OXFORD INTERNATIONAL PUBLIC SCHOOL
CLASS-XII ( SCIENCE)
HOLIDAY HOME WORK
Sl.no SUBJECT DETAILS
1. English Prepare a fileon the ideal of your lkife with creative writing and
beautiful drawings.
2 Biology Instruction—
1. Select one project for holiday homework.
2. Use filepages.(don’t use rough pages)
3. You can use this below url to get more instruction or answer
https://studyres.com/doc/1861347/investigatory-project-for-class-xii-
biology
4.important project number-5,6,7,12,13,14,15
5.after selecting the project ,send me message(your name,project name)before
starting the project.
INVESTIGATORY PROJECT FOR CLASS XII BIOLOGY
1. Project Report on Malnutrition
2. Biology Project Report on Components of Food
3. Biology Project Report on DNA Fingerprinting
4. Project Report on Pollution
5. Biology Project Report on ABO blood grouping in human beings
6. Biology Project Report on the dispersal of seeds by various agencies
7. Biology Project Report on mosquito species -major diseases caused by
it
8. Biology Project Report on Human diseases
9. Project Report on sleeping Habits in human beings
10. Biology Project Report on Manures and Chemical Fertilizers
11. Project Report on Useful Plants and Animals
12. Biology Project Report on Cancer
13. Biology Project Report on aids
14. Biology Project Report on malaria
15. BIOLOGY PROJECT REPORT ON IMMUNE SYSTEM
3
Physical
Education
Work revise all course done till now
4. Information
Practice
Do all practical question related to data frame and series in practical
5. Physics. Do the all question of N C E R T chapter 1 and 2. Also solve the question
given in page 2-9.
6. Chemistry Revise all prev. work and sir gives a separate PDF fill of problem Sheet.
7. Maths Refer to page 10-15
PHYSICS WORK SHEET
CHAPTRE -1
CHAPTRE – 2
MATHS PROB. SHEET
TARUN KUMAR ASIWAL M.Sc. CHEMISTRY , SLET , CSIR-NET , LL.B. , D.Cl. 8764333388 ,9414304777
CARBONYL COMPOUNDS 1
CARBONYL COMPOUNDS
EXERCISE # 1
Questions
based on Preparation of Aldehyde and Ketone
Q.1 The hydrolysis of benzal chloride gives–
(A) Benzyl alcohol
(B) Benzoic acid
(C) Benzaldehyde
(D) Benzophenone
Sol.[C]
CH
Cl
Cl
Benzal Chloride
Hydrolysis
HOH
CH OH
OH
–H2O
CHO
Benzaldehyde Therefore option (C) is correct.
Q.2 Ketones are first oxidation products of–
(A) Primary alcohols
(B) Secondary alcohols
(C) Both
(D) None
Sol.[B] Factual
Q.3 Calcium acetate when dry distilled gives–
(A) Formaldehyde
(B) Acetaldehyde
(C) Acetone
(D) Acetic anhydride
Sol.[C] Factual
Q.4 What is the function of BaSO4 in Rosenmund
reactionn–
(A) To stop further oxidation of aldehyde
copletely
(B) To stop further reduction of aldehyde
copletely
(C) Act as a poisonous catalyst
(D) It checks the reactivity of Pd.
Sol.[D] It checks the reactivity of Pd.
Q.5 Toluene on treatment with CrO2Cl2 gives –
(A) Chlorotoluene
(B) Benzyl chloride
(C) Benzaldehyde
(D) Benzoic acid
Sol.[C]
CH3
2CrO2Cl2
(CCl4)
CH3
CrCl2OH
OCrCl2OH
H2O
CHO
Toluene Brown Complex Benzaldehyde
This reaction is called "Etard's reaction".
Chemical Properties of Aldehyde & Ketones (Aldol Condensation, Cannizzaro reaction, Iodoform test)
Questions
based on
Q.6 HCHO with conc. alkali forms two
compounds. The change in oxidation number
would be –
(A) (0 to –2) in both the compounds
(B) (0 to +2) in both the compounds
(C) (0 to +2) in one compound and (0 to –2)
in the second compound
(D) All are correct
Sol.[C]
2.HCHO
[O]
alkali (NaOH)
22–O||
NaOCHOHCH3
HCHO
O||
NaOHC chang inoxidation
number = (0 to +2)
HCHO = CH3–OH – Changin oxidation number
= (0 to – 2)
Therefore option (C) is correct.
Q.7 C3H8O ]O[
SOH/OCrK 42722
C3H6O
.)aq(NaOHI2 CHI3,
In this reaction the first compound is –
(A) CH3CH2CH2OH
TARUN KUMAR ASIWAL M.Sc. CHEMISTRY , SLET , CSIR-NET , LL.B. , D.Cl. 8764333388 ,9414304777
CARBONYL COMPOUNDS 2
(B)
OH|
CHCHCH 33
(C) CH3OCH2CH3
(D) CH3CH2CHO
Sol.[B] CH3–
OH|
–CH CH3 ]O[
H/OCrK 722
O||
CHCCH 33
I2 + NaOH
CHI3 + CH3 –
O| |C –Na
Yellow ppct
Therefore option (B) is correct.
Q.8 CH3–CH2–CHOalkali
.Dil Product. The
product in the above reaction is–
(A) CH3–CH2COOH
(B) CH3–CH2–CH2OH
(C) CH3–CH2–
OH|CH –CH2–CHO
(D) CH3–CH2–
OH|CH – CH
|CH3
CHO
Sol.[D] CH3–CH2–
O| |C –H+CH3–
H|CH –CHO
dil.alkali
CH3–CH2–
OH|CH – CH
|CH3
–
O| |C –H
(Aldol)
–
O| |C –H
This reaction is called Aldol Reaction.
Therefore option (D) is correct.
Q.9 In the Cannizzaro reaction given below –
2Ph–CHO OH
Ph–CH2OH + PhC
the slowest step is –
(A) The attack of OH¯ at the carbonyl group
(B) The transfer of hydride to the carbonyl
group
(C) The abstraction of proton from the
carboxylic group
(D) The deprotonation of Ph - CH2OH
Sol.[B] 2Ph–CHO OH
Ph–CH2OH + PhCO2–
Mechanism
Step I :
Ph – C| |O
–H + OH Ph –
OH|
H–C|O
–H
Conjugate base of hydrate
of aldehyde.
Step II :
Intermolecular hydride ioni transfer
Ph –
OH|
–C|O
H + H– C| |O
–Ph Slow
Ph – C| |O
–OH + H–
H|
–C|O
Ph
Step III :
Ph – C| |O
–OH+NaOH Ph– C| |O
–O– +Na
+ + H2O
Ph–
H|
–C|O
H + H2O Ph –
H|
–C|OH
H
In this reaction mechanism H-transfer process
is slow and difficult :
Therefore option (B) is correct.
Q.10 Which of the following statements is wrong –
(A) All methyl ketones give a positive
iodoform test.
(B) Acetaldehyde is the only aldehyde that
gives iodoform test.
(C) All secondary alcohols give positive
iodoform test.
(D) Any alcohol that can be oxidised to an
acetyl group gives a positive iodoform test.
Sol.[C] These compound which contain CH3–
O| |C
TARUN KUMAR ASIWAL M.Sc. CHEMISTRY , SLET , CSIR-NET , LL.B. , D.Cl. 8764333388 ,9414304777
CARBONYL COMPOUNDS 3
or CH2–
OH|CH – group give a positive iodoform
test. So all secondary Alcohol can't be give
positive iodoform test therefore option (C) is
correct.
Q.11 The number of aldols formed by CH3CHO
and CH3–CH2–CHO is –
(A) 2 (B) 3
(C) 4 (D) 1
Sol.[C]
CH3
O| |C – H
+
CH3–CH2–
O| |C –H
CH3–
OH|CH CH2–
O| |C –H
CH3–CH2–
OH|CH –CH
|CH3
O| |C –H
CH3–
OH|CH –CH
|CH3
O| |C –CH3 (Cross Aldol)
CH3– CH2
OH|CH CH2 –
O| |C –H (Cross Aldol)
Four Aldols formed by CH3CHO and
CH3CH2CHO.
Therefore option (C) is correct.
Q.12 2D– C|D
= O + OH¯ Cannizzaro
X and Y
(Y is alcohol, D is deuterium)
X and Y will have structure –
(A) D –
H|C|D
– O , D – C| |O
– OH
(B) D –
D|C|D
– O , D – C| |O
– OH
(C) H – C| |O
– O , D –
D|C|D
– OH
(D) None is correct
Sol.[D] Step I :
D – C| |O
– D + OH
D –
OH|
–C| |O
D
Step II :
D –
OH|
–C|O
D + D – C| |O
– D Slow
D –
OH|C| |O
D –
D|
DC|O
Step III :
D– C| |O
–OH + NaOH D – C| |O
–O + Na
+ H2O
D–
D|
–C|O –
+ H2O D –
D|
–C|OH
D
D|
DCDOCD
|| |OHO
Therefore option (D) is correct.
Q.13
3
23
O
S)CH( A
HO B; ‘B’ is –
(A)
(CH2)4
CHO
CHO (B)
OH
OH
(C)
OH
CHO
(D)
CHO
Sol.[B]
TARUN KUMAR ASIWAL M.Sc. CHEMISTRY , SLET , CSIR-NET , LL.B. , D.Cl. 8764333388 ,9414304777
CARBONYL COMPOUNDS 4
S)CH( 22 +O3
H|C = O
H|C = O
[A]
Intermolecular
Aldol Condensation
CHO
OH
Aldol
CHO
+ H2O
OH
Therefore option (B) is correct.
Q.14 Compound (A) C7H10, on ozonolysis gives
only one compound (B)
C7H10 O3
(ii) Zn/H2O
O
H–C C
O
H
the structure (A) is
(A)
CH2=CH CH=CH2
(B)
(C)
(D)
Sol.[D]
Q.15 In Which of the following crossed aldol
condensations, only one kind of cross aldol is
formed –
(A) CH3CHO & CH3CH2CHO
(B) CH3CHO & (CH3)2CO
(C) (CH3)2CO & (C2H5)2CO
(D) C6H5CHO & CH3CHO
Sol.[D] C6H5–
O| |C –H+
H| |CH2 –
O| |C –HC6H5–
OH|CH –CH2–
O| |C –H.
LH. absent Aldol
In ths reaction only one kind of cross is formed,
because Benzaldehyde can't be have H.
Therefore option (D) is correct.
Q.16 In the given reaction final product(s) will be -
Na/liq. NH3
(excess) (A) O3, Zn
H2O (B)
CH3
(A)
CH3–C–CH2–CHO, CH2 CHO
CHO
O
(B)
CH3–C–C–CH3,
CHO
CHO
O O
(C)
O
,
C=O
CH3
CH3
(D) (CH2)3
CHO
CHO
, O
Sol.[B]
CH3–C–C–CH3,
CHO
CHO
O O
Q.17 Alkaline hydrolysis of C4H8Cl2 gives a
compound (A) which on heating with NaOH
and I2 produces a yellow precipitate of
CHI3.The compound (A) should be.
(A) CH3CH2CH2CHO
(B) CH3CH2–C–CH3
O
(C) CH3 – CH2 – CH – CH2
OH OH
(D) CH3 – CH2 – CH – CH2
OH Sol.[B] CH3CH2–C–CH3
O
Q.18 The following statement is true for
Cannizzaro reaction–
(A) The aldehyde is oxidised as well as
reduced
(B) The aldehydes not containing -
Hydrogen atoms give the reaction
TARUN KUMAR ASIWAL M.Sc. CHEMISTRY , SLET , CSIR-NET , LL.B. , D.Cl. 8764333388 ,9414304777
CARBONYL COMPOUNDS 5
(C) The reaction is not given by aldehydes
containing -Hydrogen atoms
(D) All of these
Sol.[D] All of statement are correct
Q.19 Which of the following compounds would
under go the Cannizzaro’s reaction–
(A) Acetaldehyde (B) Benzaldehyde
(C) Propionaldehyde (D) Anisole
Sol.[B] Benzaldehyde would under go the
Cannizzaro’s reaction
Q.20 Product obtained on the addition of an
aqueous alkali to benzaldehyde followed by
acid hydrolysis is–
(A) Benzoic acid (B) Benzyl alcohol
(C) Benzyl benzoate (D) All of the above
Sol.[D] All
Questions
based on Test of Aldehyde acid Ketone
Q.21 Which of the following compounds would
not form a silver mirror with Tollen’s
reagent–
(A) R.CHO (B) Ar.CHO
(C) CH3.CO.R (D) R– C– C–H
O O
Sol.[C] CH3.CO.R would not form a silver mirror
with Tollen’s reagent
Q.22 Acetaldehyde on warming with Fehling
solution gives a red precipitate of –
(A) Elemental copper
(B) Cuprous oxide
(C) Cupric oxide
(D) Mixture of all of the above
Sol.[B] Aldehyde on warming with fehling solution
give a red precipitate of cuprous oxide
RCHO+2Cu(OH)2+NaOH
RCOONa+CH2O+ 3H2O
red ppt.
Therefore option (B) is correct.
Q.23 If acetaldehyde is treated with Benedict’s or
Fehling’s solution, the following change
occurs in the system –
(A) Ag+ Agº (B) Cu
+2 Cuº
(C) Cu+2
Cu+ (D) Na
+ Naº
Sol.[C] Fehling solution
R–CHO+2Cu(OH)2 + NaOH
Cu+2
RCOONa + Cu2O + 3H2O
Cu+
Benedict's solution
RCHO + Cu+2
(complex) CH2O + Oxidaton
Red ppt product
In both of there reaction Cu2+
convert in Cu+1
Therefore option (C) is correct.
