CARBOXYLIC ACID –
ALKYL HALIDE & ARYL HALIDE –
n
1.Substitution rx on aryl halide is not easy. Why?
n
2.Why during williamson synthesis rx we can’t use aryl halide &
t-butyl halide ?
3.Why ammination of aryl halide is not easy but that of alkyl halide
is easy ?
4. Complete the following –
HBr
a. (CH3)2CH – CH=CH2
HOH
b. (CH3)3C – CH=CH2
5. Compare M.pt. & B.pt. of o-dichloro benzene & pdichlorobenzene.
6. Identify the reactant of following ethers a. 1-Proxypropane b. Ethoxybenzen
c. 2-methyl-2-methoxypropane d. 1-Methoxyethane
7. Arrange the following in the order of reactivity towards SN1 &
n
SN2 Rx as indicated below –
(a)The four isomeric bromobutanes
(b)C6H5CH2Br,C6H5CH(C6H5)Br,C6H5CH(CH3)Br,
C6H5C(CH3)(C6H5)Br
8. Explain why R – Cl is hydrolysed to R – OH slowly but the
reaction is rapid if catalytic amounts of KI are added to the
reaction mixture.
9. P – Methoxy benzyl bromide reacts faster than p-nitrobenzyl
bromide with ethnol to form ether, why?
ALCOHOL & PHENOLS
1. Write a short note on n
n
1. Oxymercuration de mercuration rx 2. Hydroboration rx
n
n
n
3. Reimer tiemann rx
4. Fries rx
5. Kolbe’s rx
2. Write the mechanism of following –
a. Conversion of alcohol to ether b. alcohol to alkene
c. Phenol to salicylic yldehyde d. Esterification
3. What happen when 2-Methylpropan-2-ol is heated with Cu at
573K.
4. Why phenol is a stronger acid than alcohol?
5. How will you convert chlorobenzene to phenol ( Dow process ).
6. Give the action of H2SO4 on ethanol .
7.How will you prepare (1) salicylic acid (2) Asprine from phenol?
8. Why symmetrical ether poessess definate dipole moment ?
9. Why di-tertbutylether & diphenylether can not be synthesise by
n
wurtz rx ?
10.Why ethers are cleaved only by acid & not by base ?
11.Orhto & para – nitrophenol are more acidic than phenols.Why?
12. Give the function of following –
a. SOCl2 b. Br2 & CS2 c. Hg(CH3COO)2 / THF
13. Complete the following –
a. 1-Methylcyclohexanol reacts with Conc. H2SO4
b. 2-Methylpropan-2-ol reacts with Cu at 573K.
14. Give the action of HI on Anisol. With mechanism?
15. Convert - (a) Acetaldehyde to isopropylalcohol.
16.Why symmitrical ether posses dipole moment?
ALDEHYDE KETONE 1.Why it is essential to control pH during the reaction of NH3
derivative on aldehydes & ketones .
2. Why aldehyde give tollens reagent test but ketones does
not ?
3. Compare the dipole moment of aldehydes & ketones .
4.Complete the following –
a. CH3- CH2 – OH to CH3CHO b. (CH3)3C – OH with Cu / 573K
c. CH3COOH with Ca(OH)2 followed by dry distillation
d. CH3CH2COOH heated with MnO
5. Why cyclohexanone form cynohydrin in good yield but
2,2,6-trimethylcyclohexanone doe s not explain ?
6. Arrange the following in the order of reactivity towords HCN –
CH3CHO , CH3COCH3 , di-terbutylketone
7. An organic compd. With molecular formula C9H10O forms 2,4n
DNP Derivative , reduces tollens reagent & undergo cannizaro rx .On
vigorous oxidation , it give 1,2-benzenedicarboxylic acid . Identify the
compd.
