chemical test for chemistry practicals.pdf

notes

1. Chemical
Test

2. Halo-alkane
&
Halo-arenes

4. R-X
AgNO
3
NH
4
OH
Cl
–
Br
–
I
–
WhitepptofAgCl
PaleyellowpptofAgBr
darkyellowpptofAgI

5. Distinguishbetweenpairofcompoundsbychemicaltest
Itcanbedistinguish byAgNO
3test.
Writeachemicaltesttodistinguishbetween chloroethaneandchlorobenzene
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ChemicalTest Chloroethane(C
2H
5Cl)
Cl
Chlorobenzene
(1) Organic compound +
aqNaOH +AgNO
3
WhiteprecipitateofAgCl
is formed which is soluble
excessofNH
4OH
Nochange

6. Writeachemicaltesttodistinguish betweenchloroethaneandbromoethane.
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Test
Chloroethane
(CH
3CH
2Cl)
Bromoethane(CH
3CH
2Br)
(1)O.C.+aqNaOH +AgNO
3Whiteppt(AgCl) Creamyyellowppt(AgBr)
(2)O.C.+NH
4OH inbothtest
tube
Whitepptget
dissolvedin NH
4OH
Yellow pptdissolvedpartly
inNH
4OH

7. Writeachemicaltesttodistinguishbetween benzylchlorideandchlorobenzene.
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Test
CH
2Cl
Benzylchloride
Cl
Chlorobenzene
(1)O.C.+aqNaOH+AgNO
3
WhitepptofAgClis
formedwhichissolubein
excessofNH
4OH
Nochange

8. Alcohol
&
Phenol

11. Lucasreagent(Conc.HCl+ZnCl
2)
Iodoformtest(I
2+NaOH)
CH
3–CH–R/H/phgivesyellowpptofCHI
3
OH
NeutralFeCl
3givesvioletppt withphenol
1
°
Alcohol 2
°
Alcohol 3
°
Alcohol
NoChange
Turbidityappears
within5minutes
Turbidityappears
immediately

12. Writea chemicaltesttodistinguishbetweenprimary,secondaryandtertiaryalcohol.
OR
Writeachemicaltesttodistinguishbetweentertiarybutylalcohol,isobutylalchol,
ethylalcohol.
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Test 1
°
Alcohol 2
°
Alcohol 3
°
Alcohol
(i)AddLucasreagent
Anhyd.ZnCl
2+Conc.HCl
NoChange
Turbidityappears
within5minutes
Turbidityappears
immediately

13. LucastestisbasedonKineticsofthereaction.3
0
alcoholsreactatthefastest
rate.
3
HC–C–OH+HCl H
3C–C–Cl+ H
2O
CH
3 CH
3
CH
3
anhydrous
ZnCl
2
CH
3
(3
0
alcohol)
CHOH+HCl
2
CHCl+HO
H
3C
H
3C
(2
0
alcohol)
H
3C
H
3C
anhydrous
ZnCl
2
CH
3CH
2OH+HCl
(1
0
alcohol)
NoChangeappears
Anhydrous
ZnCl
2

14. Distingiush betweenalcoholandphenol
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Test EthanolCH
3CH
2OH
OH
Phenol
(1)AddI
2/NaOH andheatYellowppt.ofCHI
3 Nochange
(ii)AddneutralFeCl
3 NoChange VioletColouration

15. Aldehyde , Ketone
&
Carboxylic acid

16. Ammoniacal silver nitrate solution (AgNO
3+ NH
4OH)
R—C—H
O
T. R.
R—C—R
O
T. R.
Tollen’stest can be used to
distinguish between aldehydes
and ketones
1. Tollen’sReagent
Silver mirror
[Ag(NH
3)
2OH] or[Ag(NH
3)
2] + OH
O
R—C—O + 2Ag + 2H
2O + 4NH
3R—C—H + 2[Ag(NH
3)
2] + 3OH
O

17. Compounds which react with Tollen’sreagent are :
All aldehydes (aliphatic and aromatic)
Ketones -All a-hydroxy ketones react with Tollen’sreagent
R
C R
OH
O
O
OH
(Benzoin)
Formic acid
H—C—OH
O

18. Hemi-acetals
C
H
H
OR’
OH
O
OH
C
R
HOH
2C
OR’
OH
Hemi-acetal
Hemi-acetal
a-Hydroxy hemi ketal
–R’–OH
C=O
R
HOH
2C
a-Hydroxy ketone

19. Fehling’s Solution comprises of two solutions.
Fehling A Fehling B
Aq. CuSO
4 Alkaline sodium potassium tartarate
(Rochellesalt)
HO—CH—COONa
HO—CH—COOK
Fehling A + Fehling B ---Deep blue colour
OnheatingaldehydeswithFehling’sreagent,aldehydesareoxidizedto
carboxylateionandreddishbrownppt.ofCu
2Oisobtained.
Fehling’s Solution