Q.24 For distinction between CH3CHO and
C6H5CHO the reagent used is –
(A) KCN (B) HCN
(C) NH2OH (D) PCl5
Sol.[A] Acetaldehyde react with KCN to form
acetaldehyde cynohydrine.
CH3 – CHO HCl
KCN CH3 –
OH|
–C CN.
While Benzaldehyde react with KCN to form
Benzoin.
2
O| |C –H
Benzaldehyde
KCN C| |O
– –C|OH
Benzoin Therefore option (A) is correct.
Q.25 CH3–CH=CH.CHO is oxidised to
CH3CH=CH.COOH by using–
(A) Alkaline KMnO4
(B) Selenium dioxide
(C) Osmium tetraoxide
(D) Ammonical AgNO3
Sol.[D] Factual
Q.26 Acetaldehyde and acetone can be
distinguished by all the following except–
(A) Iodine + alkali (B) Tollen’s reagent
(C) Fehling solution (D) Schiff reagent
Sol.[A] Acetaldehyde and acetone can be
TARUN KUMAR ASIWAL M.Sc. CHEMISTRY , SLET , CSIR-NET , LL.B. , D.Cl. 8764333388 ,9414304777
CARBONYL COMPOUNDS 6
distinguished by Tollen’s reagent, Fehling
solution, Schiff reagent
Q.27 Schiff’s reagent is–
(A) Magenta solution decolourised with
sulphur dioxide
(B) Magenta solution decolourised with chlorine
(C) Ammonical cobalt chloride solution
(D) Ammonical manganese sulphate solution
Sol.[A] Factual
Q.28 Which of the following does not turn schiff's
reagent to pink –
(A) Formaldehyde
(B) Benzaldehyde
(C) Acetone
(D) Acetaldehyde
Sol.[C] Aceton can't be turn schiff's reagent to pink.
Those carbonyl compound which contain
Aldehyde group give the pink colour to react
with schiff's reagent.
Therefore option. (C) is correct.
Questions
based on Other Chemical Reactions
Q.29 An aldehyde isomeric with allyl alcohol
gives phenyl hydrazone. Pick out a
ketone that too gives a phenyl
hydrazone containing the same percentage
of nitrogen –
(A) Methyl ethyl ketone
(B) Dimethyl ketone
(C) 2– Butanone
(D) 2– Methyl propanone
Sol.[B]
CH2 = CH – CH2 – OH CH3CH2 – CHO
Allyl alcohol (Propanol)
Propanole is isomer of Allyl alcohol.
CH3 – CH2
H
C = O + H2–N–NH–
CH3–CH2
H C = N–NH–
Phenyl hydrozene
Phenyl hydrazone of propane
Both of hydrazone containe same percentage of
Nitrogen. Therefore option (B) is correct.
Q.30 C2H2
4
42
HgSO
SOH Dil P (O)
QPyridine
2SOCl
R Cd)H(C 252 S
The end product in the above sequence of
reactions is –
(A) Ethylethanal
(B) 2-butanone
(C) Propanal
(D) Propanone
Sol.[B] CH2=CH2 H/HgSO4
)P(3CHOCH
]O[
)Q(O| |
OHCCH3
Pyridine
SoCl 2
)R(O| |
OCCH3
(C2H5)2Cd
CH3CH2–
O| |C –CH2–CH3
(S)
2-butanone Therefore option (B) is correct.
Q.31 ? CN,
O2H,EtOH Benzoin.
The reactant is obtained by dry distillation
of the calcium salts of the following
pairs –
(A) C6H5CH2COOH, HCOOH
(B) C6H5COOH, HCOOH
(C) C6H4 (OH)COOH, HCOOH
(D) C6H4 (NH2)COOH, HCOOH
Sol.[B]
C6H5COOH + H
O| |C OH
Benzoic acid Formic acid
C6H5CHO
C2H5OH
CN–
(Benzoin) C6H5–
O| |C –
OH|CH –C6H5
This reaction is called. "benzoin condensation".
TARUN KUMAR ASIWAL M.Sc. CHEMISTRY , SLET , CSIR-NET , LL.B. , D.Cl. 8764333388 ,9414304777
CARBONYL COMPOUNDS 7
Therefore option (B) is correct.
Q.32 The product of the reaction between acetone
and ethylene glycol is –
(A)
CH3
CH3
C
O–CH2
O–CH2
(B)
OHOH||CHCHCH 23
(C) CH3–CH–CH–CH2
O (D) None of these
Sol.[A]
CH3
CH3
C = 0 + 2
2
CH—HO|
CHOH
Ethylene glycol Acetone
CH3
CH3
C
Cyclic Ketals
O
O
CH2
CH2
Therefore option (A) is correct.
Q.33 Main product obtained by the oxidation of
C2H5COC3H7 by acidified dichromate –
(A) C3H7 COOH (B) C2H5COOH
(C) C3H8 (D) CH3COOH
Sol.[B] When oxidation of unsymmetrical Ketone occur
product is formed according to popaff's rule.
According to this rule the -CH group belongs
to the alkyl group which has high number of
carbons
CH3–CH2–CH2–
C| |O –CH2–CH3
4KMnO
OH
CH3–CH2COOH + CH3CH2COOH.
Therefore option (B) is correct.
Q.34 Following is chemoselective for the reduction
of carbonyl compounds to corresponding
alcohol –
(A) LiAlH4
(B) {(CH3)2CHO} Al / (CH3)2CHOH
(C) B2H6
(D) H2/Pt
Sol.[A] LiAlH4 is more powerful reducing agent it
reduces corbonyl compound to corresponding
alcohol
CH3–
O| |C –H 4LiAlH
CH3–CH2–OH
Aldehyde 1º Alcohol
CH3–
O| |C –CH3 4LiAlH
CH3–
OH|CH –CH3
This reagent has no effect on carbon-carbon
multiple bond
CH3 – CH = CH –
O| |C – H 4LiAlH
CH3 – CH
= CH – CH2 – OH
Therefore option (A) is correct.
Q.35
COCH3
22 NHNH)i(
KOHalcoholic)ii(product X
X is –
(A)
C=N NH2
|
CH3
(B)
CH2CH3
(C) Both correct
(D) None is correct
Sol.[B]
O| |C –CH3
22
2
NHNH)i(
OH
3
2
CH||
NH–NC
KOH.Alco
]H[4
3
2
CH|CH + NH2 – NH2
[X] Therefore option (B) is correct.
Q.36 CH3CHO 352 )HAl(OCA
ONaHC 52
B ; ‘A’ and ‘B’ are –
(A) CH3COOC2H5, CH3COCH2COOC2H5
(Tischenko) (Claisen)
(B) CH3COOH, CH3COOC2H5
(C) CH3COOH, C2H5OH
(D) All incorrect
Sol.[A]
TARUN KUMAR ASIWAL M.Sc. CHEMISTRY , SLET , CSIR-NET , LL.B. , D.Cl. 8764333388 ,9414304777
CARBONYL COMPOUNDS 8
CH3CHO 352 )HOC(Al
CH3COOC2H5
(Tischenko)
[A]
aNaHC 52 CH3COCH2COOC2H5
(Claisen)
[B]
Therefore option (A) is correct.
Q.37 Which one of the following reactions can be
used to distinguish between benzaldehyde
and benzyl alcohol –
(A) KMnO4/oxidation (B) CrO3 oxidation
(C) Sodium metal (D) Flame test
Sol.[C] Benzylalcohol react with sodium metal and
liberate hydrogen gas. While benzaldehyde can't
be react with 'Na' metal
2C6H5–CH2OH + 2Na C6H5CH2–ONa + H2
Therefore option (C) is correct.
True / False type questions
Q.38 Acetone gives iodoform test whereas HCHO
does not.
Sol. CH3
O| |C –CH3 (Acetone) contain, CH3–
O| |C – unit
so it gives iodform test whereas H–
O| |C –H can't
be contain CH3–
O| |C – or CH3–
OH|CH group so it
does not give iodoform Test. Therefore statement is True.
Q.39 wolff-Kishner reduction of acetophenone
gives toluene.
Sol.
CH3
C6H5
C = O OH/NHNH 22 CH3–CH2–C6H5
Acetophenone Ethyl Benzyne Therefore statement is false.
Q.40 Clemmensen reduction is more suitable for
reduction of those carbonyl compounds
which are sensitive to acids.
Sol. R– C| |O
–R ConHCl/HgZn
R–CH2–R This method is not used for acid sensitive
compound for Example : the given conversion can't be possible with reaction
Zn–Hg/HCl
ConHCl/HgZn
C| |O
–CH3
HO HO
CH2–CH3
CH2–CH3
HO Therefore statement is false.
Q.41 Phenylacetaldehyde on treatment with conc.
NaOH undergoes cannizzaro’s reaction.
Sol. Phenylacetaldehyde (C6H5–CH2–
O| |C –H) contain
-Hydrogen So it does not give Cannizzaro's Reaction. Therefore statement is false.
Q.42 Acetaldehyde on treatment with aqueous
NaOH solution gives sodium acetate and
ethyl alcohol.
Sol. Acetaldehyde on treatment with aqueous NaOH
solution gives Aldol compound.
CH3–
O| |C –H+
H|CH2 –
O| |C –H
NaOH
CH3–
OH|CH2 –CH2–
O| |C –H
Aldol
Therefore statement is false.
Fill in the blanks type questions
Q.43 2-Pentanone can be differentiated from
3-pentanone by ...........
Sol. Iodoform test.
Q.44 When 2-butyne is hydrated with dil. H2SO4 in
presence of HgSO4, it gives ...........
Sol. 2-butanone.
Q.45 Cannizzaro reaction is given by aldehydes
which do not have .........
TARUN KUMAR ASIWAL M.Sc. CHEMISTRY , SLET , CSIR-NET , LL.B. , D.Cl. 8764333388 ,9414304777
CARBONYL COMPOUNDS 9
Sol. -hydrogen.
Q.46 Fehling solution ‘A’ consists of an aqueous
solution of copper sulphate while Fehling
solution ‘B’ consists of an alkaline solution
of ..........
Sol. Rochelle Salt.
Q.47 Ethanol vapour is passed over heated copper
and the product is treated with aqueous
NaOH. The product is ...........
Sol. 3-hydroxy butanol.
EXERCISE # 2
Part-A (Only single correct answer type
Questions)
Q.1 In a Cannizzaro reaction the intermediate
which is the best hydride donor is –
(A) C6H5–
OH|
–C|
H
O–
(B) C6H5–
–O
|–C
|H
O–
(C)
O2N
H
O–
O–
(D)
CH3O
H
O–
O–
Sol.[D] – OCH3
is an electron releasing group this group
(– OCH3
) facilitates the release of hydride ion.
Therefore option (D) is correct.
Q.2 C6H5COCl2
4
H
BaSOPd
Intermediate
Oxidation
IntermediateonDistillatiDry
SaltCa
A
Compound (A) in above reaction sequence
is–
(A) Benzophenone
(B) Benzaldehyde
(C) Acetophenone
(D) Benzoquinone
Sol.[A] C6H5COCl 2
4
H
BaSOPb
C6H5–
O| |C –H
]O[
C6H5–
O| |C –OH
Ca-salt
Dry Distillation C6H5–
O| |C –C6H5
[A] Benzophenone
Therefore (A) is correct.
Q.3 In the reaction series –
CH3CHO 4
42
KMnO
SOH.dil P 2SOCl
Q
Heat
COONaCH3 R. The product R is –
(A) (CH3CO) 2O
(B) Cl. CH2COOCOCH3
(C) CH3COCH2COOH
(D) Cl2. CHCOOCOCH3
Sol.[A] CH3CHO + 42
4
SOH.dil
KMnO
]P[O| |
OHCCH3 2SOCl
]Q[O| |
ClCCH3
CH3COONa
+NaCl CH3 C
| |O
CH3–
O| |C
O
(R) Therefore optioin (A) is correct.
TARUN KUMAR ASIWAL M.Sc. CHEMISTRY , SLET , CSIR-NET , LL.B. , D.Cl. 8764333388 ,9414304777
CARBONYL COMPOUNDS 10
Q.4
CHO
OH
forms
COOH
OH
on reaction
with –
(A) K2Cr2O7 + H2SO4
(B) Ag2O/OH–
(C) KMnO4
(D) All of these
Sol.[B]
CHO
OH
OAg2
COOH
OH
+2Ag ppt
Therefore option (B) is correct.
Q.5 A + B forms
N
N
A and B are –
(A)
NH2
NH2
, CH3– CHO
(B) CH3CHO, CH3 - NH2
(C)
22
22
NHCH|
NHCH
,
CHO|CHO
(D) HCHO, CH3NH2
Sol.[C]
2
2
CH|CH
N–H2 + O = CH
N–H2 + O = O = CH
[A] [B]
–2H2O
2
2
CH|CH
N
N
CH
CH
Therefore option (C) is correct.
Q.6 The following preparation of caprolactum
form the oxime of cyclohexanone involves a
rearrangement
called –
NOH
NH
C = O
(A) Pinacol-Pinacolone rearrangement
(B) Claisen rearrengement
(C) Beckmann rearrangement
(D) Curtius rearrangement
Sol.[C]
N–O–H
N
C – O – H
Tautomerisation
N
C = O
This rearrangement called.
Beckmann rearrangement.
Therefore option (C) is correct.
Q.7 B 4NaBH CH=CH–CHO
/PtH2 A, then A & B are-
(A)
CH2CH2CHO,
CH=CH–CH2OH
(B)
CH2CH2CH2 OH,
CH=CH–CH2OH
(C)
CH=CH–CH2OH in both cases
(D)
CH2CH2CH2OH in both cases
Sol.[B]
CH=CH–CH2–OH 4NaBH
[B]
CH=CH–CHO Pt/H2
CH2–CH2–CH2–OH
[A]
Therefore option (B) is correct.
