RESONANCE & AROMATICITY

CENTERS : MUMBAI / DELHI /AKOLA / LUCKNOW /NASHIK / PUNE / NAGPUR / BOKARO / DUBAI # 96

FINAL LAP - 2019GENERAL ORGANIC CHEMISTRY

RESONANCE & AROMATICITY

Q.1 Which of the following statements is (are) true about resonance.(a) Resonance is an intramolecular process.(b) Resonance involves delocalization of both s and p electrons.(c) Resonance involves delocalization of p electrons only.(d) Resonance decreases potential energy of a molecule.(e) Resonance has no effect on the potential energy of a molecule.(f) Resonance is the only way to increase molecular stability.(g) Resonance is not the only way to increase molecular stability.(h) Any resonating molecule is always more stable than any nonresonating molecule.(i) The canonical structure explains all features of a molecule.(j) The resonance hybrid explains all features of a molecule.(k) Resonating structures are real and resonance hybrid is imaginary.(l) Resonance hybrid is real and resonating structures are imaginary.(m) Resonance hybrid is always more stable than all canonical structures.

Q.2 Consider structural formulas A, B and C:

(A) (B) (C)(a) Are A, B and C constitutional isomers, or are they resonance forms?(b) Which structures have a negatively charged carbon?(c) Which structures have a positively charged carbon?(d) Which structures have a positively charged nitrogen?(e) Which structures have a negatively charged nitrogen?(f) What is the net charge on each structure?(g) Which is a more stable structure, A or B? Why?(h) Which is a more stable structure, B or C? Why?

Q.3 In each of the following pairs, determine whether the two represent resonance forms of a single speciesor depict different substances. If two structures are not resonance froms, explain why.

(a) and

(b) and

(c) and

Q.4 Which of the following statement is incorrect ?(A) Contributing structures contributes to the resonance hybrid in proportion of their energies.(B) Equivalent contributing structure make the resonance very important.(C) Contributing structures represent hypothetical molecules having no real existance.(D) Contributing structures are less stable than the resonance hybrid.


FINAL LAP - 2019GENERAL ORGANIC CHEMISTRYQ.5 Which one of the following pairs of structures does not represent the phenomenon of resonance?

(A) HCCHCH||

O

2 ; HCCHCH|

¯O

2

(B) ClCHCHCH2

; ClCHCHCH2

(C) ¯OCCH)CH(||

O

23 ; OCCH)CH(|

¯O

23

(D) 323 CHCCHCH||

O

; 33 CHCCHCH|

¯O

Q.6 In which of the following lone-pair indicated is involved in resonance :

(a) (b) (c) (d)

(e) CH2= CH – 2HC (f) CH2 = CH – CH = HN

Q.7 In which of the following lone-pair indicated is not involved in resonance :

(a) CH2 = CH – HN

– CH3 (b) CH2 = CH – CH =

O

(c) CH2 = CH –

O – CH = CH2 (d) CH2 = CH – C •

•N

(e) (f)

Q.8 Which of the following is not a valid resonating structure of the other three?

(i) (a)

NH2

+ (b)

NH2+

(c)

NH2+

(d)

NH2+

(ii) (a)

3

2

CH|

¯ONHC

(b)

3

2

CH|

¯ONHC

(c)

3

2

CH|

ONCH (d)

3

2—

CH|

ONHC

Q.9 Draw the resonance forms to show the delocalization of charges in the following ions

(a) CH – C – CH3 2

O (b) H – C – CH = CH – CH2

O (c) CH2

+

(d) +

(e) O¯ (f) + NH (g) O

+



(h)

O

(i) CH3 – CH = CH – CH = CH – HC

– CH3

(j) CH3 – CH = CH – CH = CH – 2HC

Q.10 Write stability order of following intermediates:

(i) (a) 23 CHCH

(b) 33 CHCHCH

(c)

3

3

3

CH|CCH|

CH

(ii) (a) + (b) + (c) +

(iii) (a)

(b) (c)

(iv) (a)

(b) (c)

(v) (a)

(b)

(c)

(d)

(vi) (a) 23 HCCF

(b) 23 HCCCl

(c) 23 HCCBr


(i) (a) 23 CHCH

(b) 33 CHCHCH

(c)

3

3

3

CH|CCH|

CH

(ii) (a)

(b)

(c)

(iii) (a)

(b) (c)



(iv) (a) CHC (b)

CHCH2 (c)

23 CHCH

Q.12 Rank the following sets of intermediates in increasing order of their stability.

(i) (a) C6H5+ (b) p–NO2(C6H4)+ (c) p–CH3–(C6H4)+ (d) p–Cl–C6H4

+

(ii) (a)

CH2

N

O O

(b)

CH2

OMe

(c)

CH2

(iii) (a)

CH2

OH(b)

CH2

OH(c)

CH2

OH

(iv) (a)

223 CHCHCF (b) 23 CHCF (c)

3CF

(v) (a)

3CF (b)

3CCl


(i) 23 CHCH

(b) 33 CHCHCH

(c)

3

3

3

CH|CCH|

CH

(ii) (a) (b) (c)

(iii) (a) (b) (c)

(iv) (a)

CHC (b)

CHCH2 (c)

23 CHCH

(v) (a) (b) (c)

Q.14 Among the following molecules, the correct order of C – C bond lenght is(A) C2H6 > C2H4 > C6H6 > C2H2 (B) C2H6 > C6H6 > C2H4 > C2H2(C6H6 is benzene)(C) C2H4 > C2H6 > C2H2 > C6H6 (D) C2H6 > C2H4 > C2H2 > C6H6



Q.15 Which of the following is (are) the correct order of bond lengths :(A) C – C > C = C > C C > C N (B) C = N > C = O > C = C(C) C = C > C = N > C = O (D) C – C > C = C > C º C > C – H

Q.16 C1 – C2 bond is shortest in

(A) (B) (C) (D)

Q.17 In which of the following resonance is possible(A) CH2 = CH – CH2 – CHO (B) CH2 = CH – CH = O(C) CH3COCH3 (D) CH2 = CH – CH2 – CH = CH2

Q.18 Which of the following compounds have delocalised electrons ?

(A) (B)

(C) CH3CH2NH CH2 CH = CH2 (D) CH2 = CH – CH2 – CH = CH2

Q.19 Compare the C–N bond-length in the following species:

(i) (ii) (iii)

Q.20 Which of the following species are conjugated ?

(A) (B)

(C) CH = C = CH – NH2 2

: (D) All of these

Q.21 Which of the following example has both the resonating structure contributing equal to the resonancehybrid

(A) 3

3

3 CH–CHCH–CHC–CH|

3

3

3 CH–HC–CH

CH

C–CH|

(B) 23 NH–

O

C–CH||

CH – C = NH3 2|

O -

+



(C) CH – C3

O

O- CH – C3

O

O

-

(D) CH2 = CH – CH = O CH – CH = CH – O2

-+

Q.22 In which of the following pairs, indicated bond having less bond dissociation energy :(a)

BrCHCH 23 &

ClCHCH 23

(b) BrCHCHCH3 &

Br|

CHCHCH 33

(c) &

(d) &

(e) &

(f) &

Q.23 Discuss the following observations:(a) C–Cl bond in vinyl chloride is stronger than in chloroethane.(b) Carbon-carbon bond length in ethene is shorter than in CH2 = CHOCH3(c) CH3SH is stronger acid than CH3OH(d) CH3CH2NH2 is stronger base than CH2 = CHNH2.

Q.24 A canonical structure will be more stable if(a) it has more number of p bonds than if it has less number of p bonds.(b) the octets of all atoms are complete than if octets of all atoms are not complete.(c) it involves cyclic delocalization of (4n + 2) p – electrons than if it involves acyclic delocalization of (4n +

2) p – electrons.(d) it involves cyclic delocalization (4n) p – electrons than if it involves acyclic delocalizationof

(4n) p – electrons.(e) +ve charge is on more electronegative atom than if +ve charge is on less electronegative atoms.(f) –ve charge is on more electronegative atom than if –ve charge is on less electronegative atom.

Q.25 Resonance energy will be more if(a) canonical structures are equivalent than if canonical structures are non-equivalent.(b) molecule is aromatic than if molecule is not aromatic.



Q.26 In each set of species select the aromatic species.

(i) (a) (b) S

(c) +

(d) +

(ii) (a) +

(b)

(c) (d)

(iii) (a) +

(b)

(c)

(d) N

H

Q.27 Consider the given reaction:

+ 3H2 C/Pd

In the above reaction which one of the given ring will undergo reduction?

Q.28

Br

3AgNO AA

Select the correct statement about product A.(A) Product is aromatic (B) Product has high dipole moment.(C) Product has less resonance energy (D) Product is soluble in polar solvent.

Q.29 Which of the following is incorrectly orderd for resonance stability

(A)

(B)

(C) (I > II)

(D) (II = I)


FINAL LAP - 2019GENERAL ORGANIC CHEMISTRYQ.30 In which of the following pairs of resonating structures first resonating structure is more stable than

second.

(A) (B)

(C) (D)

Q.31 Hyperconjugation is possible in

(A) (B) (C) (D)

Q.32 Which of the following molecule has longest C=C bond length.(A) CH3–CH=CH–CH=CH–CH3 (B) CH2=CH–CH=CH2(C) CH3–CH=CH–CH3 (D) CH2=CH2

Q.33 In which of the following molecule (s) , the resonance effect is present ?

(A) (B) (C) (D)

Q.34 Explain why which compound is aromatic, antiaromatic or nonaromatic.

(a) O

N

isoxazole

(b) SN

1,3-thiazole

(c)

pyranO

(d)

pyrylium ionO

H+

(e)

-pyroneO

O

(f)

1,2-dihydropyridine

NH

(g)

NH2

cytosine

N

N

OH


FINAL LAP - 2019GENERAL ORGANIC CHEMISTRYQ.35 Resonance is possible/s in

(A) CH2 = 2HN

(B) CH3CH = C =

(C) (D)

Q.36 Which one of following represents different molecules?

(A) and (B) and

(C) and (D) and


(i) (a)

CH2

F

(b)

CH2CH2

ClCl

(ii) (a)

CH2

Cl

(b)

CH2

N

O O

(c)

CH2CH2

C N

(d)

CH2

(iii) (a)

O||

CHCH2

(b) 32 CHCH

Q.38 In each of the following pairs of ions which ion is more stable:

(a) (I ) C6H5– 2CH

and (II) CH2=CH– 2CH

(b) (I) CH3– 2CH

and (II) CH2 =

CH

(c) (I) and (II)

(d) (I)

33

33

CHCCH|

CHCHCH and (II)

33

33

CHCCH|

CHNCH



(e) CH2—CH=NH CH2=CH–NH

(f) CH3—C=CH–CH—CH3 CH3—C—CH=CH—CH3

O¯ O¯+

+

(g) CH3—C–CH–C—CH3 CH3—C=CH–C—CH3

O O¯O O

(h) CH3—C —C—NH2H2 CH3—C —C = NH2H2+

+

NH2 NH2

Q.39 Select the least stable resonating structure in each of the following sets of carbocation.

(i) (a) CH2 = CH — N+

O

O(b) CH2 – CH — N

++O

O

(c) CH2 – CH — N++

O

O(d) CH2 – CH = N

++O

O

(ii) (a) 2—

2 NHCHCHCHCH

(b) 2—

2 NHCHCHCHCH

(c)

2—

2 NHCHCHCHCH (d)

2—

2 NHCHCHCHCH

(iii) (a) CH2 — HC

–CH = CH – O – CH3 (b)

CH2 — CH = CH – CH =

O – CH3

(c) H2C = CH – HC

– HC

– O – CH3 (d) H2C = CH – HC

– CH = O – CH3


(i) (a) (b) (c)

(ii) (a) (b)

(iii) (a)

CH2

C

HH

H

(b)

CH2

C

HH

H (c)

CH2

C

HH

H



(iv) (a)

CH2

CH3

(b)

CH2

CH2Me

(c)

CH2

CH Me2

(d)

CH2

CMe3

(v) (a)

Me|CMe|

Me (b)

Ph|C—Ph|Ph

(c) Ph – 2CH

(d)

+

(vi) (a)

+

(b) +

(c)

+

(d) +

(e)

+

Q.41 Which one of the following molecules has all the effect, namely inductive, mesomeric and hyperconjugative?(A) CH3Cl (B) CH3–CH = CH2

(C)

O||

CHCCHCHCH 33 (D) CH2 = CH – CH = CH2

Q.42 Hyperconjugation is best described as:(A) Delocalisation of p electrons into a nearby empty orbital.(B) Delocalisation of s electrons into a nearby empty orbital.(C) The effect of alkyl groups donating a small amount of electron density inductively into a carbocation.(D) The migration of a carbon or hydrogen from one carbocation to another.

Q.43 Select the correct statement.(i) Delocalisation of s-electron is hyperconjugation.(ii) Delocalisation of p-electron is resonance.(iii) Partial displacement of s-electron is inductive effect.(A) i & iii (B) ii & iii (C) i & ii (D) i, ii, iii

Q.44 Arrange following compounds in decreasing order of electrophilic substitution.

(i)

CH3

(ii)

CH3

H—C—CH3

(iii)

CH3

CH3—C—CH3

(iv)

H

H—C—CH3

(A) i > ii > iii > iv (B) iii > iv > ii > i (C) i > iv > ii > iii (D) i > ii > iv > iii


FINAL LAP - 2019GENERAL ORGANIC CHEMISTRYQ.45 Select correct statement:

(A) –NO2 and –COOH group deactivates benzene nucleus for attack of E+ at o– and p– sites.(B) –NH2 and –OMe group activates benzene nucleous for attack of E+ at o– and p– sites.(C) –NH2 and –COOH group activates benzene nucleous for attack of E+ at o– and p– sites.(D) –NO2 and –OMe group activates benzene nucleous for attack of E+ at o– and p– sites.

Q.46 In which of the following pairs, indicated bond is of greater strength :

(a) and 22 CHCH (b)

CHCCH3 and

CHHC

(c) and ClCHCH 23

(d)

22 CHCHCHCH and

3222 CHCHCHCH

(e)

22 CHCHCHCH and 22 NOCHCH

(f) and

Q.47 Choose the more stable alkene in each of the following pairs. Explain your reasoning.(a) 1-Methylcyclohexene or 3-methylcyclohexene(b) Isopropenylcyclopentane or allylcyclopentane

(c) or

Q.48 Match each alkene with the appropriate heat of combustion:Heats of combustion (kJ/mol) : 5293 ; 4658; 4650; 4638; 4632(a) 1-Heptene (b) 2,4-Dimethyl-1-pentene(c) 2,4-Dimethyl-2-pentene (d) 4,4-Dimethyl-2-pentene(e) 2,4,4-Trimethyl-2-pentene

Q.49 Compare heat of hydrogenation (Decreasing order)

(i) &

(ii) &

(iii) &

(iv) CH2 = CH – CH & CH3 –CH = C CH3

CH3



Q.50 Compare heat of hydrogenation (Decreasing order)

(i)A B D

(ii)

Q.51 (I) Stability order and (II) heat of hydrogenation orders.

(a) (b) (c) (d)

Q.52 Among the following pairs identify the one which gives higher heat of hydrogenation :

(a) and

(b) and

(c) CH3 – CH = CH – CH3 and CH3 – CH2 – CH = CH2

(d) and

Q.53 Write increasing order of heat of hydrogenation (HOH):

(i) (a) (b)

(ii) (a) (b) (c) (d)

(iii) (a) (b) (c) (d) (e)

(iv) (a) (b) (c)



(v) (a) (b) (c) (Heat of hydrogenation per p bond)

(vi) (a) (b) (c) (HOH per benzene ring)

Q.54 Give decreasing order of heat of combustion (HOC):

(i) (a) (b) (c)

(ii) (a) (b) (c) (d)

(iii) (a) (b)

(iv) (a) (b) (c)


(i) (a) C

(b) 2HCPh

(c) +

(d)

Me|

MeCMe

(ii) (a)

CHCH2 (b)

22 CHCHCH (c) CH2 (d) CH3

(iii) (a)

Cl

CH2

(b) Cl

CH2

(c)

CH2

(iv) (a) CH2+

(b)

CH2+

(c) +



(v) (a) +

(b) +

(c) +

(vi) (a)

(b)

(c)



Q.1 (a), (c), (d), (g), (j), (l), (m)

Q.2 a = Resonance form, b= A, c = C, d= A & B, e=B&C, f = 0, g = B, h = B

Q.3 (a) is resonance form; (b) is not resonance form due to different number of l.p. and b.p.;(c) is not resonance form due to different number of l.p. and b.p

Q.4 A Q.5 D Q.6 (b), (d), (e) Q.7 (b), (d), (e) Q.8 (i) a (ii) c

Q.9 (a) 23 CHCCH|

¯O

(b) 2CHCHCHCH|

¯O

(c)

CH2 CH2CH2 CH2CH2+ +

+ +

+

(d) + (e) O¯ O O O

(f) NH+

(g) O+(h)

O¯

(i) 33 CHCHCHCHCHCHCH

(j) 23 CHCHCHCHCHCH

]

Q.10 (i) c > b > a (ii) b > c > a (iii) c > b > a (iv) b > c > a(v) d>c > b>a (vi) a < b < c

Q.11 (i) a > b > c (ii) a > b > c (iii) a > c > b (iv) a > b > c

Q.12 (i) b < d < a < c (ii) b > c > a (iii) c > a > b (iv) c > a > b(v) a > b

Q.13 (i) c > b > a (ii) c > b > a (iii) b > c > a (iv) c > b > a(v) a > c > b

Q.14 B Q.15 A, C, D Q.16 D Q.17 B Q.18 B Q.19 iii > ii > i

Q.20 D Q.21 C

Q.22 (a) I, (b) II, (c) II, (d) I, (e) I, (f) I



Q.23 (a) Due to resonance 2—

HC – CH = Cl+

(b) In CH2=CH–OCH3, there is single bond character due to resonance

2—

HC – CH = O+– CH3

(c) Conjugate base of CH3SH ie, CH3S is more stable than conjugate base of CH3OH, ie CH3O–

(d) In CH2=CH–NH2 lone pair of N is delocalized 2—

HC – CH = 2HN

Q.24 (a), (b), (c), (f) Q.25 (a), (b) Q.26 (i) a, b, d (ii) a, c (iii) b, c, d

Q.27 A Q.28 A, B, D Q.29 C

Q.30 C Q.31 A Q.32 A Q.33 A, D

Q.34 (a) A, (b) A, (c) N.A. (d) A, (e) A, (f) N.A. (g) A Q.35 A, B, C, D

Q.36 B

Q.37 (i) a < b (ii) b > c > a > d (iii) a > b

Q.38 (a) I, (b) I , (c) I I , (d) I I , (e) I I , (f) I, (g) II, (h) II Q.39 (i) c (ii) a (iii) c

Q.40 (i) a > b > c (ii) a > b (iii) c > a > b (iv) a > b > c > d(v) d > b > a > c (vi) d > e > b > a > c

Q.41 C Q.42 B Q.43 D Q.44 C Q.45 A, B

Q.46 (a) II, (b) II, (c) I, (d) I, (e) II, (f) II Q.47 (a) i , (b) i , (c) ii

Q.48 (a) 4658, (b) 4638, (c) 4632, (d) 4650, (e) 5293 Q.49 (i) 2 > 1 (ii) 2 <1 (iii) 1 < 2 (iv) 1 > 2

Q.50 (i) D > C > B > A (ii) E > C > D > B > A Q.51 (I) d > c > b > a; (II) a > b > c > d

Q.52 (a) I, (b) I, (c) II, (d) I

Q.53 (i) b > a (ii) a > b > d > c (iii) a > b > c > d > e

(iv) b > c > a (v) a > b > c (vi) a > b > c

Q.54 (i) c > b > a (ii) a > b > c > d (iii) a > b (iv) c > b > a

Q.55 (i) a » c > d > b (ii) d > b > c > a (iii) b > a > c (iv) a > c > b

(v) b > c > a (vi) b > c > a


FINAL LAP - 2019ACIDITY & BASICITY

Q.1 Write correct order of acidic strength of following compounds:(i) (a) H–F (b) H–Cl (c) H–Br (d) H–I

(ii) (a) CH4 (b) NH3 (c) H2O (d) H–F

(iii) (a) CH3–CH2–O–H (b)

3

3

CH|

HOCHCH (c) CH3–C–O–H

CH3

CH3

(iv) (a) F–CH2–CH2–O–H (b) NO2–CH2–CH2–O–H

(c) Br–CH2–CH2–O–H (d) HOCHCHNH 223

Q.2 Write correct order of acidic strength of following compounds:(i) (a) CH3COOH (b) CH3CH2OH (c) C6H5OH (d) C6H5SO3H

(ii) (a) COOH (b) COOH (c) COOH

(iii) (a) COOH|COOH

(b) CH2

COOH

COOH(c)

COOHCH|

COOHCH

2

2

Q.3 Write correct order of acidic strength of following compounds:

(i) (a) HOCCHCl||

O

2 (b)

Cl|

HOCCHCl||

O

(c)

Cl|

HOCCCl|||

OCl

(ii) (a)

F|

HOCCHCHCH||

O

23 (b)

F|

HOCCHCHCH||

O

23

(c)

F|

HOCCHCHCH||

O

222

(iii) (a) HOCCHNO||

O

22 (b) HOCCHF||

O

2

(c) HOCCHPh||

O

2 (d) HOCCHCH||

O

23




(i) (a)

O–H

NO2

(b)

O–H

Cl

(c)

O–H

CH3

(ii) (a)

O–HCl

(b)

O–H

Cl(c)

O–H

Cl

(iii) (a)

O–H

(b)

O–H

(c)

O–H

(d)

O–H


(i) (a)

O–HN

O

O

(b)

O–H

NO

O (c)

O–H

NO O

(d)

O–H

(ii) (a)

O–H

NO2

(b)

O–HN

O

O

(c)

O–H

NO2

NO2(d)

O–H

NO2

NO2NO2


(i) (a)

C–O–HO

(b)

C–O–HO

CH3

(ii) (a)

COOHCl

(b)

COOHBr



(iii) (a)

C–O–H

OMe

O

(b)

C–O–H

OMe

O

(c)

C–O–HOMe

O

(iv) (a)

C–O–H

N

O

OO

(b)

C–O–H

NO2

O

(c)

C–O–HNO2

O

Q.7 Select the strongest acid in each of the following sets :

(i) (a)

OH

CH3

(b)

OH

NO2

(c)

OH

Cl

(d)

OH

NH2

(ii) (a)

OH

NO2

(b)

OH

F

(c)

OH

CH3

(d)

OH

(iii) (a)

OHOMe

(b)

OH

OMe (c)

OH

(d)

OH

OMeQ.8 Arrange the given phenols in their decreasing order of acidity:

(I) C6H5–OH (II) F OH

(III) Cl OH (IV) O2N OH

Select the correct answer from the given code:(A) IV > III > I > II (B) IV > II > III > I (C) IV > III > II > I (D) IV > I > III > II



Q.9 Which one of the following is the most acidic?