8. What happen when Acetaldehyde is reacts with –
(a) dil. NaOH (b) Conc. NaOH (c) Zn - amalgam
9. Arrange the following in the order of nucleophilic addition
reaction – Benzaldehyde , p-tolualdehyde ,
p-nitrobenzaldehyde, acetophenone
10. How will you prepare (a) Cinnamaldehyde (b) Cinnamic acid
(c) Crotonaldehyde (d) benzoic acid from tolune (e) Lactic acid
11. What is formline.
1.Arrange the various acid dervative in the order of reactivity ?
2. Write short note on –
n
n
a. Transesterification b. Hoffman bromamide rx . c. HVZ r
3. p-fluorobenzoic acid is a weaker acid than p-chlorobenzoic
acid.Explain why?
n
4. Why carboxylic acids does not show characteristic rx of
carbonyl Group?
5. Which carboxylic acid shows the properties of both aldehyde &
alcohols?
6. How will you distinguish b/n HCOOH & CH3COOH?
COMPOUNDS CONTAINING NITROGEN
0
0
0
1.Give the order of basic strength of 1 , 2 & 3 amine in
(a) in aq. Medium (b) in non aq. Medium (c) for aromatic
amines
2.Convert –
a. methylamine to ethylamine b. Aniline to acetanilide
c. p-toludine to 2-bromo-4-methylaniline
d. Aceticacid to CH3CH2NH2 e. Aniline to N-phenylethanamide
f. Ethylamine to diethylamine
g. p-Nitroaniline to 1,2,3-tribromobenzene
h. p-Nitroaniline to 1,2,3,5-tetrabromobenzene
3. Aniline is weaker base than cyclohexylamine . Explain.
4. Give one test to distinguish b/n ethylamine from aniline .
5. How is an amide more acidic than amine ?
6.Why aniline does not under go friedel – craft reaction . Explain.
+
Hint – C6H5NH2 + AlCl3
C6H5NH2 AlCl3
7. Convert aniline to sulphanilic acid .
8. Convert nitrobenzene to p-Aminophenol in one step.
9. Why AgCl is not soluble in water but soluble in amines?
10. Arrange Following in the order of basic strength –
0 0
0
0
0
0
(a) 1 ,2 & 3 methyl amines (b) 1 , 2 & 3 ethyl amines
11. Why direct monohalogenation on aniline is not possible ?How
n
we can get monohalogenated aniline?What is the fu of
acetanilide ?
12. Explain nitration on aniline.
13. Convert – (a) Nitroethane to ethanoic acid (b) Nitroethane to
acetaldehyde
DISTINGUISHING TEST –
1.Phenol & Carboxylic acid
2. Ethanoic acid & propanoic acid
3. Ethanol & propanol
4. Propanol & propan-2-ol
5. CH3CHO & CH3CH2CHO
6. CH3CHO & C6H5CH2CHO
7. CH3CN & CH3NC
8. CH3NH2 & CH3CH2NH2
9. methanamine & N-methylmethanamine
10. CH3CH2NH2 & Aniline
11. Aniline & N-methylaniline
12. CH3NO2 & CH3 – ONO
13. Chloromethane & chlorobenzene
14. Chloroethene & 3-chloropropene ( Vinylchloride & allyl chloride )
15. Ethanol & phenol
16. Propanol & Propan-2-ol
NAME REACTIONS n
1.Reimer – tiemann rx
2. Fries rearrangement
3. Rosenmund reaction
n
4. Hoffmann rx
5. Aldol condensation
6. Hydroboration
n
7. Cannnizarro rx
8. Benzoin condensation
9. Kolbe’s reaction
10. Dow’s process
PRADEEP SHARMA, INSTITUTE OF COMPETITIVE STUDIES, SECTOR – 15 , SONEPAT
CONTACT NUMBER : 0130 – 2231322 , E – mail : pradeepsharma1976@gmail.com
b. CH3-C(OH)(CH3)-CH3
ALKYL & ARYL HALIDES n
1. Substitution rx on aryl halide is not easy because of
partial double bond between C & halogen due to
resonance.
** Show the Structure.
2. Aryl halide can’t be used in Williamson synthesis
because of partial double bond b/n C & halogen.