Q.8 Compound (A) on ozonolysis gives the
following compound
OCH|||
CHO—)(CH—C—CH—CH—CH
3
4223struct
TARUN KUMAR ASIWAL M.Sc. CHEMISTRY , SLET , CSIR-NET , LL.B. , D.Cl. 8764333388 ,9414304777
CARBONYL COMPOUNDS 11
ure of compound (A) is -
(A)
CH2–CH
CH3
CH3
(B)
CH2–CH
CH3
CH3
(C)
CH2–CH
CH3
CH3
(D)
CH2–CH2–CH
CH3
CH3
Sol.[C]
CH2–CH
CH3
CH3
Part-B (One or more than one correct
answer type Questions)
Q.9 Aldehydes and ketones give addition
reactions with –
(A) HCN
(B) NaHSO3
(C) RMgX
(D) ROH
Sol.[A,B,C,D]
Aldehydes and ketones give addition Reaction
with HCN, NaHSO3
RMgx and R–OH
C = O + HCN
C OH
CN
C = O + NaHSO3
C OH
NaSO3
C = O + RMg
C R
OMg
C = 0 + R–OH
C R
OR Therefore option (A), (B), (C) and (D) are
correct.
Q.10 Which of the following reaction are used for
detection presence of carbonyl group?
(A) Reaction with hydroxylamine
(B) Reaction with hydrazine
(C) Reaction with phenylhydrazine
(D) Reaction with semicarbazide hydrochloride
Sol.[A,B,C,D]
Main reaction of carbonyl compounds is
Nucleophilic addition.
Reaction with ammonia derivatives are
nucleophile addition Reaction.
C = O + HO – NH2 C =N – OH
Hydroxylamine Oxime
C = O + NH2 – NH2 C = N – NH2
Hydrazine Hydrozone
C = O + NH2–NH– C = N–NH–
Phenyldrazine Phenylhydrazone
C = O + NH2–NH–
O| |C –NH2 C =N–NH–
O| |C –NH2
Semicarbazide Semicarbazone
Therefore option (A), (B), (C) and (D) are
correct.
Q.11 Which of the following are examples of aldol
condensation ?
(A) 2CH3CHO NaOH.dil
CH3CHOHCH2CHO
(B) 2CH3COCH3 NaOH.dil
CH3COH(CH3)CH2COCH3
(C) 2HCHO NaOH.dil
CH3OH+ HCOOH
(D) C6H5CHO + HCHO NaOH.dil
C6H5CH2OH
Sol.[A,B] Those compound gives Aldol condensation
which contain at least one -H.
Therefore option (A) and (B) are correct.
Q.12 Benzophenone (C6H5COC6H5) does not react
with –
(A) NaHSO3 (B) CH3OH
(C) HCN (D) NH2OH
Sol.[A,B,C] Benzophenon can not be react with NaHSO3,
CH3OH, and HCN, There option (A), (B) and
(C) are correct.
Q.13 Which of the following statements about
benzaldehyde is/are true ?
(A) Reduces Tollen’s reagent
(B) Undergoes aldol condensation
(C) Undergoes Cannizzaro reaction
(D) Does not form an addition compound
with sodium hydrogen sulphite.
Sol.[A,C]
TARUN KUMAR ASIWAL M.Sc. CHEMISTRY , SLET , CSIR-NET , LL.B. , D.Cl. 8764333388 ,9414304777
CARBONYL COMPOUNDS 12
Benzaldehyde Reduces the Tollen's reagent and it's undergoes cannizaro reaction because, it
can't be contain -H atom.
CHO NaOH.dil
CH2OH
+
O| |C Na
+
Therefore option (A) and (C) are correct.
Q.14 Which of the following compounds will give
a yellow precipitate with iodine and alkali ?
(A) 2-Hydroxypropane
(B) Acetophenone
(C) Methyl acetate
(D) Acetamide.
Sol.[A,B]
Those molecules is given yellow prepcipitate
will with I2 and alkali (Haloform Reaction).
Which contain CH3–
O| |C – unit or CH3–
OH|CH – in
the molecule attached to C or H.
CH3–
OH|CH –CH3 and CH3–
O| |C –C6H5 Give
2-hydroxy propane Acetophenone
halogorm Reaction but CH3–
O| |C –OCH3 and
CH3–
O| |C –NH2 can't be give haloform reaction
therefore option (A) mol (B) are correct.
Q.15 Base-catalysed aldol condensation occurs
with –
(A) Propionaldehyde
(B) Benzaldehyde
(C) 2-Methylpropanal
(D) 2, 2-Dimethylpropanal
Sol.[A] Base-catalysed aldol condensation occurs
with Propionaldehyde
Part-B (One or more than one correct
answer type Questions)
Q.16 Under wolff-Kishner reduction condition, the
conversions which may be brought about are–
(A) Benzophenone into diphenylmethane
(B) Benzaldehyde into benzyl alcohol
(C) Cyclohexanone into cyclohexane
(D) Cyclohexanone into cyclohexanol
Sol.[A,C] wolff-Kishner reagent reduce > C = O into > CH2
> C = O OH/NHNH 22 > CH2
neBenzopheno
O| |
HCCHC 5656
OH/NHNH 22
methaneDiphenyl
HCCHHC 56256
OH/NHNH 22
Cyclohexanone Cyclohexane Therefore option (A) and (C) are correct.
Part-C (Assertion & Reasoning type
Questions)
The following questions 17 to 21 consists of
two statements each, printed as Assertion
and Reason. While answering these
questions you are to choose any one of the
following five responses.
(A) If both Assertion and Reason are true and
the Reason is correct explanation of the
Assertion.
(B) If both Assertion and Reason are true but
Reason is not correct explanation of the
Assertion.
(C) If Assertion is true but the Reason is false.
(D) If Assertion is false but Reason is true.
(E) If Assertion & Reason are false.
Q.17 Assertion : Chloral forms a stable hydrate.
Reason : On treatment with an aqueous
solution of NaOH, chloral undergoes
hydrolysis to give chloroform and sodium
formate.
Sol. [B] Hydrate of Chloral is more stable due to Intra-
molecular H-bonding.
CCl3CHO NaOH
CHCl3 + H
O| |C –ONa
–
Chloroform Sodium Formate
Therefore option (B) is correct.
Q.18 Assertion : p-Methoxybenzaldehyde is less
reactive than benzaldehyde towards
TARUN KUMAR ASIWAL M.Sc. CHEMISTRY , SLET , CSIR-NET , LL.B. , D.Cl. 8764333388 ,9414304777
CARBONYL COMPOUNDS 13
cyanohydrin towards cyanohydrin formation.
Reason : + R-effect of the methoxy group
increases the electron deficiency of the
carbonyl carbon.
Sol. [C]
CH3 O
O| |C –H
C3
S(+)
+meffe. O| |C –H
C3
S(+)
Due to +m effect on the methoxy group
increases the electron deficiency of the carbonyl
carbon.
Therefore option (C) is correct.
Q.19 Assertion : CH3– adds to >C=O group
irrevesibly but CN– ion adds reversibly.
Reason : CH3– ion is much stronger
nucleophile than CN– ion.
Sol.[A] CH3– ion is much stronger nucleophile than CN
–
ion CH3– add to >C = O group irreversible.
Or –
O| |C – + CH3 –
O|
–C|CH3
–
O| |C – + CH3 –
O|
–C|CN
(+)
(–)
–
Therefore option (A) is correct.
Q.20 Assertion : Acetaldehyde reduces Fehling’s
solution but benzaldehyde does not.
Reason : Acetaldehyde is a stronger reducing
agent than benzaldehyde.
Sol. [A] Acetaldehyde is a stronger reducing agent than
benzaldehyde so acetaldehyde reduces Fehling's
solution but benzaldehyde does not. Therefore
option (A) is correct.
Q.21 Assertion : Both Grignard reagents and
dialkylcadmium react with acid chlorides to
form tert-alcohols.
Reason : Grignard reagents are more reactive
than dialkyl cadmium.
Sol. [B] Grignard reagents react with acid chloride to
form test-alcohols. but dialkyl cadmium react
with acid chloride to form ketone. Reactivity of
organometallic compounds on the electro
positive character of the metal Mg metal is more
electropositive than so the grignard reagent
(RMgx) are more reactive than dialkyl-
cadmium.
Therefore Assertion is false but reason is true
option (D) is correct.
Part-D Column Matching
Q.22 C8H8O 2INaOH
P + Q
C| |O
–H
NaOH.Conc Q + R
CH3– C| |O
–H 2INaOH
P + S
Column A Column B
(a) P (i) NaOCH||O
(b) Q (ii) CHI3
(c) R (iii) NaOCPh| |O
(d) S (iv) Ph – CH2OH
Sol. (a)-ii, (b)-iii, (c)-iv, (d)-i
Factual
Q.23 Column A Column B
Compound Enol %
(a)
O
O
(i) 99.9
(b) CHCl2CHO (ii) 90
(c) CH3–
O| |C –CH2–
O| |C –CH3 (iii) 2.5
(d) CH3 –
O| |C – H (iv) 1
Sol. (a)-i, (b)-iii, (c)-ii, (d)-iv
Factual
TARUN KUMAR ASIWAL M.Sc. CHEMISTRY , SLET , CSIR-NET , LL.B. , D.Cl. 8764333388 ,9414304777
CARBONYL COMPOUNDS 14
EXERCISE # 3
Part-A Subjective Type Questions
Q.1 Pure HCN fails to react with aldehydes and
ketones.
Sol. HCN is weak acid and have low degree of
dissociation but in presence of a base (even
H2O), the dissociation increases appreciably to
provide appreciable CN¯ to attack >C=O bond.
Q.2 Ketones are less reactive than aldehydes.
Sol. The positive I.E. of two alkyl groups in ketones
makes the carbon atoms less positive and makes
it less reactive in comparison to aldehydes.
R
H C = O
R
H C
+ – O
–
R
R C = O
R
R C
+ – O
–
Less positive charge and thus less reactive
nature.
Q.3 Oximes are more acidic than hydroxyl amine.
Sol. NH2OH NH2O¯
Acid Base
>C=N-OH>C= N..
-
.. :
..O >
C..
– N..
= ....O
Acid Base
In oxime base, delocalization of electron (i.e.
resonance) stabilize it and thus conjugate acid
i.e. oximes are more acidic. No such resonance
exists in hydroxyl amine base (NH2O–)
Q.4 In the laboratory preparation of
acetaldehyde by oxidation of ethanol,
acetaldehyde is distilled out continuously.
Sol. During oxidation of ethanol, acetaldehyde is
formed in I step which should be distilled out,
otherwise further oxidation of acetaldehyde will
lead the formation of acetic acid.
Q.5 Although both >C = O and >C = C<
groupings have double bond they exhibit
different types of addition reactions.
Sol. The >C = C< grouping undergoes electophilic
addition while > C = O undergoes nucleophilic
addition reactions. The difference is because of
the fact that C of >C = O is more electrophilic
than carbon of > C = C < due to more
electronegative nature of O then C. Hence of
> C = O reacts with nucleophiles. The >C = C <
grouping is nucleophilic and hence adds mainly
electrophiles.
Q.6 What happens when 2, 2-Dichloropropane is
treated with aqueous NaOH and the product
treated with amalgamated zinc in presence of
conc. HCl.
Sol.
H3C
H3C C
Cl
Cl
NaOHaq CH3
CH3
C=O HClConc
)Hg(Zn
CH3
CH3
CH2
Q.7 NaHSO3 is used for the purification of
carbonyl compounds.
Sol. Carbonyl compounds form solid additive
products with NaHSO3 which are separated out.
The solid bisulphites of carbonyl compounds on
hydrolysis by dil. acid regenerate original
carbonyl compound and thus this property is
used for the purification of carbonyl compounds
as well as for their separation.
Q.8 Protic acids, Lewis acids and bases all
activate carbonyl group.
Sol. Protic and Lewis acids increase the partial
positive charge on the carbonyl carbon and
hence activate the group.
>C = O + H+ > C
– OH
Bases activate the -methylene component of
the carbonyl compounds by converting them
into carbanions.
R - CH2 – CHO + : B R – HC
– CHO + BH
Note : Proton can be expelled only from -
carbon because such carbanion can stabilize due
to resonance.
Q.9 Differentiate between acetophenone and
benzophenone.
Sol. Acetophenone (C6H5. CO.CH3) gives haloform
test while benzophenone (C6H5. CO. C6H5) not.
Q.10 How will you prepare Diethyl ketone from
propionic acid : (2 steps).
Sol. CH3 .CH2 .COOH 2Ca(OH)
TARUN KUMAR ASIWAL M.Sc. CHEMISTRY , SLET , CSIR-NET , LL.B. , D.Cl. 8764333388 ,9414304777
CARBONYL COMPOUNDS 15
(CH3 .CH2COO)2Caondistillati
dry
CH3.CH2 .COCH2 .CH3
Q.11 An organic compound (A) with molocular
formula C3H6O is not easily oxidised. It gives
on reduction C3H8O (B) which on reaction
with HBr gives a bromide (C). Grignard
reagent obtained from (C) reacts with (A) to
give C6H14O (D). Identify compounds from
(A) to (D).
Sol. CH3COCH3 (Acetone) (A) CH3CHOHCH3 (Isopropyl alcohol); (B) CH3CHBrCH3 (Isopropyl bromide) (C) (CH3)2CH–C–(OH)(CH3)2(2,3-Dimethyl butanol-2)
(D)
Q.12 Two organic compounds ‘A’ and ‘B’ with
molecular formula, C3H6O, react with HCN
in different manner to produce, (C) and (D)
respectively on subsequent hydrolysis of (C)
and (D) gives optically active substances.
