Solution Slot - 1 (Chemistry)
Page # 1
ACIDITY & BASICITY EXERCISE - 1
HINT & SOLUTON
Q.1 (i)
(a) H – F
(b) H – Cl
(c) H – Br
(d) H – I
– Cl– F Br I– In group we see the size factor more the size, more stable the anion, more the acidity. d > c > b > a (acidic strength)
CH4
(ii)
(a) – CH3
(b)
H–F
H2O
NH3
–H+
–H+
(c)
NH2
–H+
(d)
OH–
–H +
F–
In period we see the E.N. factor more the electron negativity, more stable the anion, more the acidity. –
–
–
–
E.N. order ⇒ F > OH > NH2 > CH3 Acidity order = d > c > b > a
CH3 (iii)
CH3 – CH2 – O – H
(a)
(b) CH3 – CH – O – H (c)
CH3 – C – O – H
CH3
CH3
– +I CH3 – CH2 – O
More stable
CH3 – CH – O
CH3
–
CH3 2(+I) effect
CH3 – C – O– CH3 Three + I group less stable
(acidic strength)a > b > c (iv)
⇒ F – CH2 – CH2 – O– ⇒ NO2 – CH2 – CH2 – O– ⇒ Br – CH2 – CH2 – O–
(a) F – CH2 – CH2 – O – H (b) NO2 – CH2 – CH2 – O – H (c) Br – CH2 – CH2 – O – H ⊕
⊕
⇒ N H3 – CH2 – CH2 – O–
(d) N H3 – CH2 – CH2 – O – H ⊕
–I order = – N H3 > –NO2 > – F > –Br acidic strength = d > b > a > c Q.2(i)
(a) CH3 – COOH
(b) CH3 – CH2 – OH
CH3 = C – O– || O
–H+
–H+
– +I CH3 – CH2 – O – O
(c) C6H5OH
–H+
C6H5 – O– (Phenoxide ion)
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Solution Slot - 1 (Chemistry)
O
(d) C6H5SO3
O S
Page # 2
– O
[–I order – C – R > Ph] O
(More the resonating structures, more the stability) d>a>c>b
(ii)
(a)
COOH
(b)
–H+
COOH
– C –O 3
sp
COOH
(c)
–H+
–H+
C–O
C – O–
O
O sp (more-I)
O
2
sp
–
Stability = C > b > a acidity = c > b > a CH2 – COOH
COOH
(a)
COOH
–H+
–H+ COO
–H+
CH2
COOH (iii)
CH2 – COOH
COOH
–
–
COO
(b)
–
CH2 – C – O
(c)
O
CH2
COO–
–
CH2 – C – O
COO–
O O
Q.3
(a)
O
Cl – CH2 – C – O – H
–H+
Cl – CH2 – C – O - (one -I group)
O (b)
O
Cl – CH2 – C – O – H
–H+
Cl Cl (c)
Cl O
Cl
Cl – C – C – O – H Cl
– Cl – CH2 – C – O (two -I grop)
–H+
O –
Cl – C – C – O (Three –I groups) Cl (most stable)
Ans. c > b > a
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Solution Slot - 1 (Chemistry)
Page # 3 O (ii)
(a)
O
–H+
CH3 – CH2 – CH – C – O – H
CH3 – CH2 – CH – C – O
F (b)
F
–H+
CH3 – CH – CH2 – C – O – H
CH3 – CH – CH2 – C – O
O
F
O
F
O
(c)
–H+
CH2 – CH2 – CH2 – C – O – H
CH2 – CH2 – CH2 – C – O –
F
O
F
In this type of case are see
rule.
