Iit Chemistry Paper I
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IIT-JEE - 2009 Paper - I CHEMISTRY CODE-1
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PART - I SECTION - I Straight Objective Type This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C), and (D) for its answer, out of which ONLY ONE is correct .
1..
The Henry’s law constant for the solubility of N 2 gas in water at 298 K is 1.0 × 105 atm. The mole fraction of N 2 in air is 0.8. The number of moles of N 2 from air dissolved in 10 moles of water at 298K and 5 atm pressure is Sol: C Acc. to Henry’s law P = K H .x
PAGE
B) 4.0 × 10 −5
A) 4.0 × 10 −4
C) 5.0 × 10−4
D) 4.0 × 10 −6
[Partial pressure of N 2 = 0.8 × 5 = 4]
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4 = 10 × x 5
x = 4 × 10−5
x = 4 × 10 −5 =
4 × 10−5 =
x 10
moles of N 2 x = Total moles 10 + x
(10 + x ≅ 10)
x = 4 × 10−4
0.8 moles of N 2 - 1 mole of air 4 × 10 −4 moles of N 2 - ?
4 × 10−4 × 1 1 = × 10−3 2 8 × 10 −1 = 5 × 10 −4
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The correct acidity order of the following is OH
OH
(I)
(II)
COOH
COOH
(III)
CH 3 (IV)
Cl
PAGE
A) (III) > (IV) > (II) > (I)
B) (IV) > (III) > (I) > (II) C) (III) > (II) > (I) > (IV) D) (II) > (III) > (IV) > (I) Sol: A
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The reaction of P4 with X leads selectively to P4O6 . The X is A) Dry O2
B) A mixture of O2 and N 2
C) Moist O2
D) O2 in the presence of aqueous NaOH
Sol: B N 2 acts as a diluent and prevents further oxidation.
4.
Among cellulose, poly(vinyl chloride), nylon and natural rubber, the polymer in which the intermolecular force of attraction is weakest is A) Nylon B) Poly (vinyl chloride) C) Cellulose Sol: D
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D) Natural Rubber
Natural rubber is an elastomer.
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5.
Given that the abundances of isotopes
54
Fe, 56 Fe and
atomic mass of Fe is A) 55.85 B) 55.95
57
Fe are 5%, 90% and 5%, respectively, the
C) 55.75
D) 56.05
Sol: B Average mass =
5 × 54 + 90 × 56 + 5 × 57 = 55.95 100
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The IUPAC name of the following compound is
OH
CN Br
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A) 4-Bromo-3-cyanophenol
B) 2-Bromo-5-hydroxybenzonitrile
C) 2-Cyano-4-hydroxybromobenzene D) 6-Bromo-3-hydroxybenzonitrile Sol: B
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CN is the principal functional group. 7.
Among the electrolytes Na2 SO4 , CaCl2 , Al2 ( SO4 )3 and NH 4 Cl , the most effective coagulating agent for Sb2 S3 sol is A) Na2 SO4
B) CaCl2
C)
Al2 ( SO4 )3
D) NH 4 Cl
Sol: C Based upon Hardy-Schulze rule. 8.
The term that corrects for the attractive forces present in a real gas in the van der Waals equation is A) nb
B)
an 2 V2
C) −
an 2 V2
D) -nb
Sol: B
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SECTION - II Multiple Correct Choice Type This section contains 4 multiple choice questions. Each questions has 4 choices (A), (B), (C) and (D) for its answer, out of which ONE OR MORE is/are correct. 9.
The compound(s) formed upon combustion of sodium metal in excess air is(are) A) Na2 O2 Sol: A, B
10.
B) Na2 O
C) NaO2
D) NaOH
PAGE
The correct statement(s) about the compound H 3C ( HO ) HC − CH = CH − CH (OH )CH 3 X is(are) A) The total number of stereoisomers possible for X is 6
B) The total number of diastereomers possible for X is 3
C) If the stereochemistry about the double bond in X is trans, the number of enantiomers possible for X is 4
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D) If the stereochemistry about the double bond in X is cis, the number of enantiomers possible for X is 2 Sol: C, D In cis-configuration meso compound is possible.