** Resonance structure
t-butyl halide can’t be used in Williamson’s
0
synthesis because of stable 3 carbocation which
leads to elimination.
3.Hint : Ammination of aryl halide is not easy due to
partial double bond.
4. a. CH3-C (Br)(CH3)-CH2-CH3
b. CH3-C(OH)(CH3)-CH(CH3)-CH3
5. Comparision of B.pt & M.pt –
B.pt : o-dichlorobenzene > p-dichlorobenzene
M.pt : p-dichlorobenzene > o-dichlorobenzene
6. a. The reactants of 1-Proxypropane are
CH3CH2CH2ONa + CH3CH2CH2Cl
b. The reactants of Ethoxybenzene are
sodium phenoxide & chloroethane.
c. The reactants of 2-methyl-2-methoxypropane are
CH3-C(ONa)(CH3)-CH3 + CH3Cl
d. The reactants of 1-Methoxyethane are
CH3ONa + CH3CH2Cl or CH3CH2ONa + CH3Cl
7.
8. In Notes
9. p-Methoxy benzylbromide reacts faster than pnitrobenzyl bromide with ethanol to form ether
because e- releasing –OCH3 group increases the
bondlength of carbon-halogen bond,increases the
rate whereas e-withdrawing NO2 group decreases
the bondlength & hence decreases the reactivity.
10. In Notes
11. In Notes
ALCOHOLS & PHENOLS –
1. In Notes
2. In Notes
3. CH3-C(OH)(CH3)-CH3 Cu ,573 K (CH3)2C=CH2
4. In Notes (Based upon resonance)
5.
–Cl 2NaOH,
-ONa H+/HOH
-OH + NaOH
623K,300atm
6. In Notes
7. a. Salicylic acid – Reimer Tiemann or Kolbe’s rxn
b. Phenol
Salicylic acid (CH3CO)2CO Aspirine
8. Symmetrical ether possess irregular geometry,in
which resultant of individual bond moment cancel
the effect of each other.
0
9. Di-tert-butylether can not be prepared because of 3
0
halide.we know that in 3 halide,there is partial
double bond b/n C & halogen .
10. Lone pair of OH- repelled by lone pair of oxygen of
ether,hence rxn of ether is not catalysed by base.
+
But the charged H can attack on lone pair of
n
oxygen of ether,hence rx of ether are catalysed by
acid.
11. Ortho & para –nitrophenol are more acidic than
phenol because of –I & -R effect of NO2 gp.Nitro
gp. when present at o & p-position stablise
conjugate base & increases the acidic strength.
12. a. SOCl2 :- -OH gp. to –Cl gp.
b. Br2 & CS2 :- Monohalogenation of Phenol
Br
-OH Br2 ,CS2
-OH + Br-OH
c. Hg(CH3COO)2/THF :- OxymercurationDemercuration , Markovnikoff’s addition of H2O
13. a.
OH
Conc. H2SO4
14.
HI
–OCH3
(CH3)2C=CH2
-OH + CH3I
H+ + I-
H–I
O-CH3
Cu ,573 K
H
O-CH3
H+
OH
+
CH3 +
I-
15.CH3CHO
CH3MgBr
H+ /HOH
Conc. H2SO4
CH3-C(OH)(CH3)-CH3
-HOH
(CH3)2C=CH2
CH3I
CH3-CH(OH)-CH3 Cu /573KCH3COCH3
B2H6 , H2O2
CH3MgBr,H+/HOH
(CH3)2-CH-CH2OH
16. Same as in Ques.14
17. Same as in Ques.8
ALDEHYDES & KETONES –
1. In Notes
2. Aldehydes give Tollen’s reagent test because
Aldehydes due to high reactivity are easily oxidized
by weak oxidizing agent like tollen’s reagent.