(A) (B) (C) (D) CH2=CH–CH3

Q.10 Which one of the following phenols will show highest acidity?

(A) (B) (C) (D)

Q.11 Which of the following is weakest acid?

(A) (B) (C) (D)

Q.12 The correct pKa order of the follwoing acids is :OH OHOHO HO

(I) (III)(II)O OOO OO

(A) I > II > III (B) III > II > I (C) III > I > II (D) I > III > II

Q.13 Arrange pH of the given compounds in decreasing order:(1) Phenol (2) Ethyl alcohol (3) Formic acid (4) Benzoic acid(A) 1 > 2 > 3 > 4 (B) 2 > 1 > 4 > 3 (C)3 > 2 > 4 > 1 (D) 4 > 3 > 1 > 2

Q.14 Arrange acidity of given compounds in decreasing order:(I) CH3–NH–CH2–CH2–OH (II) CH3–NH–CH2–CH2–CH2–OH

(III) OHCHCHN)CH( 2233

(A) III > I > II (B) III > II > I (C) I > II > III (D) II > I > III

Q.15 Consider the following compound

(A) (B) (C) CCOOHCH||

O

3(D)

Which of the above compounds reacts with NaHCO3 giving CO2



Q.16 Say which pka belong to which functional group in case of following amino acids :

(i) cysteine : 1.8, 8.3 & 10.8

(ii) glutamic acid : : 2.19, 4.25, 9.67

Q.17 Record the following sets of compounds according to increasing pKa ( = – log Ka)

(a) , , cyclohexane carboxylic acid.

(b) 1-butyne, 1-butene, butane(c) Propanoic acid, 3-bromopropanoic acid, 2-nitropropanoic acid(d) Phenol,o-nitrophenol, o-cresol(e) Hexylamine, aniline, methylamine

Q.18 Explain which is a stronger acid.

(a) CH3CH3 & BrCH2NO2 (b) 33 CHCCH||

O

& CNCHCCH||

O

23

(c) CH3 – CHO & CH3 – NO2

Q.19 Explain which is a weaker acid.

(a)

OH

O=C–CH3

or (c)

OH

O=C–CH3

or

(b) or

Q.20 Which of the following would you predict to be the stronger acid ?

(a) or

(b) CH3 – CH2 – CH2 – OH or CH3 – CH = CH – OH(c) CH3 – CH = CH – CH2 – OH or CH3 – CH = CH – OH



BASICITYQ.1 Write increasing order of basic strength of following:

(i) (a) F (b) Cl (c) Br (d) I

(ii) (a) 3CH (b)

2NH (c) OH (d) F

(iii) (a) R–NH2 (b) Ph–NH2 (c)

O||

NHCR 2

(iv) (a) NH3 (b) MeNH2 (c) Me2NH (d) Me3N (Gas phase)

(v) (a) NH3 (b) MeNH2 (c) Me2NH (d) Me3N (in H2O)

Q.2 Write increasing order of basic strength of following:

(i) (a) NH

O

(b) NH

(c) N

Me

(ii) (a)

NH2

(b)

NH2

(c) NH

(iii) (a) O N2

N

(b) Me

N

(c) F

N

(iv) (a)

NH2

NH3

(b)

NH2

Cl(c)

NH2

CH3

(d)

NH2

H


(i) (a) CH3–CH2– 2HN

(b) CH3–CH= HN

(c) CH3–C

N

(ii) (a)

O||

HNCCH 23

(b) CH3–CH2– 2HN

(c)

HN||

HNCCH 23

(d)

HN||

HNCHN 22

(iii) (a) NH

(b)N

(c)

NH2



(iv) (a)NH–C–CH3

O

(b)

NH2

(c)

NH–CH CH2 3–

(v) (a)

NH2

N

MeMe

O O

(b)

NH2

NMeMe

O O


(i) (a)

NO2

NH2

(b)

CN

NH2

(c)

OMe

NH2

(d)

NH2

NH2

(ii) (a)

NH2CH

HH

(b)

NH2

CH

HH

(c)

NH2

CHH

H

(iii) (a)

NO2

NH2

(b)

NH2

NO2

(c)

NH2

NO2

(iv) (a)

NH2

(b)

NH2CH

HH

(c)

NH2

(v) (a)

NMe2

OMe

(b)

NMe2

OMe(c)

NOMe

Me Me

Q.5 Select the strongest base in following compound :

(i) (a) N

(b)

N

H

(c)N

O

H

(d)N

S

H



(ii) (a)

NH2

(b) NH

(c)N

(d)N

H

(iii) (a) N

N

H

(b) N

H

(c)N

CH3

(d)N

H

(iv) (a) N¯Li+

(b) N

H

(c)N

H

(d)N

Me

Q.6 Arrange the following compound in decreasing order of their basicity.(i) (a) H2C = CHNa (b) CH3CH2Na (c) CH3CH2ONa (d) HC CNa

(ii) (a) NH2 (b) CH – NH22 (c) NH2

NO2

(d) C – NH2

O(iii) (a) HO¯ (b) NH3 (c) H2O

Q.7 Basicity order in following compound is :

N

N

NH

O CH3

CH3

CH2 – NH – C – CH 3H2N– C – CH 2b

d

a

c

CH3

CH3

(A) b > d > a > c (B) a > b > d > c (C) a > b > c > d (D) a > c > b > d

Q.8 Consider the following bases:(I) o-nitroaniline (II) m-nitroaniline (III) p-nitroanilineThe decreasing order of basicity is:(A) II > III > I (B) II > I > III (C) I > II >III (D) I > III > II

Q.9 Consider the basicity of the following aromatic amines:(I) aniline (II) p-nitroaniline (III) p-methoxyaniline (IV) p-methylanilineThe correct order of decreasing basicity is:(A) III > IV > I > II (B) III > IV > II > I(C) I > II > III > IV (D) IV > III > II > I



Q.10 Which one of the following is least basic in character?

(A) (B) (C) (D)

Q.11 In each of the following pair of compounds, which is more basic in aqueous solution? Give an explanationfor your choice:

(a) CH3NH2 or CF3NH2 (b) CH3CONH2 or H2N NH2

(c) n-PrNH2 or CH3CN (d) C6H5N(CH3)2 or 2,6-dimethyl-N-N-dimethylaniline(e) m-nitroaniline or p-nitroaniline

Q.12 From the following pair, select the stronger base:(a) p-methoxy aniline or p-cyanoaniline (b) pyridine or pyrrole(c) CH3CN or CH3CH2NH2

Q.13 Choose the member of each of the following pairs of compunds that is likely to be the weaker base.(a) H2O or H3O (b) H2S, HS–, S2– (c) Cl–, SH–

(d) F–, OH–, NH2–,

3CH (e) HF, H2O, NH3 (f) OH–, SH–, SeH–

Q.14 Explain which compound is the weaker base.

(a) or (b) CH2 = CH – CH = CH – CH2– or CH2 = CH – CH2

–

(c) OHCCO||||

OO

or OHCCHO||||

OO

(d) or

Q.15 Rank the following amines in increasing basic nature.

(a)

(i) (ii) (iii) (iv)

(b)

(i) (ii) (iii) (iv)Q.16 Arrange the basic strength of the following compounds.(a) OH– CH3COO– Cl–

(i) (ii) (iii)(b) CH C– CH2 = CH– CH3CH2

–

(i) (ii) (iii)(c) CH2 = CHCH2NH2 CH3CH2CH2NH2 CH C – CH2NH2

(i) (ii) (iii)



Q.17 Arrange the basic strength of the following compounds.

(a)

(i) (ii) (iii)

(b)

(i) (ii) (iii)

(c)NH2 NH2 NH2

H3C O2N(i) (ii) (iii)

Q.18 Arrange the following compounds in order of increasing basicity.

(a) CH3NH2, CH33NH , CH3NH— (b) CH3O—, CH3NH—, CH3

2CH

(c) CH3CH = CH—, CH3CH22CH , CH3CC—

Q.19 Rank the amines in each set in order of increasing basicity.

(a)NH2 NH2 N

H

(b)

HN NH N

(c) N NN N H HH

Q.20

Pyrimidine Imidazole PurineAmong the following which statement(s) is/are ture:(A) Both N of pyrimidine are of same basic strength(B) In imidazole protonation takes places on N–3.(C) Purine has 3 basic N.(D) Pyrimidine imidazole and purine all are aromatic



ACIDITY

Q.1 (i) d>c>b>a (ii) d>c>b>a (iii) a > b > c (iv) d>b>a>c

Q.2 (i) d > a > c> b (ii) c > b > a (iii) a > b > c

Q.3 (i) c > b > a (ii) a > b > c (iii) a > b > c > d

Q.4 (i) a > b > c (ii) a > b > c (iii) d > b > c > a

Q.5 (i) c > a > b > d (ii) d > c > a > b Q.6 (i) b > a, (ii) b > a, (iii) c > b > a, (iv) c > a > b

Q.7 (i) b (ii) a (iii) b

Q.8 C Q.9 B Q.10 C Q.11 B Q.12 B Q.13 B Q.14 A

Q.15 A, B, C, D

Q.16 (i) cysteine : (ii) glutamic acid :

Q.17 (a) 3<2<1; (b) 1<2<3; (c) 3<2<1; (d) 2<1<3; (e) 2<3<1

Q.18 (a) 2; (b) 2; (c) 2 Q.19 (a) 2; (b) 2; (c) 2 Q.20 (a) 2; (b) 2; (c) 2

BASICITY

Q.1 (i) a > b > c > d (ii) a > b> c> d (iii) a > b> c (iv) a < b< c< d(v) c > b> d> a

Q.2 (i) a < b < c (ii) c > a > b (iii) b > c> a (iv) c > d> b > a

Q.3 (i) a > b > c (ii) d > c > b > a (iii) b > c > a (iv) c > b > a(v) b > a

Q.4 (i) d > c > b > a (ii) c > b > a (iii) b > a > c (iv) a > b > c(v) c > a > b

Q.5 (i) d (ii) b (iii) a (iv) a

Q.6 (i) b > a > d > c (ii) b > a > c > d (iii) a > b > c

Q.7 B Q.8 A Q.9 A Q.10 A

Q.11 (a) i, (b) ii, (c) i, (d) ii, (e) i Q.12 (a) i, (b) i, (c) ii

Q.13 (a) 2; (b) 1; (c) 1; (d) 1; (e) 1; (f) 3 Q.14 (a) 2; (b) 1; (c) 2; (d) 2

Q.15 (a) 3<2<1<4; (b) 1<2<3<4 Q.16 (a) 1>2>3; (b) 1<2<3; (c) 3<1<2

Q.17 (a) 2<1<3; (b) 1<2<3; (c) 2 > 1 > 3 Q.18 (a) 2<1<3; (b) 1<2<3; (c) 3<1<2

Q.19 (a) 2 > 1> 3, (b) 1 > 2> 3, (c) 1 > 3 > 2, Q.20 A, B, C, D


FINAL LAP - 2019ISOMERS

Exercise-1

Q.1 Which will show geometrical isomerism.

(A) (B) (C) (D)

Q.2 Configuration of is

(A) R (B) S (C) D (D) L

Q.3 is diastereomers of

(A) (B) (C) (D)

Q.4 Give the correct order of initials T or F for following statements. Use T if statement is true and F if it isfalse.I. Me–CH=C=C=CH–Br is optically active.II. All optically active compound are chiral.III. All chiral pyramidal molecules are optically inactive.

IV CH3–CH2–CH2–COOH and

COOH|

CHCHCH 33 are positional isomers.

(A) TTTF (B) FTFT (C) FTFF (D) TFTT

Q.5 From the following four structures select:(a) The optically active isomers (b) Optically inactive isomers(c) Enantiomers pairs (d) Distereomer pairs

(I) (II) (III) (IV)


FINAL LAP - 2019ISOMERSQ.6 Discuss the optical activity of the following two compounds and also label them as polar and non-polar.

(I) (II)

Q.7 Consider the following six structures:

(I) (II)

(III) (IV)

(V) (VI)

The stereochemical relationship between : (a) I and II, (b) III and IV, (c) II and III, (d) I and V, (e) IV and (VI)

Q.8 Indicate the stereo centres in the following molecule and total number of stereoisomers in the followingmolecule.

Q.9 For each of the following pair, deduce the stereochemical relationship, i.e., whether they are enantiomers,diastereomers or identical.

(a) and (b) and

(c) and



(d) and

Q.10 For each of the following molecules, indicate, whether they are chiral, achiral or meso compound:

(I) (II) (III) (IV)

(V) (VI) (VII)

Q.11 For each of the following pair of structures, indicate, if the compounds are identical, constitutional isomers,enantiomers, distereomers, diffrerent.

(a) and (b) and

Q.12 How many stereoisomer may have this natural occuring compound.

(A) 8 (B) 16 (C) 64 (D) 128

Q.13 How many optically active stereoisomers are possible for butane-2, 3-diol(A) 1 (B) 2 (C) 3 (D) 4

Q.14 Identify whether each of following pair of compounds are identical or enantiomers, diastereoisomer orconstitutional isomers.

(a) and (b) and

(c) and (d) and

(e) and (f) and



(g) (h)

(i) (j)

(k) (l)

(m) and (n) and

(o) and (p) and

Q.15 How many stereoisomers are possible for the molecule shown below:

Q.16 Discuss the type of isomerism exhibited by the following pairs:

(a) and

(b) and



(c) and

(d) CH3– CH2– CH2– CN and

CN|

CHCHCH 33

(e) and

Q.17 Draw resonance structure of the amides shown below and select them which are stereoisomeric:

(a) (b) (c)

(d)

Q.18 Which of following will exhibit geometrical isomerism?(A) 1-phenyl-2-butene (B) 3-phenyl-1-butene(C) 2-phenyl-1-butene (D) 1,1-diphenyl-1-propene

Q.19 How many optical isomers are possible for butane-2, 3-diol(A) 1 (B) 2 (C) 3 (D) 4

Q.20 MeCH=CH–CH=C=CH–CH=CH2Total number of geometrical isomers possible for above compounds are:(A) 16 (B) 8 (C) 4 (D) 2

Q.21 total stereoisomers of the given compound is___________.

Q.22 Which point on the potential energy diagram in represented by Newmann projection shown?


FINAL LAP - 2019ISOMERSQ.23 Truxillic acid can exist theretically in five stereosiomeric form all of which are known an optically inactive

explain,where a = – CO2H, b = – C6H5

Q.24 The most stable enol-form of compound is

(A) (B) (C) (D)

Q.25(a) A cyclobutandicarboxylic acid exist in two stereo-isomeric forms in which one is polar but non-resolvablewhile other is non-polar but resolvable into enantiomers. Deduce structures of all these compounds.

(b)Select resolvable compounds.

(i) (ii) (iii)

3CH|

OSPh

(iv) (v) MeCHBrCH2Me (vi)

(vii) (viii) (ix) BrDHNMe 2

(x)

OH|

MeCHCHMeCH 22 (xi) (xii) C=C=CH2

Q.26 Which of the Newman projections shown below represents the most stable conformation about theC1–C2 bond of 1-iodo-2-methyl propane?

(A) (B) (C) (D)


FINAL LAP - 2019ISOMERSQ.27 Select the pair of enantiomer and diastereomers out of the following:

Q.28 In each of the following molecules, indicate the presence of a centre of chirality with an asterisk (*)

(a) (b) (c) (d) (e)

Q.29 Choose the correct option for the following molecule in view of chemical bonding

(A) non-planar (B) 0 (C) A & B both (D) = 0

Q.30 Match the column-I with column-II. Note that column-I may have more than one matching optionsin column-II.

Column–I (stability) Column-II (Reason)

(A) > 3CH (P) Inductive effect

(B) < –3CH (Q) Resonance

(C) > H3C– 2CH (R) Hyperconjugation

(D) > (S) steric hindrance

Q.31

is dimethyl derivative of a compound 'A' and is fairly stable but most of the molecules of 'A'

gets converted into another compound 'B' on keeping. What are the structures of 'A' and 'B'. Explain thereason of conversion A into B.



Exercise-21. Which of the following are isomers

(A) I, II only (B) III, IV only (C) I, II, III (D) All

2. No. of functional groups present in the following compound is :

(A) 5 (B) 7 (C) 6 (D) 8

3. Quinine is the most important alkaloid obtained from cinchona bark. It’s molecular formula is C20H24N2O2. Itmay contain(A) 5 double bond & 6 ring (B) 6 double bond & 4 ring(C) 6 double bond & 3 ring (D) 7 double bond & 5 ring

4. In the structure of 4-Isopropyl-2,4,5-trimethylheptane, number of 10, 20 & 30 H’s are respectively(A) 18, 5, 4 (B) 21, 4, 3 (C) 18, 4, 3 (D) 21, 5, 4

5. How many 1° amines are possiple for the molecular formula C7H9N which contain benzene ring also(A) 1 (B) 4 (C) 3 (D) 2

6. Find out the total number of geometrical isomers of the following compound(CH3)CH = C = C(CH3)(C2H5)(A) 0 (B) 2 (C) 3 (D) 4

7. Which of the following will not show geometrical isomerism

(A) (B) (C) (D)

8. How many products are formed by monochlorination of

CH3

CHCH3

CH3H C3

.

(A) 13 (B) 12 (C) 15 (D) 14

9. Which will show geometrical isomerism

(A)

CH2

CH3(B)

CHCH3

CH3(C)

CHCH3

(D)

CHCH3CH3


FINAL LAP - 2019ISOMERS10. How many stereoisomers are possible for Gorlic acid (C18H30O2)

CH=CH

CH -CH2 2

CH(CH ) CH=CH(CH ) COOH2 6 2 4

(A) 6 (B) 4 (C) 8 (D) 2

11. Molecular formula of Anthanthrone

O

O

is

(A) C22H8O2 (B) C21H10O2 (C) C23H10O2 (D) C22H10O2

12. ==

==

O How many geometrical isomers the above compound will posses

(A) 2(4-3/2) (B) 26 (C) 25-22 (D) 25

13. Structure of Biotin, (C10H16N2O3S) is as follows

No. of chiral carbon & D.U. (degree of unsaturation) for Biotin are(A) 4,3 (B) 3,3 (C) 4,4 (D) 3,4

14. Which of the following pairs are nonsuperimposable over each other ?

I H OH

CH3

C H2 5

HHO

CH3

C H2 5

II CH3-O-CH3 CH3-CH2-OH

III

IV CH3-CH2-CHO CH3-CH2-CH2-CHO

V and

(A) I, II, III, IV (B) I, V (C) I, II, III (D) I, II, IV


FINAL LAP - 2019ISOMERS15. Which statement is correct about the following structures

BrClH

OCH3

CH3

C H2 5

I

BrCl

HH O3C

CH3

C H2 5

II

Br

Cl

HH O3C

CH3

C H2 5

III(A) I & III are structural isomers(B) I & II & I & III are enantiomers(C) I, III are enantiomers and I, II are structural isomers(D) I, II & III are stereoisomers

16. Indicate the number of chiral carbon atoms in the following molecule.

HO(A) 6 (B) 7 (C) 8 (D) 9

17. 3, 5-Dimethylcyclohexanol has total stereoisomers(A) 3 (B) 4 (C) 8 (D) 12

18. 2, 4-Dimethylcyclohexanol has total stereoisomers(A) 4 (B) 3 (C) 8 (D) 12

19. In the number of stereoisomers are

(A) 26 (B) 25 (C) 24 (D) 23

20. Ph – CH2 – CH2 – COOH PdRe

Br2 PhCH2CHBrCOOH

The product is(A) racemic mixture (B) pure (+) (C) pure (–) (D) none of these

21. and are

(A) enantiomers (B) diastereomers (C) mesomers (D) all of these

22. can have

(A) geometrical isomer (B) optical isomers (C) epimer (D) none of these



23.

Which is more stable(A) A > B (B) A = B (C) A < B (D) none

24. In the given conformation shown below

orientation of I and Br atoms are(A) (a, e) (B) (e, e) (C) (e, a) (D) (a, a)

25. In the above conformation Br, Et are (geometrical relationship)(A) cis (B) trans (C) both (D) none of these

26. In the above conformation Me, Et are(A) trans (B) cis (C) both (D) none of these

27. Between &

(A) I is more stable than II (B) II is more stable than I(C) I & II are equally stable (D) can’t say about their relative stability

28. 1-Bromo-2-chlorocyclohexane has several conformations. The stablest one has(A) (1e – Br, 2e – Cl) (B) (1e – Br, 2a – Cl) (C) (1a – Br, 2a – Cl) (D) (1a – Br, 2e – Cl)

29. Among the 1, 3-dibromo cyclohexane conformations, the stablest one is(A) (1a – Br, 3a – Br) (B) (1e – Br, 3e – Br) (C) (1a – Br, 3e – Br) (D) (1e – Br, 3a – Br)

30.

Between I, II the stable one is(A) I (B) II (C) equally stable (D) none of these

31. Between and the stable one is

(A) I (B) II (C) both are equally stable (D) can’t be predicted

32. is

(A) optically active (B) optically inactive (C) meso in nature (D) none of these



33. and are

(A) diastereomers (B) conformational enantiomers(C) configurational enantiomers (D) racemic form

34. Between and the relationship is

(A) conformational enantiomer (B) identical(C) conformational diastereomer (D) none of these

35. and are

(A) enantiomers (B) identical (C) optically inactive (D) all of these

36. The compound

(A) is opticaly acitve (B) is a meso compound(C) has zero dipole moment (D) has all groups in cis form

37. Which name is correct for

(A) cis-(a)-2-Decalol (B) cis-(e)-2-Decalol (C) trans-(a)-2-decalol (D) trans-(e)-2-Decalol

38.