11.
The compound(s) that exhibit(s) geometrical isomerism is(are) A) [ Pt (en)Cl2 ]
B) [ Pt (en) 2 ]Cl2
C) [ Pt (en) 2 Cl2 ]Cl2
D) [ Pt ( NH 3 ) 2 Cl2 ]
Sol: C, D
12.
The correct statement(s) regarding defects in solids is(are) A) Frenkel defect is usually favoured by a very small difference in the sizes of cation and anion
PAGE
B) Frenkel defect is a dislocation defect
C) Trapping of an electron in the lattice leads to the formation of F-center D) Schottky defects have no effect on the physical properties of solids Sol: B, C
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SECTION - III Comprehension Type This section contains 2 groups of questions. Each group has 3 multiple choice questions based on a paragraph. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct. Paragraph QParagraph for Question Nos. 13 to 15.No. 13 to 15 A carbonyl compound P, which gives positive iodoform test, undergoes reaction with MeMgBr followed by dehydration to give an olefin Q. Ozonolysis of Q leads to a dicarbonyl compound R, which undergoes intramolecular aldol reaction to give predominantly S.
PAGE −
1. O3 1. MeMgBr 1. OH → Q → R → P S 2. Zn , H O 2. ∆ 2. H + , H O 2
2
3. H 2 SO4 , ∆
13.
The structure of the carbonyl compound P is
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A) O
Me
Me
O
Me
FeCl3 + K 4 [ Fe(CN )6 ] → Fe4 [ Fe(CN )6 ] O
Prussian blue
C)
D) O
Et
Me
Sol: B
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The structures of the products Q and R, respectively, are O H
Me
A) Me
Me
COMe , Me
Me O
PAGE H
B)
CHO
Me
Me
,
Me
Me
O
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C)
CHO
Me
,
Et
Et
Me O
Me
CH 3 CHO
D) Me
, Me
Et
Sol: A
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The structure of the product S is O
B)
A)
O
Me
Me
O
Me
O
PAGE
C)
Me
D)
Me
Me
Sol: B Solution for 13 to 15
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. MeMgBr 1 → +
H 2 SOH →
2. H H 2 O
O
⊕
HO
HO
(P)
⊕
H
PAGE (Q)
O3 / ZnH 2O
O
H
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O
OH −
O (S)
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Paragraph for Question Nos. 16 to 18 p-Amino-N, N-dimethylaniline is added to a strongly acidic solution of X. The resulting solution is treated with a few drops of aqueous solution of Y to yield blue coloration due to the formation of methylene blue. Treatment of the aqueous solution of Y with the reagent potassium hexacyanoferrate(II) leads to the formation of an intense blue precipitate. The precipitate dissolves on excess addition of the reagent. Similarly, treatment of the solution of Y with the solution of potassium hexacyanoferrate (III) leads to a brown coloration due to the formation fo Z. 16.
The compound X is A) NaNO3 C) Na2 SO4 Sol: D
17.
The compound Y is A) MgCl2
PAGE B) NaCl
D) Na2 S
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C) FeCl3
D) ZnCl2
Sol: C
FeCl3 + K 4 [ Fe(CN )6 ] → Fe4 [ Fe(CN )6 ] Prussian blue
18.