"DNP" Distance
Power Number
Acidic strength = a > b > c O
O
(iii)
(a)
– NO2 – CH2 – C – O O
–H+
NO2 – CH2 – C – O – H O
(b)
O F – CH2 – C – O O–
–H+
F – CH2 – C – O – H O
(c)
O
Ph – CH 2 – C – O – H
– Ph – CH2 – C – O
–H+
O
(d)
O CH3 – CH2 – C – O–
–H+
CH3 – CH2 – C – OH
:OH
(b)
NO2 (strongel acid) ↓
:OH
(c)
:
(a)
:
:
Acidic Strength : a > b > c > d
Q.4(i)
–
:OH
CH3
Cl
↓
↓
– O
– O
– O
NO2 –I
Cl –I
CH3 +I
(most stable conjugate base) Acidic strength = a > b > c
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–
Solution Slot - 1 (Chemistry) :
Page # 4
:
:
(ii)
Cl
(a)
:OH
:O
:O – H (b)
(c)
Cl
Cl
:O – H
(a)
(b)
:
:OH (iii)
:O – H
:
:
:
Inductive effect by distance decreases. Acidity order a > b > c
:O – H
(c)
–H+
–H+
– O
– O
(d)
–H+
–H+
– O
– O
CH3 CH3
:
stability of Conjugate base : d > b > c > a acidity order : d > b > c > a
:
:
:O – H N Q.5(i)
:O – H
P O
(a)
(b)
(–M) Hydrogen Bonding
N
:
:OH
:O – H O (–I)
(c)
N O
O
(d)
O
(–M)
(a)
NO2 –M
:O – H N (b)
:O – H NO2
O O
H–Bonding (–M)
:
:O – H
(ii)
:
:
:
Acidity order = c > a > b > d
:OH O2N
NO2
(d)
(c)
NO2 –I
NO2 –I
Acidity order = d > c > a > b
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Solution Slot - 1 (Chemistry)
Page # 5 O C–O–H
O C–O–H CH3
–H+ Q.6(i)
(a)
O
(b)
–
C–O
–H+
O C–O
–
CH3 Due to ortho effect COOH goes out of plane So max. acidity
b>a
(ii)
COOH Cl
(a)
(b)
COOH Br
O
.. C–O .. – H
Br is more bulky then Cl ortho effect
O
C–O
–
(b)
C–O
+I
O Me (+M) Acidity order c > b > a
(iv)
O
O
O
(a)
(c)
–
O Me
–
C–O O Me Ortho effect max. acidity
O
O
O
C–O–H
C–O–H
C – OH NO2
(a)
(b)
N O
C–O–H .. O .. Me
.. O .. Me
:O .. Me
(iii)
O
.. C–O .. – H
(c)
–I
NO2
Ortho effect
O
–M –I
Acidity order = c > a > b
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Solution Slot - 1 (Chemistry)
Page # 6
OH Q.7(i)
OH
OH
(a)
(b)
OH
(c)
CH3
(d) Cl
NO2 –M –I
NH2
Strongest acid ⇒ (b)
OH (ii)
OH
OH
(b)
NO2 –I is more
OH
(c)
(d)
CH3
F
So maximum acidity.
OH
OH O Me
(iii)(a)
OH
OH
(b)
(c)
(d)
O Me
+M
O Me –M
–I acidity ∝ with drawing tendency of group. so (b) max. acidity.
Q.8
(I) C6H5 – OH
(II)
F
(III)
OH
+M > –I
Cl
OH
(iv) O2N
OH
–M
–I > M
IV > III > II > I
Q.9
(A)
(b)
(c)
(d) CH2 – CH – CH3
The conjugate base of compound (b) is most stable due to aromaticity (i.e.
OH (A)
CH 3 NO2 –I
)
CH3
CH3 Q.10
–
(B)
H3C O2N –I
OH CH3
(C)
H3C
OH
H3C
NO2 –M
OH (D)
H 3C NO2 –I
Maximum acidity (C). Ans. (C)
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Solution Slot - 1 (Chemistry)
Page # 7 COOH
COOH Q.11 (A)
COOH
(B)
(C)
(D)
COOH OH
(ortho effect)
OH +I
OH +M
(most acidic) Compound (B) is less acidic due to (+M) effect of –OH group because it is present on pera position.
OH
OH
OH
HO
(II)
Q.12 (I)
–
(III) O
O
O
O
O
O
–
–O ⊕
–
–
⊕
O 4 eq. Aromatic R.S.