3. In GOC
Cu / 573 k or PCC or collin’s reagent
4. a. CH3CH2OH
CH3CHO
b. (CH3)3C-OH Cu / 573 K
(CH3)2C=CH2
c.CH3COOH Ca(OH)2(CH3COO)2Ca dry ,dil NaOHCH3COCH3
d. CH3CH2COOH MnO , Heat CH3CH2COCH2CH3
5. 2,2,6Trimethylcyclohexanone does not form
cynohydrin in good yield because of three methyl gp.
at α-position w.r.t >C=O .Thus nucleophilic attack by
CN- does not occur due to steric hindrance.No such
steric effect is present in cyclohexanone.
6. Reactivity towards HCN –
CH3CHO > CH3COCH3 > (CH3)3C-CO-C(CH3)3
+I
+I
+I
+I
+I
Steric hindrance
NO2 _
7.C2H5-CHO 2,4-DNP C2H5 -CH=N-NH-NO2
[ O]
COOH-
cannizaro ,
conc. NaOH
CH2OH-
8. a. 2CH3CHO
b. CH3CHO
c. CH3CHO
-COOH
-C2H5 + COONa-
dil.NaOH
conc. NaOH
Zn-Hg /HCl
-C2H5
H+ ,Heat
CH3CH(OH)CH2CHO
CH3CH=CHCHO
CH3CH=CH-CH=CH-CH=…….
CH3-CH3
9. The order of nucleophilic addition reaction :NO2-
-CHO >
-CHO >CH3-
-CHO >
-COCH3
- I α Reactivity , + I α 1 / Reactivity
10.
a .
–CHO + CH3CHO dil. NaOH
-CH(OH)CH2CHO
-CH=CHCHO
CINNAMALDEHYDE
Heat
b.
-CHO +(CH3COO)2
CH3COONa
H+/HOH
-CH(OH)CH2COOH
-CH(OH)CH2COOCOCH3
-CH=CHCOOH Cinnamic acid
Heat
c. CH3CHO + CH3CHO dil.NaOH CH3CH(OH)CH2CHO ∆ , H+
CROTONALDEHYDE CH3CH=CHCHO
d.
–CH3
KMnO4 / H+
-COOH
e. CH3CHO + HCN CH3CH(OH)CNCH3CHOHCOOH
11. Formalin is 35-40% aq. soln of HCHO.
PRADEEP SHARMA, INSTITUTE OF COMPETITIVE STUDIES, SECTOR – 15 , SONEPAT
CONTACT NUMBER : 0130 – 2231322 , E – mail : pradeepsharma1976@gmail.com
4. Aniline will give dye test and ethylamine does not.
-NH2 NaNO2 , HCl
-N2+Cl-
CARBOXYLIC ACID :1.
2. Transestrification :- The conversion of one ester into
other, when reacts with suitable alcohol.
CH3COOCH3 + C2H5OH
CH3COOC2H5 + CH3OH
Methyl acetate
Ethyl acetate
Hof’fmann Bromamide :- It is the conversion of amide
into amine having one carbon atom less.
R-CO-NH2
Br2 , 3 KOH
R-NH2 + KBr + K2CO3
HVZ :- Halogen in red P to acid leads to replacement of
α-Hydrogen.
Cl ,Red P
Cl ,Red P
CH3COOH 2
Cl CH2-COOH 2
Cl2CH-COOH
3. In m-fluoro benzoic acid,+R effect of F is more than +R
of Cl in m-Chlorobenzoic acid because in F ,e- transfer
from 2p orbitals of F to 2p-orbitals of Carbon,which is
easy as compared to 3p of Cl to 2p of carbon.
4. Carboxylic acids does not show characteristic rxn of
carbonyl Group because carboxylic acid does not
contain true carbonyl gp. due to Resonance.
O
O- +
- C-OH
-C = O-H
5. The carboxylic acid which shows the properties of
both aldehyde & alcohols is Formic acid , HCOOH.
6. Difference b/n HCOOH & CH3COOH CH3COOH on reduction gives ethyl alcohol & also
give Idoform test.show the rxn….
HCOOH gives Tollen’s reagent test.
7. In Notes
8. a. CH2=CH2 1.Cl2 , 2. KCN, 3. dil.HCl ,4. 2Cl2 / RED p ,AgOH aq.