Select the correct statement/s(A) All are identical compounds (B) III and IV are identical and meso compounds(C) I and II are optically active compounds(D) I and II are configurational isomers while II and IV are conformational isomers

39. Draw the most stable conformation of 1-Chloropropane. Which statement is correct about thisconformation.(A) It is the most polar form (B) It has maximum torsional strain(C) It has minimum steric strain (D) A and C both


FINAL LAP - 2019ISOMERS40. The optically active alkyl chloride is

(A) (B) (C) (D)

41. Compound A (C6H12) does not absorb H2 in presence of Ni. It forms two monochloro isomers on photochemi-cal chlorination. Its structure can be

(A) (B) (C) (D)

42. Which of the following molecule/s is / are chiral :

(A)

322

323

CHCHCH|

CHNCHCH

(B) (C) (D)

I II III IV(A) I (B) I, II (C) III, IV (D) All

43. Which of the following species will be optically active :

(A) (B) 733

52

HCNCH|

HC

(C) (D)

44. How many spatial orientations are possible in the following compound ?

(A) 2 (B) 4 (C) 6 (D) 8

45.

CHO|

HCHO|

OHCH2

represents the Fischer projection formula :

(A) D (B) L (C) d (D) l

46. Hydrogenation of the adjoining compound in the presence of poisoned palladium catalyst gives.

(A) an optically active compound (B) an optically inactive compound(C) a racemic mixture (D) a diastereomeric mixture

47. Mention the relationship between compounds of following pairs (isomers (type), Homologs, identicals)(I) (II)

(A)



(B)

(C)

(D)

(E)

48. Which of the following is a meso isomer,

(A) (B) (C) (D)

49. How many optically active products can be formed in the following reaction

(A) 2 (B) 4 (C) 6 (D) 8

50. Which of the following compound has zero dipole moment in one of the stable conformations

(A) HO – CH2 – CH2 – OH (B) CH3 – CHCl – CHBr – CH3

(C) d-CH3 – CHCl – CHCl – CH3 (D) meso-CH3 – CHCl – CHCl – CH3

51. Which of the following are identical compounds

(A) I, II (B) I, III (C) I, IV (D) II, III

52. The following two compounds are

(A) conformational isomers (B) optical isomers(C) structural isomers (D) identical compounds


FINAL LAP - 2019ISOMERS53. Establish the relationship between the following pairs, as enantiomers, identicals, diastereomers or struc-

tural isomers.

(i) (ii)

(iii) (iv)

54. Observe the structures of compounds A, B, C and D. Write the relationship between the given pairs ofcompounds.

(I) A and B are .......... (II) A and C are ............(III) A and D are ......... (IV) C and D are ..........

55. Compounds with molecular formula C2H7N can be(A) Functional isomers and metamers (B) Only functional isomers(C) Functional and positional isomers (D) Chain, positional, functional and metamers

56. Total number of isomers having molecular formula C3H4O and only functional group can be,

(A) 1 (B) 2 (C) 3 (D) 4

57. How many products are formed by monochlorination of (Ignore stereoisomers)

(A) 4 (B) 2 (C) 5 (D) 3

58. The total number of 4° (quaternary) carbon atoms present in cholestanol is

(A) 3 (B) 8 (C) 4 (D) 2

59. Total number of phenol isomers of the compound C8H10O is(A) 10 (B) 9 (C) 8 (D) 6

60. The following two compounds I and II are ...................... isomers.

and

(A) Geometrical (B) Geometrical and positional(C) Positional (D) Geometrical, optical and positional


FINAL LAP - 2019ISOMERS61. Which of the following will not show geometrical isomerism

(A) (B) (C) (D)

62. Fill the correct type of isomerism(A) CH3 – CH2 – CH2 – CH2 – NH2 (B) CH3 – CH2 – NH – CH2 – CH3

(C) CH – CH – N – CH3 2 3

CH3

(D) CH3 – CH2 – CH2 – NH – CH3

(E) CH – CH – CH – NH3 2 2

CH3

(i) A & B are__________ (ii) B & C are__________(iii) A & C are__________ (iv) B & D are__________(v) A & E are__________ (vi) C & E are__________

63. For molecular formula C5H10 :x Total number of possible structural isomers.y How many of them can show geometrical isomerism.

(A) 10, 2 (B) 9, 2 (C) 9, 1 (D) 8, 4

64. Which of the following carbonyl compound will give two product after reaction with NH2OH :

(1) (2) (3) (4)

(5) (6) (7)

(A) 2, 3 , 4, 5, 6, 7 (B) 1, 2, 3, 5, 6, 7 (C) 2, 3, 4, 5, 6 (D) 1, 3, 4, 6, 7

65. Which of the following will show geometrical isomerism.

(1) (2)

(3) (4)

(5) (6)

(A) 1, 2, 3 (B) 1, 2, 5 (C) 1, 2, 4 (D) 3, 5, 6



66.

Incorrect statement for the above structure :

(A) I, II & III have CnH2n-2 general formula (B) I , II & III have same empirical formula(C) I, II & III have same molecular formula (D) I , II are identical and homologue of compound III.

67. Which of the following will show geometical isomer.

(A) (B) (C) (D)

68. Total number of geometrical isomers in the following compound,

Ph–CH=CH– –CH2–CH= –OH

(A) 2 (B) 4 (C) 8 (D) 16

69. Which of the following will not show geometrical isomerism.

(1) (2) (3)

(4) (5) (6)

(7) (8) (9)

(10) (11) (12) = CH – CH = CH – Ph

(13) (14)

(A) 1, 2, 5, 6, 8, 9, 14 (B) 3, 6, 7, 8, 9, 11, 14 (C) 3, 6, 10, 11, 12, 13(D) Except 2, 10, 13

70. Which of the following is pair of geometrical isomerism :

(A) and



(B) and

(C) and

(D) and

71. Write down total number of geometrical isomers after reaction with NH2OH.

(1)

(2)

72. Indicate the type of isomers for the following compounds :

(i) (ii)

(iii) (iv)

(v) (vi)

(vii)

73. In which option all compounds can show geometrical isomerism.

(A)



(B) , C2FClBrI.

(C) , Ph – CH = N – OH, Ph2N2

(D) CH3–CH = N–OH , H2C = CH–CH = CH–CN.

74. Consider the following four statementsS1 : Geometrical isomers are not mirror image isomer.S2 : A compound having double bond (restricted bond) always show geometrical isomerismS3 : Acyclic compounds having single bond does not show geometrical isomerismS4 : Cyclodecene can exist in cis & trans form

(A) T F F T (B) T T F T (C) F F F T (D) T F T T

75. Which will not show geometrical isomerism

(A) (B) (C) ClCH = CHCl (D) Ph—N=N—Ph

76. Which of the following compound posses plane of symmetry -

(A) (B) (C) (D)

77. Which of the following compounds are optically active ?

(A) (B) (C) (D)

78. Well known pain killer Nurofen is an ibuprofen how many stereoisomers it would have ?

If we have a racemic mixture of ibuprofen which one of the following can be used to resolve the isomers.

(A) 4, CH3 – CH2 – OH (B) 8,

Ph|

OHCCH|Ph

3 (C) 4,

Ph|

OHCCH|D

3 (D) 8, CH3 – CH2 – OH


FINAL LAP - 2019ISOMERS79. Which of the following pairs can be separated by fractional distillation :

(A) H

MeMe

H

O

NH

NH and H

MeH

Me

O

NH

NH

(B)

H

H

OH

OH

COOH

COOH

and

HOOC

HO

H

COOH

OH

H

(C) CH –C–O–C H3 3 7

O

and CH –O–C–C H3 2 5

O

(D)

HCOOH

NH2

CH3CH3

H

and

HCOOH

NH2

CH3CH3

H

80. Arrange following in decreasing order of percantage enol content.

(i) (ii) (iii) (iv)

(A) I > II > III > IV (B) II > I > III > IV (C) II > III > I > IV (D) III > II > IV > I

81.||O

Ph Ph

HCH3

OH / H O2

-

Number of isomeric products (m) Fractionaldistillation Number of fractions (n)

values of (m) and (n) are(A) 2, 1 (B) 1, 1 (C) 2, 2 (D) 1, 2

82. Which of the following statement is/are correct about below newmann projection :

(A) I and II are functional isomers. (B) II and III are functional isomers(C) II and III are metamers (D) I and III are metamers



Exercise-1

Q.1 (A, B, C) Q.2 (A, C) Q.3 (D)Q.4 (C)

(II) Cyclic chiral pyramidal molecules are optically active.

(IV) CH3–CH2–CH2–COOH and

COOH|

CHCHCH 33 are chain isomers.

Q.5 (a) II and IV are chiral, hence optically active, (b) I and III are achiral, posses plane of symmetry, henceoptically inactive, (c) There is no enantiomer pair, both II and IV are identical structure, (d) I and II, IIand III, I and III are pair of distereomers.

Q.6 Compound I is optically inactive since it contain a plane of symmetry. Compound II is enantiomeric sinceit does not contain plane of symmetry, hence chiral. Also compound I is polar while II is non polar.]

Q.7 (a) Both I and II are optically, but they are not mirror image of one another, hence, they are distereomers(b) Distereomers ,(c) Enantiomers,(d) I and V are enantiomers,(e) IV and VI are distereomers]

Q.8 [8] Q.9 (a) Enantiomers, (b) Identical, (c) Identical, (d) Identical]

Q.10 (I) Achiral, (II) Achiral, (III) Meso, (IV) Meso, (V) Meso, (VI) Chiral, (VII) Chiral

Q.11 (a) Distereomers, (b) Identical and meso

Q.12 (C) Q.13 (B)

Q.14 (a) Enantiomer, (b) Constitutional, (c) Constitutional, (d) Identical, (e) Identical,(f) Diastereomer, (g) Identical, (h) Enantiomer, (i) Enantiomer, (j) Enantiomer,(k) Diastereomer, (l) Identical, (m) Enantiomer, (n) Enantiomer, (o) Enatiomer, (p) Identical

Q.15 This compound has two chiral carbon, and a double bond capable of showing geometrical isomerismgiving rise to total eight different configurations possible for the molecule as shown below:

Q.16 (a) Both are similar structures, (b) They are positional isomers, (c) They are enantiomers, (d) chain ,(e) Metamers

Q.17 (a)

(b)



(c)

(d) ]

Q.18 (A) Q.19 (C) Q.20 (D)

Q.21 [4] Q.22 [B]

Q.23 All isomer contain symmetry] Q.24 (D)

Q.25(a) The compound must be 1,2-cyclobutan-dicarboxylic acid since all other constitutional isomers are non-resovable.

(b) (i) R, (ii) R, (iii) R, (iv) R, (v) R, (vi) R, (vii) N.R, (viii) R, (ix) N.R, (x) N. R., (xi) N.R. (xii) N.R

Q.26 (C)

Q.27 Enantiomer (A,B) (C,D); Diastereomer (A,C)(A,D);(B,C)(B,D)

Q.29 (D) Q.30 (A) P,Q,R; (B) P; (C) P, R; (D) Q

Q.31 A is and it tautomerise into B as it is more stable because it is aromatic compound

B is


FINAL LAP - 2019ISOMERSExercise-2

1. (C) 2. (C) 3. (B) 4. (B)

5. (B) 6. (A) 7. (A) 8. (D)

9. (B, D) 10. (B) 11. (D) 12. (D)

13. (D) 14. (A) 15. (C) 16. (C)

17. (B) 18. (C) 19. (C) 20. (A)

21. (B) 22. (B) 23. (C) 24. (A)

25. (B) 26. (B) 27. (A) 28. (A)

29. (B) 30. (B) 31. (A) 32. (A)

33. (B) 34. (C) 35. (B) 36. (B)

37. (C) 38. (B, C) 39. (D) 40. (C, D)

41. (C) 42. (C) 43. (C) 44. (D)

45. (A) 46. (B)

47. (A) Chain isomer(B) Metamers(C) Metamers(D) Metamers(E) Homologs

48. (B) 49. (C) 50. (D) 51. (D)

52. (C)

53. (i) Identical (ii) Structural (positional) isomers (iii) Identical (iv) Identical54. (I) Functional isomers (II) Homologs

(III) Homologs (IV) Metamers

55. (B) 56. (C) 57. (D) 58. (D)

59. (B) 60. (C) 61. (D)

62. Functional isomer Functional isomerFunctional isomer MetamerPosition isomer

63. (A) 10, 2

(1) C = C – C – C – C (2) C – C = C – C – C (3) C–C–CC|C

(4) C–C–CC|C

(5) C–CC–C

|C

(6) (7) (8) (9) (10)

64. (C) 65. (D) 66. (D) 67. (A, C)


FINAL LAP - 2019ISOMERS68. (C) 69. (B) (B)

70. (A, B, D)

71. 8, 16

72. (i) Functional isomers (ii) Functional isomers (iii) Position isomers (iv) Functional isomers(v) Position isomers (vi) Chain isomers (vii) Position isomers

73. (D) 74. (D) 75. (A) 76. (C)

77. (D) 78. (C) 79. (C)

80. (A, C) 81. (A) 82. (C) 83. (A, C)


FINAL LAP - 2019REACTION MECHANISM

Write the structure of more stable rearranged form of the following carbocations:

(1) CH3CH2CH2+ (2) 323 CHCHCH)CH(

(3)

+

3 3 3(CH ) CCHCH

(4) (CH3CH2)3CCH2+ (5) (6)

(7)

OCHCH|CHCHHC

23

222

(8) (9) +

(10) (11)

OHCH

CHCCCH

H

3

33||

|

(12)

33

32

33

CHCH||

CHCHHCCCCH||CHCH

(13) (14) (15)

(16)

OH|

33

33

CHCCCH||CHCH

(17) (18) (19)

(20) 23 HCCHCHCH

(21) (22) (23)

(24) (25) (26) (27) (28)

(29) (30) (31) (32)


FINAL LAP - 2019REACTION MECHANISM

(6) (7) OCHCH|

CHCHCH

23

23

(8)

(10) (11)

OHCH||

CHCCCH|H

3

33

(12)

33

32

33

CHCH||

CHCHCHCCCH||CHCH

(13) (14) (15)

(16) (17) (18) (19)

(20) 23 CHCHHCCH

(21) (22)

(23) OCCH3

(24) (25) (26)

(27) (28) (29)

(30) (31) (32)


FINAL LAP - 2019REACTION INTERMEDIATES

EXERCISE-I

Q.1 2-chloropentane on halogenation with chlorine gives 2,3, dichloropentane. What will be the structure offree radical species formed in the reaction?(A) Planar (B) Trigonal planar (C) Square planar (D) Pyramidal

Q.2 The correct order of rate of Wurtz recation.

(I) CH2–F ether

Na CH2–CH2

(II) CH2–Cl ether

Na CH2–CH2

(III) CH2–Br ether

Na CH2–CH2

(IV) CH2–I ether

Na CH2–CH2

(A) I > II > III > IV (B) II > I > III > IV(C) IV > III > II > I (D) In all rate of Wurtz reaction is same

Q.3KCOCHCH

|KCOCHCH

23

23

iselectrolys (A) (Major)

Major product (A) of above reaction

(A) (B) (C) (D)

Q.4 Consider the following reaction –

O

OEt

3NH.liq

ether/Na reagent

s'Fenton224 OHFeSO P

the major product P is:

(A) O OH

(B) O O



(C) OO H

(D) OHO H

Q.5 Peroxide.CCl

NBS

4

ether/Na (X)

X is

(A) (B)

(C) (D) None of these

Q.6 Find out the correct order of rate of reaction towards allylic substitution.

(I) CH3–CH = CH2 (II) CH3–CH2–CH=CH2 (III) 23

3

CHCHCHCH|

CH

(A) I > II > III (B) II > I > III(C) III > II > I (D) III > I > II

Q.7 What will be the major product, when 2-methyl butane undergoes bromination in presence of light?(A) 1-bromo-2-methyl butane(B) 2-bromo-2-methyl butane(C) 2-bromo-3-methyl butane(D) 1-bromo-3-methyl butane

Q.8 Which can not be the possible product of the given reaction

O||

OAgCCHCH 23

,CClBr

4

2 product(s)

(A) CH3 – CH2 – Br (B)

O||

CHCHOCCHCH 3223

(C) CH3 – CH2 – CH2 – CH3 (D) CH3 – CH2 – CH3


FINAL LAP - 2019REACTION INTERMEDIATESQ.9 Pick the correct statement for monochlorination of R-secbutyl Bromide.

Me

Et

HHBr C300

Cl2

(A) There are five possible product ; four are optically active one is optically inactive(B) There are five possible product ; three are optically inactive & two are optically active(C) There are five possible product ; two are optically inactive & three are optically active(D) None of these

Q.10 Correct order of rate of photochlorination for following compounds is:

CH3–CH3 CD3–CD3

3

33

3

CH|

CHCCH|HC

(I) (II) (III)(A) II < I < III (B) I < II < III (C) III < I < II (D) II < III < I

Q.11

,CCl

NBS

4Allylic brominated products

Find out the incorrect statement.(A) It gives total 9 allylic brominated products(B) All allylic brominated products are optically active(C) Substrate has 7 allylic hydrogens(D) NBS gives Br2 constantly to reaction mixture.

Q.12 Which of the following carbocation is most stable?

(A) (B)

(C) (D)

Q.13 Which carbocation is least likely to form as an intermediate?

(A) C)HC( 356 (B) (C) (D) HCCH2


FINAL LAP - 2019REACTION INTERMEDIATESQ.14 For the reactions

(I) + Clr , o1H (II) + Clr , o

2H

(III) + Cl , o

3H (IV) + Cl , o4H

The correct decreasing order of enthalpies of reaction for producing carbocation is

(A) o1H > o

2H > o3H > o

4H (B) o4H > o

1H > o2H > o

3H

(C) o3H > o

2H > o1H > o

4H (D) o2H > o

1H > o4H > o

3H

Q.15Br

(I), which is not the correct statement

(A) I is more soluble than bromocyclopropane(B) I gives pale yellow ppt. on addition with AgNO3(C) I is having lower dipole moment than bromocyclopropane

(D) I is more ionic than

Q.16 Which one of the following carbocation would you expect to rearrange.

(A) (B) (C)

(D)

Q.17 How many 1,2-Shifts are involved during the course of following reaction:

(A) 1 (B) 2 (C) 3 (D) 4

Q.18 How many 1,2-Shifts are involved during the course of following reaction:

42SOH.conc

(A) 1 (B) 2 (C) 3 (D) 4



Q.19 H (X).

Product (X) is

(A) (B) (C) (D)

Q.20 Among the given compounds, the correct dehydration order is:

(I) (II) (III) (IV)

(A) I < II < III < IV (B) II < III < IV < I(C) I < III < IV < II (D) I < II < III = IV

Q.21 C5

H

P. The product P is:

(A) (B) (C) (D)

Q.22 Rate of dehydration when given compounds are treated with conc. H2SO4.

(P) (Q) (R) (S)

(A) P > Q > R > S (B) Q > P > R > S (C) R > Q > P > S (D) R > Q > S > P

Q.23 42SOH X

X is

(A) (B) (C) (D)



Q.24 H major product is

(A) (B) (C) (D)

Q.25 OH3 A, A is

(A) (B) (C) (D)

Q.26H

(A) (B) (C) (D)

Q.27

42SOH.ConcA.

Product A is:

(A) (B) (C) (D)

Q.28 How many products are obtained in the given reaction :

Et

Et Ph

Ph

OHHO +

CH3

CH3 Ph

Ph

OHHO 42SOH

(A) 1 (B) 2 (C) 3 (D) 4



Q.29 Which of the following is not correct about P2.O

OH

Mg

2 P1

H P2

(A) It is a spiro compound (B) It is a Ketone(C) It can show tautomerism (D) It is an alkene

Q.30

33

223

3

CHCH||

NHCH—C—CHCH|CH

2HNO (X) (major)

Major product of above reaction is

(A) (B)

(C) (D)

Q.31 Which will dehydrate at fastest rate by H3PO4:-(A) 2-methyl butane-2-ol (B) 3-methyl butane-2-ol(C) Butane-1-ol (D) 2-methyl butane-1-ol

Q.32 What is the order of reactivity with HBr.

(a) (b) (c)

(A) a > b > c (B) b > a > c(C) c > b > a (D) b > c > a

Q.33

CH3

DH

CH = CH3

4CCl

HBr

What is stereochemistry of product?(A) Racemic mixture (B) Optically inactive(C) Diastereomers (D) Meso product


FINAL LAP - 2019REACTION INTERMEDIATESQ.34 In the given reaction:

2Br [X]

[X] is:

(A) (B) (C) (D)

Q.35 Select the incorrect statement about the product mixture in the following reaction :

4

2

CCl

Br Products

(A) it is optically active (B) it is racemic mixture(C) it is a resolvable mixture (D) it is a mixture of erythro compounds


FINAL LAP - 2019REACTION INTERMEDIATESEXERCISE-IIQ.1 Which of the following can be produced by Wurtz reaction in good yield.

(A) (B) (C) (D)

Q.2 Select true statement(s):(A) Instead of radical substitution, cyclopropane undergoes electrophilic addition reactions in sun light.(B) In general, bromination is more selective than chlorination.(C) The 2,4,6-tri-tert, butylphenoxy radical is resistant to dimerization.(D) The radical-catalysed chlorination, ArCH3 ArCH2Cl, occurs faster when Ar = phenyl than whenAr = p-nitrophenyl.

Q.3 Choose all alkane that give only one monochloro derivative upon reaction with chlorine in sun light.

(A) (B) (C) (D)

Q.4

h/CCl

NBS

4 HBr

(X) + (Y) enantiomeric pair

(A) (B) (C)

Br

Br(D)

Br

Br

Q.5 Select correct statement about the product (P) of the reaction:

42 CCl/Br P

(A) P is optically inactive due to internal compensation(B) P is optically inactive due to the presence of plane of symmetry in the molecule(C) The structure of P can have three optical isomers possible.(D) P can have four possible optical isomers.