20. Aromatic aldehydes (ex: benzaldehyde) do not react with fehlingsolution.
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R—C—H
O
F.R.
R—C—R
O
F.R.
R—C—H + 2Cu
2+
+ 5OH
O
R—C—O + Cu
2O + 3H
2O
O

21. When methyl ketones react with halogen in the presence of base oxidation
of methyl ketones take place.
R—C—CH
3+3X
2+ 4NaOH
O
Salt of carboxylic acid
R—C—ONa+ CHX
3
O

(Haloform)
CHCl
3–Chloroform (liquid)
CHBr
3–Bromoform(liquid)
CHI
3–Iodoform(yellow solid)
10. Oxidation of methyl ketone (haloform reaction)

22. Ketones Methyl ketones undergo haloform reaction
R—C—CH
3
O
Aldehydes Only acetaldehyde can undergo haloform reaction
H
3C—C—H
O
Alcohols Alcohols of the following type also undergo haloform
reaction
—CH—CH
3
OH
a
NaOH + X
2is also a mild oxidizing agent
H
3C—CH—CH
3
OH
H
3C—CH
2—OH
NaOX
NaOX
H
3C—C—CH
3
O
H
3C—CHO {Give iodoform test}
Haloform Reaction

23. Chemical
Test
O
||
H–C–H
O
||
CH
3–C–H
O
||
CH
3–CH
2–C–CH
O
||
CH
3–C–CH
3
2, 4-DNP Test
O.C. +2,4-DNP
Red/
orange ppt
formed
Iodoform
Test
O.C. +
(NaOH + I
2)
Yellow ppt
of iodoform is
formed
Tollen’sTest
O.C. +
Tollen’s
reagent
Silver
mirror is
formed
Fehling’s Test
O.C. +
Fehling (A+B)
Solution
Reddish
brown ppt
is formed
O
||
C
H
O
||
C
CH
3

24. Test Acetaldehyde CH
3CHO O
||
Acetone H
3C–C–CH
3
(1) Add Fehling’s
reagent
Brown red precipitateNo change
CH
3CHO + 2Cu
+2
+ 5OH
–
CH
3COO + Cu
2O(red brown) + 3H
2O
(2) Add Tollen’stestSilver mirror No change
(i)
Difference between Aldehyde and Ketone
CHEMICAL TEST

25. Test Ethanal (CH
3CHO) Benzaldehyde (C
6H
6CHO)
(1) Add I
2/NaOH Yellow iodoform (CHI
3)No change
(2) Add Fehling’s
reagent
Red ppt of Cu
2O No change
(ii)

26. (iii)
Test Pentan-3-one Pentan-3-one
(1) Add I
2/NaOH and heat O
||
CH
3 –C –CH
2CH
2CH
3
Yellow ppt. (CHI
3)
O
||
CH
3CH
2 –C –CH
2CH
3
No change
O
||
4NaOH + 3I
2+ CH
3 –C –CH
2CH
2CH
3CHI
3+ CH
3CH
2CH
2COONa + 3Nal + 3H
2O
(Yellow ppt.)
Chemical Test

27. Test Acetophenone Benzophenone
(1) Add I
2/NaOH Yellow iodoform (CHI
3)No change
Yellow ppt No change
COCH
3
C
O
COCH
3
4NaOH + 3I
2+ CHI
3+
COONa
+Nal+ 3H
2O
(iv)
Chemical Test

28. Test Formic acid (HCOOH)Acetic acid (CH
3COOH)
(1) Tollen’sreagentSilver mirror(Ag) No Change
HCOOH+AgNH
32
+
+2ഥO→CO
2+Ag+2H
2O+2NH
3
(i)
Difference between Different-Different Carboxylic Acid

29. (i)
Test Phenol (White solid)Benzoic acid (White solid)
(1) Add neutral FeCl
3Violet color No Change
(2) Add NaHCO
3(aq)No change Effervescence of CO
2
C
6H
5COOH + NaHCO
3C
6H
5COONa + CO
2+ H
2O
OH COOH
Difference between Phenol and Carboxylic Acid

30. Test CH
3COCH
3 acetoneAcetic Acid CH
3COOH
(1)Add I
2/NaOH + heat Yellow iodoform
(CHI
3)
No change
4NaOH + 3I
2+ CH
3COCH
3CHI
3+ CH
3COONa + 3Nal + 3H
2O
(2)Add NaHCO
3 No change Effervescence of CO
2
CH
3COOH + NaHCO
3CH
3COONa + H
2O + CO
2
(ii)
Difference between Ketone and Carboxylic Acid

31. (ii)
Test Acetaldehyde CH
3CHO Acetic acid CH
3COOH
(1)Add Fehling’s
reagent
Red ppt. of Cu
2O No change
(2)Add Tollen’sreagentSilver mirror (Ag) No change
(3)Add NaHCO
3 No change Effervescence of CO
2