Sol. CH3CH2CHO; CH2 = CH CH2OH; (A) (B) CH3CH2CH(OH)CN; CH3 – CH(CN)CH2OH; (C) (D)
CH3CH2 HC*
(OH)COOH;
(E)
CH3 – HC*
(COOH) CH2OH
(F)
Q.13 An organic compound (A) adds Br2 to give
C5H8Br2O. It does not react with Tollen’s
reagent but enters into reaction with phenyl
hydrazine. Ozonolysis of (A) gives an
acetaldehyde and C3H4O2 which readily loses
CO to form acetaldehyde. What is (A) ?
Sol. CH3CH=CHCOCH3
(A)
Q.14 An organic compoud (A), C2Cl3HO, reduces
Tollen’s reagent and on oxidation gives a
monocarboxylic acid (B), C2Cl3HO2. (B) on
distillation with sodalime gave a sweet
smelling liquid (C) containing 89.12% Cl.
What are (A), (B) and (C).
Sol. CCl3CHO, CCl3COOH, CHCl3
(A) (B) (C)
Q.15 An organic compound (A) C9H10O is inert to
Br2 in CCl4. Vigorous oxidation with hot
alkaline permanganate yields benzoic acid.
(A) gives a precipitate with semicarbazide
hydrochloride and with 2,4-dinitrophenyl
hydrazine. Write all possible structures of
(A).
Sol. C6H5COCH2CH3; C6H5CH2COCH3;
(A) (B)
C6H5CH2CH2CHO; C6H5CH(CH3)CHO
(C) (D)
Q.16 An organic compound having molecular
formula C5H10O exists in two chain isomers,
(A) and (B). Isomer (A) undergoes the
Cannizzaro’s reaction to give 2, 2 - dimethyl
propanoic acid and 2, 2-dimethylpropanol-1.
Compound (B) in the presence of dilute alkali
undergoes aldol condensation to from 3-
hydroxy-2-propyl heptanal. Give graphic
representations of (A) and (B).
Sol. (CH3)3CCHO CH3CH2CH2CH2CHO
(A) (B)
Q.17 Two organic compounds (A) and (B) have
same empirical formula CH2O. Vapour
density of (B) is (A) twice the vapour density
of (A). (A) reduces Fehling solution but does
not react with NaHCO3. Compound (B)
neither reacts with NaHCO3 nor reduces
Fehling solution. What are (A) and (B) ? Also
report an isomer of (B) if it reacts with
NaHCO3.
Sol. (A) HCHO (Formaldehyde)
(B) HCOOCH3 (Methyl methanoate)
Q.18 An organic compound (A) with molecular
formula C3H6O is oxidised by Fehling
solution and gives silver mirror with Tollen’s
reagent. (A) gives on reduction C3H8O (B)
with hydrobromic acid followed by heating
with Mg (B) gives (C), which reacts with
ethylene oxide to give C5H12O (D). On
oxidation (D) gives C5H10O2 (E). Idientfy
compounds from (A) to (E).
Sol. CH3CH2CHO; CH3CH2CH2OH;
(A) (B)
TARUN KUMAR ASIWAL M.Sc. CHEMISTRY , SLET , CSIR-NET , LL.B. , D.Cl. 8764333388 ,9414304777
CARBONYL COMPOUNDS 16
CH3CH2CH2MgBr; CH3CH2CH2CH2CH2OH;
(C) (D)
CH3CH2CH2CH2COOH
(E)
Q.19 Write the structure of the product (s) formed
in each case.
(a)
C| |O
CH2CH3
Cl
22
2
ClCH
Cl
(b)
O
CH3
OH;NaOH
SHCHHC
2
256
(c)
O
CH3
+ LiCu (CH3)2 OH.2
etherdiethy l.1
2
(d)
O
+C6H5CHONaOH
,waterethanol
Sol. (a)
Cl|
CCHCH| |O
3Cl
(b)
SCH2C6H5
O
CH3
(c)
CH3
O
CH3
(d)
CHC6H5
O
Q.20 Suggest reasonable structure for compounds
A and B.
O
O
CH3
+ CH2=CH– C| |O
–CH3
Methanol
KOH A
O
O
CH3
CH2 CH2C| |O
CH3
Benzene
])CH(OC[Al 333 B
Sol. (A)
O
O
CH3
CH2CH2 (CO)CH3
(B)
O
O
CH3
Q.21 Identify the reagents appropriate for each step
in the following syntheses -
(a)
OH
OH
HC| |O
CH2CH2CH2CH2C| |O
H
CH| |O
(b)
CH3
CH(CH3)2
CH3C| |O
CH2CH2
23
2
)CH(CH|
CHCHCH| |O
(CH3)2CH 3CCH| |O
Sol. (a)
OH
OH
HIO4 H– C| |O
–CH2–CH2–CH2– C| |O
–H
NaOH/
C| |O
H
(b)
TARUN KUMAR ASIWAL M.Sc. CHEMISTRY , SLET , CSIR-NET , LL.B. , D.Cl. 8764333388 ,9414304777
CARBONYL COMPOUNDS 17
CH3
CH(CH3)2
O3 CH3– C| |O
–CH2–CH2–
23
2
)CH(CH|
CHCH C| |O
–H
NaOH/
C| |O
–CH3
H2O/Zn
Aldol
Condensation
(CH3)2CH
Q.22 Outline reasonable mechansims for each of
the following reactions.
(a)
CH2CH2CH2CH2Br
O
benzene
)CH(KOC 33
O
(76%)
(b) C6H5C| |O
C| |O
C6H5 + C6H5CH2C| |O
CH2C6H5
KOH
O
C6H5
C6H5 C6H5
C6H5
(91 - 96%)
(c) C6H5CH2C| |O
CH2CH3 + CH2=
56
56
HC|
HCCC| |O
OHCH
NaOCH
3
3
O
C6H5 CH3
C6H5
C6H5
(51%)
Sol. (a)
CH2CH2CH2CH2Br
O
benzene
)CH(OK 33
CH2CH2CH2CH2
O
H
O
–h
(b)
C6H5
O| |C
O| |C
C6H5
+
C6H5 CH2
O| |C
CH2 C6H5
KOH
Alodal condensation
C
C6H5
HO
C
C C
C6H5
O| |C
C6H5
H OH
C6H5
Alodal
–2H2O
C
C6H5
C
C6H5
C6H5 C6H5
O| |C
(c)
TARUN KUMAR ASIWAL M.Sc. CHEMISTRY , SLET , CSIR-NET , LL.B. , D.Cl. 8764333388 ,9414304777
CARBONYL COMPOUNDS 18
C6H5
C
CH2 O| |C
C6H5
+
CH2
C6H5 O| |C
CH2 CH3
Na2CO3
CH3OH
CH2
C6H5
O| |C
C|
HC 56
C–OH
C–H
CH3
C6H5
H|
CH
–H2O
C|
HC 56
CH2 C–C6H5
C–CH3 CH
C6H5 O| |C
Part-B Passage based Question
Passage - I (Q. 23 to 27)
Aldehyde without –H atoms on reaction with
concentrated alkali solution undergo an
oxidation-reduction (disproportionation)
reaction. One half of the aldehyde is reduced to
a 1° alcohol and the other is oxidised to a
carboxylic acid. This is known as Cannizzaro
reaction.
2–C–H
O
NaOH%50
– CH2OH +
–C–O–Na
O
(no –H)
Two steps are involved in this reaction.
(i) Attack of OH ion on the carbonyl group of
one aldehyde yields an oxyanion.
H–C–H + OH
O
H–C–H
O
OH
(oxyanion)
(ii) As a result of hydride transfer, the final
product is formed
H–C–H
O
OH
C = O H
H
+
H–C + CH3O
O
OH
Proton
exchange
H–C–ONa + CH3OH
O
– + Na
H–C + CH3OH
O
O
Intramolecular Cannizzaro’s reaction is also
possible
CHO
CHO
NaOH CH2OH
COONa
Q.23 The aldehyde which not shows Cannizzaro’s
reaction is -
(A) HCHO (B) C6H5CHO
(C) CCl3–CHO (D) (CH3)3C. CHO
Sol.[D] The aldehyde which has not -H shows
Cannizzaro’s reaction.
Q.24 The product formed in the following reaction
will be -
CHO
CHO
NaOH product -
(A) CH2OH
CH2OH (B)
CH2OH
COONa
(C)
COONa
COONa (D) All of these
Sol.[B] CHO
CHO
NaOH CH2OH
COONa
Q.25 Mixture of C6H5CHO and HCHO is treated
with NaOH then cannizzaro’s reaction
involves-
(A) Oxidation of HCHO
(B) Reduction of HCHO and Oxidation of
C6H5CHO
(C) Reduction of C6H5OH
(D) Both A & C
Sol.[D] Reduction of HCHO and Oxidation of
C6H5CHO, Reduction of C6H5OH
TARUN KUMAR ASIWAL M.Sc. CHEMISTRY , SLET , CSIR-NET , LL.B. , D.Cl. 8764333388 ,9414304777
CARBONYL COMPOUNDS 19
Q.26 Product (P) in the following reaction
Ph–C–CHO
O
OHHC
HO
52
(P)
(A) Lactic acid (B) Man delic acid
(C) Salicylic acid (D) Malonic acid
Sol.[C] Ph–C–CHO
O
OHHC
HO
52
Ph–C–COOH
O
Q.27 Identify A, B, C in following reaction
C8H6O2OHHC
HO/CH2
52
2 C8H10O2 3PBrC8H8
Br2
(A) (B) (C)
(A) CHO
CHO
CH2.OH
CH2.OH
CH2.Br
CH2.Br
(B) CHO
CHO
OH
OH
OH
Br
Br
Br
(C) CHO
CHO
CH3 OH
OH CH3
Br
Br
(D) All of the above
Sol.[D] All of the above
Passage - II (Q. 28 to 32)
Most common reaction for aldehyde &
ketones are nucleophilic addition reactions.
The reactivity of the carbonyl group arises
from the electronegativity of the oxygen atom
and the resulting polarisation of the C=O
bond. The electrophilic carbon atom is sp2
hybridised and flat, leaving it relatively
undinderd and open to attack from either face
of the double bond. As a nucleophile attacks
the carbonyl group, the carbon atom changes
hybridisation from sp2 to sp
3. The electrons o
the bond are forced out to oxygen atom to
form an alkoxide anion, which protonates to
give the product. A carbonyl group is a weak
base and it can become protonated in an
acidic solution. Protonated carbonyl group is
strongly electropilic, inviting attack by a
weak nucleophilic.
In most cases aldehydes are more reactive
than ketones towards nucleophilic addition.
This reactivity can be explained by electronic
effect and steric effect.
Q.28 In the following reaction the nucleophile that
attacks the ketone
33 CHCCH||O
4NaBH
OHCH–CH 23
33 CHCHCH|OH
(A) B – ··
(B) CH3–CH2– O – ··
(C) H – ··
(D) NaBH4
Sol.[C] Step I
CH3
CH3
C = O + NaBH4
CH3–C–O–BH3Na
|
CH3
|
H
Nucleophile
attack
CH3 –C–OBH3Na +
CH3 |
H
|
Step II
CH3 –
H|
–C|CH3
O BH3Na+B + CH3
CH3 C=O
CH3 –
H|
–C|CH3
O – B Na
4
H2O B(OH)3+NaOH+Cl3– –CH
|CH3
OH In this reaction the Nucleophile is H
–.
Therefore option (C) is correct.
Q.29 Which of the following reactions is feasible
favourable.
(A) CH3–CH=O+H2O H
CH3–CH OH
OH
TARUN KUMAR ASIWAL M.Sc. CHEMISTRY , SLET , CSIR-NET , LL.B. , D.Cl. 8764333388 ,9414304777
CARBONYL COMPOUNDS 20
(B) 33 CHCCH||O
+H2O H
OH
|CHCCH
|OH
33
(C)
C| |O
+H2O H
OH|C|OH
(D)
O + H2O H OH
OH
Sol.[D] Steric hindrance and mere alkyl substitutents
make carbonyl compounds less reactive towards
any nucleophile, electron-with drawing groups
and small rings make them more reactive. More
is the angle strain in the cyclic ketones, more is
it's reactivity for nucleophilic addition reactions.
So
O is more reactive than other
carbonyl compound. Which given in option.
Therefore option (D) is correct.
Q.30 What will be the product in the following
reactional
CH3– C| |O
–CH2– C| |O
–H COOHCH
BH)COOCH(Na
3
33 P
(A) CH3– C| |O
–CH2 –
H|
–C|
OH
H
(B) CH3– CH|
OH
–CH2– C| |O
–H
(C) CH3– CH|
OH
–CH2–CH2–OH
(D) CH3– CH|
OH
–CH2–COOH
Sol.[A]
CH3COCH2
H
C=O + + –
H|B
(CH3)3
CH3COCH2
H
C=O +
H|B
(CH3COO)3
CH3COCH2
H
C=OB
H
(CH3COO)3
CH3
O| |C –CH2–
H|
H–C|
OH
H–O–H
therefore option (A) is correct.
Q.31 Correct reactivity order towards nucleophilic
addition –
(A) HCHO > CH3–CHO > C6H5–CH2–CHO
(B) HCHO > CH3– C| |O
–CH3 > CH3–CHO
(C) HCHO > CH3– C| |O
–CH3 > C6H5–CHO
(D) HCHO > C6H5–CH2–CHO > CH3CHO
Sol.[D] Reactivity towards Nucleophilic addition a
groupofpowerI
1
– I group
So the reactivity of different carbonyl compounds in decreasing order.