Q.6 Products formed when HCl adds to 2,4- hexadiene is:(A) 4-chloro-2-hexene (B) 2-chloro-3-hexene(C) 2-chloro-4-hexene (D) 1-chloro-2-hexene

Q.7 In the given reaction

C7H12 (A) HBr

(A) can be

(A) (B) (C) (D)


FINAL LAP - 2019REACTION INTERMEDIATESQ.8 Which of following reaction product are diastereomer of each other.

(A) 4

2

CCl

Br (B)

H)ii(

NaCN)i(

(C) 4CCl

HBr (D)

Et|

PhCHCHCHCH3 peroxide

HCl

Q.9 Which of the following can be formed during this reaction?

OH OH3

(A) OH

(B) OH

OH

(C) O

(D) O

Q.10 Each of the compounds in column A is subjected to further chlorination. Match the following for them.Column A Column B

(A) CHCl2–CH2–CH3 (P) Optically active original compound

(B) CH2Cl–CHCl–CH3 (Q) Only one trichloro product

(C) CH2Cl–CH2–CH2–Cl (R) Three trichloro product.

(D) CH3–CCl2–CH3 (S) Four trichloro product

(E)

33

33

CHCH||

CHC—CCH||

ClCl

(T) Atleast one of the trichloro product is

optically active.

(U) Two trichloro products.


FINAL LAP - 2019REACTION INTERMEDIATESEXERCISE-III

Q.1 Identify P1 to P8.

CH3 – CH = CH – CH2 CH2 CH3

D T

H H

BrH P1 + P2

CH3 – CH =CH –CH2–

D|

CHCH 2 CH3

T

H BrH P1 + P2 + P3 + P4

CH3 – CH = CH –

TD||

CHCHCHCHCH 322 BrH P1 + P2 + P3 + P4 + P5 + P6 + P7 + P8

Q.2 Give the product of the following reaction.

(i) O

OH

Mg

2 A (ii)

CH=OCH=O OH

Mg

2 B

(iii) EtCMe||

O

OH

Mg

2

42SOH C (iv) MeCPh

||O

OH

Mg

2

42SOH D

Q.3 Identify missing products in the given reaction sequence.CH3 – CH2 – CH3 hv/Br2 (A) KOHaq (B)

42SOH (C)

4

2

CCl

Br (D) = ?

Q.4 Find out the total no. of products (including stereo) in the given reaction :CH3

NBS, CCl4

Peroxide, Products.

Q.5 With the help of following data show HBr exhibits the peroxide effect.H1

0/kJ mol–1 H20/kJ mol–1

H–X X

+ CH2 = CH2 X CH2 – C

H2 XCH2– C

H2 + H–X XCH2CH3 + X

HCl –67 + 12.6HBr –25.1 – 50.2HI +46 –117.1

Q.6 Addition of small amount of (C2H5)4Pb to a mixture of methane and chlorine, starts the reaction at140°C instead of the usual minimum 250°C. Why?

Q.7 On chlorination, an equimolar mixture of ethane and neopentane yields neopentyl chloride andethyl chloride in the ratio 2.3 : 1. How does the reactivity of 1° hydrogen in neopentane compare withthat of a 1° hydrogen in ethane?



Q.8 It required 0.7 g of a hydrocarbon (A) to react completely with Br2 (2.0 g) and form a non resolvableproduct. On treatment of (A) with HBr it yielded monobromo alkane (B). The same compound (B) wasobtained when (A) was treated with HBr in presence of peroxide. Write down the structure formula of(A) and (B) and explain the reactions involved.

Q.9 Give product(s) in each of the following reactions .

(a) CH3 –

3CH|CH – CH2 – CH2– CH3 hv/Br2 (A) (major)

(b) + NBS 256 )COOHC( (B)

(c) CH3 – CH2 – CH = CH2 + Me3COCl hv (C) + (D)

(d) C6H5 – CH2 – CH2 – CH3

3

3

3

CH|

/ClOCCH|

CH

(E) (major)

Q.10 We saw that acid-catalyzed dehydration of 2,2-dimethyl-cyclohexanol afforded 1,2-dimethylcyclohexene.To explain this product we must write a mechanism for the reaction in which a methyl shift transforms asecondary carbocation to a tertiary one. Another product of the dehydration of 2,2-dimethylcyclohexanolis isopropylidencyclopentane. Write a mechanism to rationalize its formation.

heatH

+ C(CH3)2

2,2-Dimethylcyclohexanol 1,2-Dimethylcyclohexene Isopropylidenecyclopentane

Q.11 (a) Write a reasonable and detailed mechanism for the following transformation.

42SOH

.conc + H2O

(b) HOH/H

-Terpeniol


FINAL LAP - 2019REACTION INTERMEDIATESQ.12 Assuming that cation stability governs the barrier for protonation in H – X additions, predict which

compound in each of the pairs in parts (a) & (b) will be more rapidly hydrochlorinated in a polar solvent.(I) (II)

(a) CH2 = CH2 or

(b) or

Q.13 Choose the member of the following pairs of unsaturated hydrocarbons that is more reactive towardsacid-catalysed hydration and predict the regiochemistry of the alcohols formed from thi s compound.

(a)

(I)

or

(b) or

(c) or

Q.14 Give product in the following reaction.

(i) NH2

HCl

NaNO2 A (ii) NH2 HCl

NaNO2 B

(iii)

CH2NH2

HCl

NaNO2 C (iv)

NH2

HCl

NaNO2 D

(v) NH2OH

HCl

NaNO2 E

Q.15 What are the products of the following reactions ?

(a) PhCH = CHCH3 + HBr A (b) + HI B

(c) + HBr Peroxide C (d) + HCl D


FINAL LAP - 2019REACTION INTERMEDIATESQ.16 Complete following reaction:

(a) HCl

(b)CH3

4

2

CCl

Br

(c) 2

2

CS

Cl

Total number of products obtained in this reaction is ?

Q.17

OH

H Write the mechanism.

Q.18 Compare rate of reaction towards pinacol pinacolone rearrangement.

)I(

OHOH||

Ph—C—C—Ph||

CHCH 33

)II(

OHOH||

Ph—C—C—CH||PhCH

3

3

)III(

OHOH||

CH—C—C—CH||CHCH

33

33

Q.19 Calculate the percentage of products i.e. P1, P2 & P3, if reactivity for chlorination with 1°H, 2°H & 3°His 1 : 3.8 : 4.5 respectively.

+ Cl2 nationMonochlori

h

Cl

ClCl

(P1) (P2) (P3)

+ +

Q.20 Write all the monochlorinated products (including stereo) of isohexane.



EXERCISE-IQ.1 B Q.2 C Q.3 C Q.4 B Q.5 C Q.6 C Q.7 B

Q.8 D Q.9 D Q.10 A Q.11 B Q.12 C Q.13 C Q.14 B

Q.15 C Q.16 B Q.17 D Q.18 C Q.19 A Q.20 A Q.21 D

Q.22 C Q.23 D Q.24 C Q.25 B Q.26 D Q.27 D Q.28 B

Q.29 B Q.30 C Q.31 A Q.32 B Q.33 C Q.34 D Q.35 A

EXERCISE-IIQ.1 B, D Q.2 B, C, D Q.3 A, B, D Q.4 B, D

Q.5 A, B, C Q.6 A, B Q.7 A, B, C Q.8 A, B, D

Q.9 A, B, D Q.10 (A) S,T (B) P,S,T (C) U, (D) Q, (E) T,U

EXERCISE-III

Q.1 P1 =

TDBr|||

CHCCHC)CH(CCH|||

HHH

32223 P2 =

TDH|||

CHCCHC)CH(CCH|||HHBr

32223

P3 =

THBr|||

CHCCHC)CH(CCH|||HDH

32223 P4 =

THH|||

CHCCHC)CH(CCH|||

HDBr

32223

P5 =

HHBr|||

CHCCHC)CH(CCH|||TDH

32223 P6 =

HHH|||

CHCCHC)CH(CCH|||

TDBr

32223

P7 =

HDH|||

CHCCHC)CH(CCH|||

THBr

32223 P8 =

HDBr|||

CHCCHC)CH(CCH|||THH

32223

Q.2 (i) (A)

OH

OH(ii) (B)

OH

OH

*

*

RR, SS, RS (3 products)



(iii) (C) CH – C – C –3

CHEtEt

3

||O

(iv) (D) CH – C – C –3

ChPhPh

3

||O

Q.3 CH3 – CH2 – CH3 hv/Br2

)A(CHCHCH

|rB

33 KOHaq

)B(CHCHCH

|OH

33

)D(BrBr||

CHCHCH 32 4

2

CCl

Br )C(

CHCHCH 32 H2SO4

Q.4 9 Q.7 1.15 times more reaction Q.8 A =

Q.9 (a) A:

3

3223

CH|

CHCHCHCCH|Br

(b) B:

(c) C:

Cl|

CHCHCHCH 23 D: ClCHCHCHCH 23

(d) E:

Cl|

EtCHPh

Q.11 (b) H

H/HOH

Q.12 (a) II, (b) I Q.13 (a) II, (b) I, (c) II ;

Q.15 (a)

Br|

EtCHPh , (b) Me2C(I)–Et, (c) , (d)

Q.16 (a)CH3 CH3

CH3

CH3

H HCl

Cl+

(diastereoisomers)



(b) + (diastereoisomers)

(c) H HH HCl Cl

Cl Cl

Me Me

Et Et

+ (Enantiomers)

Q.17

OH OH2

H

–H

–H O2

H

Q.18 More stable the carbocation, more will be the rate of reaction. II > I > III

Q.19 P1 = 13.26 %, P2 = 66.96 %, P3 = 19.82 % Q.20 8 products


FINAL LAP - 2019GRIGNARD REAGENT

Q.1 The order of reactivity of alkyl halide in the reaction R–X + Mg RMgX is(A) RI > RBr > RCl (B) RCl > RBr > RI (C) RBr > RCl > RI (D) RBr > RI> RCl

Q.2 Br–CH2–CC–CH2–BrOEt

)excess(

Mg

2

BrMg–CH2–CC–CH2–MgBr

ProductThe major product is:(A) Br–Mg–CH2–CC–CH2–Br (B) Cyclobutyne(C) —)CHCCCH(— n22 (D) CH2 = C=C=CH2

Q.3 On conversion into Grignard followed by treatment with ethanol, how many alkyl halides (excludingstereoisomers) would yield 2-methyl butane.(A) 2 (B) 3 (C) 4 (D) 5

Q.4 Which of the following reacts with Grignard reagent to give alkane?(A) nitro ethane (B) acetyl acetone (C) acetaldehyde (D) acetone

Q.5 How many litres of methane would be produced when 0.595 g of CH3MgBr is treated with excess ofC4H9NH2(A) 0.8 litre (B) 0.08 litre (C) 0.112 litre (D) 1.12 litre

Q.6 How many litres of ethene would be produced when 2.62 g of vinyl magnesium bromide is treated with224 ml of ethyne at STP.(A) 0.224 litre (B) 0.08 litre (C) 0.448 litre (D) 1.12 litre

Q.7

MgBr OH

+ AA

(A) (B) (C) (D)

O – Ph

Q.8 In which of the following reactions 3°alcohol will be obtained as a product.

(A) MgBr (excess) + ClCH||

O

H

(B) PhMgBr (excess) + ClCCH||

O

3 H



(C) CH3MgBr (excess) + 33 CHCOCCH||||

OO

H

(D) CH3MgBr (excess) + EtOCCl||

O

H

Q.9 ether

Mgequivalent1 X OD2 Y; Y is is

(A) (B) (C) (D) None of these

Q.10 Compounds are shown with the no. of RMgX required for complete reaction, select the incorrect option(A) CH3COOC2H5 1(B) CH3COCl 2(C) HOCH2COOC2H5 3

(D) 4

Q.11 What will be the order of reactivity of the following carbonyl compounds with Grignard's reagent?

(I) C = OH

H(II) C = O

H

CH3

(III) C = OCH3

CH3(IV) C = O

Me C3

Me C3(A) I > II > III > IV (B) IV > III > II > I(C) II > I > IV > III (D) III > II > I > IV

Q.12 Carbonyl compound (X) + Grignard reagent (Y) Me PhOH

EtX , Y will be

(A) PhCEt||

O

, Me Mg Br (B) PhCMe||

O

, Et Mg Br

(C) EtCMe||

O

, Ph Mg Br (D) PhCEt||

O

, Et Mg Br



Q.13 (R) - 2-Bromooctane

H)iii(CO)ii(

Mg)i(

2

X

X is

(A) (B)

(C) A and B both (D) None of these

Q.14 In which one of the following reaction products are not correctly matched in(A) RMgX + CO2

H)2(Carboxylic acid

(B) RMgX + C2H5OH Alkane(C) RMgX + CH3CH2Cl Alkene

(D) RMgX + Cl O Ether

Q.15 The number of moles of grignard reagent consumed per mole of the compound

is:

(A) 4 (B) 2 (C) 3 (D) 1

Q.16 Select the correct statement:(A) 1,4-dibromobutane react with excess of magnesium in ether to generate di-Grignard reagent.(B) 1,2-dichlorocyclohexane treated with excess of Mg in ether produces cyclohexene.(C) Vicinal dihalides undergo dehalogenation to give alkene when heated with Zn dust or Mg.(D) 1,3-dichloropropane by treatment with Zn dust or Mg forms cyclopropane.

Q.17 CH3–CH=CH2 2Br

etherDry

Mg ClNH

CHCCH||O

4

33

H

End product of above reaction is

(A) 322

2

CHCCHCHCH||CH

(B)

3

32

CH|

CHCCHCHCH

(C)

3

322

CH|

CHCCHCHCH|OH

(D)

3

222

CH|

OHCHCHCHCHCH



Q.18 MgBr + H–C–Cl

O

product.

(A) 3CHC||

O

(B) CH2CH=O

(C) CH = O (D)

OCH|

CHCH 3

Q.19Br

OOH|||

OH.3CHCHCHCH.2

ether/Mg.1

323

Product (s)

Select the product from the following

I : II :

OOH|||

CHCHCHCH 23 III :

(A) III (B) I, III (C) I, II (D) II, III

Q.20 5252 HOCCOHC||

O

MgBrCH2 3 A. Product A formed

(A) is ethyl acetate(B) further react with CH3MgBr/H2O+ to give acetone(C) further react with CH3MgBr/H2O+ to give t-butyl alcohol(D) Can give pinacol when treated with Mg followed by H2O

Q.21 Order of rate of reaction of following compound with phenyl magnesium bromide is:

IClCMe

||O

II

HCMe||O

III

EtOCMe||O

(A) I > II > III (B) II > III > I (C) III > I > II (D) II > I > III

Q.22 Select the correct order of decreasing reactivity of the following compounds towards the attack ofGrignard reagent(I) Methyl benzoate (II) Benzaldehyde (III) Benzoylchloride (IV) Acetophenone(A) II > III > I > IV (B) III > IV > II > I(C) III > II > IV > I (D) II > IV > I > III

Q.23 O CH MgX3

NH Cl4Product is

(A) Enantiomer (B) Diastereisomer (C) Meso (D) Achiral



Q.24 Nucleophilic addition of Grignard reagent cannot occur in

(A) 33 CHCCCH||||

OO

(B) 323 CHCCHCCH||||

OO

(C) 3223 CHCCHCHCCH||||

OO

(D) O

O

Q.25

O||

ClCHCHCCHCH 2223 MgBrCH3 A, A is

(A)

OH|

ClCHCHCCHCH|

CH

2223

3

(B)

O||

CHCHCHCCHCH 32223

(C) (D)

Q.26 32223 CHCOCHCHCCHCH||||

OO

OH)ii(

)molone(MgBrCH)i(

3

3 A, A formed in this reaction is

(A)

3

32223

CH|

CHCOCHCHCCHCH|||

OOH

(B) 3223 CCHCHCCHCH||||

OO

(C) (D)

OHOH||CCHCHCCHCH||

CHCH

3223

33

Q.27 PhCH3 hCl2 (A)

etherMg (B)

ClNHCHOCH

4

3 (C)

(A) CH – CH – CH2 3

|OH

(B)

CH3

CH – CH3

|OH

(C)

CH3

CHCH3HO

(D)

CH3

C – CH3

CH3

OH



Q.28 Select the correct order of reactivity towards Grignard reagent for nucleophilic attack.

(A) RCR||

O

> HCR||

O

(B)

O||

HCCHCl 2 >

O||

HCCHCH 23

(C) OCCH||

O

3 NO2 < OCCH||

O

3

(D) ORCR||

O

> 2NRCR||

O

Q.29 In the reaction sequence:

H/OH)ii(

CuCl/MgBrCH)i(

2

3 (X) Major + (Y)

(X) & (Y) respectively are

(A)

OH

CH3

,

CH3 OH

(B) ,

(C) , (D) ,

Q.30C

OHO

O

SHH

HCC

(1)

(4)

(3)

(2)

)moles2(RMgX

Deprotonation will occur from the following positions:(A) 1,2 (B) 1,3 (C) any two positions (D) 1,4

Q.31 Which of the following reacts with 4 moles of RMgX.

(A)

COCl

CHO(B)

CH –Cl2

COOEtOH

ROC



(C) O

O

H

EtO

(D) O

O

CHOCH=CH2

Et

RN

Q.32Br

Mg (A) OH/H)ii(

CO)i(

2

214

(B) 3NaHCO (C) gas

Product C is(A) CO (B) 14CO2(C) CO2 (D) A mixture of 14CO2 and CO2

Q.33 2CH3MgBr

H)ii(

ONHCH)i( 23

(A) CH3–O–NH–CH3 (B) CH3–NH–CH3(C) CH3–NH2 (D) CH3–OH

Q.34

O

OH/H)ii(

MgBrCH)i(

2

3

(A)

OH

CH3(A) The product is optically active(B) The product contains plane of symmetry(C) The product shows geometrical isomerism.(D) The product shows optical isomerism.

Q.35 Which of the following is incorrect.

(A) Cl OC H2 5

C

O

)eq1(

MgXCH3 523 HOCCCH||

O

(B) OC H2 5

OC H2 5

OC H2 5

CH3–C)eq1(

MgXHC 52 523 HOCCCH||

O

(C) CH3MgX + C||

S

= S OH3 SHCCH

||S

3

(D) CH3MgX + C||

O

= O OH3 OHCCH

||O

3


FINAL LAP - 2019GRIGNARD REAGENT Q.36 Which of the following reacts faster with RMgX.

(A) BrCR||

O

(B) HCR||

O

(C) OEtCR||

O

(D) 2NHCR||

O

Q.37 CH3MgBr + CH2=CH C||

O

H OH3 Product (1, 4 addition). It is

(A)

3

2

CH|

HCCHCH|OH

(B)

OH|

CHCHCHCH 32

(C) CH3CH2CH2CHO (D) none

Q.38

O

Me

ClNH)ii(

PhMgBr)i(

4 Product.

Products in this reaction will be(A) Stereoisomers (B) Enantiomer (C) Diastereomers (D) Geometrical isomers

Q.39 .)eq1(MgBrCH3 ?

The product is:

(A) (B) (C) (D)

Q.40 CH2 = C = O

)equi2(MgBrCH)ii(

Br)i(

3

2 C4H8O

(A) (B) (C) (D) All of these



Q.41 RMgX OH)ii(

CNCH)i(

3

3 (A) ClNH.aq)ii( 4

RMgX)i( (B) will be

(A) 1° ROH (B) 2° ROH (C) 3° ROH (D) Alkene

Q.42 CH3—CH—CH2

O

OH

MgClCH

2

3

(A)

3

23

CH|

OHCHCHCH (B)

OH|

CHCHCHCH 323

(C)

3

33

CH|

CHCHCH (D) HO – CH2 – CH2 – CH2 – CH2 – OH

Q.43 The reaction of 1 mole each of p-hydroxy acetophenone and methyl magnesium iodide will give

(A) CH4 + IMgO C—CH3

O(B) CH –O3 C—CH3

O

(C) CH –C3 OH

OMgI

CH3

(D) CH O3 C—CH3

OMgI

Q.44 (i) O

+ Ph Mg Br 1r Ph CH2 CH2 OH

(ii) O

+ Ph Mg Br 2r Ph CH2 CH2 CH2 OH

(A) r2 > r1 (B) r1 > r2 (C) r1 = r2 (D) r1 = 2r2

Q.45 How many moles of Grignard reagent will be required by one mole of given compound?

C – OEt

C – Cl

O

OCH –CH2 2

Cl

SHHO

(A) 7 (B) 6 (C) 8 (D) 5

Q.46 Consider the given organometallic compound.(I) (CH3)2Hg (II) (CH3)2Zn (III) (CH3)2Mg (IV) CH3LiThe correct decreasing order of ionic character is(A) I > II > III > IV (B) II > I > III > IV (C) I > III > II > IV (D) IV > III > II > I



Q.47 CH3CH = 3CHCCH||

O

(i) CuI,CH MgBr3

(i) CH MgBr3

(ii) H O3+

(ii) H O3+

P

Q

(A) P is CH3CH =

Me|

MeCCH|

OH

Q is CH3CH =

Me|

MeCCH|

OH

(B) P is

3

323

CH|

CHCCHCHCH||

O

Q is CH3CH =

Me|

MeCCH|

OH

(C) P is

OH|

)CH(CCHCHCH 233 Q is

O||

CHCCHCH)CH( 3223

(D) P is

O||

CHCCHCH)CH( 3223 Q is

OH|

)CH(CCHCHCH 233

For Q. No.48 to Q. No. 50Consider the given reaction and answer the following questions

COOCH3

OCH3

SH||OO

O

)excess(

MeMgBr Products

Q.48 No. of RMgX consumed in the reaction is(A) 4 (B) 5 (C) 6 (D) 7

Q.49 How many product will be fromed in given reaction (excluding stereo)(A) 2 (B) 3 (C) 4 (D) 5

Q.50 Which of the following reaction will give the same Hydrocarbon formed as one of the product in theabove reaction.(A) EtMgBr + Me – OH (B) PhMgBr + Me – OH (C) MeMgBr + Ph – OH (D) MeMgBr + CH3 – CHO

Q.51 Compare the two methods shown for the preparation of carboxylic acids:

Method 1: RBr etherdiethyl

Mg RMgBr

OH)ii(

CO)i(

3

2 RCO2H

Method 2 : RBr NaCN RCN heat

HCl,OH2 RCO2H



Which of the following statements correctly describes this conversion?