The compound Z is A) Mg 2 [ Fe(CN )6 ]
B) Fe[ Fe(CN ) 6 ]
C) Fe4 [ Fe(CN )6 ]3
D) K 2 Zn3 [ Fe(CN )6 ]2
Sol: B
FeCl3 + K3 [ Fe(CN )6 ] → Fe [ Fe(CN )6 ] Brown
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SECTION - IV Matrix - Match Type This section contains 2 questions. Each question contains statements given in two columns, which have to be matches. The statements in COLUMN-I are labelled A, B, C and D, while the statements in COLUMNII are labelled p, q, r, s and t. Any given statement(s) in COLUMN-II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following examples:
19
Match each of the compounds in Column I with its characteristic reaction(s) in Column II. Column I Column II A) CH 3CH 2CH 2CN B) CH 3CH 2OCOCH 3
PAGE
p) Reduction with Pd − C / H 2 q) Reduction with SnCl2 / HCl
C) CH 3 − CH = CH − CH 2 OH
r) Development of foul smell on treatment with chloroform and alcoholic KOH
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D) CH 3CH 2CH 2CH 2 NH 2
(DIBAL-H) t) Alkaline hydrolysis Sol: A → p, q, s, t ; B → p, s, t; C → p; D → r
20.
Match each of the diatomic molecules in Column I with its property/properties in Column II. Column I Column II A) B2
p) Paramagnetic
B) N 2
q) Undergoes oxidation
C) O2−
r) Undergoes reduction
D) O2
PAGE
s) Bond order ≥ 2 t) Mixing of ‘s’ and ‘p’ orbitals.
Sol: A → p, q, r; B → q, r , s; C → p, q, r , s; D → p, q, r
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PART - I SECTION - I Straight Objective Type This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C), and (D) for its answer, out of which ONLY ONE is correct .
1..
The Henry’s law constant for the solubility of N 2 gas in water at 298 K is 1.0 × 105 atm. The mole fraction of N 2 in air is 0.8. The number of moles of N 2 from air dissolved in 10 moles of water at 298K and 5 atm pressure is Sol: C Acc. to Henry’s law P = K H .x
PAGE
B) 4.0 × 10 −5
A) 4.0 × 10 −4
C) 5.0 × 10−4
D) 4.0 × 10 −6
[Partial pressure of N 2 = 0.8 × 5 = 4]
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4 = 10 × x 5
x = 4 × 10−5
x = 4 × 10 −5 =
4 × 10−5 =
x 10
moles of N 2 x = Total moles 10 + x
(10 + x ≅ 10)
x = 4 × 10−4
0.8 moles of N 2 - 1 mole of air 4 × 10 −4 moles of N 2 - ?
4 × 10−4 × 1 1 = × 10−3 2 8 × 10 −1 = 5 × 10 −4
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The correct acidity order of the following is OH
OH
(I)
(II)
COOH
COOH
(III)
CH 3 (IV)
Cl
PAGE
A) (III) > (IV) > (II) > (I)
B) (IV) > (III) > (I) > (II) C) (III) > (II) > (I) > (IV) D) (II) > (III) > (IV) > (I) Sol: A
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The reaction of P4 with X leads selectively to P4O6 . The X is A) Dry O2
B) A mixture of O2 and N 2
C) Moist O2
D) O2 in the presence of aqueous NaOH
Sol: B N 2 acts as a diluent and prevents further oxidation.
4.
Among cellulose, poly(vinyl chloride), nylon and natural rubber, the polymer in which the intermolecular force of attraction is weakest is A) Nylon B) Poly (vinyl chloride) C) Cellulose Sol: D
PAGE
D) Natural Rubber
Natural rubber is an elastomer.
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5.
Given that the abundances of isotopes
54
Fe, 56 Fe and
atomic mass of Fe is A) 55.85 B) 55.95
57
Fe are 5%, 90% and 5%, respectively, the
C) 55.75
D) 56.05
Sol: B Average mass =
5 × 54 + 90 × 56 + 5 × 57 = 55.95 100
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The IUPAC name of the following compound is
OH
CN Br
PAGE
A) 4-Bromo-3-cyanophenol
B) 2-Bromo-5-hydroxybenzonitrile
C) 2-Cyano-4-hydroxybromobenzene D) 6-Bromo-3-hydroxybenzonitrile Sol: B
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CN is the principal functional group. 7.