1 acidity ∝ Pk a
OH
O
–
O
O
O
O
⊕
O
–
⊕
O–
–
–
⊕ –
⊕ O O Non aromatic
acidity (ka) order = I > II > III Pka order = III > II > I
Q.13
(1) Phenol
(2) Ethyl alcohol
OH
(3) Formic acid
CH 2 – OH
(4) Benzoic acid COOH
H – C – OH O
pH ∝
1 1 ∝ acidity stability of conjugate Base
COOH– –
Stability of conjugate base = H – C – OH >
CH2 – O–
O– >
>
O Q.14
(I) (II)
CH3 – NH – CH2 – CH2 – OH CH3 – NH – CH2 – CH2 – O– CH3 – NH – CH2 – CH2 – CH2 – OH CH3 – NH – CH2 – CH2 – CH2 – O–
(III)
(CH3)3 N – CH2 – CH2 – OH
⊕
⊕
N( C H3)3 – CH2 – CH2 – O– Strong –I effect
Acidity order = III > I > II Q.15
Compound which are stronger acid than H2CO3 gives CO2 with NaHCO3. → Compound (A), (B), (C), (D) all are stronger acid than H2CO3.
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Solution Slot - 1 (Chemistry) 16.
Page # 8
(i) Cysteine :
COOH
HS NH2 1 Acidity ∝ Pk a
Acidity order = –COO – > –S – > –NH – pka order = – NH2 > – S > – COOH 10.8 8.3 1.8 (ii) glutomic acid : HOOC
COOH 2
1
NH2
acidity : – COOH > – NH2
.. acidity of –COOH (2) > – COOH (1) due to – I effect to –NH2 order of acidity : – COOH (2) > – COOH (1) > – NH2 order of Pka : – NH2 > – COOH (1) > – COOH (2)
17.
(a)
COOH
OH
OH
,
,
O–
COO–
acidity stablity of CB =
COOH
acidity =
<
OH
>
O–
<
OH
>
Pka order = 3 < 2 < 1
(b) 1 - butyne, 4 3 2 CH3 − CH2 − C
1-butene,
butane
1 CH
sp 2
sp
sp 2
acidity order = 1 > 2 > 3 Pka order = 1 < 2 < 3
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Solution Slot - 1 (Chemistry)
Page # 9 (c) Prophanoic acid, C − C − C − OH
3-Bromo propanoic acid
2-nitro propanoic acid
3 3 1 C − C − C − OH
2 1 C ? C ? C ? OH
O
acidity 3>2>1 pka order 3 < 2 < 1 (d) Phenol, O-nitrophenol, OH
||
NO2 O (–I)
O
Br –I
O-cresol
OH
OH NO2
CH3 +I +H
–M –I acidity order = 2 > 1 > 3 PKa order = 2 < 1 < 3 (e)
–
(1) CH3 − CH2 − CH2 − CH2 − CH2 − NH2 → CH − CH − CH − CH − CH − CH − NH 3 2 2 2 2 2 –
NH2
NH +
–H (2)
+
–H
(3)
– NH
acidity ∝ stability of conjugate Base stability order = 2 > 3 > 1 acidity order = 2 > 3 > 1 Pka = 2 < 3 < 1
O ||
18.
(a) CH3 − CH2 − Br & CH3 − NO2 (1)
+I
–
CH3 − CH − Br
–
(2) CH2 − NO2
–I (2) is stronger acid
(2) > (1)
&
(b) CH − C − CH 3 3
CH3 − C − CH2 − CN
(C) CH3 – CHO
CH3 – NO2
|| O
O || (1) CH2 − C − CH3 –
–
(2) CH3–C–CH–CN || –I O (2) > (1)
–
(1) CH2–C–H || – O (2) CH –NO 2 2 (2) > (1)
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Solution Slot - 1 (Chemistry)
Page # 10
OH
OH
acidity ∝
as (2)
19.