(A) Both method 1 and method 2 are appropriate for carrying out this conversion.(B) Neither method 1 nor method 2 is appropriate for carrying out this conversion.(C) Method 1 will work well, but method 2 is not appropriate.(D) Method 2 will work well, but method 1 is not appropriate.

Q.52 Which of the given compound can not show acid-base reaction with Grignard reagent.

(A) CH3–NO2 (B) (C) Cl (D)

Q.53 (a)

O

O

O

OClCl

Cl

SHOMe

C

)excess(

RMgCl Number of moles of Grignard reagent consumed.

(b) Number of mole of Grignard consumed in given molecule.(When Grignard reagent is in excess)

OO

OONH2

NHCl



Q.1 A Q.2 D Q.3 C Q.4 A,B Q.5 C

Q.6 C Q.7 A Q.8 B,C,D Q.9 D Q.10 A

Q.11 A Q.12 A,B,C Q.13 C Q.14 C Q.15 A

Q.16 A,C,D Q.17 B Q.18 C Q.19 C Q.20 C,D

Q.21 A Q.22 C Q.23 A Q.24 B,D Q.25 C

Q.26 C Q.27 A,B,C Q.28 B,D Q.29 B Q.30 A

Q.31 D Q.32 C Q.33 C,D Q.34 B,C Q.35 B

Q.36 D Q.37 C Q.38 A,C,D Q.39 D Q.40 A

Q.41 C Q.42 B Q.43 A Q.44 B Q.45 A

Q.46 D Q.47 C Q.48 C Q.49 C Q.50 C

Q.51 C Q.52 D Q.53 (a)-7, (b)-7


FINAL LAP - 2019OXIDATION & REDUCTION

OXIDATION OF ALKENES, ALCOHOLS & CARBONYL COMPOUNDS

(I) OXIDATION OF ALKENES

R–CH=CR2 R–CH—CR2

R–CH—CR2

OsO4

Cold dil.alkalineKMnO4

H2OOH

OH

OH

OH* Cold dil. alkaline KMnO4 is called as Bayer’s reagent.* Overall syn addition* Given by alkenes & alkynes* Benzene & Cyclopropane can not give this reaction.If we use acidic KMnO4 or warm KMnO4 or too concentrated KMnO4 the oxidative cleavage ofGlycol occurs resulting in mixture of Carboxylic acids & Ketones.

R–CH = CR2

4KMnO,H RCOOH + R2C = O

Hot acidic KMnO4, Hot acidic K2Cr2O7 & hot acidic NaIO4 gives same result with alkene. The effectis similar to that of oxidative ozonolysis on alkenes.

Preilschaive reaction :Epoxidation of alkenes is reaction of alkenes with peroxyacids.

CH2=CH2 + HOOCCH||

O

3 CH –CH2 2

O + OHCCH

||O

3

With the decrease in nucleophilicity of double bond, rate of reaction decreases. With the decrease in ewithdrawing substituents in leaving group, rate decreases.

R RRR

CH2 CH2–OHCH2–OH

–H+

H O2 –CHCH2

CH HO–CHCHO O–H

dil.H SO2 4

H+

+ +

H O2

RCO H3

RCO H3O

OH

OH

OH

OH

HCO H3

H O3

R

CH2

CHO

R

CH2

O212

AgCH +



(II) OXIDATION OF ALCOHOLSOxidising agents

(1) Cu / 300°C (or Red hot Cu tube) (2) H /KMnO , 4 (Strong oxidising agent)(3) H /K Cr O , 2 2 7 (Strong oxidising agent) (4) PCC (Pyridinium chloro chromate)

HN CrO3 Cl or N CrO3 + HCl

(5) Collin’s reagent (6) Sarett reagent (i.e. PCC in CH2Cl2)

(N

(2 mol) + CrO3 + CH2Cl2)N

+ CrO3 + HCl + CH2Cl2

(7))dichromatePyridinium(

PDC

2

Cr O2 7

N

(8) Jones reagent (H2CrO4 in Anhydrous acetone)

or CrO3 + H2SO4 in acetone.Sufficiently mild so that it oxidises alcoholswithout oxidising or rearranging double bonds(8 or 9)

(9) TsCl + DMSO + NaHCO3 (10) MnO2-Oxidises only allylic or benzylic–OH.i.e.

RCH2OH ClTs RCH2OTss3NaHCO

DMSO RCHO 1° Allylic or benzylic OH 2MnO Aldehyde

R2CHOH ClTs R2CH–OTss3NaHCO

DMSO R2CO 2° Allylic or benzylic OH 2MnO Ketone

R3COH ClTs R3C–OTss3NaHCO

DMSO × No effect on 3° ROH and on Carbon-carbon

multiple bond.(11) Periodic cleavage (12) NBS

A similar oxidation is obtained incase of HIO4known as periodic cleavage. (13) Openaur oxidation

R–CH–OH

R– CH — O – I = O

R–CH

R C–OH2

R C–O–H2

+ +

+

HO–I=O

O

O

O

O

O

HIO3

R C=O2

R – CH – R

(R CH–O) 2 2

R CH–OH + 2

R – C – R

CH –C – CH3 3OH

O

O

Al(OCMe )3 3

Al(OCMe )3 3

Me COH + Al(OCHR )3 2 3

Al–O

3R C=O + (Me C–O)2 2 3

3Me C2

O CR2H

AlBut reaction is only observed for Vic-diols. Oxidation of alcohol with aluminium tertiary

butoxide is Openaur oxidation.OH

Al (OCMe )3 3

O

acetone


FINAL LAP - 2019OXIDATION & REDUCTION Different oxidising agents are used to oxidise alcohols in corresponding carbonyl compounds and carboxylicacids.

e.g. (I)alcohol1

OHCHR 2

agent

oxidisingmild

O||

HCR (Aldehyde) eg. 1,4,5,6,7,8,9,12

(II)alcohol2

'RCHR|

OH

agent

oxidisingmild 'RCR||

O

(Ketone) eg.1,2,3,4,5,6,7,8,9,12,13

(III)alcohol1

OHCHR 2

agent

oxidisingstrong

O||

OHCR eg. 2,3

(IV)

3

3

3

CH|

OHCCH|

CH

C300Cu

CH3

CH2

CH3

C Dehydration takes place.

(V) Double bond or Tripple bond is not affected by 1,4,5,6,7,8,9,10(VI) No effect on 3° alcohol by 2,3,4,5,6,7,8,9,10,12,13

(III) OXIDATION OF CARBONYL COMPOUNDS.1. RCHO + [Ag(NH3)2]OH RCOOH + NH3 + Ag

Aldehyde acts as reducing agent, they can reduce mild oxidizing agents like Tollen’s Reagent. Tollen’stest Gentle Heating for 20 to 25 mins.

2. Fehling’s SolutionsFehling’s A Fehling’s B

COONa

COOK

OHOHH

Haq. CuSO4 Alk. solution of Roschellesalt (sodium potassium tartrate)It act’s a carrier for Cu2+ as it make reversible complex with Cu2+

This test is also used is Blood and Urine test.

RCHO + Cu2+ OH2 ).pptred(OCuRCOO

CuRCOOH

2–

3. Benedict’s solutionSodium Citrate + NaOH + NaHCO3 + CuSO4

RCHO + Cu2+ OH2 ).pptred(OCuRCOO

CuRCOOH

2–

4. RCHO + HgCl2 + H2O RCOOH + 2HCl + Hg2Cl2

RCHO + Hg2Cl2 + H2O RCOOH + 2HCl + 2Hg greyish solution

5. Schiff’s ReagentSchiff’s Reagent is aq. solution of following base decolourised by passing SO2.Aldehyde restore pink colour of Schiff’s reagent.



NH2

C

NH2

SO2 RCHO

Colourless solution (Schiff’s Reagent)

p-Rosaniline HydrochlorideMagenta colour (Fuschin)

RCOOH + Pink colour

NH2Cl+

Ketons are not easy to oxidize so they do not give these 5 tests. These five tests can be used to distinguishaldehyde and ketones. Both gives 2,4 DNP test

KETONES ARE DIFFICULT TO OXIDIZEKetones can be oxidized from their enolic form at high temperature with very strong oxidizing agent.Oxidation of ketones is sometimes governed by Popoff’s rule. According to this rule carbonyl groupremains with the smaller alkyl group. More electron rich alkene will be easy to oxidized.

Me – C – Me

O

MeCOOH + CO + H2 2O[O]

Allylic oxidationSeO2 is a selective oxidizing agent with converts –CH2– group adjacent to carbonyl group into carbonylgroup.The reagent, in general, oxidises active methylene and methyl groups to ketonic and aldehydicgroups respectively.

CCH||

O

2 2SeO

CC||||

OO

; 3CHC||

O

2SeO CHOC||

O

Double bonds, triple bonds and aromatic rings may also activate the methylene group.The methylene ormethyl group to the most highly substitued end of the double bond is hydroxylated according to theorder of preference of oxidation CH2 > CH3 > CH groups.

CH3= CH–CH3 2SeO 22 CHCHCH|

OH

2° C – H > 1° C – H > 3° C – HRate of reactivity orderCH3 – CH = CH – CH2–CH3 2SeO

OH|

CHCHCHCHCH 323

CH3 HO–CH2

CH – CH3CH – CH3

SeO2

CH3CH3

SeO2

HO



Q.1 (i) CH2 = CH2 4/H KMnO

(ii) CH3–CH= CH2 4/H KMnO

(iii) 4/H KMnO

(iv) 4/H KMnO

(v) 4/H KMnO

(vi) 4/H KMnO

(vii) 4/H KMnO

(viii) 4/H KMnO

(ix) 4/H KMnO

(x) 4/H KMnO

(xi) 4/H KMnO

(xii) C10H10 4/H KMnO

COOHC|

COOHCCCHOOC

Q.2 A to F alkenes with minimum possible carbon.

(i) A 4/H KMnO

MeCOOH as the only product

(ii) B 4/H KMnO

Oas the only organic product

(iii) C 4/H KMnO

MeCH2COOH as the only organic product

(iv) D 4/H KMnO

O

O

(v) E 4/H KMnO

O||

COOHCCCCCCCCHOOC||

O

(vi) F 4/H KMnO

acetone + ethanoic acid

Q.3 (i)

mCPBA\hydrolysis

1% alkalineKMnO4

(A)

(B)

(ii) mCPBA

(iii) mCPBA

Mehydrolysis (iv)

mCPBAhydrolysis

Me Me

H HC = C



(v) Me

MeH

H mCPBAhydrolysis

C = C (vi) Me

PhH

HC = C

Ag2O or 2Ag + O2

12

Q.4 (i) CH3– CH2 – CH2 – OH

,H/KMnOor

¯,OH/KMnO

4

4 ?

K Cr O / H , 2 2 7

or conc. HNO3, ?

(ii) 323 CHCHCHCH|OH

H,OCrKor

H,KMnO

722

4 ?

(iii)

HO

HOOH

OH(1)

(10)

(2) or (3)

(4) or (5) or (6)

??

?

?

or (7) or (8) or (9)

Q.5 (i) CH2 = CH – (CH2)3 – CH2 – OH PCC

(ii) C6H5 – CH = CH – CH2 –OH PCC

(iii) OHCHCHCHCHCH|OH

2223 PCC (A)

)ii(

NaOHDil)i( (B)

(iv)

CH OH2

OHPCC (v) CH2 = CH – CH2–OH 2MnO ?

(vi) CH O3

CH O3

CH–CH –CH –OH2 2

OH

Acetone

MnO2 ?

(vii) OHCHCCHCCH|

CH

2

3

Acetone

MnO2 ? (viii) 356 CHCHHC|OH

4

2

CCl

MnO ?



(ix)

3

2256

CH|

OHCHCHCHCHCHHC 3NaHCOTsClDMSO ?

Q.6 (i) HO Acetone

butoxidetertAluminium ?

(ii) 322 CHCHCHCHCH|

OH

butoxidetertAluminium

(iii)

OH

nebenzoquinop

butoxidetertAluminium

?

Q.7 Which one of the following alcohols are oxidised by MnO2?

(A) C6H5 – CH2 – CH2–OH (B) 322 CHCHCHCHCH|

OH

(C) 33 CHCHCHCHCH|

OH

(D) CH3–CH2 – CH2 –OH

Q.8

(i)

OH|

OHCHCHMe 2 4HIO (ii)

OHOH||

Et—CH—CMe2 4HIO

(iii)OH

OH 4HIO (iv)

OH|

OHCHCHCHCHHO 222 4HIO

(v)

OHOH||

CHCHCH—CH 322 4HIO (vi)

OHOHOH|||

CHCHCH—CH 32 4HIO

(vii)

OHOHOHOH||||

CHCHCH—CH 22 4HIO (viii)

OHO|||

MeCHCMe 4HIO

(ix)

OO||||

MeCCMe 4HIO

Q.9 Which will give the Tollen test.

(i) O OH

H (ii) O OMe

(iii) 2CHCR|||

OHO

(iv) HO

HO



Q.10 (a) C – C – C –C

O[O] (b) Me CH–C–Me2

O[O]

(c) Me C–C–Me3

O[O] (d)

O [O]

Q.11 (a) CH –CHO3

SeO2

(b) Me CO2

SeO2

(c) C – C – C – CSeO2C – C – C –C

O

P1 mCPBA P2 LAH P3

(d)O SeO2

(e) SeO2

(f) CH – CH = CH 3 2Acrolein? (g)

CH3 C

2 step(1) step etard

HO

(h) CH –C – H3 H – C – C – H

O OSeO2

conc. NaOH H /+ P1 P2

O

Q.12 How will you differentiate HCHO and PhCHO ?

Q.13 How will you differentiate HCHO and MeCHO ?



Reducing agents and their role

Group Product LAH in LiAlH(OCMe3)3 NaBH4 LiAH4 B2H6 H2+

ether in THF in EtOH +AlCl3 in THF catalyst

–CHO –CH2OH + – + + + +

>C=O >CH–OH + – + + + +

–CO2H –CH2OH + – – + + +

–CO2R –CH2OH + – – + + +

–COCl –CH2OH + +* + + – +

–CONH2 –CH2NH2 + – – + + +

(RCO)2O RCH2OH + – – + + +

–CN –CH2NH2 + – – + + +

>C=NOH –CH2NH 2 + – – + – +

>C=C< >CH–CH< – – – – + +

–CC– –CH=CH– – – – + + +

1° RX RH + – – + – +

* Product is RCHO


FINAL LAP - 2019OXIDATION & REDUCTION Q.1 How many alkene on catalytic reduction give corresponding products.

(i) (A) Pt/H2 n-butane (ii) (B) Pt/H2 Iso-pentane

(iii) (C) Pt/H2 Neo-pentane (iv) (D) Pt/H2 Cyclopentane

(v) (E) Pt

H2

Q.2 Give the expected major product for each reaction, including stereochemistry where applicable.

(a) CH3–CH2–CH=CH2 Pt

H2 P1 (b)D2Pt P2

(c)H3C

CH3

C = CH

H

Ni

D2 P3 (d)Me

Me 2H/Ni P4

(e) excessexcessH /Pt2

H /Pt21 eq.

P5

P6

Q.3 (i) CH2 = CH–CH2 – CH = O excessexcessH /Pt2

H /Pt21 eq.

P1

P2

(ii) CH2=CH–CH2–CN excessexcessH /Pt2

H /Pt21 eq.

P3

P4

Q.4 Identify the product?(i) 4NaBH Me–CHO 4LiAlH (ii) 4NaBH Me2CO 4LiAlH

(iii) 4NaBH Me–COCl 4LiAlH (iv) 4NaBH Me–COOEt 4LiAlH

(v) 4NaBH Me–COOH 4LiAlH (vi) 4NaBH Me–COOMe 4LiAlH

(vii) 4NaBH Me–CONH2 4LiAlH (viii) 4NaBH Me–CONH–Me 4LiAlH

(ix) 4NaBH Me–CONMe2 4LiAlH (x) 4NaBH Me–CH=NH 4LiAlH

(xi) 4NaBH CH3–CH = CH2 4LiAlH

Q.5 Give product in following reactions.

(i) 4NaBH O 4LiAlH (ii) 4NaBH OH–N 4LiAlH

(iii) 4NaBH ON

H

4LiAlH (iv) 4NaBH

N 4LiAlH

(v) 4NaBH OO

4LiAlH (vi) 4NaBH OO

O

4LiAlH

(vii) 4NaBH Me–CO–N 4LiAlH (viii) 4NaBH CH2=CH–CHO 4LiAlH

(ix) 4NaBH Ph–CH=CH–CHO 4LiAlH (x) 4NaBH

O 4LiAlH


FINAL LAP - 2019OXIDATION & REDUCTION Q.6 Give product in following reactions.

(a) H

O

HIPdRe (b)

O

OHCHO

HIPdRe

(c)

COCH3

CH2OHHO HIPdRe


(i)N O

LAH (ii)

(iii)

AlCl3

OCH – CH – CH3 2

LAH

LAH

(iv)

H CCOO3 COOCH3

4LiAlH (A) + (B) + (C)

(v)

O

O


(i)

CHO

OCOOEt

NaBH4 (ii) C—OEt

O

LAH (A) NaOI (B)

(iii) LiAlH , D O4 2

LiAlH4

H O2



Q.9

(a) O=C—CH3 HCl

)Hg(Zn (A) NBS (B) KOH

.alc (C) 22OR

HCl (D)

(b)

O

,HCl

)Hg(Zn(c) 3CHCPh

||O

HCl

)Hg(Zn

(d)

O

OH,HCl

)Hg(Zn

2 (e)

O

heat,KOH)ii(

NNHH)i( 22

(f)O

O O

N H2 4

Zn(Hg)

KOH, heat

HCl(excess)

(excess)

Q.10 Suggest appropriate reagents for following conversion.

(i)

O

OC—H

?

OH

OC—H

(ii)O

O

C—OEt ? O

OH

C—H

H

(iii)

O O

OH

OH

A

B

C



OXIDATIONQ.1 (i) CO2 (ii) CH3–COOH + CO2

(iii) O + CO2 + HOOC – C – C – C–COOH (iv) O + CH3–COOH

(v)COOHCOOH (vi)

COOHO

(vii) O

O(viii) 2CH3–COOH + CO2

(ix)COOH|COOH

+ COOHCH

|COOHCH

2

2

(x)

O

O+

COOHCH|

COOHCH

2

2

(xi)HOOC

COOH

O

O

(xii)

CHC|

CH|

CHCCHCHCHCCH

2

22

Q.2 (i) Me–CH=CH–CH3 (ii)

(iii) Me–CH2–CH=CH–CH2–CH3 (iv)

(v) (vi)

Q.3 (i) OHOH

OHOH

(A) (B) (ii) O

(iii)OH OH

OHOH

Me Me(iv)

Me Me

Me Me

OHOHHO HHHH HO

(v)

Me

Me

OHOH

HH (vi)

Me H

PhHC—C

O



Q.4 (i) CH3–CH2–COO , CH3–CH2–COOH

(ii) 323 CHCHCCH||

O

(iii) (1) O

OCHO

(2) or (3)

O O

OH OH

O O

O O

COOH

O O

(4) or (5) or (6) or (7) or (8) or (9) O

OHO

CHO

(10) O

OHHO

HO

Q.5 (i) CH2 = CH – (CH2)3 – CHO (ii) Ph – CH = CH – CHO

(iii) (A) CHOCHCHCCH||

O

223 (B)

O

(iv)

O

CHO(v) H2C = CH–CHO

(vi)MeO

MeO

C–CH –CH –OH2 2

O

(vii) CHOCCHCCH|

CH3

(viii) 3CHCPh||

O

(ix)

3

2

CH|

CHOCHCHCHCHPh

Q.6 (i)

O

(ii) 322 CHCCHCHCH||

O

(iii)CHO

Q.7 C



Q.8 (i) Me–CHO + HCHO (ii) O + Et – CHO

(iii)CHOCHO (iv) HO–CH2–CH2–CHO + HCHO

(v) HCHO + CH3– CH2–CHO (vi) HCHO + HCOOH + CH3–CHO

(vii) 2HCHO + 2HCOOH (viii) Me–COOH + Me–CHO

(ix) 2 Me–COOH

Q.9 (i), (iii), (iv)

Q.10 (a) 2C–COOH

O[O] (b)Me CH–C–Me Me CO + MeCOOH2

O(c)Me CH–C–Me Me C–COOH + CO + H O3 2 2

O

(d)COOHCOOH

Q.11 (a)CH –CHO CH – CHO

O

SeO (b) Me – C – C – H

O

SeO

O

(c) C – C – C – CSeO2C – C – C –C

O

CCCC||||

OO

mCPBA CCOCC||||

OO

LAH CH3–CH2–OH

(d)O O

O

SeO(e)

HO

(f) CH – CH = CH 3 2

H – C – CH = CH 2CH – CH = CH 2 2

Acrolein?