H– O| |C
–H > I
56HC
– CH2–CHO > I3CH
CHO
Therefore option (D) is correct.
Q.32 Choose the correct statement about
cannizzarro’s reaction –
(A) Rate determining step involves
nucleophilic attak of _
OH
(B) Rate determining step involves the
nucleophilic attack of carbanion
(C) Rate determining step involves the
nucleophilic attack of hydride ion
(D) All of the above
Sol.[C] According to Reaction mechanism of Cannizzaro reaction
Step I
R–
O| |C –H + OH
R–
OH|
OH–C|O
Step II
TARUN KUMAR ASIWAL M.Sc. CHEMISTRY , SLET , CSIR-NET , LL.B. , D.Cl. 8764333388 ,9414304777
CARBONYL COMPOUNDS 21
R–
OH|
H–C|O–
+H – C| |O
–R Slow R– C| |O
–OH+H–
H|
R–C|
O
attack of hydride Step III
R– C| |O
–OH+NaOH R– C| |O
–O– + Na
+ H
R–
H|
–C|O
H + H2O R–
H|
–C|OH
H
Therefore in two reaction mechanism.Read
determining step (slow step) involves the nucleophilic
attack of hydride. So option (C) is correct.
Passage –III (Q. 33 to 37)
Aldehydes and ketons are specially
susceptible to nucleophilic addition because
carbonyl group C = O
is polar (due to
electronegativity difference between carbon
and oxygen).
C = O
–
Positive charge on carbon makes it reactive
towards the nucleophile. This addition is
catalysed by acid.
Reactivity of carbonyl compound towards
nucleopilic addition increases with increase in
the electron deficiency at carbonyl carbon.
Thus, (– I.E.) groups increase while (+ I.E.)
groups decrease the reactivity of carbonyl
compound.
Q.33 Which among the following is most reactive
to give nucleophilic addition ?
(A) FCH2CHO (B) CICH2CHO
(C) BrCH2CHO (D) ICH2CHO
Sol.[B] CICH2CHO is most reactive to give
nucleophilic addition
Q.34 Select the least reactive carbonyl compound
for nucleophilic addition -
(A)
C6H5–C–C6H5
CH3OH
O
(B)
C6H5–C–CH3
CH3OH
O
(C)
C6H5–C–H
CH3OH
O
(D)
CH3–C–H
CH3OH
O
Sol.[B]
C6H5–C–CH3
CH3OH
O
Q.35 Which one of the carbonyl compound is most
reactive towards NaCN/
H
(A) PhCHO
(B) CH3.O
CHO
(C) CH3
CHO
(D) NC
CHO
Sol.[A] PhCHO carbonyl compound is most
reactive towards NaCN/
H
Q.36 The most reactive compound towards
cyanohydrine formation is
(A) PhCHO
(B) NO2
CHO
(C) HO
CHO
(D) PhCH2CHO
Sol.[D] PhCH2CHO is the most reactive
compound towards cyanohydrine formation
Q.37 The increasing order of the rate of HCN
addition to compounds a – d is -
(a) HCHO (b) CH3COCH3
(d) PhCOCH3 (d) PhCOPh
(A) a < b < c < d (B) d < b < c < a
(C) d < c < b < a (D) d < a < c < b
Sol.[B] d < b < c < a
Passage –IV (Q. 38 to 40)
An organic compound (A) with molecular
formula C5H8 when treated with Na in liq.
NH3 followed by treatment with n-propyl
TARUN KUMAR ASIWAL M.Sc. CHEMISTRY , SLET , CSIR-NET , LL.B. , D.Cl. 8764333388 ,9414304777
CARBONYL COMPOUNDS 22
iodide yields C8H18(B). When treated with
dil. H2SO4 containing Hg2+
, (A) gives the
ketone C5H10O (C). On oxidation with
alkaline KMnO4, (B) gives two isomeric
acids C4H8O2 (D & E).
Q.38 Compound (C) may give following reaction -
(A) Cannizaro (B) Haloform
(C) Both (D) None
Sol.[B] Compound (C) is CH3–CH2–
O| |C –CH3.
and 2-pentanone
Therefore option (B) is correct.
Q.39 When (A) is treated with BH3/H2O2, HOH the
compound formed contains following group -
(A) an aldehyde
(B) a ketone
(C) an alcohol
(D) acarboxylic group
Sol.[A] CH3–CH2–CH2–CCH 223 OH/BH
OH|CHCHCHCHCH 2223
CH3–CH2–CH2–CH2–CHO
Pentanaldehyde
Therefore option (A) is correct.
Q.40 (D) & (E) on decarboxylation with soda lime
produces -
(A) same alkanes (B) different alkanes
(C) Isomeric alkanes (D) can't be predicted
Sol.[A] CH3–CH2–CH2–
O| |C –OH NOH/CaO
CH3CH2–CH2–CH3
Same alkane is formed therefore option (A) is
correct.
Passage - V (Q. 41 to 45)
An organic compound A(C11H14O) reacts
with 1 mole of Br2 to give C11H14Br2O. A on
treatment with aluminium isopropoxide in
presence of excess of P-benzoquinone to give
B with molecular formula C11H12O. B also
reacts with 1 mole of Br2 to C11H12Br2O. B
reduces Tollen’s reagent and when reacted
H2O2 followed by periodic acids produces
two compounds C(C8H8O) & (C3H4O3)D
reduces ammonical silver nitrate solution
forming malonic acid. C upon permaganate
oxidation gives E (C8H6O4) which is soluble
in aqueous alkali and on heating it give its an
hydride C with molecular formula C8H4O3.
Q.41 What is A in the above passage –
(A)
CH3
CH=CH–CH2–CH2–OH
(B) CH3
CH=CH–CH2–CH2–OH
(C) CH2–OH
CH=CH–CH2–CH3
(D) none of these
Sol.[B] A CH3
CH=CH–CH2–CH2–OH
Therefore option (B) is correct.
Q.42 Identify B
(A)
OH|
CHOCHCH–CH|OH
2
(B)
OHOH||
CHOCHCHCH 2
CH3
(C)
CH=CH–CH2–CHO
CH3
(D)
O||
CHC 3
CH3
Sol.[C] B
CH=CH–CH2–CHO
CH3
Therefore option (C) is correct.
Q.43 What is C in the given passage –
(A)
CHO CH3
(B)
CH2–OH CH3
(C)
OCH3
CH3
(D)
CHO
CHO
TARUN KUMAR ASIWAL M.Sc. CHEMISTRY , SLET , CSIR-NET , LL.B. , D.Cl. 8764333388 ,9414304777
CARBONYL COMPOUNDS 23
Sol.[A] C CH3
O| |C –H
Therefore option (A) is correct.
Q.44 What is D –
(A) CH3– C| |O
–COOH (B)
CHO|
COOHCH2
(C) CH2
COOH
CH2OH
(D) CH2
CHO
CHO
Sol.[D] HO–
O| |C – CH2 –
O| |C – H
Therefore option (B) is correct.
Q.45 What is E –
(A)
COOH
COOH
(B) CH3
COOH
(C)
O
C| |O
O| |C
(D)
COOH
COOH
Sol.[A] E
C| |O
–H
O| |C –H
Therefore option (A) is correct.
TARUN KUMAR ASIWAL(12 years experienced ) M.Sc. CHEMISTRY , B.Ed. , NET , SET -2012 , LL.B. , D.Cl,
8764333388 CARBOXYLIC ACID & IT'S DERIVATIVES 1
CARBOXYLIC ACIDS & IT'S DERIVATIVES
EXERCISE # 1
Questions
based on Preparation of carboxylic acid and
derivatives
Q.1 Give the order of decarboxylation of the
following acid –
CH3COOH ; CH2 =CH–CH2 – COOH ;
I II
CH2(COOH)2 ;
COOH O2N
NO2
NO2
III IV
(A) I > II > III > IV (B) III > IV > II > I
(C) IV > III > II > I (D) I > III > II > IV
Sol.[C]
CH3 – C – O CH3 + CO2
O unstable (I)
CH2=CH–CH2 –C–O
O Stable due to
conjugation (II)
CH2=CH–CH2 +CO2
H–O–C–CH2 –C – O
O
stabilised due to
resonance
HO–C –CH2
O O
C – O O2N
NO2
NO2 O
+CO2 O2N
NO2
NO2
more stable IV
Therefore stability of carbanion -decarboxy-
lation IV > III > II > I.
Option (C) is correct.
Q.2 The end product Y in the sequence of
reaction:
RX CN
X NaOH
Y is –
(A) An alkene
(B) A carboxylic acid
(C) Sod. salt of carboxylic acid
(D) A ketone
Sol.[C] RX CN
R–CN NaOHD2 R–
O| |C –H+NH3
NaOH
R–
O| |C –O
–Na
+
Sodium Salt of carboxylic acid.
Q.3 The oxidation of toluene with hot KMnO4
gives –
(A) Benzoic acid (B) Benzaldehyde
(C) Benzene (D) Benzyl alcohol
Sol.[A]
CH3
]O[
KmNOHot 4
CH3
+ H2O
Benzoic acid
Q.4 Identity the product Z in the series :
CH3CN OHHC/Na 52 X 2HNO
Y OH/KMnO4 Z
(A) CH3CHO
(B) CH3CH2CONH2
(C) CH3COOH
(D) CH3CH2NHOH
Sol.[C] CH3COOH
Questions
based on Chemical properties of carboxylic acids
and derivatives
Q.5 Which of the following acid does not form
anhydride –
(A) CH3COOH
(B) CH-COOH
CH-COOH
(C)
CH2COOH
CH2COOH
(D) CH-COOH
HOOC-CH Sol.[D] CH-COOH
HOOC-CH
does not form anhydride
TARUN KUMAR ASIWAL(12 years experienced ) M.Sc. CHEMISTRY , B.Ed. , NET , SET -2012 , LL.B. , D.Cl,
CARBOXYLIC ACID & IT'S DERIVATIVES 2
Q.6 Formic acid and formaldehyde can be
distinguished by treating with –
(A) Benedict’s solution (B) Tollen’s reagent
(C) Fehling’s solution (D) NaHCO3
Sol.[D] Formic acid is strong acid than formaldehyde
formic acid react with NaHCO3 and liberated
CO2 but formaldehyde does not react with
NaHCO3.
Therefore option (D) is correct.
Q.7 The product formed when adipic acid is
heated–
(A) COOH
(B)
C=O
(C)
O
O
O
(D)
COOH
COOH
Sol.[B]
O
CH2–CH2–C–OH
CH2–CH2–C–OH
O
O
+ CO2 + H2O
Adipic acid Cyclo pentanone
Adipic acid undergo dehydration as well as
decarboxylation to give cyclopentanone when
its heated.
So option (B) is correct.
Q.8 Ethyl acetate react with hydrazine to form -
(A) Acetamide and urea
(B) Acetic acid hydrazide and ethanol
(C) Hydroxamic acid and ethanol
(D) Ethyl isocyanate
Sol.[B]
O
CH3–C–OC2H5 + NH2–NH2
O
CH3–C–NH–NH2 + C2H5OH
Acetic acid
hydrazide
Ethanol
Therefore option (B) is correct.
Q.9 X and Y in the following reaction sequence
are–
X 52OP
R– C||O
–NH2OH
Br2 Y
(A) R–CN, RCH2NH2
(B) RNH2, RCN
(C) RCN, RNH2
(D) RCN, RCH2NH2
Sol.[C] H2O+R–CN 52OP
R– C| |O
–NH2 OH
Br2 R–NH2
[X] [Y]
Therefore option (C) is correct.
Q.10 Benzoic acid gives benzene on being heated
with X and phenol gives benzene on being
heated with Y. Therefore X and Y are
respectively –
(A) Soda lime and copper
(B) Zinc dust and sodium hydroxide
(C) Zinc dust and soda lime
(D) Soda lime and zinc dust
Sol.[D]
COOH
)elimsoda(
CaO/NaOH
+ CO2
OH
dustZn
+ ZnO
Therefore X and Y are respectively sodalime
Zn-dust.
Option (D) is correct.
Q.11 COOH
is converted into
CHO
by–
(A) (i) LiAlH4 (ii) Cu/
(B) (i) Cu/ (ii) LiAlH4
(C) (i) Ag2O (ii) Cu/
(D) (i)SOCl2(ii)LiAlH4
Sol.[A] COOH
4LiAlH
OH
TARUN KUMAR ASIWAL(12 years experienced ) M.Sc. CHEMISTRY , B.Ed. , NET , SET -2012 , LL.B. , D.Cl,
CARBOXYLIC ACID & IT'S DERIVATIVES 3
Cu /
CHO
Therefore option (A) is correct
Q.12
O
.O OHCH3 Y, Y is –
(A)
OH COOCH3
CH2 – CH2 (B)
OCH3 COOH
CH2 – CH2
(C) Both are correct (D) None is correct
Sol.[A]
O
.O
3OCHH
O
.OH C – OCH3
[Y]
Therefore option (A) is correct.
Q.13 End product of this conversion
CH3– C| |O
–CH2CH2CH2CO2H H/OH.2
NaBH.1
2
4 is –
(A)
O
O
CH3
(B)
O
O
CH3
(C)
O
O
(D) CH3–
OH
CH –CH2 – CH2CO2H
Sol.[A] CH3– C| |O
–CH2CH2CH2COOH 4NaBH
CH3–C–CH2–CH2–CH2–C=O
O H H–O
H+/H3O+
O
O
CH3
Therefore option (A) is correct.