– M effect
O = C – CH3 –M So (2) is weaker acid
OH
OH
(b)
as (2)
O = C – CH3
CH3 +H
–M So (2) is weaker acid
SH
–
OH
S
O
–
or
(c)
S
–
Stability of C.B. =
O
–
.>
Acidity ∝ stability of C.B.
OH So
is weaker acid
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Solution Slot - 1 (Chemistry)
Page # 11
COOH
O
or
+ || O N
O || – C – OH
O 20.
COO
–
O2N
COO
–
–M acidity ∝ – M effect so (2) is more acidic (b) CH3 − CH2 − CH2 − OH ↓ CH3 – CH2 – CH2 – O –
or
So (2) is stronger acid (c) CH3 – CH = CH – CH2 – OH ↓ (d) CH3 – CH = CH – CH2 – O –
CH3 – CH = CH – OH ↓ CH3 – CH = CH – O – Showing Resonance so more acidic or
CH3 – CH = CH – OH
(2) CH3 – CH = CH – O – Show Resonance so more acidic
so (2) is stronger acid
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Solution Slot - 1 (Chemistry)
Page # 12
EXERCISE - 2 Q.1
(i)
HINT & SOLUTON
1 Basicity ∝ Stability of anion Stability order F¯ < Cl¯ < Br¯ < I¯ (size) Basicity order F¯ > Cl¯ > Br¯ > I¯
(ii)
Stability order CH3− < NH2− < OH− < F − (E.N.) Basicity order CH3− > NH2− > OH− > F −
(iii)
(iv)
.. Pn – NH2 .p. delocalised (weak base)
.. R – NH2 .p. localised (more basic)
.. R – C – NH2 O .p. more delocalised (least base)
basicity order a > b > c In gas phase basicity ∝ (+I) effect. Basicity order
NH3 < Me +I
NH3 < Me +I
NH
Me < Me +I +I
N
Me +I
Me +I (v)
In H2O, 3°-amine is less basic than 1°-amine due to less solvation. Basicity order : 2°–amine > 1°–amine > 3°–amine > NH3
O Q.2
(i)
.. NH
NH
.p. localised .. NH2
.. N Me
.p. delocalised .. NH2
(ii)
NH .. .p. localised (2° amine)
1°–amine Basicity c > a > b
.. N (iii)
.. N
O2N Strong-I Basicity ∝
Me +I-effect
.. N F Weak –I effect
1 (−I) effect
∝ (+I) effect Basicity b > c > a
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Solution Slot - 1 (Chemistry)
Page # 13 .. NH2
(iv)
.. NH2
+
(a) NH3 Strong-I-effect of + –NH3, least basic
(b) Cl Weak –I effect of –Cl, less basic
.. NH2
.. NH2
(c) CH3 +I effect of –CH3, most basic
.. Q.3(i) (a) CH3 CH2 NH2 sp3 Least electronegative. most basic order a > b > c
(ii)
(d) No effect of any group. Weaker than C, but stronger than a & b compound. .. (b) CH3 – CH = NH sp2
.. (c) CH3 – C ≡ N
sp most electronegative least basic.
.. (b) CH3CH2NH2
.. (a) CH3 – C – NH2
Lone pair localised
O least basic, l.p. of nitrogen is delocalised. .. (c) CH3 – C – NH2
.. .. (d) NH2 – C – NH 2
NH .. More basic due to delocalisation of only one l.p.
NH Most basic, due to more delocalisation of l.p.
Basicity order d > c > b > a
(iii)
(b)
(a)
NH .. * * *
(c)
N ..
NH .. 2
Compound 'a' is least basic, because l.p. of Nitrogen will more take participate in resonance due to aromaticity. Compound 'b' is most basic, because its l.p. is localised. Compound 'c' is more basic than 'b' because its l.p. is delocalised.
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Solution Slot - 1 (Chemistry)
Page # 14
O .. NH – C – CH3 (iv)
l.p. of nitrogen is in conjugation with benzene as well as – C – CH3 , least basic.
O .. NH2 l.p. of nitrogen is in conjugation only with benzene. it is more basic than compound 'a'.