OH
H – C - COOH
H – C – COOH
OH
CH3Cl ,AlCl3 anhy.
KMnO4 / H+
b.
–CH3
-COOH
HO
-N2+Cl- +
HO
-N=N-

Orange Dye
5. In amide,conjugate base is stabilized by resonance.
CH3CONH2
CH2-CO-NH2 + H+
.. _
_
:O: ..
:O: ..
: CH2-C- NH2
CH2=C – NH2
In amine + R & +I effect of NH2 gp. , make amine basic.
6. In Notes
7.
–NH2 conc. H2SO4
8.
–NO2
-NH3+HSO4-
electricity , acid
SO3H-
-NH2
-NHOH Rearrang. OH-
-NH2
9. With amines AgCl form soluble complex due to the
presence of lone pair.
AgCl + 2 NH3
[ Ag(NH3)2 ]Cl
No such complex formation takes place with H2O.
10. Notes –
In Non-aqueous –
..
..
a. CH3NH2 < (CH3)2-NH < (CH3)3N
b. C2H5NH2 < (C2H5)2NH < (C2H5)3N
In aqueous solution –
a. (CH3)2NH > CH3NH2 > (CH3)3N
b. (CH3)2NH > (CH3)3N > CH3NH2
11. Monohalogenation of aniline is not possible directly
because of high e- releasing power of –NH2 which
result in tri substitution. _Br
-NH2 Br2 , HOH Br-NH2
Br
Monohalogenation can be done with the help of
acetanilide.Acetanilide decreases the e- releasing
power of NH2 gp.
-NH2 + ClCOCH3
-NHCOCH3
Acetanilide
-NHCOCH3
COMPOUNDS CONTAINING
NITROGEN :-
b.
c. CH3-
1. NaNO2 / HCl ,2. SOCl2 , 3. KCN alc.
1. (CH3CO)2O
-NH2
d. CH3COOH
CH3CH2NH2
CH3COCl
-NHCOCH3
f.
CH3-CH2-NH2
g. NO2-
CH3-
1. NH3 2. Sn + HCl
–NH2
CH3CH2Cl a lc.
-NH2
Br
(CH3CH2)2NH
-NH2 1. NaNO2 / HCl , 2. H3PO3
Br-
-NH2 Br2 /HOH
Br
- NH2
Br
h.
3.
-NHCOCH3
+
-NH2
Br
H / HOH
-CH3COOH
Br-
-NH2
CH3CH2NH2
-NHCOCH3
1. CH3COCl,2. Br2 /HOH , 3. H+/HOH
e.
-NHCOCH3 +BrBr
+
-
–NH2
-
H / HOH
-CH3COOH
1. In Notes
2. a. CH3NH2
Br2
..
–NH2 , lone pair of aniline is delocalized over
entire benzene ring & are not available for donation
due to resonance.Resonating str. --No such delocalization is present in cyclohexyl amine,
hence e- density is more localized & can be easily
donated & stronger is basic character.
12. Nitration of aniline –
-NH2 conc. HNO3 /,H2SO4 NO2( 51%)
+
-NH2 +
-NH2
NO2(47%)
-NH2
NO2 (2%)
The formation of meta-product is due to the formation of
anilinium ion in highly acidic medium.
-NH2 + H+
-NH3+ Anilinium ion.
13. (a) CH3CH2NO2 dil.HCl/HOH CH3COOH
(b) CH3CH2NH2 TiCl3 ,Mc-murry reagent CH3CHO
14. SO3H-NH2 consists of two gp.-one acidic
-SO3H & other basic –NH2 gp. Hence possess two
value of constants, Ka & Kb .Further,existence of
Zwitter ion shows two values of Ka & Kb.
_
SO3H-NH2
SO3-NH3+
PRADEEP SHARMA, INSTITUTE OF COMPETITIVE STUDIES, SECTOR – 15 , SONEPAT
CONTACT NUMBER : 0130 – 2231322 , E – mail : pradeepsharma1976@gmail.com
DISTINGUISHING TEST :We have to write the reactions for both the compds. for
test,if possible.