Q.14
COOH + NaHC*O3
—
COONa + CO2
C* is with in the product –
(A) CO2
(B)
COONa
(C) Both
(D) None
Sol.(A)
C–O–H + Na O–C–OH
O
*
O
Sod. Bicarbonate
C–O–Na + H–O–C–O–H
O
*
O
CO2 + H2O
Therefore option (A) is correct.
Q.15 Identify the product A in the following
reaction
COOH CH2
3 COOH
CH3COOH + A
(A) CO2 (B) CH3CHO
(C) CH3OH (D) None of these
Sol.[A] COOH CH2
3 COOH
CH3COOH + CO2
Therefore option (A) is correct.
Q.16 A carboxylic acid (x) & (y) on heating with
P2O5 (x) give corresponding anhydride but
(y) remain unaffected. Carboxylic acid (y)
is-
(A) CH3COOH (B) HCOOH
(C) both (A & B) (D) CH3CH2COOH
Sol.[B] Carboxylic acid (y) is HCOOH
Q.17 Malonic acid and succinic acid are
distinguished by -
(A) Heating (B) NaHCO3
(C) Both A & B (D) None of these
Sol.[A] By Heating
Q.18 Which of the following statement is correct -
(A) Nucleophilic additions to the carbon-
oxygen double bond is a characteristic
reaction of aldehyde and ketones.
TARUN KUMAR ASIWAL(12 years experienced ) M.Sc. CHEMISTRY , B.Ed. , NET , SET -2012 , LL.B. , D.Cl,
CARBOXYLIC ACID & IT'S DERIVATIVES 4
(B) Carboxylic acid and their derivative are
characterized by nucleophilic substitution
reaction.
(C) In both case initial step involves a
nucleophilic attack on the carbonyl
carbon but in second step aldehyde and
ketone accept a proton to yield additional
product whereas in the case of acyl
compound leaving group is ejected to
form substitutional product.
(D) All are correct
Sol.[D] Factual
Q.19 The relative order of reactivity of acyl
derivatives is -
(A) R – C||O
–Cl > R – C||O
– O – C||O
– R >
R – C||O
– NH2 > R – C||O
– OR'
(B) R – C||O
–Cl > R – C||O
– OR' >
R– C||O
– O – C||O
– R > R – C||O
– NH2RBT
(C) R – C||O
– Cl > R – C||O
– O – C||O
– R >
R – C||O
– OR' > R – C||O
– NH2R
(D) None of the above
Sol.[C] R – C||O
– Cl > R – C||O
– O – C||O
– R >
R – C||O
– OR' > R – C||O
– NH2R
Q.20 What is the correct order of alkaline
hydrolysis of different ester CH3COOCH3
[rate = r1] , CH3COOC2H5
[rate = r2] CH3COOC3H7 [rate = r3] -
(A) r1 >r2 > r3 (B) r1 < r2 < r3
(C) r1 < r2 > r3 (D) r1 > r2 < r3
Sol.[A] r1 >r2 > r3
Q.21 Which of the following esters cannot undergo
claisen self condensation -
(A) CH3CH2CH2CH2COOC2H5
(B) C6H5COOC2H5
(C) C6H11CH2COOC2H5
(D) C6H5CH2COOC2H5
Sol.[ B] C6H5COOC2H5 esters cannot undergo claisen
self condensation
Q.22
2O
)A(
C
O
C
O
O
elimsoda/OH2
COOH
Oxidizing agent (A) used is
(A) K2Cr2O7 / H+ (B) AlK . KMnO4
(C) Chromic Acid (D) V2O5
Sol.[D] Oxidizing agent (A) used is V2O5
Q.23 In the reaction
C6H5NH2 Cº50
HCl/NaNO2
(A) KCN
CuCN (B)
OH/H 2 (C)
the product (C) is -
(A) C6H5CH2NH2 (B) C6H5COOH
(C) C6H5OH (D) None of these
Sol.[B] Product (C) is C6H5COOH
Q.24 Benzoic acid on treatment with hydrazoic
acid (N3H) in the presence of concentrated
sulphuric acid gives
(A) Benzamide
(B) Sodium benzoate
(C) Aniline
(D) C6H5CON3
Sol.[C] Factual
Q.25 Number of cross products in the given
reaction: (Without considering stereoisomers)
CH3COOC2H5 + C6H5–CH2–COOC2H5
OHHC
ONaHC
52
52
(A) One (B) Three
(C) Two (D) Four
Sol.[C] Two
Q.26 Arrange following compounds in decreasing
order of reactivity for hydrolysis reactions -
(I) C6H5COCl
(II)
COCl NO2
(III)
COCl CH3
TARUN KUMAR ASIWAL(12 years experienced ) M.Sc. CHEMISTRY , B.Ed. , NET , SET -2012 , LL.B. , D.Cl,
CARBOXYLIC ACID & IT'S DERIVATIVES 5
(IV)
C–Cl OHC
O
(A) II > IV > I > III (B) II > IV > III > I
(C) I > II > III > IV (D) IV > III > II > I
Sol.[A] II > IV > I > III
Q.27 In the given reaction sequence :
CH3–CH2–OH
H)ii(
/HO/KMnO)i( 4 (A)
/NH)ii(
SOCl)i(
3
2 (B) KOH/Br2 (C) will be -
(A) Methylamine
(B) Eltylamine
(C) Propylamine
(D) Acetamide
Sol.[A] (C) will be Methylamine
Q.28 Which optically active compound on
reduction with LiAlH4 will give optically
inactive compound ?
(A) CH3–CH–COOH
OCH3
(B) CH3–CH2–CH–COOH
OH
(C) CH3–CH2–CH–COOH
CH2OH
(D) CH3–CH–CH2–COOH
OH Sol.[C] CH3–CH2–CH–COOH
CH2OH
Q.29
CNH2
O
52OP
W
OH
MgBrCH
3
3 X
22 I,)OH(Ca
.)yellowppt( Y
Z will be
(A)
C–CH3
O
(B)
COOH
(C)
C
O
(D)
Sol.[C] Z will be
C
O
True / False type questions
Q.30 Acetonitrile on hydrolysis with a dilute
mineral acid gives acetone.
Sol.
leAcetonitri
NCCH3 H/HOOH
acidAcetic
NHCOOHCH 33
Acetonitrile on hydrolysis with a dilute mineral
acid give acetic acid, not an acetone.
Therefore statement is false.
Q.31 Esterfication involves the heating of a
carboxylic acid with an alcohol in presence of
a protonic acid (H2SO4 or HCl gas) as
catalyst.
Sol. In esterfication reaction :
O
CH3–C–OH + C2H5OH
acid
alcohol
H2SO4
or HCl(gas)
O
CH3C–OC2H5 + H2O
ester
Therefore statement is true.
Fill in the blanks type questions
Q.32 Kolbe’s electrolysis of potassium succinate
gives CO2 and ........
Sol. Ethylene
Q.33 In Hoffmann-bromamide reaction, the
migration of an alkyl or aryl group occurs
from ....... to ..... atom.
Sol. Carbon, nitrogen
Q.34 Ethanenitrile on hydrolysis gives ...........
Sol. Ethanoic acid
Q.35 Hell-Volhard Zelinsky reaction involves the
replacement of an ....... atom from the alkyl
group of a monocarboxylic acid by a .....
atom.
Sol. -hydrogen, halogen
TARUN KUMAR ASIWAL(12 years experienced ) M.Sc. CHEMISTRY , B.Ed. , NET , SET -2012 , LL.B. , D.Cl,
CARBOXYLIC ACID & IT'S DERIVATIVES 6
EXERCISE # 2
Part-A (Only single correct answer type
Questions)
Q.1 X HI
Pred CH3COOH 4LiAlH
Y. What
does not true for X and Y –
(A) X is hydrocarbon of general formula
CnH2n+2 while Y belong to alkanol
(B) X can be obtained by reducing
CH3CH2Cl while Y by its hydrolysis
(C) X gives positive litmus test but Y does
not
(D) X and Y both belong to different
homologeous series
Sol.[C] CH3–
CH3HI
Pred CH3COOH 4LiAlH
CH3CH2OH
[X] [Y]
Y give positive litmus test but 'X' does not.
Therefore option (C) is correct.
Q.2 The product of which of the following
reaction is capable of changing orange colour
of 272OCr to green colour of Cr
+3 –
(A) CH3(COOH)2
(B) CH3CN OH3
(C) HCN OH3
(D) CH3CONH2 OH2
Sol.[C] HCN OH3 H–COOH +
4NH
Formic acid is strong acid, that its capable of
changing orange coloure of 272OCr to green
colour Cr+3
.
Q.3 In the reaction sequence –
CH3CO
CH3CO O
X
CH3CONH2
Y
CH3C N Z
CH3COOC2H5
(A) NaOH, PCl5, Na + alcohol
(B) NH3, P2O5, aqueous ethanol/H+
(C) NH3, NaOH, Zn + NaOH
(D) NH3, Conc. H2SO4, aqueous methanol
Sol.[B] CH3CO
CH3CO
O
NH3
CH3–C–NH2 + CH3COOH
X
O
P2O5 (Y)
CH3C N + H2O
Aqueous ethanol
CH3COOH + C2H5OH
CH3–C–OC2H5
O
Therefore option (B) is correct.
Q.4 OHC–CH2–CH2 –CH2–CH2–OH
is converted into
O
O by –
(A) (i) KMnO4 (ii) H+, H
(B) (i) Na2Cr2O7 (ii) H+,
(C) (i) Ag2O (ii) H+,
(D) All of these Sol.[C] OHC–CH2–CH2 –CH2–CH2–OH
Ag2O
HOOC–CH2–CH2–CH2–CH2–OH
O
O
H+/
Therefore option (C) is correct.
Q.5
.O
O .O
O
, on saponification of the given
ester is formed –
(A) OH OH
CH2 – CH2
and OHC–CHO
(B) OH–CH2–CH2–OH and OHC–COOH
(C)
OH OH
CH2 – CH2
and HOOC–COOH
(D) HO–CH2–CH2–COOH and HCOOH
TARUN KUMAR ASIWAL(12 years experienced ) M.Sc. CHEMISTRY , B.Ed. , NET , SET -2012 , LL.B. , D.Cl,
CARBOXYLIC ACID & IT'S DERIVATIVES 7
Sol.[C]
CH2
CH2 +
O O H H–O
O H H–O
C
C
O
—
.O
O .O
O
Therefore option (C) is correct.
Q.6 A sweet smelling ester, with molar mass 116,
on hydrolysis produces a carboxylic acid and
an alcohol. Alcohol give positive iodoform
reaction which of the following formula could
correspond to above statements –
(A)
CH3–C–O–CH2–CH2–CH2–CH3
O
(B) CH3CH2–O–COC3H7
(C) CH3–O–COC4H9
(D)
CH3–CH2–C–O–CH2–CH2–CH3
O
Sol.[B]
O
C3H7 C–OCH2CH3 OH3
O
C3H7 C–OH + CH3CH2OH
Alcohol give positive
Iodoform reaction
Therefore option (B) is correct.
Q.7 YOHCH
NaBH
3
4
O
O
O
4LiAlHX
X and Y are –
(A)
O
O
OH
in both case
(B) OH
OH
in both case
(C) OH
OH
OH
,
O
O
OH
(D) Formation of A and B is not possible
Sol.[C] LiAlH4 is powerful, non selective reducing
agent. It reduces a wide range of functional
group –COOH, COOR, –CN, –
O| |C –NH2
But NaBH4 is more specific and selective for
the reduction of aldehydes and ketones. It has
no effect on ester nitro group, C = C etc.
O
O
OH
OHCH
NaBH
3
4
O
O
O
4LiAlH OH
OH
OH
[Y] [X]
Therefore option (C) is correct.
Q.8 CH3–CH2–COOH
423 SOH.ConcNaN
X
by reaction R1
CH3–CH2–COOH
P/Br2 Y
by reaction R2
Which is correct alternate –
(A) X Y
CH3CH2NH2
Br
CH3 – CHCOOH
R1 R2
Schmidt HVZ
(B) X Y
CH3CH2CONH2 CH3CH2COBr
R1 R2
HVZ Schmidt
(C) X Y
CH3CH2NH2 CH3CH2COBr
R1 R2
HVZ Schmidt
(D) None of these
Sol.[A] CH3CH2COO
423 SOH.ConcNaN
CH3CH2NH2 + N2 + CO2
[x]
This reaction is known as Schmidt reaction
[R1]
CH3CH2COOH
P/Br2 CH3
Br
|CHCOOH + HBr
[Y]
This reaction is known as Hell-Volhard-
zelinsky (HVZ) reaction.
Therefore option (A) is correct.
TARUN KUMAR ASIWAL(12 years experienced ) M.Sc. CHEMISTRY , B.Ed. , NET , SET -2012 , LL.B. , D.Cl,
CARBOXYLIC ACID & IT'S DERIVATIVES 8
Q.9 In the given reaction :
COOH
OH
3PCl
[X]
[X] will be -
(A)
COOH
Cl
(B)
COCl
OH
(C) O
C O
O
(D)
COCl
Cl
Sol.[B]
COCl
OH
Q.10
O
COOH
CH2COOH
X, X is –
(A)
O
C
CH2C
O
O
O (B)
O
COOH
CH3
(C)
O
CH2COOH
(D) None
Sol.[C] X is
O
CH2COOH
Q.11
Br
H C
CH3
O O
is reacted with
OH)iii(
CO)ii(etherMg)i(
3
2 Z.
The final product Z is –
(A)
COOH
H C
CH3
O O
(B)
COOMgBr
H C
CH3
O O
(C)
COOH
H C
CH3
O
(D)
COOH
COOH
CH3
Sol.[C]
Br
H C
CH3
O O
etherMg)i(
MgBr
OMgr C
CH3
O
H
OH
CO
3
2
COOH
H C
CH3
O
Therefore option (C) is correct.