.. NH – CH2CH3 due to extra +I effect of –CH2CH3 group, compound 'c' is most basic than a & b.. order c > b > a
CH3
NH2
NH2 CH3
(v)
Me
No2
Me NO2
Compound a is least basic than b, due to ortho effect of two –CH3 group.
NH 2
NH2
NH2
NH2
Q.4(i)
–M
NO2
–M
CN
+M
Order of –M Order of +M
–NO2 > –CN –NH2 > –OMe
Basicity ∝ +M ∝
1 −M
OMe
+M
NH2
∴ Basicity order d > c > b > a
NH2 Ortho effect (ii)
NH2
NH2
CH3 CH3
+I
CH3
+H
Compound C is most basic due to +H effect of –CH3 group. But compound a is least basic due to ortho effect of –CH3 group. Basicity order c > b > a
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Solution Slot - 1 (Chemistry)
Page # 15 NH2
NH2 (iii)
(a)
NH2
(b)
Ortho effect
NO2
(c)
NO2
–M
+I
NO2
Order b > a > c
NH2 Me
NH2
NH2
C – Me
CH3 (iv)
Me Ortho effect, less basic
Greater ortho effect, least basic.
Order a > b > c
Me
N
Me
Me
N
Me
Me
N
Me SIR
OMe
(v)
OMe
+M OMe
–I
Order c > a > b
Q.5(i)
(a)
.. N2 sp
(b)
.. N H sp & delocalisation of l.p. 2
S
(c)
(d) N
N 2
H sp
3
H sp
In compound 'd', hybridisation of N is sp3. i.e. less electronegative and –I effect of S is less. (ii)[b] In compound b, N atom is sp3 hybridised, and l.p. of nitrogen is localised. (iii) Compound a is most basic because charge density on double bonded nitrogen is very high due to resonance. (iv) In compound a, the e– charge density is very high due to –ve charge.
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Solution Slot - 1 (Chemistry)
Basicity ∝
Q.6(i)
Page # 16
1 E.N. of anion
Order of E.N. CH3CH2 < CH2 = CH– < HC ≡ C– < CH3CH2O– (ii)
.. NH2
(a)
l.p. is in conjugation only with benzene ring.
l.p. localised, most basic
NH2
(c)
.. CH2 – NH2
(b)
.. C – NH2
(d)
O
NO2 Ortho effect
l.p. is in conjugation with –C– group. Which is most O with drawing.
Order : b > a > c > d
O Q.7
CH3
Nb
CH3 CH2 – NH – C – CH3
H2N – C – H2C c
NH2 a
CH3
N d
CH3
a is most basic due to high charge density. b is more basic than d and c due to ortho effect. c is least basic because l.p. of nitrogen is conjugated with – C – group.. O
Order : a > b > d > c
NH2
NH2
NH2
NO2 Q.8
I least basic due to ortho effect
II
(I)
NO2 –M III
NH2
NH2 Q.9
NO2 –I
NH2
NH2
(III)
(II)
NO2 Order : III > IV > I > II
–M
(IV) OCH3
+M
CH3
+H
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Solution Slot - 1 (Chemistry)
Page # 17
Q.10
.. N H This lone pair very much delocalised because of getting Aromaticity so do not donate lone pair so it is least basic.
Q.11 (a)
CH3NH2
or
CF3NH2
F
.. CH3 – NH2
.. C – NH2
F
+I
F –I
have more tendensity to donate the pair of e–.
NH (b)
CH3CONH2
or
(c)
In aquous solution 2° are more basic because of solvation energy. 2>1 n-Pr NH2 or CH3CN .. .. CH3 – CH2 – CH2 – NH2 CH3 – C ≡ N
H2N
3
sp 'N' 1>2 (d)
NH2
sp Nitrogen more E.N weak basic
C6H5N (CH3)2 or 2,6-di methyl-N, N-dimethyl aniline.