32. Amine

33. Writethechemicaltesttodistinguishbetween
pre./sec./ter.amine.
ItcanbedistinguishbyHinsbergreagent.Benzenesulphonylchloride
(C
6H
5SO
2CI),whichisalsoknownasHinsberg'sreagent,reactswith
primaryandsecondaryaminestoformsulphonamides.
(i)Thereactionofbenzenesulphonylchloridewithprimaryamineyields
N-ethylbenzenesulphonylamide.Thehydrogenattachedtonitrogenin
sulphonamideisstronglyacidicduetothepresenceofstrongelectron
withdrawingsulphonylgroup.Hence,itissolubleinalkali.
N-Ethylbenzenesulphonamide(soluble in alkali)
S–Cl+H–N–C
2H
5
O
O H O
S–N–C
2H
5+HCl
O
H
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34. (ii)Inthereactionwithsecondaryamine,N,N-diethyl–
benzenesulphonamideisformed.SinceN,N-diethylbenzene
sulphonamidedoesnotcontainanyhydrogenatomattachedto
nitrogenatom,itisnotacidicandhenceinsolubleinalkali.
N-N-Diethylbenzenesulphonamide
(insoluble in alkali)
S–Cl+H–N–C
2H
5
O
O
C
2H
5
O
S–N–C
2H
5+HCl
O
H
H
3C
(iii)Tertiaryaminesdonotreactwithbenzenesulphonylchloride.
Thistestisusedforthedistinctionofprimary,secondaryand
tertiaryaminesandalsofortheseparationofamixtureof
amines.However,thesedaysbenzenesulphonylchlorideis
replacedbyp-toluenesulphonylchloride.
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35. Writethechemicaltesttodistinguishbetweenaliphaticand
aromaticamine.
(a)Primaryaliphaticaminesreactswithnitrousacidtoformaliphatic
diazoniumsaltswhichbeingunstable,liberatenitrogengas
quantitativelyandalcohols.Quantitativeevolutionofnitrogenis
usedinestimationofaminoacidsandproteins.
R-NH
2+ HO-N=0 R–OH + N
2+ H
2O
NaNO
2+ HCl HNO
2+ NaCl
R −NH
2 + HNO
2 [R−N
2Cl] ROH + N
2+ HCl
NaNO
2+ HCl
+−
H
2O
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36. (b)Aromaticaminesreactwithnitrousacidatlowtemperature
(273-278K)toformdiazoniumsalts
BenzenediazoniumChloride

Orange dye if formed
b–nepthol+ NaOH
C
6H
5−NH
2  C
6H
5−N
2Cl +NaCl +2H
2O
NaNO
2+ 2HCl +−
275 −278K

37. Test
Primary
Amine
Secondary
Amine
Tertiary
Amine
Hinsberg
reagent
C
6H
5SO
2Cl
White ppt
soluble in
NaOH
White ppt
soluble in
NaOH
No
Change
(i)
Difference between Primary, Secondary and Tertiary Amine

38. Test
Methanamine
(CH
3NH
2)
Dimethyl Amine
(CH
3NHCH
3)
(1)
Add Hinsberg
reagent C
6H
5SO
2Cl
White ppt
(soluble in NaOH)
White ppt insoluble
in NaOH (aq.)
(2)
Add CHCl
3+
KOH(alc.)
Four odourof
isocyanide
No Change
(ii)
CH
3NH
2+ CHCl
3+ KOH CH
3N ≡C + 3KCl + 3H
2O
–+

39. (iii)
Test Aniline N-methyl-aniline
(1)
Add Hinsberg
reagent
(C
6H
5SO
2Cl)
White ppt solution in
NaOH
White ppt insoluble in NaOH
(2)
Add CHCl
3+
KOH(alc)
Foul odourof
isocyanide
No Change
NH
2
NHCH
3
C
6H
5SO
2Cl + H
2NC
6H
5C
6H
5SO
2–NHC
6H
5C
6H
5SO
2–N –Na
OH
NaOH
y 
C
6H
5
NH
2
+ HCCl
3+ 3KOH 
N ≡C
+ 3KCl + 3H
2O

40. (i)
Test Aniline
Methylamine
(CH
3NH
2)
(1) Add Br
2/H
2O
Colourof Br
2(aq.)
decolourised
No change
NH
2
NH
2
+ 3Br
2(aq) 
NH
2
Br
Br
Br
+HBr
(Colourless)
Difference between Aliphatic and Aromatic

41. (ii)
Test
Cyclohexylamine (Aniline)
(1)AddNaNO
2and
HClandthenadd
phenolinpresence
ofabase
N
2gasisliberated
andnocolourationis
observed.
Stablediazoniumsalt
isformedwhichreacts
withphenoltoform
orangedye.
NH
2NH
2