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More stable A , greater dissociation, stronger acid. No acidic proton electron donating benzene ring present. ethyl group present allows intensifies delocalisation of of cannot act as negative charge negative charge on an acid! and destabilizes Oalkoxide anion O of into benzene ring stabilizing the phenoxide ion
negative charge is dispersed over 2 electronegative oxygen atoms in 2 equivalent resonance structures causing the anion to be greatly stabilised
Presence of benzene enables delocalisation of negative charge in the anion into the benzene ring, ring, stabilising the anion
More stable A , greater dissociation, stronger acid. Compare stability of A , taking CH 3COO as reference: alkyl groups are elctron donating by inductive effect. It intensifies intensifies the negative charge & destabilises the anion. CH 3CH2CH2CH2COO is only slightly less stable than CH3COO due to longer alkyl chain (effect of alkyl chain length is minimal) OH is electron withdrawing by inductive effect. It helps to disperse the negative charge & stabilises the anion. HOCH 2COOH is more acidic than CH 3COOH. the presence of 3 Cl atoms (electron withdrawing by inductive effect) disperses the negative charge extensively. CCl3COO is greatly stabilised & CCl 3COOH is the strongest acid among the 4.
2.
All three compounds compounds react with with Na(s) to give effervescence effervescence of of H 2(g) and the + corresponding Na salts which are soluble in water. E
CH2OH
H3C
+
+
OH
Na (s)
COOH
+
F
-
Na(s)
+
CH2O Na + 1/2 H2
-
H3C
+
O Na + 1/2 H2
-
Na (s)
+
COO Na + 1/2 H2
G Both E and F have no reaction with aqueous Na 2CO3. G reacts with aq. Na 2CO3 to give effervescence of CO 2(g) and Na + salt which is soluble in water.
2
COOH
+ Na2CO3 (aq)
2
-
+
COO Na + CO2
G 1
E has no reaction with aqueous NaOH. Both F and G react with NaOH to give the corresponding Na + salts which are soluble in water.
+
H3C
OH
-
H3C
NaOH (aq)
+
O Na + H2O
F -
+ NaOH (aq)
+
COO Na + H2O
COOH
G Add aqueous Na2CO3 to the sample. If effervescence of CO 2 is detected, then the sample is G. If not, then the sample is either E or F, which can be differentiated by adding aqueous NaOH. Only F is soluble in aqueous NaOH. 3. C 6H and C 6H [acid strength] 5 COOH 5OH Test: Add Na2CO3 (aq) [or NaHCO3 (aq)] Observations: Effervescence observed. No effervescence observed Colourless gas (CO 2) evolved formed white ppt with aq. Ca(OH) 2 Equation: 2 CH3COOH + Na2CO3 2 CH3COO Na+ + CO2 + H2O
(b) HOCH 2CH and ClCH 2 CH 2COOH [acyl chloride vs acid] 2COCl Test: Add AgNO3(aq) Observations: White fumes No white fumes White ppt (AgCl) No ppt observed Equation: HOCH2CH2COCl + H2O HOCH2CH2COOH + HCl Cl (aq) + Ag+(aq) AgCl (s)
(c)C 6H and C 6H and C 6H 5COCl 5COBr 5 CH 2 Cl [acyl chloride vs acyl bromide vs alkyl halide] Test: Add AgNO3(aq) Observations: White fumes White fumes White fumes White ppt (AgCl) Off-white ppt (AgBr) White ppt (AgCl) only when heated Equation: C6H5COX + H2O C6H5COOH + HX C6H5CH2Cl + H2O + C6H5CH2OH + HCl X (aq) + Ag (aq) AgX (s) X= Cl, Br Cl (aq) + Ag +(aq) AgCl (s)
(d) CH3COOCH2CH2CH3 and CH3COOCH(CH3)2 Test: 1) Heat with NaOH (aq) 2) Add aq. I2 Observations: No yellow ppt formed yellow ppt formed Equation: CH3COOCH2CH2CH3 +OH CH3COO + HOCH2CH2CH3 CH3COOCH(CH3)2 +OH CH3COO + HOCH(CH3)2 HOCH(CH3)2 + 4I2 + 6OH CH3COO + CHI3 + 5I + 5H2O
4.
(a)
dilH SO , aq.KMnO
2 4 CH3CH2OH + [O] CH3COOH + H2O heat
CH3COOH
SOCl2 room temperature
CH3COCl + SO2 + HCl
2
O =
(b)
(c)
CH3CH2OH + CH3CH2COCl (from a)
CH3CH2OH + [O]
C
H
CH3C-O-CH2CH3
K2Cr 2O7(aq), dil H2SO4, heat with immediate distillation
CH3CHO + H2O
O CH3
room temperature
+
OH
KCN
HCN
CH3
C
CN
H
OH CH3
C CN H
OH
dil HCl + 2H2O + H+
(d) CH3CH2CH2OH + 2 [O]
CH3
C COOH
heat
+ NH4+
H
KMnO4 (aq), dil H2SO4, heat
CH3CH2COOH + H2O
room temperature
CH3CH2COOH + SOCl2
CH3CH2COCl + SO2 + HCl O
+CH3CH2COCl + Na O
5.
C O
+ NaCl
COCl
COOH
+
room temperature SOCl2
COCl
+ 6.
CH3CH2
NH2CH2COOH
room temperature
+
HCl
+
CONHCH 2COOH
+
SO2
HCl
F undergoes neutralization with NaOH(aq)
F
contains an acidic fn gp F is either a phenol or a carboxylic acid F undergoes condensation reaction with 2,4-DNPH but does not undergo oxidation with Fehling’s solution F is a carbonyl compound that is not an aliphatic aldehyde F is either an aromatic aldehyde or a ketone F undergoes electrophilic substitution with Br 2 water to give G.
3
F
contains a strongly activated benzene ring F is a phenol Since G has 2 H atoms substituted with 2Br atoms G is a 2- or 4-substituted phenol F gives a positive iodoform test to yield H F is a methyl ketone H is a carboxylic acid with 1 C less H undergoes neutralization with both NaOH(aq) and Na2CO3(aq). H is a carboxylic acid (confirmed) H is obtained by alkaline hydrolysis of J followed by acidification. H is 1,2-disubstituted. OH
O C OH
H
is OH
OH
O
Br
C
O C
OH CH3
O C OH
CH3 Br
F
G
H
[1] per correct structure
Equations: [F + NaOH] OH
O-Na+ O
O C
CH3
C
+ NaOH
CH3
+ H2O
[F + 2,4-DNPH] OH
C
OH
O2N
CH3
CH3 C
O
+
H2NNH
NO2
O2N
NNH
NO2 + H2O
[F + Br 2(aq)] OH
O C
OH CH3
Br
O C
+ 2Br 2
CH3
+ 2HBr
Br
[F + aq. alkaline I2]
4
OH
O-
O C
O C
CH3 + 3I2 + 5OH-
O-
+ CHI3 + 3I- + 4H2O
[acidification after iodoform test] [acidification after J + NaOH] O-