H3C
N
H3C
CH3
H3C
.. CH3 N CH3
Goes out of plane So more basic
Ortho effect So 2 > 1 (e)
m-Nitro aniline or
p-nitro aniline
.. NH2
.. NH2
NO2 –I So more basic
NO2 –M 1>2
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Solution Slot - 1 (Chemistry)
Q.12 (a) p-methoxy aniline
Page # 18
or
p-cyano aniline
NH2
NH2
OCH3 +M So more basic 1 > 2 (b)
Puridine
or
CN –M Pyrrole
.. N
N .. Localised lone pair (c)
CH3C ≡ N or ↓ sp nitrogen so less basic
Q.13 (a) (b) (c)
.. CH3CH2 N H2 ↓ sp3 Nitrogen
2>1
.. or H3O+ H 2O (2) +ve charged spcie is weak base. H2S, –SH, S2– (1) Neutral so weak base More –ve charge more donating tendency. Cl–, –SH
(1)
(d)
Delocalised lone pair so least basic.
H
S
F– OH– – CH3 NH2–
Cl E.N. ↑ size small
More sable less basic NH2– CH3– OH– F– Size ↓
F–
(f)
..
(e)
(1) So more stable less basic .. .. .. HF .. , H2O , NH3 'F' more electronegative so do not donate the pair of e– so less basic. – – – OH SH SeH Big size –ve charge stable, So weak base. Basicity ∝
1 Stability
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Solution Slot - 1 (Chemistry)
Page # 19 .. NH2
.. NH2
Q.14 (a)
or
NO2 –M so more tendency to donate the pair of e– of NH2 in the ring. So weak base – CH2 = CH – CH2– (b) CH2 = CH – CH = CH – CH 2 or long resonance So–ve charge delocalised So weak base
O (c)
O
O
O
–O – C – C – OH or HO – C – C – OH Anions are more Basic than Neutral. So (2) is a weak base. .. OH
.. OH CH3 +I
CF3 –I
or
Strong Base
Weak Base
.. NH 2
.. NH2
CH3 Q.15(a) (i)
.. NH2
.. NH2
(ii)
(iii)
NO2 –M (iv)
Localised l.p. of N
Ortho effect 3 (iii) Basicity = (i) < (ii) < (iii)
(i)
2
(c)
sp .. CH2 = CH – CH2 – NH2
(i)
3
(ii) (iii)
sp .. CH3 – CH 2 – CH2 – NH 2 .. CH ≡ C – CH2 – NH 2 sp
Basic strength : 3 < 1 < 2
.. NH – C6H5
NH2 Q.17(a) (i)
NH2
(ii)
(iii)
More resonance More delocalised so least basic (ii) < (i) < (iii) NH2
NH2
NH2 Cl
(b)
(ii)
(i) Ortho effect
(iii) Cl –I
More
Cl Less
–I
1 3
NH2 O2N +M
Q.18 (a) CH3NH2 (Neutral), CH3+NH3 (+ve charged), CH3NH– (–ve charged) 2 2 Basicity order = 3 < 1 < 2
394,50 - Rajeev Gandhi Nagar Kota, Ph. No. : 93141-87482, 0744-2209671 IVRS No : 0744-2439051, 52, 53, www. motioniitjee.com ,
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Solution Slot - 1 (Chemistry)
Page # 21
.. NH2
H
NH2
N
Q.19 (a) More resonance least basic
2>1>3
H
N–H
(b)
N ..
N: Localised l.p.
l.p. Delocalised
1>2>3
(c)
H–N
.. N–H
N
N–H
Participate in resonance so least basic
5 6
4
Q.20 N 1
7
5
4
6
N
5
N
4
N1
8
N
N
3
:N – H 3
1
2
2
Pyrimidine Localised lone pair of N, 6πe–,aromatic
Imidazole Protonation takes place on N–1
– 1
9
N
2
3 H Purine Three basic 'N' N1, N3, N7
+
N
N–H 3
2
–
6πe aromatic So (A), (C) & (D) are aromatic.
394,50 - Rajeev Gandhi Nagar Kota, Ph. No. : 93141-87482, 0744-2209671 IVRS No : 0744-2439051, 52, 53, www. motioniitjee.com ,
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