Q.12 Y
2
3
Br,KOD)ii(
,HN)i( COOH
2
3
Br,KOH)ii(
,ND)i(,
X, what are X and Y –
(A) X is
NH2 ; Y is
ND2
(B) X is
ND2 ; Y is
NH2
(C) both
ND2
(D) both
NH2
Sol.[A]
COOH 3ND)i(
C–NH2
O
+ H2O
C–N
O
H
H
• •
+
OD
C–N–H
O
Br–Br
• •
C–N
O
Br
H
• •
OD
+
TARUN KUMAR ASIWAL(12 years experienced ) M.Sc. CHEMISTRY , B.Ed. , NET , SET -2012 , LL.B. , D.Cl,
CARBOXYLIC ACID & IT'S DERIVATIVES 9
C – N – Br
O
• •
tarrangemenRe
N–C–O + Br
• •
N = C = O + OD • •
N–C=O • •
OD
DO + CO2
ND2
ND – C – O
O • •
D–OD
[Y]
C–OH
O
/ND3
C–ND3
O
C–ND3
O
+ 4KOH + Br2
NH2 + 2Br + K2CO3 + 2D2O
Therefore option (A) is correct.
Q.13 An ester (I) with molecular formula C9H10O2
was treated with excess of CH3MgBr and the
complex so formed was treated with H2SO4 to
give an olefin (II). Ozonolysis of (II) gave a
ketone with molecular formula C8H8O which
show positive iodoform test. The structure of (I)
is –
(A) C6H5COOC2H5 (B) C6H5COOC6H5
(C) H3COCH2COC6H5
(D) p– CH3O – C6H4– COCH3
Sol.[A]
O
C6H5COC2H5 + CH3MgBr C6H5–C–OC2H5
CH3
OMgBr
CH3MgBr H3O
+
Con. H2SO4 C6H5–C–CH3
CH3
OH
C6H5–C–CH2
CH3
O3
C6H5–C–CH2O
CH3
O
[II]
Rotophenon
Therefore option (A) is correct.
Q.14
NH2 NH2
CH2 – CH2
on reaction with diethyl oxalate
form –
(A) NH2CH2NH– C| |O
COOC2H5
(B)
O N
H
H
N O
(C) H2N–CH2–CH2–N–COOH
H
(D) H2N – CH2 – CH2 – N =
OC2H5
C–COOC2H5
Sol.[B]
CH2
CH2 +
O NH–H C2H5–O
NH–H C2H5–O
C
C
O
—
O N
H
H
N O
+ 2C2H5OH
Therefore option (B) is correct.
Q.15 Give the structure of the expected product of
the following reaction
CH3
O
O
+ CH3NH2 — (X)
(A)
CH3
O
NCH3
(B)
CH3
O
NCH3 HO
(C) CH3–NH– C||O
– CH2CH2–
OH
CH –CH3
(D) None
TARUN KUMAR ASIWAL(12 years experienced ) M.Sc. CHEMISTRY , B.Ed. , NET , SET -2012 , LL.B. , D.Cl,
CARBOXYLIC ACID & IT'S DERIVATIVES 10
Sol.[C]
CH3
O + CH3–NH–H
O
CH3–NH–C–CH2–CH2–CH–O–
CH3
H+
CH3–NH–C–CH2–CH2–CH–OH
CH3
O
Therefore option (C) is correct.
Q.16 Which carboxylic acid (X) of equivalent mass
of 52g / eq loses CO2 when heated to give an
acid (Y) of equivalent mass of 60g/eq.
(A) CH3COOH
(B) CH2(COOH)2
(C) HOOCCH2CH2COOH
(D) HOOC(CH2)3COOH
Sol.[B] The carboxylic acid (X) is CH2(COOH)2
Q.17 59 g of amide obtained from the carboxylic
acid RCOOH, on heating with alkali gave 17g
of ammonia. The acid is –
(A) HCOOH (B) CH3COOH
(C) CH3CH2COOH (D)C6H5COOH
Sol.[B] The acid is CH3COOH
Q.18
COOH
ROH/NH/Na 3 ? Product is –
(A)
COOH
(B)
COOH
(C)
COOH
(D)
COOH
Sol.[B]
COOH
ROH/NH/Na 3
COOH
Part-B (One or more than one correct
answer type Questions)
Q.19 Which one of the following compounds will
give HVZ reaction ?
(A) COOH
(B)
COOH
(C) COOH (D)
COOH
Sol.[B,C]
COOH
COOH , of the compounds
will give HVZ reaction
Q.20 Phenol and benzoic acid may be distinguished
by their reaction with –
(A) Aqueous NaOH
(B) Aqueous NaHCO3
(C) Neutral FeCl3
(D) Aqueous NH3
Sol.(B), (C) Phenols are weaker acids than carboxylic
acid. It is because of this reason that phenols do
not decompose carbonates and bicarbonates
evolving CO2. This test distinguishes phenols
and carboxylic acid.
C6H5OH 3FeClNeutralBlue or violet colour
Benzoic acid does not react with Neutral FeCl3.
Therefore option (B) and (C) are correct.
Q.21 In the given reaction :
O
R– C – OH ]X[
O
R – C – O – CH3
[X] will be -
(A) CH2N2 (B) CH3OH/H
(C) MeCOOH (D) Me2SO4
Sol.[A,B,D] X will be CH2N2, CH3OH/H
, Me2SO4
Q.22 Which of the following statements are true
about HCOOH ?
(A) It is a stronger acid than CH3COOH
(B) It forms formyl chloride with PCl5
(C) It gives CO and H2O on heating with
conc. H2SO4
(D) It reduces Tollen’s reagent
Sol.[A,C,D]
(A) HCOOH is a strong acid than CH3COOH
(B) HCOOH + PCl5 HCOCl + PoCl3 + HCl
Unstable
TARUN KUMAR ASIWAL(12 years experienced ) M.Sc. CHEMISTRY , B.Ed. , NET , SET -2012 , LL.B. , D.Cl,
CARBOXYLIC ACID & IT'S DERIVATIVES 11
So the formyl chloride does not form.
(C) HCOOH
42SOH.ConcCO + H2O
(D) Formic acid reduces to tollen's reagent
Therefore option (A), (C) and (D) are correct.
Q.23 The best methods to reduce cyclohex-2-en-1-
one to cyclohex-2-en-1-ol is are –
(A) PCC, CH2Cl2
(B) MPV reduction
(C) NaBH4
(D) Ni/H2
Sol.[B]
O
4NaBH
OH
Cyclohex-2- ene-1-one Cyclohex-2-ene-1-ol
NaBH4 is more specific and selective for the
reduction of aldehydes and ketones. It has no
effect on C = C, C = C nitro group ester.
Therefore option (C) is correct.
Q.24 Which of the following on oxidation with
alkaline KMnO4 followed by acidification
with HCl would give benzoic acid ?
(A) Toluene (B) Ethylbenzene
(C) o-Xylene (D) p-Xylene
Sol.[A, B]
CH3
H/KMnO4
COOH
Tolune Benzoic acid
CH2–CH3
H/KMnO4
COOH
Ethyl benzene Benzoic acid
Therefore option (A) and (B) are correct.
Q.25 Nitrogen gas is evolved when HNO2 reacts
with –
(A) C6H5COCH3 (B) C6H5CONH2
(C) CH3CH2NH2 (D) NH2CONH2
Sol.[B,C,D]
C6H5
O| |C –NH2 2HNO
C6H5COOH + N2 + H2O
Benzoic acid
CH3CH2NH2 + 2HNOCH3CH2OH + N2 H2O
NH2CONH2 + HNO2 2N2 3H2O + CO2
Therefore option (B), (C) and (D) are correct.
Q.26 Which of the following on reduction with
LiAIH4 will give ethyl alcohol ?
(A) (CH3CO)2O
(B) CH3COCl
(C) CH3CONH2
(D) CH3COOC2H5
Sol.[A,B, D]
(A)
CH3C
O
O CH3C
O
4LiAlH 2CH3CH2OH
Ethyl alcohol
(B) CH3–
O| |C –Cl 4LiAlH
C2H5OH+LiCl+AlCl3
(C) CH3CH2–
O| |C –NH2 4LiAlH
CH3CH2CH2 NH2+H2O
(D) CH3COOC2H5 4LiAlHCH3CH2OH + CH3OH
So option (A), (B) and (D) are give the ethyl
alcohol by reduction with LiAlH4.
Q.27 Ethyl acetate can be prepared by –
(A) Reaction of ethyl alcohol with acetic acid
in presence of dry HCl gas
(B) Treatment of acetaldehyde with
Al(OC2H5)3
(C) Reaction of acetyl chloride with ethyl
alcohol in presence of acidic medium
(D) Reaction of acetic anhydride with ethyl
alcohol in presence of acidic medium
Sol.[A,B,C,D]
(A) C2H5OH+CH3–
O| |C –OH2
HCH3
O| |C –OC2H5
(B)
O
CH3C + O = C – CH3
H
H
352 )HOC(Al
C
Acetaldehyde O
CH3C–O–CH2–CH3
Ethyl acetate
This reaction is know as
TARUN KUMAR ASIWAL(12 years experienced ) M.Sc. CHEMISTRY , B.Ed. , NET , SET -2012 , LL.B. , D.Cl,
CARBOXYLIC ACID & IT'S DERIVATIVES 12
Tischenko reaction.
(C) CH3
O| |C –Cl + C2H5OH CH3
O| |C –OC2H5 + HCl
Ethyl acetate
(D) (CH3CO)2O+C2H5OHCH3
O| |C –OC2H5+CH3COOH
Ethyl acetate
Therefore option (A),(B),(C) and (D) are correct.
Q.28 The formula C4H4O4 can represent –
(A) A cyclic ester of dibasic acid
(B) A cis-dibasic acid
(C) A trans-dibasic acid
(D) An , -unsaturated dibasic acid which
on heating forms a monobasic acid
Sol.[A,B,C,D]
Formula C4H4O4 represent.
(A) A cyclic ester of dibasic acid
CH2–O–C
O
CH3–O–C
O
(B) A cis-dibasic acid
H–C–COOH
H–C–COOH
Maleic acid
(C) A trans-dibasic acid
H–C–COOH
COOH–C–H
Pumaric acid
(D)
HC CH
COOH COOH
CH2=CH–COOH+CO2
Monobasic acid
Therefore option (A),(B),(C) and (D) are correct.
Part-C (Assertion & Reasoning type
Questions)
The following questions 29 to 30 consists of
two statements each, printed as Assertion
and Reason. While answering these
questions you are to choose any one of the
following five responses.
(A) If both Assertion and Reason are true and
the Reason is correct explanation of the
Assertion.
(B) If both Assertion and Reason are true but
Reason is not correct explanation of the
Assertion.
(C) If Assertion is true but the Reason is false.
(D) If Assertion is false but Reason is true.
(E) If Assertion & Reason are false.
Q.29 Assertion : Nitration of benzoic acid gives
m-nitrobenzoic acid.
Reason : carboxylic group increases the
relative electron-density at meta-position.
Sol.[C] –C| |O
–OH, is an electron attracting group thus
cause deactivation of benzene ring particularly
at o, p that m-position, thus electrophilic reagent
can attack at m-position.
Therefore assertion is true but the reason is
false.
So option (C) is correct.
Q.30 Assertion : CH3COCH2COOC2H5 will give
iodoform test.
Reason : It contains CH3CO–group linked to
a carbon atom. Sol.[D] Ethyl acetoacetate (CH3
O| |C –CH2COOC2H5) does
not give iodogorm test, although it contains
CH3–
O| |C
group attached to carbon (methylene
group). This is due to active nature of
methylene group at which iodination occurs and
not on methyl group of CH3CO unit.
Therefore assertion is false but reason is true so
option (D) is correct.
Part-D Column Matching
Q.31 Column-A Column-B (a) H.V.Z. reaction (i) potassium salt of
Monocarboxylic acid
(b) Oxidation of Formic (ii) replacement of
acid -hydrogen by
halogen
(c) –COOH group (iii) H2O + CO2 (d) Salt in Kolbe’s (iv) deactivating
electrolytic reaction
TARUN KUMAR ASIWAL(12 years experienced ) M.Sc. CHEMISTRY , B.Ed. , NET , SET -2012 , LL.B. , D.Cl,
CARBOXYLIC ACID & IT'S DERIVATIVES 13
Sol. (A)ii, (B)iii, (C)iv, (D)i ,
Factual
Q.32 Column-A Column-B
(a) Clemmenson (i) ,-unsaturated
reduction compound
(b) Cannizaro’s reaction (ii) red brown
precipitate
(c) Aldol condensation (iii) aldehyde
+ Zn–Hg/HCl
(d) Fehling solution (iv) aldehydes
containing no
-hydrogen
(e) Claisen-Schmidt (v) carbonyl
reaction compounds having
-hydrogen
Sol. (A) iii, (B)iv, (C)v, (D)ii, (E) i
Factual
Q.33 Column-A Column-B
(Reagents reacting with (Product formed)
PhCH2COOH) (a) CH3MgBr (i) PhCH2COCl
(b) PCl5 (ii) PhCH2COOCH3
(c) NH3, followed by (iii) CH4
heating
(d) CH3OH in the presence(iv) PhCH2CONH2
of conc. H2SO4
Sol. (A) iii, (B)i, (C)iv, (D)ii
Factual
Q.34 Column-A Column-B
(a) PhCONH2 (i) B2H6/AcOH/H2O
PhCH2OH
(b)
C6H5
O
O
(ii) LiAlH4
CH2OH–(CH2)2–CHOH–C6H5
(c) C6H5CH=CH–COOH (iii) H2/Pd + BaSO4
C6H5–CH=CH2OH
(d) CH3COClCH3–CHO (iv) None
Sol. (A) iv, (B)ii, (C)iv, (D)iii,
Factual
Q.35 Column-A Column-B
(a) RCN reduction
(i) 1° Amine
(b) RCN OH)ii(
MgBrCH)i(
2
3 (ii) Alcohol
(c) RNC hydrolysis
(iii) Ketone
(d) RNH2 2HNO (iv) Acid
Sol. (A) i, (B)iii, (C)i, iv, (D)ii Factual
TARUN KUMAR ASIWAL(12 years experienced ) M.Sc. CHEMISTRY , B.Ed. , NET , SET -2012 , LL.B. , D.Cl,
CARBOXYLIC ACID & IT'S DERIVATIVES 14
EXERCISE # 3
Part-A Subjective Type Questions
Q.1 Fluoro acetic acid is stronger than chloro
acetic acid.