SeO2

MnO2

OOH

(g)

SeO2

MnO2

CH3 C

CH2

2 step(1) step etard

H

OH

O

(h) CH –C – H3 H – C – C – H

O OSeO2

conc. NaOHO

HCOOH+ CH OH3

HCOONa+ CH OH3

H / +

Q.12 F and B test Q.13 Iodoform test


FINAL LAP - 2019OXIDATION & REDUCTION REDUCTION

Q.1 (i) cis & trans 2-butene & 1-butene; (ii)C|

CCCC C|

CCCC C||

CCCC

(iii) zero (Neo-pentane can not be prepared by catalytic hyrogenation of alkene); (iv) One

(v) Including optical = 4 , Excluding optical = 3

Q.2 (a) (b)

D

D

(c)

CH3 CH3

CH3 CH3

HHH

HDDD

D(d)

CH3

CH3

H

H

(e)

2 G.I.

*

Q.3 (i) P1 CH3–CH2–CH2–CH2–OH

P2 CH2=CH–CH2–CH2–OH

(ii) P3 CH3–CH2–CH2–CH2–NH2

P4 CH2=CH–CH2–CH = NH

Q.4 (i) MeCH2–OH, Me–CH2OH (ii) Me2CH–OH, Me2CH–OH

(iii) MeCH2–OH, Me–CH2OH (iv) No reaction , Me–CH2–OH + EtOH

(v) No reduction, MeCH2–OH (vi) No reaction, MeCH2–OH + MeOH

(vii) No reduction, MeCH2NH2 (viii) No reaction, Me–CH2–NH–Me

(ix) No reduction, MeCH –N2

Me

Me

(x) No reaction, Me–CH2–NH2

(xi) No reaction, No reaction

Q.5 (i) OH, OH (ii) OHNH , OHNH

(iii) No reaction, NH

(iv) No reaction, NH



(v) No reaction, CH2OHOH

(vi) No reaction, CH2OHOH

(vii) No reaction, Me–CH –N2 (viii) CH2=CH–CH2–OH, CH2=CH–CH2–OH

(ix) Ph–CH=CH–CH2–OH, Ph–CH2–CH2–CH2–OH

(x)HO

, HO

Q.6 (a) (b) CH3

(c)

Q.7 (i)N

(ii)OH CH OH2

(iii) CH – CH – CH3 3 , CH –CH – OH3 2 CH2

OH

(iv)

HO CH2OH

(A)

+ MeOH (B) + Et–OH (C)test

NaOI By only (C) (v)CH OH2

OH

Q.8

(i)

CHO CH OH2CH OH2

CH OH2COOEtCH3CH3

OHOH OCOOEt CHCH

NaBH4 + EtOH

(ii) CH2OH + EtOH NaOI +Iodoform test

(iii) OD OHH


FINAL LAP - 2019OXIDATION & REDUCTION Q.9

(a)

C—CH3

O

HCl

)Hg(Zn

CH2—CH3

NBSBr—C—CH3

H

KOH.alc

CH=CH2

22OR

HCl

Cl—CH–CH3

(b) (c) Ph–CH2–CH3 (d) (e)

(f)O O

,

Q.10 (i) Protection Glycol C=OR

H+

HO—CH2

HO—CH2

CR O

OH cyclic-acetal

Glycol is used for protection of aldehyde and Ketone.O

OC—H

CH –OH2

CH –OH2

H

O

O

OCH

OH

LiAlH

3

4

OH

OC—H

(iii) (A) Ni will reduce alkene, aldehyde and all it is not specific ;(B) NH2 – NH2 / H2O2 ;(C) LiAlH4


FINAL LAP - 2019NAME REACTIONS

(1) ALDOL CONDENSATIONThe -hydrogen of carbonyl compounds are acidic due to the fact that the anion (enolate ion) is stabilizedby resonance.

CH – C – H2||O

|H OH (base)

CH = C – H2 |O

CH – C – H2 ||O

(enolate anion) (enolate anion)(3 Hydrogens)

Base catalysed AldolIn aqueous base, two acetaldehyde molecules react to form a -hydroxy aldehyde called aldol. Thereaction is called Aldol condensation. The enolate ion is the intermediate in the aldol condensation ofaldehyde and ketone. Acetaldehyde for instance, forms a dimeric product aldol in presence of a dilutebase (10% NaOH).

2CH3CHO C5

HOH/HO

OH|

CHOCHCHCH 23 OH2

CH3 – CH = CH – CHO

-hydroxy butyraldehyde (Aldol)Mechanism :

H–C–CH2 H–C–CH2–CH–CH3

H–C–CH–CH–CH3

H–C–CH3

H–C=CH2

O O

O O

O

O

O

O

OHOH

HH–O–H

HOHH–C=CH–CH–CH3

–OH

H–C–CH=CH–CH3

rds

OH

Aldols are stable and may be isolated. They, however can be dehydrated easily by heating the basicreaction mixture or by a separate acid catalyzed reaction. Thus if the above reaction is heated theproduct is dehydrated to 2-butenal (crotonaldehyde).


FINAL LAP - 2019NAME REACTIONSAcid catalysed AldolIn acid catalysed aldol condensation enol form of carbonyl is the nucelophile in place of enolate.Mechanism :

CH –C–CH3 3

CH –C–CH3 3 CH –C–CH3 3

CH –C–CH3 3

CH –C=CH3 2

CH –C=CH3 2

CH3

CH3

CH3

CH3CH3H

CH –C=CH–C–CH3 3

CH –C–CH=C–CH3 3

CH –C—CH –C–CH3 2 3

CH –C—CH –C–CH3 2 3CH –C—CH–C–CH3 3

O

O O–H

O–H

OH

O–H

O

O

O–H

OO

OH

OHOH2

H / H O,+2

H+

( unsaturated carbonyl compound)

+

–H O2

–H

H

(Aldol)Q.1 Write the product and mechanism for given reactions.

(I)

O||

HCCHCH 23 NaOHDil (A) (B)

(II) O NaOHDil (C)

(III) C6H5 – CH2– CHO NaOHDil (D)

(IV) O NaOHDil (E) (F)

(V)

O||

CHCHC 356 NaOHDil (G) (H)

Q.2 Identify the intramolecular aldol product?

(I) 3223 CHCCHCHCCH||||

OO

NaOHDil (A) (B)

(II) 322223 CHCCHCHCHCHCCH||||

OO

NaOHDil (C)

(III) 32223 CHCCHCHCHCCH||||

OO

NaOHDil (D)

(IV)

O

O

NaOHDil (E)

(F)



(V) 3222223 CHCCHCHCHCHCHCCH||||

OO

NaOHDil (G)

Q.3 Find out the total number of possible aldol products (including and excluding stereo products)

(I) CH3–CHO + CH3–CH2–CHO C5

HOH/NaOH

(II) C6H5 –CHO + CH3–CHO C5

HOH/NaOH

(III) CH3–CHO +

O||

CHCCH 33 C5

HOH/NaOH

(IV) CH3–CH = O +

O||

CHCCHCH 323 C5

HOH/NaOH

(V) C6H5 – CHO +

O||

CHCCHCH 323 C5

HOH/NaOH

Q.4 Identify the structure of substrate?

(I) A C5

NaOHDil

OH

O

A = ?

(II) )mole2(A

C5

NaOHDil

3

323

CHO|||

CH—C—CHCCH|OH

A = ?

Q.5 Complete reaction sequence:

(i) OH/Zn)ii(

)eq1(O)i(

2

3 (a) )ii(

OH)i(

(b)

(ii)

O

)iii(

CHOHC)ii(

OH)i(

56

Product

Q.6 Complete the retro aldol and retro aldol condensation reactions :

(i) (X) + (Y)

OH CH3 CH–CH3

O OH

(ii) (X) OH

CH3

O

OH



(iii) [X] )ii(

OH)i(

O

(2) CLAISEN CONDENSATION

Esters undergo SNAE Reaction, when attacked by a Nu� generated by the interaction of a base(usually base related to the Alkoxy anion of ester) with one of the molecule of ester and this Nu� attackson another molecule. The reaction over all is considered as condensation of esters known as claisenester condensation.

ORCMe2||

O

ionAcidificat)ii(RONa)i(

ORCCHCMe||||

OO

2

(-keto ester)Mechanism :

ORCCH||

O

3 ROH

RO

RO

Na

Na

Na

(rds)

Me–C=CH–COOR Me–C–CH=C–OR

O OO

Acidification

ORCCHCMe||||

OO

2

Some times, when two ester groups are present within the molecule then the condensation occursintramolecule then cyclization caused thus is known as Dieckmann cyclization or Dieckmann'scondensation.

Q.1 MeCOOEt ionAcidificat)ii(

EtOK)i( AA

Q.2 EtCOOMe ionAcidificat)ii(

MeOK)i( B

Q.3 MeCOOMe + EtCOOMe ionAcidificat)ii(

MeOK)i( C

Q.4 + CH3COOC2H5 ionAcidificat)ii(

ONaHC)i( 52 D + D’

Q.552

52

HCOOC|

HCOOC + C6H5CH2COOC2H5

ionAcidificat)ii(

ONaHC)i( 52 E



Q.6ionAcidificat)ii(

PhONa)i( F

Q.7 C H – N2 5

CH CH COOEt2 2

CH CH COO2 2 Et

ionAcidificat)ii(

ONaHC)i( 52 G OH3 H (Piperidone derivative)

Q.8 ionAcidificat)ii(

EtOHinNaOEt)i( I

Q.9 NaOEt J OH3 K

Q.1052

52

HCOOC|

HCOOC +

COOC H2 5

COOC H2 5

Diethyl- , - DimethylGlutarate

ionAcidificat)ii(

ONaHC)i( 52 L OH3 M


FINAL LAP - 2019NAME REACTIONS(3) PERKIN CONDENSATION

In Perkin reaction, condensation has been effected between aromatic aldehydes and aliphatic acidanhydrides in the presence of sodium or potassium salt of corresponding acid of that anhydride, to yield, -unsaturated aromatic acids.

C6H5–CHO +

OO||||

CHCOCCH 33

,OH/H)ii(

,COONaCH)i(

2

3 C6H5 – CH = CH – COOH + CH3COOH

Mech.

Q.1 H3C CHO

,OH/H)ii(

,COONaCHCH/O)COCHCH()i(

2

23223 A

Q.2

OH)ii(

,COONaCH/O)COCH()i(

3

323 B C

Q.3

,OH/H)ii(

,COONaCH/O)COCH()i(

2

323 D

Q.4

O||

HCHC 56 +

OH)ii(

,SuccinateSodium)i(

3E

Q.5 + F ,OH)ii(MeCOONa)i(

3


FINAL LAP - 2019NAME REACTIONS(4) KNOEVENAGEL REACTION

Reaction of active methylene group with aldehyde & Ketones is known as knoevenagel reaction.

RCHO+piperidineor

Pyridine RCH = C(COOR)22

2

CO,)ii(

H/OH)i(

COOHCHCHR

Z can be

CHO, COOMe, CN, NO2, RC||

O

, SOR, SO2R, SO2OR etc.

Mechanism:

NR3 + H2C(COOR)2 23 )COOR(HCHNR

OH2

R–CH = C(COOR)2 2

2

CO,)ii(

H/OH)i(

R – CH = CHCOOH

High reactivity of the methylene group of the active methylene compound prevents self-condensation ofthe aldehyde.

Q.1 CH3CHO + H2C(COOR)2

,OH)ii(

Pip)i(

3A

Q.2 C6H5CHO + H2C(COOR)2

,OH)ii(

Pip)i(

3 B

Q.3acidGlyoxalicCHOHOOC + H2C(COOR)2

,OH)ii(

Pip)i(

3 C

Q.4 HCPh||

O

+ NHEt2 D

Q.5 HCPh||

O

+ CH3–NO2

2115 NHHCn E

Q.6 +

,COOHCH

NHCCH||

O

3

23 F

Q.7 + HCPh||

O

NaOEt

G


FINAL LAP - 2019NAME REACTIONS(5) REFORMATSKY REACTION

-halo esters when treated with Zn in gives organometallic halo ester which provides the attacking Nu�for the another reactant, which is a carbonyl compound. When Nu� attacks on carbonyl compound itgives an intermediate which upon acidic hydrolysis followed by heating, results in formation of -unsaturated acid. The overall reaction is known as Reformatsky reaction.

Ph – CHO +

Br|

OEtCCH||O

2

/OH

Zn

3 Ph – CH = CH – COOH

Benzaldehyde Cinnamic acidMechanism :

Br|

COOEtCH2 Zn CH – COOEt |ZnBr

2Ph–CH

||O

Ph–CH–CH –COOEt2

|O

ZnBr

H O3

EtOH + Zn + Ph–CH–CH –COOH2

Br

OH

|OH

/–H O2

Ph–CH=CH–COOHCinnamic acid

Q.1 C6H5COCH3 + BrCH2COOEt

,OH.2

Zn.1

3 AA

Q.2

,OH)iii(

PhCHO)ii(Zn)i(

3B

Q.3 + CH3CHO ,OH)ii(

Ether/Zn)i(

3

C

Q.4 + BrCH2COOEt OH.2

Zn.1

3 D

Se E

Q.5O||

CHOCHCHCHCHCHCEtO|Br

2222

OH)ii(

OEt/Zn)i(

3

2 FSe G

Q.6

,OH)iii(OCH)ii(

OEt/Zn)i(

32

2 H



Q.7 (1) Zn + BrCHCH = CH – COOCH3

(2) H O (3) , – H O3 2

I

(1) SOCl2(2) CH Li3

J (1) Zn + BrCH COOEt2

(2) H O (3) , – H O3 2

K

CH OH2

Vitamin A1

LiA

lH4

(6) CANNIZARO REACTIONThis reaction is given by aldehyde having no - hydrogens in the presence of conc. NaOH/ or KOH/.

HCH||

O

KOH OCH

||O

K+ + CH3OH

Mechanism :

OH

O O

H HH–C–H C r.d.s O–CH3 + H–O–C–H

||O

OHCH3 + O–C–H

||O

In the presence of a very strong concentration of alkali, aldehyde first forms a doubly charged anion (I)from which a hydride anion is transferred to the second molecule of the aldehyde to form acid and analkoxide ion. Subsequently, the alkoxide ion acquires a proton from the solvent.

H–C

H–C

H–C–H H–C–H

H–C–H

O

O

O O

O H

H

O

OH

H

O

O

OH OH

(I)

(I)+ C = O + H–C–O

H

H

HCH O2 HCH OH2H O2

From solvent + OH

Q.1 Which of following will not undergo Cannizaro reaction

(A) (B) (C) (D) Cl3C–CHO

Q.2 ClC–Ph||

O

4

2

BaSOPd

H

Ph – CHO

KOH,

HCHO

(C) + (D)

Product (C) & (D) are

(A) Ph–CO2H, Ph–OH (B) –2CO–Ph , –

2HCO

(C) Ph–CH2OH, –2CO–H (D) –

2CO–Ph , CH3OH



Q.3CHO|CHO

NaOH.conc (A) H (B)

Product (B) is

(A) H2C=CH–CO2H (B) (C)

O||

COCH||COCH||

O

2

2

(D) H2C=C=O

Q.4 (i) HCHO NaOD

(ii) DCHO NaOD

Q.5 (i) Ph–CHO DOD/DO

(ii) Ph–CHO OH

18

Q.6 (i) Ph–CHO + HCHO KOH

(ii)

OO||||

HCCPh KOH

Q.7 (i) MeCH2 – CHO KOH

(ii) Me2CH–CHO KOH.conc

Q.8 (i) Me–NO2 + )excess(

HCHO KOH (ii) Me–CHO +)excess(

HCHO KOH


FINAL LAP - 2019NAME REACTIONS(7) BENZIL-BENZILIC REARRANGEMENT OR BENZILIC ACID REARRANGEMENT

The base catalysed reaction of 1, 2-diketones to a salt of -2-hydroxy carboxylic acid is known asBenzilic acid rearrangement, this reaction is mainly applicable when aryl group is present on both carbonylcarbons.

Ph – C – C – Ph||O

||O

(Benzil)

H)ii(

¯OH)i( HO – C – C – Ph

||O

|

|

Ph

OH

(Benzilic acid)

Mechanism :

Ph – C – C – Ph||O

||O

OH

Ph – C – C – Ph|O

||O

|OH

HO – C – C – Ph||O

|

|

Ph

O

HO – C – C – Ph||O

|

|

Ph

OH

(Benzilic acid)

H O – C – C – Ph||O

|

|

Ph

OH

Q.1 CH3 C – C – C H6 5

|| ||O O

OH

Q.2 O C – C O||O

||O

(Furil)

H,2

NaOH.1

Q.3 562 HCCCCHPh||||OO

(i) OH

(ii) HOH / H

Q.4 5656 HCCHCHC|||

OHO

Areagent

Fenton (i) OH

(ii) HOH / H

B.

Q.5O

O (i) NaOH

(ii) H O3

Q.6

O

O

(i) NaOH

(ii) H O3


FINAL LAP - 2019NAME REACTIONS(8) MICHAEL ADDITION

Michael Addition : unsaturated carbonyl compound undergo michael reaction with compoundshaving active methylene. Many different nucleophile can add to unsaturated carbonyl compound.When the nucleophile is an enolate the addition reaction has a special name MICHAEL REACTION.Mechanism :

R – C – CH – C – R2

||O

OH (base)

– H O2

||O

R – C – CH – C – R||O

||O R – CH= CH – C – R

||O

unsaturatedcarbonyl compound

HOH– OH

CH – CH – CH = C – RCH – CH – CH – C – R2

O

RR

R–CR–C

R–CR–C

OOO

OO

Q.1 CH – C – CH – C – CH + CH = CH – CH3 2 3 2 || || ||O O O

OH

– H O2

?

Q.2 33 CHCCHCHCH||

O

+ CH2

COOEt

COOEt OCHCH 23 ?

Q.3 23 NHCCHCHCH||

O

+ 3223 CHOCCHCCHCH||||

OO

OCH3 ?

Q.4 323 OCHCCHCHCHCH||

O

+ NCCHCCH||

O

23 OCH3 ?

Q.5 CH–CH3

O + CH2

COOR

COOR

,OH)ii(

NaEtO

3

?


FINAL LAP - 2019NAME REACTIONS(9) TISCHENKO REACTION / TISCHENKO CONDENSATION

(1) This reaction takes place between two molecules of aldehydes.(2) Reaction is catalysed by aluminium ethoxide.(3) This is a two-step reaction, i.e., redox reaction followed by ester formation. Thus this reaction is

extension of Cannizzaro reaction.

R – CHO + R – CHO

Al)OHC( 352 RCOCHR

||O

2 Mechanism :

R – C = O + Al(OC H )2 5 3

|H

R–C=O–Al (OC H )2 5 3

H

R–CH=O

R–CH = O – C – O – Al(OC H )2 5 3|

|

R

H

Hydride ion transfer

R – CH – O – C – R + Al(OC H )2 2 5 3

||O

Ester

Q.1 CH3 – CHO ?Al)OHC( 352

Q.2 CH3–CH2–CHO ?Al)OHC( 352

Q.3 OHC – CH2–CH2 – CH2 – CHO ?Al)OHC( 352

Q.4 CH3–CH2 – CH2 – CH2 – CHO ?Al)OHC( 352


FINAL LAP - 2019NAME REACTIONS(10) COREY HOUSE SYNTHESIS

Reaction of Gilman’s reagent with alkyl halide gives alkane as one of the product which is known asCorey House synthesis,

[R Cu] Li2 R – X' R – R + RCu + LiX'

Gilman reagent(Lithium dialkyl cuprate)

To obtained good yield of alkane, the alkyl halide R’–X must be either a methyl halide, a primary alkylhalide, or a secondary cycloalkyl halide.(Best CH3X, if possible)

Preparation of Gilman reagent:

R – X XLi

Li2

um)iLith(alkylmole)(2

LiR XLi

,XCu

reagentGilman

2 LiCu][R

Complete the following reaction sequence :

Q.1 CH3– I OEt

Li

2

(A) CuI (B)

ICHCHCHCHCH 22223 (C)

CH3CH2CH2CH2BrOEt

Li

2

(D)CuI

(E)

BrCHCHCHCHCH 22223 (F)

Q.2 (A) hCl2 (B) Li (C)

CuCl (D)

ClCH3

3

323

CH|

CHCHCHCH

Q.3 CH3 – CH2 – Br

,CuBr)ii(

Li)i( (A) ClCH3 (B)

Q.4 I

CuI)ii(Li)i(

(A)

BrCHCH 23 (B)

Q.5 CH3 – CH2 – Br Li (A) CuBr

Cl|

CHCHCH 33

(ii)

(i)

H

O

(C)

(D)

(B)

Q.6 How can you prepare following compounds using corey house synthesis.

(i) CH3 (ii) CH2 = CH – CH2 – CH3

(iii) CH2



Q.7 (CH3)2CuLi + ClHCCHCH 2

14

2

Q.8 (CH3)2CuLi + CH – CH – CH3 2

O

(A) H (B)

Q.9 (CH3)2CuLi +

||O

(A) H (B)

Q.10 (CH3)2CuLi +

O||

HCCHCH2 (A) H (B)

Q.11 Which of the combination of Gilman’s reagent & alkylhalide will be most suitable for preparation offollowing compounds.

(i) CH3 – CH2 – CH3 (ii)

3

33

3

CH|

CHCCH|

CH

(iii)

3

3

3

CH|

PhCCH|CH

(iv) (v)

Q.12 Identify the product of following reactions:

(a) 14

CuLi)CH(ClCHCCH||

CH

2323

2

(b)


FINAL LAP - 2019NAME REACTIONS(11) OZONOLYSIS

The reaction of alkene or Alkyne with ozone (O3) followed by hydrolysis is known as ozonolysis.It is two types : (I) Reductive ozonolysis In presence of reducing agent

(II) Oxidative ozonolysis In presence of oxidising agentReducing agents : Zn, H2O or Zn, CH3COOH or (CH3)2S or (Ph)3P etc.Oxidising agents : H2O2 or

O||

HOOCR or Ag2O etc.

Example 1 :

C=C C=OC CR RRR’ R’O

O OR RRH H

O3 Zn H2O/

H2O2

step I–70°C

Reductiveozonolysis

Oxidativeozonolysis

+ R’–C–H

R–C–R R’–C–OH

O

O O

+

[SCT- Cut the double bondand paste two oxgen atomsand vice versa]

Mechanism :

C=OR’

H

R–C–R R’–C–H

O O

+

C

C

R

R’

R

H

+ +O

O

O

C

C

R R

H

H H

O

OO

OO

R

R’R C

CO

OR

R

R’O=C

R’

R—C—O

R

OC

O

Note : In case of oxidative ozonolysis aldehyde (not ketone) further undergoes oxidation which givesacid as product.

Q.1 Give the product of the following reaction.

(i) H2C = CH2 OH/Zn)ii(

O)i(

2

3 (ii) CH3–CH = CH2 OH/Zn)ii(

O)i(

2

3

(iii) 23

3

CHCCH|

CH

OH/Zn)ii(

O)i(

2

3 (iv) OH/Zn)ii(

O)i(

2

3

(v) OH/Zn)ii(

O)i(

2

3 (vi)H

OH/Zn)ii(

O)i(

2

3

(vii) H2C=CH–CH2–CH=CH–CH3 OH/Zn)ii(

O)i(

2

3


FINAL LAP - 2019NAME REACTIONSQ.2 Find out the structure of reactant.