1. Phenol & CH3COOH Phenol give white ppt with Br2 / H2O.But CH3COOH
does not.
_Br
-OH + Br2 HOH Br-OH 2,4,6-Tribromophenol
Br
(white ppt )
CH3COOH + Br2 HOH
No ppt
 Phenol can be distinguish from acid by Azo-dye
test.
2. CH3COOH & CH3CH2COOH – On redn ethanoic acid
give ethanol which give iodoform test & Propanoic acid
n
on red give propanol & does not give iodoform test.
CH3COOH Ni/Pt CH3CH2OH I2 + NaOH , NaOH CHI3 + CH3COONa
15. CH3CH2OH & C6H5-OH :- Ethanol give Iodoform test.
But phenol does not.
(Phenol gives white ppt with Br2 water but ethanol does not.)
16. CH3CH2CH2OH & CH3-CH(OH)-CH3 :- Propan-2-ol
give Iodoform test and Propanol does not.
CH3-CH(OH)-CH3 NaOH + I2 CH3COCH3 3NaOI
I3C-CO-CH3 NaOH CHI3 + CH3COONa
Yellow ppt.
NAME REACTIONS – From Notes
Yellow ppt
3. Ethanol & Propanol – As above ,Iodoform test.
4. Propanol & Propan-2-ol Propan-2-ol gives Iodoform test.
5. CH3CHO & CH3CH2CHO CH3CHO give Iodoform test.
6. CH3CHO & C6H5CH2CHO –
CH3CHO give Iodoform test.
n
0
7. CH3CN & CH3NC – CH3CN on red give 1 Amine &
n
0
CH3NC on redNC on red give 2 Amine.
0
0
1 & 2 amines can be distinguished by Hinsberg reagent
test.
O
CH3CN Na+ C2H5OH CH3CH2NH2 -SO2Cl
-S-NHCH2CH3
O
KOH
Dissolve in KOH
O
Na+ C2H5OH
- SO2Cl
CH3NC
CH3-NH-CH3
-S-N(CH3)2
O
KOH
Not soluble in KOH.
8. CH3NH2 & CH3CH2NH2 –
CH3CH2NH2 gives Iodoform test
CH3NH2 NaNO2 , HCl CH3OH  does not give iodoform test
CH3CH2NH2NaNO2 , HClCH3CH2OH I2 + NaOHCHI3 + CH3COONa
Yellow ppt
9. CH3NH2 & CH3NHCH3 –
CH3NH2 gives Hinsberg reagent test.
n
10. CH3NO2 & CH3ONO - Nitro on red give amine which
can be distinguish by Hinsberg reagent test & other
can’t be distinguish.
CH3NO2 Sn + HCl CH3NH2
CH3ONO Sn + HCl CH3OH
0
11.
–NH2 &
-NHCH3 : - Aniline is a 1 amine &
N-methyl aniline is a 20 amine.Hence can be
distinguished by Hinsberg reagent test.
12. CH3CH2NH2 &
-NH2 – Aniline give azo dye test
whereas CH3CH2NH2 does not.
-NH2
NaNO2 , HCl
+
-N2 Cl
-
-OH
-N=N-OH
Orange dye
14. CH2=CH-Cl & Cl-CH2-CH=CH2 :- 3-Chloropropene on
reaction with aq. KOH follo wed by AgNO3 give white
ppt of AgCl.No such ppt is observed in case of
CH2=CH-Cl.
aq. KOH
Cl-CH2-CH=CH2
OH-CH2-CH=CH2 + KCl
AgNO3
AgCl + KNO3
White ppt
13. CH3Cl &
-Cl - Chlorobenzene will not give above
test .but chloromethane give this test.
PRADEEP SHARMA, INSTITUTE OF COMPETITIVE STUDIES, SECTOR – 15 , SONEPAT
CONTACT NUMBER : 0130 – 2231322 , E – mail : pradeepsharma1976@gmail.com