Sol. F–CH2–C–OH F–CH2 –C–O
CH2–N=N
O O
(strong –I power)
–I
Cl–CH2–C–O–H Cl–CH2 –C–O
O O
Anion stabilised by
–I group.
Therefore fluoro acetic acid is stronger than
chloroacetic acid.
Q.2 C – O bond length in formic acid are 1.23 Aº
and 1.36 Aº but in sodium formate both
carbon and oxygen bonds have same value
i.e. 1.27 Aº.
Sol. In sodium formate both carbon and oxygen
bonds have same value because in formate ion
equal resonating structure is formed.
H–C
O–
O
H–C
O
O–
H–C
O
O
–
–
But resonance does not present in formic acid.
Therefore C–O bond length in formic acid are
different.
Q.3 The C – G bond in acid derivative is expected
to be shorter and stronger than alkyl
derivative.
Sol. In acid derivative
O
R–C–G
xÅ
Resonance present
O
R–CH2–G
yÅ no Resonance
O
R – C – G R–C=G
O
Double bond
characters are
present
Therefore C–G bond in acid derivative is
expected to be shorter and stronger than alkyl
derivative.
Q.4 Ester (A) 5PCl B + C
KOH.aq
(CH3)2CHCH2OH
B + C4
2
BaSO
Pd/H (D) 4LiAlH
(CH3)2CHCH2OH
Sol. (A) (CH3)2CHCOOCH2CH(CH3)2 ;
(B) (CH3)2CHCOCl ;
(C) (CH3)2CHCH2Cl
Q.5
COOH
COOH
COOH
A
B
Sol. (A)
COOH
COOH
; (B)
C
C
O
O
O
Q.6
COOH
COOH
4LiAlH
(A) HCl
(B)
)C(
CH2COOH
CH2COOH
BaO
(D) + CO2 + H2O
Sol. (A)
CH2OH
CH2OH
;
(B)
CH2Cl
CH2Cl
;
(C) CN–/H3O
+;
(D)
C=O
Q.7 A OAgMoist 2 B
]O[C
D
(HCOO)2Ca
G2
52
SOCl,2
OHHC/Na H
I
CH3CN OH3 X
A KCN
J OH3
X
Sol. (A) CH3Cl ; (B) CH3OH ;
(C) HCOOH ; (D) Ca(OH)2 ;
(G) HCHO ; (H) CH3Cl ;
(I) KCN ; (J) CH3CN ;
(X) CH3COOH
TARUN KUMAR ASIWAL(12 years experienced ) M.Sc. CHEMISTRY , B.Ed. , NET , SET -2012 , LL.B. , D.Cl,
CARBOXYLIC ACID & IT'S DERIVATIVES 15
Q.8 HC CHexcess
COOHCH3 (A)
B + C
HOH
CH3COOH
Sol. (A) CH3CH(OOCCH3)2 ;
(B) CH3CHO ;
(C) (CH3CO)2O
Q.9 On standing in dilute aqueous acid,
compound A is smoothly converted to
mevaionolactone.
Me
CH2COOH
Me
O O
)2(
OH)1( 3
OH
CH3
O
O
Suggest a reasonable mechanism for this
reaction. What other organic product is also
formed ?
Sol.
Me
CH2COOH
Me
O O H2O+
H2C
HO
CH2
OH
CH3 O
CH3
Intermol ester
OH
CH3
O
O
Q.10 Outline reasonable mechanisms for each of
the following reactions.
(a)
O O
+ BrMgCH2CH2CH2CH2MgBr
OH.2
THF.1
2
HO CH2CH2CH2OH
(b)
O S H2NCH2CH2
eoustanspon
O
HS
N
H
Sol. (a)
O O
Br–Mg–CH2 CH2–Mg–Br
+
CH2 – CH2
O
OH . CH2 CH2
CH2 – CH2
OH
OH
CH2 CH2
(b)
O S HNCH2–CH2
• •
• •
H
• •
HS
O–
N–H HS
O
N–H
Q.11 An organic compound A (C4H9NO) on
treatment with bromine and alkali form
another compound B (C3H9N) on treatment
with sodium nitrite and dilute hydrochloric
acid. B yield C (C3H8O). C can be oxidized to
D (C3H6O). Which can also synthesised from
methyl
Sol. (A) (CH3)2CHCONH2; (B) (CH3)2CHNH2;
(C) (CH3)2CHOH; (D) CH3COCH3
Q.12 An organic compound A (C8H8O) an
treatment with NH2OH. HCl give B and C.
Compound B and C can be converted into D
and E respectively by treatment of H2SO4.
Compound B, C, D and E are all isomer of
molecular formula C8H9NO. When D is
boiled with KOH, an oil F (C6H7N) separate
out. F react rapidly with CH3COCl to give
back (D). On other hand. E on boiling with
alkali followed by acidification give a white
solid G (C7H6O2). Identify A to G.
Sol. (A)
–C–CH3
O
; (B)
–C=N–OH
CH3
;
TARUN KUMAR ASIWAL(12 years experienced ) M.Sc. CHEMISTRY , B.Ed. , NET , SET -2012 , LL.B. , D.Cl,
CARBOXYLIC ACID & IT'S DERIVATIVES 16
(C)
–C = N
CH3 OH
;(D)
–NH–C–CH3
O
;
(E)
–C–NH–CH3
O
; (F) –NH2
;
(G)
–C–OH
O
Q.13 An organic compound (A) of formula C3H6O
from a monoxime with hydroxylamine and
form iodoform on heating with I2 and NaOH
and sodium acetate. Compound A on reaction
with NaCN and dil H2SO4 gave product B.
Which on hydrolysis produced (C).
Compound (C) on heating gave (D), which on
decarboxylation gave (E) of formula C3H6.
Compound (E) on ozonolysis gave one
molecule of acetaldehyde and one of
methanal. What are A to E ?
Sol. (A) CH3–CO–CH3;
(B)(CH3)2C(OH)CN ;
(C) (CH3)2C(OH)COOH ;
(D) CH2=C–(CH3)–COOH ;
(E) CH3–CH=CH2
Q.14 Compound (A) with an empirical formula,
C7H9N on diazotization give a product which
undergoes reaction with Cu2Cl2 & HCl to
give a compound (B). B on oxidation give a
compound (C). Compound (A) on treatment
with Br2/H2O form C7H6NBr3 (D). Give the
structural formula of A, B, C & D.
Sol.
CH3
NH2
CH3
Cl
COOH
Cl
CH3
NH2
Br
Br Br
Part-B Passage based Question
Passage - I (Q. 15 to 17)
-keto acids are highly unstable acids and
undergo decarboxylation on heating at 90º–
100ºC Via formation of cyclic transition state.
Decarboxylation of -keto acids involves
both free acids and the carboxylate ion. Loss
of CO2 from carboxylate ion yields the
carbanion which is stabilised by resonance.
O
• •
R–C–CH2–C–O
O
R – C – CH2 + CO2
O • •
O • •
R–C=CH2
This carbonion is formed faster than the
simple carbanion i.e. alkyl carbanion that
would be form a simple carboxylate ion
(RCOC–) because it is more stable. It is more
stable of course due to accommodation of the
negative charge by the keto group.
Q.15 Which of the following on heating gives
acetone –
(A)
CH3–C–CH2–C–OH
O O
(B)
CH3–CH2–C–CH2–C–OH
O O
(C)
CH3–C–CH–C–OH
O O
CH
(D) All
Sol.(A)
CH3–C–CH2–C–OH
O O
CH3–C–CH2+CO2 • •
O
CH3–C=CH2
O–
H2O/H+
CH3–C=CH2
OH
CH3–C–CH3
OH
Therefore option (A) is correct.
TARUN KUMAR ASIWAL(12 years experienced ) M.Sc. CHEMISTRY , B.Ed. , NET , SET -2012 , LL.B. , D.Cl,
CARBOXYLIC ACID & IT'S DERIVATIVES 17
Q.16 Arrange the following in the increasing order
of ease of decarboxylation.
I.
CH3–C–CH2–C–OH
O O
II.
CH3–C–CH2–CH2–C–OH
O O
III.
COOH
NO2 O2N
NO2
IV.
CH2–C–OH
O
(A) I < II < III < IV (B) II < I < IV < III
(C) II < IV < I < III (D) IV < III < II < I
Q.17 What will be the final product in the
following reaction :
CH3–C–H
O
5PClA
KCNB
NaOH
OH2 C HCl
D
E
(A)
CH3–C–NH2
O
(B)
CH3–CH2–C–OC2H5
O
(C) CH3–CH2–CH3
(D)
CH3–CH2–C–OH
O
Sol.(D)
CH3–C–H
O
5PCl Cl CH3CH
Cl [A] KCN
CN CH3CH
CN [B]
H2O/NaCl
COONa CH3CH
COONa [C]
NaCl COOH CH3CH
COOH [C]
CH3–CH2–COOH+CO2
Therefore option (D) is correct.
Passage - II (Q. 18 to 20)
Claisen condensation takes place between
two molecules of ester in presence of strong
base like R–ONa in alcohol or NaH in ether.
Two ester molecules may be same or
different. The basic need of this reaction is
the presence at least one -Hydrogen, the
reaction is termed as crossed claissen
condensation.
This reaction is an example of nucleophilic
substitution reaction. In this reaction one
molecules of ester behaves as substrate and
other molecule having -Hydrogen atom
behaves as nucleophile.
The product of the reaction is -Ketoester.
Step (i)
H
CH2–C–OC2H5
• •
O
C2H5O—
CH2–C–OC2H5
O
–
Step (ii)
CH2–C–OC2H5
O
+ CH2–C–OC2H5
O • •
CH3–C–OC2H5
O
CH2–C–OC2H5
• •
O
CH3–C–OC2H5
O
CH2–C–OC2H5
• •
O
CH3–C–CH2–C–OC2H5
O O
This reaction also takes place between ester
and ketone having atleast one -hydrogen.
Ketone behaves as nucleophile. In diesters
intramolecular claissen condensation gives a
cyclic -ketoester.
Q.18 What is X in the following reaction.
C–OC2H5
O
C–OC2H5
O
+
O
H)ii(
OHHC/ONaHC)i( 5252X
(A)
CH2–C–OC2H5
O
O
TARUN KUMAR ASIWAL(12 years experienced ) M.Sc. CHEMISTRY , B.Ed. , NET , SET -2012 , LL.B. , D.Cl,
CARBOXYLIC ACID & IT'S DERIVATIVES 18
(B)
C–OC2H5
O
O
(C)
C–C–OC2H5
O
O
O
(D)
C–C–OC2H5
O
O
O
Sol.(C)
O
H + C2H5O
O
+C2H5OH
O
C–OC2H5
O
C–OC2H5
O
+
O
C–OC2H5
O–
C–OC2H5
O
C–C–OC2H5 + C2H5O
O
O
O
Therefore option (C) is correct.
Q.19
CH2–CH2–C–OC2H5
O
CH2–CH2–C–OC2H5
O
OHHC
ONaHC
52
52 A.
What is A–
(A)
CH2–C–OC2H5
O
O
(B)
C–OC2H5
O
O
(C)
C–OC2H5
O
O
(D)
C–CH3
O
O
Sol.(C)
CH2–CH–C–OC2H5
O
CH2–CH2–C–OC2H5
O
H
C2H5O
CH2–CH–C–OC2H5
O
CH2–CH2–C–OC2H5
O
C–OC2H5
O
O
–C2H5O
CH–C–OC2H5
O–
O
C
CH2
CH2 CH2
OC2H5
Therefore option (C) is correct.
Q.20 What will be the product in the following
reactions –
CH2–CH2–C–OC2H5
O
CH2–CH2–C–OC2H5
O
CH3–CH2–N
OHHC
ONaHC
52
52 Product
(A) CH3–CH2–N O
(B) CH3–CH2–N O
COOC2H5
(C) CH3–CH2–N
COOC2H5
(D)
CH3–CH2–N
O
C–OC2H5
O
Sol.(B)
CH2–CH–C–OC2H5
O
CH2–CH2–C–OC2H5
O
CH3–CH2–N
H
C2H5O–
TARUN KUMAR ASIWAL(12 years experienced ) M.Sc. CHEMISTRY , B.Ed. , NET , SET -2012 , LL.B. , D.Cl,
CARBOXYLIC ACID & IT'S DERIVATIVES 19
CH2–CH–C–OC2H5
O
CH2–CH2–C–OC2H5
O
C2H5–N
CH2–CH–C–OC2H5
O
CH2–CH2
C2H5–N
C–O–
OC2H5
C2H5–N O
C–OC2H5
O
Therefore option (B) is correct.