(i) X OH/Zn)ii(

O)i(

2

3

O||

HCCHCH 23 (ii) X OH/Zn)ii(

O)i(

2

3 O + OH

H

(iii) X OH/Zn)ii(

O)i(

2

3 O +

O

H(iv) X

OH/Zn)ii(

O)i(

2

3

O

H +

H

O

(v) X OH/Zn)ii(

O)i(

2

3 O (vi) X OH/Zn)ii(

O)i(

2

3

O

O

+ CH3

H3CC=O +

H

H C=O

(vii) X OH/Zn)ii(

O)i(

2

3

O

O(viii)

1210HCX

OH/Zn)ii(

O)i(

2

3 OO

O + HCHO

(ix)1812HC

X OH/Zn)ii(

O)i(

2

3

O

+ O=C=O + OO

+ HCHO

(x))X(

HC 46OH/Zn)ii(

O)i(

2

3 C3H2O3 (xi) 1812HC

X OH/Zn)ii(

O)i(

2

3 O O

O

+ HCHO

Q.3 Give the ozonolysis product of the following.

(i) X Zn

)(O3 3 H

HCO

O

(ii) OH/Zn

O

2

3 ?

How many species.

(iii) OH/Zn

/O

2

3

How many species are found.

Q.4 How mnay ozonoids are possible in given reaction.

(i) CH2 = CH2 OH/Zn)ii(

O)i(

2

3 (ii) CH3 – CH = CH – CH3 OH/Zn)ii(

O)i(

2

3

(iii) CH3 – CH = CH2 OH/Zn)ii(

O)i(

2

3


FINAL LAP - 2019NAME REACTIONSQ.5 The reactants that lead to product (a) and (b) on ozonolysis are

HCHO

(a) (b)

(A) (B)

(C) (D)

Q.6 Which starting material should be used to produce the compound shown below?

? OHZn

O

2

3

(A) (B) (C) (D)

Q.7 This compound on ozonolysis gives which of the following compounds

(I) (II)

OO||||

CHOCCHCCOHC||

O

2

(III)

CHO|CHO (IV)

O||

OCHCHCCOHC||

O

2

(A) I, II, III & IV (B) I, III, IV (C) I, II, III (D) I, II, IV


FINAL LAP - 2019NAME REACTIONS(12) OXYMERCURATION-DEMERCURATION

OMDM is a hydration process of alkene according to Markovnikov’s rule with no rearrangement of cyclicmercuinium ion. When reaction with that reagent is complete, sodium borohydride and hydroxide ion are addedto the reaction mixture.

R–CH = CH2 ¯HO,NaBH)ii(

THF,OH,)OAC(Hg)i(

4

22

OH|

CHCHR 3

(i)Hg(OAc) , MeOH, THF 2

(ii) NaBH , HO¯4

OMe|

CHCHR 3

Mechanism for oxymercuration:

CH CH = CH + Hg — OAc 3 2 CH CH — CH3 2 CH CHCH3 2—Hg—OAc

CH CHCH3 2—Hg—OAc

H2O

OAcOAc

¯OAc

Hg+

OH

OH

+

H

+ AcOH

+ AcO¯

Sodium borohydride (NaBH4) converts the carbon-mercury bond into a carbon-hydrogen bond. Becausethe reaction results in the loss of mercury, it is called demercuration.

OH|CHCHCH 23 —Hg—OAc

¯HO

NaBH4

OH|CHCHCH 33 + Hg + AcO¯

Q.1 OH,NaBH)ii(

OHCH)OCOCH(Hg)i(

4

33 Q.2 OH,NaBH)ii(

OH,)OAc(Hg)i(

4

22

Q.3

CH3

¯HO,NaBH)ii(

OHCH,)COOCF(Hg)i(

4

323 Q.4

OH,NaBH)ii(

HOCH,)OAc(Hg)i(

4

32

Q.5OH

OH,NaBH)ii(

OH,)OAc(Hg)i(

4

22 Q.6

C CH

OH,NaBH)ii(

OH,)OAc(Hg)i(

4

22

Q.7 How could each of the following compounds be synthesized from an alkene by OMDM?

(i)

OH

(ii)

OCH CH2 3

(iii)

OH|

CHCCHCH|CH

323

3

(iv)

3

323

3

OCH|

CHCCHCH|CH


FINAL LAP - 2019NAME REACTIONS(13) HYDROBORATION-OXIDATION

Hydroboration has been developed by brown as a reaction of tremendous synthetic utility because alkylboranes are able to undergo a variety of transformation. Hydroboration is a one step, four centre, cisaddition process in accordance with Markovnikov’s rule but after oxidation it seems to be appear toviolate Markovnikov’s rule.

CH3CH = CH2 THF,BH)i( 3

(CH3CH2CH2)3 B

OH¯ ,H O , H O2 2 2

CH COOH3

AgNO3

NH2Cl

Cl2

Hydroboration oxidation

Hydroboration reduction

Dimerisation,

CH CH CH OH23 2

CH CH CH23 2

CH CH CH33 2

CH CH CH23 2 –NH2

CH CH CH23 2 –Cl

CH CH CH2 2 3

Mechanism of Hydroboration:

CH CH=CH3 2

H—BH2

CH3—C—C—H

H

H

H

BH2

+

—

More stable transition state

ealkylborananBH|

CHCHCH

2

223 2 (CH3CH2CH2)3 B

Mechanism of oxidation :HOOH + HO¯ HOO¯ + H2O

R–B R—B—OR RO—B—OR

RO—B¯— OR

RO—BROH + RO—B

R—B—O—OH

R R OR

OR

OROR

OH

O—HO¯

R

R R

+ O—OH

repeat the twopreceding stepstwo times

repeat the twopreceding stepstwo times

+ OH¯ HO¯

RO¯ +3 ROH + BO33–



Q.1 CH2 = CH2 22

3

OH¯,HO)ii(

BH)i( Q.2

AcOH/OH)ii(

THF,BH)i(

2

3

Q.3 22

3

OH¯,HO)ii(

BH)i( Q.4

AcOH/OH)ii(

THF,BH)i(

2

3

Q.5(i)BD , THF3

(i)BH , THF3

(i) BT ,THF3

(ii)CH COOH, H O3 2

(ii) CH COOD, D O3 2

(ii) CH COOD, D O3 2

Q.6

C CH

HO/OH)ii(

THF,BH)i(

22

3

Q.7

CH = CH2

ClNH)ii(

THF,BH)i(

2

3 Q.8

CH = CH2

,AgNO)ii(

THF,BH)i(

3

3

Q.9

CH = CH2

2

3

Cl)ii(

THF,BH)i( Q.10 CH3–C CH

,AgNO)ii(

THF,BH)i(

3

3


FINAL LAP - 2019NAME REACTIONS (14) RIEMMER TIEMENN REACTION

It is believed that chloroform and hydroxide ion react to produce an electron deficient intermediatedichloro carbene : CCl2 (DCC)

HHO— H2O + : :CCl2 + Cl

Treatment of phenol with DCC in basic medium introduces an aldehyde group, onto the aromatic ring.This reaction is known as Reimmer Tiemenn reaction.

H)iii(,CCl:)ii(

HO)i(

2 +

The Reimmer–Tiemenn reaction involves electrophilic substitution on highly reactive phenoxide ring. Theelectrophilic reagent is dichloro carbene : CCl2

+ :CCl2 HO—

H

o-product is major product because :(i) there are two o-positions available as compared to one para.(ii) o-product is more stable due to the formation of six membered chelate formation.

If 1 ° amines (aliphatic and aromatic) react with DCC in basic medium it yield isocyanide or carbylamine.This reaction is known as carbylamine reaction:

The product is known as isocyanide & it is a foul smelling substance.

Q.1 Step marked by is :(A) Aromatization reaction (B) intramolecular acid base reaction(C) both of the above (D) none of the above

Q.2 Addition of NaCl in aqueous solution will make Riemmer Tiemann reaction:(A) slower (B) faster (C) no effect on rate (D) can't be predicted



Q.3 If is used instead of during Riemmer Tiemenn reaction the product formed is:

(A) (B) (C) (D)

Q.4 If CCl4 is used in place of CHCl3 during Riemmer Tiemenn reaction, the product formed is:

(A) (B) (C) (D)

Q.5 o/p ratio in Riemmer Tiemenn reaction follows following order, if NaOH & KOH are used one by oneas base(A) NaOH = KOH (B) NaOH > KOH(C) NaOH < KOH (D) can't be decided on the basic of information given here

Q.6 Which of the following will not give carbylamine reaction(A) t-butyl amine (B) aniline (C) sec. butylamine (D) N-methyl methanamine

Q.7 Correct order of rate of carbylamine reaction for following compounds is:

(I) (II) (III)(A) I > II > III (B) II > I > III (C) II > III > I (D) III > I > II

Q.8 Rate of carbylamine reaction for Me–NH2 will be :(A) CHF3 > CHCl3 (B) CHCl3 = CHF3 (C) CHCl3 CHF3 (D) CHF3 < CHCl3



(1) ALDOL CONDENSATION

Q.1 (I) (A)

OCH|||

HCCHCHCHCH|OH

3

23

(B)

OCH|||

HCCCHCHCH

3

23

(II) (C)

O OH

(III) (D)

OOH|||

HCCHCHCHPh|Ph

2

(IV) (E)

O OH

(F)

O

(V) (G)

OOH|||

PhCCHCPh|

CH

2

3

(H)

O||

PhCCHCPh|

CH3

Q.2 (I) (A)HO

O

(B)

O

(II) (C)

CH3 OHC—CH3

O

(III) (D)CH3

OH

O

(IV) (E)

O

OH(F)

O

(V) (G)CH3

C—CH3

O


FINAL LAP - 2019NAME REACTIONSQ.3 Excluding Stereo Including Stereo

(I) 4 12(II) 2 4(III) 4 6(IV) 6 16(V) 4 12

(I)*

CHOCHCHMe|

OH

2 = 2

3CHOH||

CHOCHCHMe**

= 4

OH|

CHOCHCHEt *2

= 2

3CHOH||

CHOCHCHEt ** = 4

(II)*

CHOCHCHPh|

OH

2 = 2

*CHOCHCHMe

|OH

2 = 2

(III)*

CHOCHCHCH|

OH

23 = 2

3

23

CH|

CHOCHCCH|

OH

= 1

3

323

CH|

COCHCHCCH|

OH

= 1

*COCHCHCHCH

|OH

323 = 2

(IV)

3

33

CH|*

COCHCHCHCH *|OH

= 4



*CHOCHCHCH

|OH

23 = 2

332

33

CHCHCH||*

COCHCH———CCH *|OH

= 4

*CHCOCHCHCH

|OH

323 = 2

32

3223

CHCH*|

CHCOCHCHCHCH|

OH

= 2

32

23

CHCH*|

CHOCHCCH|

OH

= 2

(V)

3

3

CH|*

COCHCHCHPh *|OH

= 4

*CHCOCHCHCHPh

|OH

322 = 2

Et*|

EtCOCHCMe|OH

2 = 2

3

3

CHEt||*

COCH–CH—CMe *|OH

= 4

Q.4 (I)

||

||

O

O

(II)

O||

CHCCH 33

Q.5 (i) (a) C – H||O

CH – CH – C – H2 2 ||O

(b) C – H||O

(ii) CH—C H6 5

O



Q.6 (i) X = ||O

CH3

Y =

O||

HCCH3

(ii) X = 3223 CHCCHCHCCH||||OO

(iii) X = CHOCHCHCHCCH||O

2223

(2) CLAISEN CONDENSATION

Q.1 OEtCCHCMe||||

OO

2 Q.2

)S/R(CH|

OMeCCHCEt||*||

OO

3

Q.3 4 products Q.4 , CH COCH COOC H3 2 2 5

Q.5 Q.6

O

OC—OPh*

(R/S)

Q.7 (G)

COOEt

OC H – N2 5, (H)

Q.8 Q.9 (J) C–O Et

C–O EtO

O

O

(K) O

COOH

COOH

Q.10 (L) O

O

OO

C–OC H2 5

C–OC H2 5

(M)

O

O

O

Oenol is more stable

H

(3) PERKIN CONDENSATION

Q.1 H3C

3CH|

COOHCCH Q.2OH

CH–CH –COOH2

(B)

O

(C)

O

|OH



Q.3 CH=CH–COOH Q.4 OH

OH

O

O

H C2

C

C

C H – CH – CH6 5

OH

Q.5 MeCOCMe||||

OO

(4) KNOEVENAGEL REACTIONQ.1

acidCrotonicCOOHCHCHCH3 Q.2

acidCinnamicCOOHCHCHHC 56

Q.3acidMaleic

COOHCHCHHOOC Q.4 Ph – CH=

Q.5 Ph – CH = CH – NO2 Q.6

Q.7 = CH–Ph

(5) REFORMATSKY REACTION

Q.1 C H –C=CH–COOH6 5 |CH3(E/Z)

Q.2

)Z/E(CH|

COOHCCHPh

3

Q.3 PhCH–CH3

OOH

Q.4 (D)

HO CH COOH2

, (E)

Q.5 (F)

COOHOH

(G)

COOH

Q.7 (I) , (J) ,

(K) ,



(6) CANNIZARO REACTION

Q.1 D Q.2 C Q.3 B

Q.4 (i) CH3OD + HCOONa (ii) DCH2OD + DCOONa

Q.5 (i) PhCH2OD + PhCOONa (ii) Ph–CH2OH + PhCOONa18

Q.6 (i) Ph–CH2OH + HCOOK (ii)

OOH|||

OKCCHPh

Q.7 (i)

Me|

HCCCHMeCH||

O

2 (ii) Me2CH – CH2OH + Me2CHCOOK

Q.8 (i)OHOH

HO

NO2 + CH OH

+ HCOOK3 (ii)

OH

OH

OH

HO

+ HCOOK

(7) BENZIL-BENZILIC REARRANGEMENT OR BENZILIC ACID REARRANGEMENT

Q.1 CH3 C – COO|

OH

|C H6 5

Q.2 OC

O

|OH

(Furilic acid)

|COOH

Q.3 Ph – CH – C2|C H6 5

COOH

OHQ.4 (A) PhCCPh

||||OO

(B)

Ph|

COOHCPh|OH

Q.5OH

COOHQ.6

OH

COOH

(8) MICHAEL ADDITION

Q.1 CH – C – CH – C – CH3 3

|CH – – CH2 CH2

||O

|| ||O O

Q.2CH –CH–CH –C–CH3 2 3

CH –C–OCH CH2 3)( 2

O

O



Q.3

CH –CH–CH –C–NH3 2 2

CH – CH – C–CH–C–O–CH3 2 3

O

O O Q.4

CH –CH –CH–CH –C–O–CH3 2 32

CH –C–CH–C N3

O

O

Q.5CH–CH – COOH2

CH3

O

(9) TISCHENKO REACTION / TISCHENKO CONDENSATION

Q.1 523 HCOCCH||

O

Q.2 32223 CHCHCHOCCHCH||

O

Q.3

O

O

Q.4322222223 CHCHCHCHCHOCCHCHCHCH

||O

(10) COREY HOUSE SYNTHESISQ.1 (A) CH3Li, (B) (CH3)2CuLi, (C) CH3–CH2CH2CH2CH2CH3, (D)CH3CH2CH2CH2Li,

(E) (CH3CH2CH2CH2)2CuLi , (F) CH3CH2CH2CH2– CH2CH2CH2CH2CH3

Q.2 (A) CH3–CH2–CH3 (B)

3

23

CH|

ClCHCHCH

(C)

3

23

CH|

LiCHCHCH (D)

CuLi

CH|

CHCHCH

33

23

Q.3 (A) LiCu]CHCH[ 223 (B) CH3 – CH2 – CH3

Q.4 (A) 2

Cu Li

(B) CH –CH2 3

Q.5 (A)

LiCHCH 23

(B) LiCu]CHCH[ 223

(C) 33 CHCHCH|Et

(D) Et

O



Q.6 (i) ClCH2

3CuLi]Ph[ Ph–CH3

(ii) [CH =CH–CH ] CuLi2 2

2

CH –Cl3 CH2–CH–CH2–CH3

(iii)

2

Ph – CH – Cl2[Ph] CuLi Ph–CH2–Ph

Q.714

223 CHCHCHCH

Q.8 (A)

LiO|

CHCHCHCH 323

(B)

OH|

CHCHCHCH 323

Q.9 (A) CH3

O Li

(B) CH3

O

Q.10 (A)

OLi|

HCCHCHCH 23

(B)

O||

HCCHCHCH 223

Q.11 (i) CuLi]CHCH[ 223 + CH3 – Cl (ii) CuLi]CMe[ 23

+ CH3 – Cl

(iii) C–Me

MePh

2

CuLi + Me–Cl (iv) CuLi + Me–Cl

(v) CuLi + Me–Cl

Q.12 (a)

214

323

HC||

CHCHCCH (b)

OLi|

CHCHCHCCHCH|CH

323

3

(11) OZONOLYSIS

Q.1 (i) OH

H +

H

HO=C (ii) HCCH

||O

3 + H

H C=O (iii)CH3

CH3

O +

H

H C=O



(iv)

O O

H H (v) O O

H

(vi) O O

+

O O

H (vii) H

H C=O +

HH||

OCCHCO 2 +

H|

CHCO 3

Q.2 (i) H H

(ii) H

H(iii)

H

(iv)

H

H (v) (vi)

(vii) (viii) (ix) C=C—C=CH2

CH3

CH3

(x) (xi)

Q.3 (i) (ii)

O

O

H

(iii) C

C

CH3

CH3

O

O +

C

C

CH3

H

O

O

+ C

CH

H O

O

Q.4 (i) 1(ii) 3(iii) 6

Q.5 (A, B, C) Q.6 (B) Q.7 (C)

(12) OXYMERCURATION-DEMERCURATION

Q.1OCH3

Q.2

OH

Q.3

H C3 OCH3



Q.4

OCH3

Q.5 O Q.6

COCH3

Q.7 (i) OH,NaBH)ii(

OH,)OAc(Hg)i(

4

22

OH

(ii) OH,NaBH)ii(

OHCHCH,)OAc(Hg)i(

4

232

OCH CH2 3

(iii) OH,NaBH)ii(

OH,)OAc(Hg)i(

4

22

OH|

CHCCHCH|CH

323

3

(iv) OH,NaBH)ii(

OHCH,)OAc(Hg)i(

4

32

3

323

3

OCH|

CHCCHCH|CH

(13) HYDROBORATION-OXIDATION

Q.1 CH3CH2OH Q.2HH

OHH

Q.3

CH3

H

H

OH , CH3

H

H

OH

] Q.4CH3CH3

H

OH

,

CH3

CH3

HOH

]

Q.5 (i)

HD||CHCHCH 23 , (ii)

DH||CHCHCH 23 , (iii)

CH D2 CH D2

CH3 CH3

H HT T ]

Q.6

CH2CHO

Q.7

CH –CH –NH2 2 2

Q.8 (CH )2 4

Q.9

CH –CH –Cl2 2

Q.10 Z Z, E E, Z E of CH3–CH = CH–CH=CH–CH3



(14) RIEMMER TIEMENN REACTION

Q.1 (C)

Q.2 (A) Common ion effect will decrease concentration of 2ClC

Q.3 (A) Abnormal Riemmer Tiemann reaction

Q.4 (D)

Q.5 (B) Due to bigger size of K+ it cannot fit in chelate that well.

Q.6 (D) D is 2° amine

Q.7 (B) Based on order of nucleophilicity

Q.8 (D) Order of electrophilicity 2ClC

> 2FC


FINAL LAP - 2019BIOMOLECULES, POLYMERS & POC

EXERCISE-I

BIOMOLECULESQ.1 The commonest disaccharide has the molecular formula

(A) C10H18O9 (B) C10H20O10 (C) C18H22O11 (D) C12H22O11

Q.2 On complete hydrolysis of starch, we finally get(A) Glucose (B) Fructose(C) Glucose and fructose (D) Sucrose

Q.3 The term anomers of glucose refers to(A) Isomers of glucose that differ in configurations at carbons one and four (C–1 and C-4)(B) A mixture of (D)-glucose and (L)-glucose(C) Enantiomers of glucose(D) Isomers of glucose that differ in configuration at carbon one (C–1)

Q.4 Which of the following is an example of aldopentose?(A) Erythrose (B) Ribose (C) Fructose (D) Dihydroxyacetone

Q.5 Glucose and fructose form(A) Same osazone (B) Same acid on oxidation(C) Same alcohol when reduced (D) Different osazone

Q.6 The change of optical rotation of glucose solution with time is refered to as:(A) Mutarotation (B) Inversion (C) Specific rotation (D) Autorotation

Q.7 To become a carbohydrate a compound must contain at least:(A) 2 carbon atoms (B) 3 carbon atoms (C) 4 carbon atoms (D) 6 carbon atoms

Q.8 Methyl--D-glucoside and methyl--D-glucoside are(A) Epimers (B) Anomers(C) Enantiomers (D) Conformational diastereomers

Q.9 The correct name of 'sucrose' is(A) -D-glucopyranosyl--D-fructofuranoside(B) -D-glucopyranosyl--D-fructofuranoside(C) -D-glucopyranosyl--D-fructofuranoside(D) -D-glucopyranosyl--D-fructofuranoside

Q.10 Which one of the following is laevorotatory(A) Glucose (B) Sucrose (C) Fructose (D) None of these

Q.11 Which of the following is a disaccharide(A) Lactose (B) Starch (C) Cellulose (D) Glucose



Q.12 The two functional groups present in a typical carbohydrate are(A) – OH and – COOH (B) – CHO and – COOH(C) > C = O and – OH (D) – OH and – CHO

Q.13 The monomer of cellulose is(A) Fructose (B) Galactose (C) Glucose (D) None of these

Q.14 Glucose is a(A) Monosaccharide (B) Disaccharide (C) Trisaccharide (D) Polysaccharide

Q.15 Which pair is different for reaction with Fehling solution:(A) Glucose, Fructose (B) HCHO, CH3CHO(C) CH3COCH3, C6H5CHO (D) Glucose, Sucrose

Q.16 Glucose contains in addition to aldehyde group(A) One secondary OH and four primary OH groups(B) One primary OH and four secondary OH groups(C) Two primary OH and three secondary OH groups(D) Three primary OH and two secondary OH groups

Q.17 Glucose reacts with excess of phenyl hydrazine and forms(A) Glucosazone (B) Glucose phenyl hydrazine(C) Glucose oxime (D) Sorbitol

Q.18 Which set of terms correctly identifies the carbohydrate shown

HOH C2

H

H

OH H

OH

OH

H

O

1. Pentose 2. Hexose3. Aldose 4. Ketose5. Pyranose 6. Furanose(A) 1, 3 and 6 (B) 1, 3 and 5 (C) 1, 4 and 6 (D) 2, 3 and 6

Q.19 Hydrolysis of sucrose is called:(A) Inhibition (B) Saponification (C) Inversion (D) Hydration

Q.20 Which of the following pentoses will be optically active

ICOHH|

HCOH|

HOCH|

HCOH|CHO

2IICOHH|

HOCH|

HCOH|

HCOH|CHO

2IIICOHH|

HCOH|

HCOH|

HCOH|CHO

2

(A) All (B) II and III (C) I (D) II



Q.21 The secondary structure of a protein refers to(A) -helical backbone (B) Hydrophobic interactions(C) Sequence of -amino acids (D) Fixed configuration of the polypeptide backbone

Q.22 A tripeptide is written as Glycine-Alanine-Glycine. The correct structure of the tripeptide is

(A) NH2

O

NH

CH3

O

NH

CH3

COOH

(B) NH2

O

NHO

NH

CH3

COOH

CH3

(C) NH2

O

NHO

NH COOHCH3

(D) NH2

O

NHO

NH COOH

CH3

CH3

Q.23 Which compound can exist in a dipolar (zwitter ion) state(A) C6H5CH2CH(N = CH2) COOH (B) (CH3)2CH·CH(NH2)COOH(C) C6H5CONHCH2COOH (D) HOOC·CH2CH2COCOOH

Q.24 Which of the following statement is incorrect for maltose.(A) It is a disaccharide (B) It undergoes mutarotation(C) It is a reducing sugar (D) It does not have hemiacetal group.

