Hydroxyl group R–OH

The first 3 tests (i) – (iii) given on the right are quite general for most alcohols AND other substances too!

(ii) Mix the dry sample with a little phosphorus(V) chloride and test as above.

(iii) Warm with a little ethanoic acid and a few drops of conc. sulphuric acid. Pour into water and stir gently.

(ii) as for (i) but no ester smell!

(iii) You should get a 'pleasant' characteristic smell of an ester.

An ester and hydrogen chloride are formed

(ii) R–OH + PCl₅ ==> R–Cl + POCl₃ + HCl

a chloro compound and hydrogen chloride are formed.

(i) and (ii) Ag⁺_(aq) + Cl^–_(aq) ==> AgCl_(s) from the hydrogen chloride fumes dissolved in water.

(iii) CH₃COOH + ROH ==> CH₃COOR + H₂O

Note: Alcohols are distinguished from carboxylic acids (pH < 7) by being neutral pH 7 - alcohols do not change blue litmus red, or create effervescence with a pinch of sodium hydrogencarbonate.

RCH₂OH, R = H, alkyl or aryl (NOT a phenol).

(ii) is not a good test on its own, since so many other readily oxidisable organic compounds will give the same reaction, though following it up by testing for an aldehyde gives it much more validity.

(ii) Distil with potassium dichromate(VI) and mod. conc. H₂SO_4(aq) 

Not very convenient !!!

(ii) If product distilled off immediately an aldehyde odour can be detected and the solution colour changes from orange to green.

(ii) R–CH₂OH + [O] => R–CHO + H₂O

or the full works!

3R–CH₂OH + Cr₂O₇^2–  + 8H⁺ ==>

2Cr³⁺ + 3R–CHO + 7H₂O

The orange dichromate(VI) ion is reduced to the green chromium(III) ion. If the organic product is collected you could test for an aldehyde.

R₂CHOH, R = alkyl or aryl.

(ii) is not a good test on its own, since so many other reducible organic compounds will give the same reaction, though following it up by testing for a ketone gives it much more validity.

(ii) Distil with K₂Cr₂O₇/H₂SO_4(aq) 

Not very convenient !!!

(ii) If product distilled off immediately a ketone odour can be detected and the solution colour changes from orange to green.

R₂CHOH + HCl => R₂CHCl + H₂O

(ii) R₂CHOH + [O] => R–CO–R + H₂O

or the full works!

3R₂CHOH + Cr₂O₇^2–  + 8H⁺ ==>

2Cr³⁺ + 3R–CO–R + 7H₂O

The orange dichromate(VI) ion is reduced to the green chromium(III) ion. If the organic product is collected you could test for an aldehyde.

R₃COH, R = alkyl or aryl.

(ii) Distil with K₂Cr₂O₇/H₂SO_4(aq) Not very convenient !!!

(ii) No aldehyde or ketone readily formed, no change from orange to green.

R₃COH + HCl => R₃CCl + H₂O

(iii) Stable to modest oxidation.

The formation of CHI₃, triiodomethane (or old name 'iodoform'.

triiodomethane, melting point 119^oC.

–CHOH–CH₃

BUT, it is also given by the aldehyde ethanal CH₃CHO and all ketones with the 2–one structure R–CO–CH₃  ('methyl ketones')

Its a combination of halogenation and oxidation and is not a definitive test for anything, it just indicates a possible part of a molecules structure.