1-propanol= colourless 3. 2-propanol= colourless After:

 Tube 2 : HOCH2CH2CH3(aq)+Na(s) NaOCH2CH2CH3(aq) + H2(g)

1. Ethanol+Na+PP The priority of rectivity Purpke solutio(+)n and the  Tube 3 : is : CH CHOHCH CH (aq)+ Na(s)  3 2 3 solution is hot and there is  Ethanol CH3CHONaCH2CH3(aq) + H2(g) bubles, time needed to react  1-propanol is 14,44 s  2-propanol 2. 1-Propanol+Na+PP Purple solution(++), the solution is hot and the is bubles. Time to react is 20,45 s 3. 2-Propanol+Na+PP Purple solution (++), there is buble and the time needed to react is 47,16 s

N o

Procedure

4 2 mL reagent lucas

1butha nol

2butha nol

sychlo hexan ol

 Shaked  Note the time until the solution turbid or separated into 2 layers  Noted the resut

result

Tetriary buthyl alcohol

Result

Hypothesis

Before : Lecas reagent is colourless

 C4H9OH(aq) + HCl(aq) ≠

   

1-buthanol = colourless 2-buthanol=colourless Sychlohexanol=colourless Tetriary buthyl alcohol= colourless

After : Lucas reagent+1-buthanol= colorless

 CH3CHOHCH2CH3 + HCl  CH3CHOClCH2CH3 + H2O

 C6H11OH + HCl  C6H11OCl + H2O

Time to react (x) Lucas reagent+2-buthanol= colourles  (CH3)3COH + HCl (CH3)3CCl +H2O Time to react (x) Lucas reagent + Sychlohexanol = Colorless, time to react (x) Lucas reagent+ tert-buthyl alcohol= colourless

Coclution

N o 5

Procedure 0,1 g fenol  Added with aquadest 2

Result

Hypothesis

Conclution

Before : Fenol= orange Brome= yellow H2O = colourless

C6H5OH + Br2  C6H2OHBr3

Reaction of fenol and bromine water produce 2,4,6-tribromo fenol that difficult to dissolved in water. And the yellow solution occure is Brom

ml  Added with bromine

After : Fenol+H2O+Bromine water = yellow solution

water  Shaked untill yellow colour disappear Result

6 1-2 drops of fenol

1-2 drops of resoisinol

1-2 drops of 2propanol

 Poured into test tube  Dilute with some water  Added with 1-2 drops of iron (III)

Before: FeCl3 :yellow H2O : colourless 1.Fenol: orange 2.resorsinol : brown 3. 2-propanol: colourless After



chloride  Shaked  Notetd the result

result

 

Fenol+H2O+FeCl3= purple solution Resorsinol +H2O+ FeCl3= colourless solution then become clear yellow 2propanol+H2O+FeCl3=Clea r yellow

FeCl3 will form a complex compound which will produce different colour if react with hydroxyls

 FeCl3 will only react with fenol and resorsinol because they have OH that reactive  2-propanol not react with FeCl3 because it less reactive