Q.25 Identify the correct statement about lactose.(A) It consists of one galactose and one glucose unit(B) Mutarotation is not possible(C) Anomeric carbon of galactose is attached to 4' carbon of glucose which is –1, 4'-glycoside bond.(D) Lactase is not used to cleave the –1, 4'-glycoside bond.

Q.26 Which of the following carbohydrates would be most abundant in the diet of strict vegetarian?(A) Amylose (B) Glycogen (C) Cellulose (D) Maltose

Q.27 Naturally occuring (+) - sucrose is :(A) -D-glucopyronoside--D-fructofuranoside(B) -D-glucopyronoside--D-fructofuranoside(C) -D-glucopyronoside--D-fructofuranoside(D) -D-glucopyronoside--D-fructofuranoside



Q.28 D-Ribose when treated with dilute HNO3 forms.

(A)

HO

HO

O

O

HHH

OHOHOH , Achiral (B)

OH

HO

O

O

HH

H

OHHO

OH , Chiral

(C)

HO

O

O

HH

H

OHHO

HO

OH , Achiral (D)

OH

HO

O

O

HHH

OHOHOH

, Chiral

Q.29 Consider the given process.

H

HH

HOOHOH

CH OH2

H–C–OH

CHO

(I)

OH¯

H O2 H

HH

HOOHOH

CH OH2

C–OH

CHOH

(II)

OH¯

H O2 H

HH

HOOHOH

CH OH2

HO–C–H

CHO

(III)and identify the incorrect statement.(A) Configuration at C–2 is lost on enolisation(B) I and III are epimers(C) Proton transfer from water to C–1 converts ene diol to an aldose.(D) D-glucose can isomerise to D-fructose through enol intermediate.

Q.30 Glucose HI/P)ii(

OH/HCN)i( 3

P

P is :(A) n- heptanoic acid (B) 2-methyl hexanoic acid(C) n-heptane (D) 2-methyl hexane

Q.31 When methyl D-glucopyranoside is treated with HIO4 how many moles of HIO4 are consumed with permole of the sugar ?(A) 2 (B) 3 (C) 4 (D) 5

Q.32 The configuration of the C-2 epimer of D-glucose is-(A) 2R, 3S, 4R, 5S (B)2S, 3S, 4R, 5R(C) 2S, 3R, 4S, 5R (D)2R, 3S, 4R, 5R

Q.33 Mutarotation involves(A) Racemisation (B)Diastereomerisation(C) Optical resolution (D)Conformational inversion



Q.34 Same osazone derivative is obtained in case of D-glucose, D-Mannose and D-Fructose due to(A) the same configuration at C-5(B) the same constitution.(C) the same constitution at C-1 and C-2(D) The same constitution and same configuration at C-3, C-4, C-5 and C-6 but different constitutionand configuration at C-1 and C-2 which becomes identical by osazone formation.

Q.35NaBH4

NaBH4

(P)

(R)

D(–) –Erythrose

D(–) –Threose Which of the following statement is correct about P and R ?(A) Both are optically active(B) Both are optically inactive(C) P is optically inactive and R is optically active(D) Neither P nor R has asymmetric carbon.

Q.36 Which of the following compounds will not show mutarotation:(A) -D (+) glucopyranose (B) -D(+) glucopyranose(C) Methyl--D-glucopyranoside (D) -D(+) galactopyranose

Q.37 Amylose and cellulose both are linear polymers of glucose. The difference between them is:(A) Amylose has (1 4') linkage and cellulose has (1 4') linkage(B) Amylose has (1 4') linkage and cellulose has (1 4') linkage(C) Amylose has (1 4') linkage and cellulose has (1 6') linkage(D) Amylose has (1 4') linkage and cellulose has (1 6') linkage

Q.38 Glycogen on hydrolysis gives:(A) Lactose and Glucose (B) Only Glucose(C) Glucose and Fructose (D) Glucose and Maltose

Q.39 An example of disaccharide made up of two unit of the same monosaccharide.(A) Maltose (B) Sucrose (C) lactose (D) None

Q.40 The colour of the precipitate formed when a reducing sugar is heated with Fehling solution is:(A) Brown (B) Red (C) Blue (D) Green

Q.41 Osazone formation involves only 2 carbon atoms of glucose because of(A) Oxidation (B) Chelation (C) Reduction (D) Hydrolysis

Q.42 Methyl--D-glucose and methyl --D-glucose are(A) Epimers (B) Anomers (C) Enantiomers (D) Constitutional isomers

Q.43 Hydrolysis of lactose with dilute acid yield(A) Equimolar mixture of D-glucose and D-glucose.(B) Equimolar mixture of D-glucose and D-galactose.(C) Equimolar mixture of D-glucose and D-fructose.(D) Equimolar mixture of D-galactose and D-galactose.



Q.44 Celluose is a straight chain polysaccharide composed of(A) D-glucose units joined by -glycosidic linkage(B) D-glucose units joined by -glycosidic linkage(C) D-galactose units joined by -glycosidic linkage(D) D-galactose units joined by -glycosidic linkage

Q.45 - amino acid when heated with BaO forms -(A) - unsaturated acid (B) - unsaturated amine(C) Carboxylic acid (D)Amine

Q.46 The pH of the solution containing following zwitter ion species is NH3

R

COO¯

H

(A) 4 (B)6 (C) 8 (D)9

Q.47 Peptide linkage is -

(A) –C – O –

O(B)–C – NH2

O

(C) –C – NH–

O

(D)–C – NH–NH2

O

Q.48 The monomer of nucleic acids are held together by(A) Phosphoester linkage (B) Amide linkage(C) Glycosidic linkage (D) Ester linkage

Q.49 Test used to identify peptide linkage in protein is:(A) Biuret (B) Ninhydrin test (C) Molisch test (D) 2,4-DNP test

Q.50 Which one of the following structures represents the peptide chain:

(A) –N–C–C–C–N–C–C–N–C–C–C–

H H

H

O O

(B) –N–C–N–C–NH–C–NH–

H

HO

O

(C) –N–C–C–C–C–NH–C–C–C

H

(D) –N–C–C–N–C–C–N–C–C–

H H H

O O OQ.51 Among the following L-serine is:

(A)

COOH

H

H N2 CH OH2 (B)

COOH

H

NH2

HOCH2



(C) COOHH

NH2

CH OH2

(D)

COOH

HH N2

CH OH2

POLYMERS

Q.52 The monomer that undergo radical polymerisation most easily is

(A) CH2=CH2 (B) C6H5CH=CH2 (C) CH2=CMeMe

(D) CH3–CH=CH2

Q.53 Select the incorrect statement about Nylon 2-nylon-6.(A) It is a copolymer.(B) It is biodegradable.(C) It is an alternating polyamide.(D) It is made up of

2

3

NH|

COOHCHCH and H2N(CH2)5COOH.

Q.54 The polymer formed as a result of following sequence of reaction is

2HC CH 2HC CH polymerisationpolymerisationHCl (1 eq.)HCl (1 eq.)Cu Cl2 2Cu Cl2 2

NH Cl4NH Cl4(A)(A) (B)(B) (C)

(A) Saran (B) PVC (C) Neoprene (D) Chloroprene

Q.55 The monomer that can undergo radical, cationic and anionic polymerisation with equal ease -(A) Me–C=CH2

Me

(B) Ph–CH=CH2 (C) CH2 = CH2 (D) CH2=CH–CN

Q.56 Protein is a polymer of:(A) Glucose (B) Terephthalic acid (C) Amino acid (D) Glycol

Q.57 Enzymes are:(A) Minerals (B) Oils (C) Fatty acids (D) Proteins

PRACTICAL ORGANIC CHEMISTRYQ.58 Consider the reaction-

(i) CHCl +NaOH3 P (major) + Q (Minor)(ii) H O3

+

OH

Mixture of P and Q can be best separated by -(A) Steam distillation (B) Vacuum distillation(C) Fractional distillation (D) Crystallisation



Q.59 Which of the following compound(s) will give blue colour when it is converted into Lassaigne's extractand FeSO4 is added followed by FeCl3.

(I) (II) (III) NH2–OH (IV) 22 NHCNHNH||

O

(A) I and IV (B) IV only (C) I, III & IV (D) I, II, III & IV

Q.60 Select reagent which is used in laboratory to differentiate 1°, 2° and 3° amines from each other:(A) NaOH, I2 (B) PhSO2Cl (C) CHCl3 , KOH (D) CS2, HgCl2



EXERCISE IIBIOMOLEUCLES

Q.1 Carbohydrates may be :(A) Sugars (B) Starch(C) Polyhydroxy aldehyde/ ketones (D) Compounds that can be hydrolysed to sugar

Q.2 Select the correct statement:(A) S-glyceraldehyde is also known as L-glyceraldehyde(B) The configuration of the stereocenter most distant from the carbonyl group determines whether amonosaccharide is D or L.(C) Glucose and all naturally occurring sugars are D-sugars(D) D-erythrose and D-threose are diastereomers.

Q.3 Select the incorrect statement.(A) Monosaccharide are insoluble in organic solvents like diethyl ether.(B) Anomers of a cyclic monosaccharides differ in the position of the OH group at the hemiacetalcarbon.(C) D-ribose the OH group used to form the five membered furanose ring is located on C4.(D) Aldopentoses and ketohexoses form pyranose rings in solution.

Q.4 The peptide bond is a key feature in :(A) Vitamins (B) Proteins (C) Nucleotides (D) Polypeptides

Q.5 Which is not correct about monosaccharides.(A) Optically active polyhydroxy carbonyl compounds.(B) Fructose & glucose can not be distinguish by Br2/H2O(C) Glucose and mannose are anomers(D) Fructose & glucose can be distinguish by Fehling solution

Q.6 Which of the following is disaccharides ?(A) Lactose (B) Sucrose (C) Cellulose (D) Maltose

Q.7 Select the correct statement.(A) Proteins upon hydrolysis gives -amino acid only.(B) Except glycine, all other naturally occuring -amino acids are optically active.(C) In fibrous proteins polypeptide chains are held together by hydrogen and disulphide bonds.(D) Fats upon hydrolysis gives -amino acids

Q.8 Select the correct statement.(A) Coiling of polypeptide chain form fibrous protein.(B) Quarternary structure of protein also exist.(C) Lysine is an amino acid with basic side chain.

(D) The absolute configuration of NH3 – CH(CH2OH)COO¯ (L-serine) is S.

Q.9 Select the correct statement.(A) All proteins are polyamides formed by joining amino acids together.(B) All L-amino acids except cysteine have the S-configuration.(C) Proline is 1° amine consisting 6 membered ring.(D) Proline is a 2° amine consisting of five membered ring.



Q.10 Select the correct option.(A) Isoelectric point is the pH at which an amino acid exists primarily in its neutral form.(B) Isoelectric point is the average of pK a values of -COOH amino -

3NH groups [valid only forneutral amino acid](C) Glycine is characterised by two pKa values.(D) For neutral amino acid the concentration of zwitter ion is maximum at its isoelectric point.

Q.11 Amino acids are synthesised from.(A) -Halo acids by reaction with NH3.(B) Aldehydes by reaction with NH3 and cyanide ion followed by hydrolysis.(C) Alkyl halides by reaction with the enolate anion derived from diethyl acetamidomalonate & hydrolysis.(D) Alcohols by reaction with NH3 and CN¯ ion followed by hydroysis.

Q.12 Which of the following carbohydrates developes blue colour on treatment with iodine solution ?(A) Glucose (B) Amylose (C) Starch (D) Fructose

Q.13 Which of the following are correct.(A) Br2/H2O can be used to differentiate between aldose & ketose.(B) All monosaccharides are reducing sugar(C) Osazone formation destroys the configuration about C2 of an aldose but not effect the rest ofmolecule(D) Mono saccharides undergoes mutarotation.

Q.14 Match the column :Column I Column II

(A) Sucrose (P) Two acetals(B) Maltose (Q) No hemiacetal(C) Lactose (R) 1,4'-glycosidic bond(D) Cellulose (S) Hydrolysis product is glucose

POLYMERSQ.15 Select the correct statement.

(A) High density polythene is a linear polymer.(B) Low density polythene is a branched chain polymer.(C) Chain growth polymers are also known as addition polymer.(D) Step growth polymer is also known condensation polymer.

Q.16 Select the correct statement.(A) Elastomers have the weakest intermolecular forces.(B) Buna-N is an elastomer with crosslinks.(C) Some fibres have crystalline nature.(D) Thermoplastic polymers have stronger intermolecular forces than fibres



Q.17 Match the column.Column I Column II

(A)

Cl|

)CHCHCCH( n22 (P) Thermoplastic polymer

(B)n4262 )C)CH(CN)CH(N(

||||||OOHH

(Q) Thermosetting polymers

(C) n2 )CHCH(|Cl

(R) Fibres

(D)

OH OH

(S) Elastomer


(A) Teflone (P) Natural Polymer

(B) Natural Rubber (Q) Condensation

(C) Terylene (R) Linear Polymer

(D) Proteins (S) Co-Polymer

POC

Q.19 + AC O2AcONa (P)

CHO

Before isolating (P) unreacted Ph–CHO is removed first. Select the correct statement.

(A) P is cinnamaldehyde(B) Removal of PhCHO is done by passing steam into the mixture(C) Removal is done by simple distillation(D) P is cinnamic acid.



Q.20 Match the columnColumn I Column II(component of mixture) (Reagent)

(A) Crystalline Na2CO3 + Sodium citrate + CuSO4(aq. sol.) (P) Fehling solution

(B) CuSO4 + Rochelle Salt + NaOH(aq. sol.) (Q) Nesseler's Reagent

(C) 10% - naphthol in alcohol (R) Benedict's solution

(D) HgCl + KI + KOH (aq. sol.) (S) Molisch's Reagent



EXERCISE III

Q.1 The pKa values for the three acidic group P,Q,R are 4.3, 9.7 and 2.2 respectively

HOOC–CH –CH –COOH2

NH3

(R) (P)

(Q)

+

Calculate the isoelectric point of the amino acid ?

Q.2 Statement 1 : Bromine water changes glucose to gluconic acid.Statement 2 : Bromine water acts as oxidising agent.(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C) Statement-1 is true, statement-2 is false.(D) Statement-1 is false, statement-2 is true.

Comprehension (3 to 5)Consider the following reversible process for a reaction of D-glucose.

HO

HO

O

OH OH

CH OH2

D-glucose( or form)

H [X]

CH OH3

H+

H+

Y

Z

-D-isomer

-D-isomer

Q.3 The structure of intermediate [X] is

(A) HO

HO

O

OH

CH OH2

(B) O

O

OH OH

CH OH2

(C)

OCH OH2HO

HO O(D)

O

OHHO

HOOH

CH OH2

Q.4 Select the correct option.

(A) Y is HO

HO

O

OHOMe

HCH OH2

(B) Z is HO

HO

O

OHH

OCH3

CH OH2

(C) Y is O

OHOH

HOCH3

CH OH2

(D) Z is HO

HO

O

OHH

OCH3

CH OH2



Q.5 Which of the following is positional isomer of -D glucose?

(A)

CH OH2O

HOHO

OH

OCH3

(B)

CH OH2O

HOHO

OHOCH3

(C)

CH OH2

OHO

OH

OH

OH (D)

CH OH2

OHO

OHOCH3

OH

Comprehension (Q.6 to Q.8)

The monomer G

(G = Me or Cl) when treated with Zieglar - Natta catalyst undergo polymerisation

in the manner given below -Zieglar-Natta

cis-poly-1,3-butadiene

Catalyst nn

Q.6 The Zieglar-Natta catalyst is(A) TiCl4 (B) R3Al (C) R3 Al / TiCl4 (D) R3 B/TiCl4

Q.7 The polymer obtained when monomeric unit used is CH2=C–CH=CH2

Cl

(A) Neoprene (B) Stilbene (C) Styrene (D) Chloropicrin

Q.8 Which of the following statement is not true considering the process given above.(A) The general class of polymer formed is known as homopolymer(B) The polymer obtained is stereoregular(C) Buna–N can be prepared using above process(D) Synthetic rubber can be formed by above process using 1,3- butadiene.

Q.9 Match the column :Column I Column II(Carbohydrate) (Properties)

(A) Starch (P) Mutarotation(B) Sucrose (Q) Non reducing sugar(C) Lactose (R) -glycosidic bond(D) Maltose (S) -glycosidic bond

(T) Reducing sugar(U) Hemiacetal



Q.10 Match the column :Column I Column II

(A)

OO

OOH

OH

HO

HOHO

HCH OH2

CH OH2

H

OH (P) -glycoside bonds

(B)

OO

OOHOH

OH

OHOH

HO

HOCH 2

HOCH2

HH

(Q) Reducing sugar

(C)

O

OHO

HO H

HO

CH OH2

HOCH2

HO

OH

O

CH OH2

(R) Forms enediol intermediate

(D)

O OO OH

OH OH

HO

HO HO

HOCH2 HOCH2

(S) -glycoside bond


(A) Addition polymer (P) Buna-S(B) Condensation polymer (Q) Buna-N(C) Homopolymer (R) Polythene(D) Copolymers (S) Nylon 6,6



Q.12 Match the columnColumn I Column II(Functional group) (Test used or complex formed during confirmatory test)

(A) Aldehydic (P) [(C6H5O)6Fe]–3 (violet)

(B) Phenolic (Q)

CH — C–O–

CH — C–O–

CuOO

OO(blue)

(C) Alcohol (R)

RCH(OH)OSONH

RCH(OH)OSONHC NH

violet red

(S) (ROH)2Ce(NO3)4 (Red)

(T) Molisch's Test

Q.13 Read following statements.(a) Protein on complete hydrolysis give.(b) Hormons belong to the class of(c) Carbohydrates are stored in the body as(d) The excess of glucose is stored in the liver asWhich set of terms correctly identifies the statements given above.(i) Lipids (ii) -amino acid (iii) Peptides (iv) Fats(v) Proteins (vi) Cellulose (vii) Glycogen (viii) Peptones



EXERCISE-I

Q.1 D Q.2 A Q.3 D Q.4 B Q.5 A Q.6 A Q.7 B

Q.8 B Q.9 A Q.10 C Q.11 A Q.12 D Q.13 C Q.14 A

Q.15 D Q.16 B Q.17 A Q.18 A Q.19 C Q.20 A Q.21 D

Q.22 C Q.23 B Q.24 D Q.25 A Q.26 C Q.27 A Q.28 A

Q.29 C Q.30 A Q.31 A Q.32 B Q.33 B Q.34 D Q.35 C

Q.36 C Q.37 B Q.38 B Q.39 A Q.40 B Q.41 B Q.42 B

Q.43 B Q.44 B Q.45 D Q.46 B Q.47 C Q.48 A Q.49 A

Q.50 D Q.51 C Q.52 B Q.53 D Q.54 C Q.55 B Q.56 C

Q.57 D Q.58 C Q.59 B Q.60 B

EXERCISE II

Q.1 A,B,C,D Q.2 A,B,C,D Q.3 A,B,C Q.4 B,D Q.5 B,C,D

Q.6 A,B,D Q.7 A,B,C Q.8 B,C,D Q.9 A,B,D Q.10 A,B,C,D

Q.11 A,B,C Q.12 B,C Q.13 A,B,C,D Q.14 (A) Q,S (B) S (C) R,S (D) R,S

Q.15 A,B,C,D Q.16 A,B,C Q.17 (A) S, (B)R, (C)P, (D) Q

Q.18 (A) R (B) P,R (C) Q,S (D) P,Q,S Q.19 B,D Q.20 (A)R (B) P (C) S (D) Q

EXERCISE III

Q.1 3.25 Q.2 A Q.3 A Q.4 D Q.5 C Q.6 C Q.7 A

Q.8 C Q.9 (A) Q,S (B) Q,S (C) P,R,T,U (D) P,S,T,U

Q.10 (A) P,Q,R (B) P (C) P (D) P,Q,R Q.11 (A) P,Q,R (B) S, (C) R, (D) P,Q,S

Q.12 (A) Q,P (B) P, (C) S (D) T Q.13 2577

Date post:	07-Feb-2022
Category:	Documents
Upload:	others
View:	9 times
Download:	0 times

RESONANCE & AROMATICITY

Documents