Among the electrolytes Na2 SO4 , CaCl2 , Al2 ( SO4 )3 and NH 4 Cl , the most effective coagulating agent for Sb2 S3 sol is A) Na2 SO4
B) CaCl2
C)
Al2 ( SO4 )3
D) NH 4 Cl
Sol: C Based upon Hardy-Schulze rule. 8.
The term that corrects for the attractive forces present in a real gas in the van der Waals equation is A) nb
B)
an 2 V2
C) −
an 2 V2
D) -nb
Sol: B
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SECTION - II Multiple Correct Choice Type This section contains 4 multiple choice questions. Each questions has 4 choices (A), (B), (C) and (D) for its answer, out of which ONE OR MORE is/are correct. 9.
The compound(s) formed upon combustion of sodium metal in excess air is(are) A) Na2 O2 Sol: A, B
10.
B) Na2 O
C) NaO2
D) NaOH
PAGE
The correct statement(s) about the compound H 3C ( HO ) HC − CH = CH − CH (OH )CH 3 X is(are) A) The total number of stereoisomers possible for X is 6
B) The total number of diastereomers possible for X is 3
C) If the stereochemistry about the double bond in X is trans, the number of enantiomers possible for X is 4
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D) If the stereochemistry about the double bond in X is cis, the number of enantiomers possible for X is 2 Sol: C, D In cis-configuration meso compound is possible.
11.
The compound(s) that exhibit(s) geometrical isomerism is(are) A) [ Pt (en)Cl2 ]
B) [ Pt (en) 2 ]Cl2
C) [ Pt (en) 2 Cl2 ]Cl2
D) [ Pt ( NH 3 ) 2 Cl2 ]
Sol: C, D
12.
The correct statement(s) regarding defects in solids is(are) A) Frenkel defect is usually favoured by a very small difference in the sizes of cation and anion
PAGE
B) Frenkel defect is a dislocation defect
C) Trapping of an electron in the lattice leads to the formation of F-center D) Schottky defects have no effect on the physical properties of solids Sol: B, C
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SECTION - III Comprehension Type This section contains 2 groups of questions. Each group has 3 multiple choice questions based on a paragraph. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct. Paragraph QParagraph for Question Nos. 13 to 15.No. 13 to 15 A carbonyl compound P, which gives positive iodoform test, undergoes reaction with MeMgBr followed by dehydration to give an olefin Q. Ozonolysis of Q leads to a dicarbonyl compound R, which undergoes intramolecular aldol reaction to give predominantly S.
PAGE −
1. O3 1. MeMgBr 1. OH → Q → R → P S 2. Zn , H O 2. ∆ 2. H + , H O 2
2
3. H 2 SO4 , ∆
13.
The structure of the carbonyl compound P is
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A) O
Me
Me
O
Me
FeCl3 + K 4 [ Fe(CN )6 ] → Fe4 [ Fe(CN )6 ] O
Prussian blue
C)
D) O
Et
Me
Sol: B
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The structures of the products Q and R, respectively, are O H
Me
A) Me
Me
COMe , Me
Me O
PAGE H
B)
CHO
Me
Me
,
Me
Me
O
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C)
CHO
Me
,
Et
Et
Me O
Me
CH 3 CHO
D) Me
, Me
Et
Sol: A
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The structure of the product S is O
B)
A)
O
Me
Me
O
Me
O
PAGE
C)
Me
D)
Me
Me
Sol: B Solution for 13 to 15
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. MeMgBr 1 → +
H 2 SOH →
2. H H 2 O
O
⊕
HO
HO
(P)
⊕
H
PAGE (Q)
O3 / ZnH 2O
O
H
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O
OH −
O (S)
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Paragraph for Question Nos. 16 to 18 p-Amino-N, N-dimethylaniline is added to a strongly acidic solution of X. The resulting solution is treated with a few drops of aqueous solution of Y to yield blue coloration due to the formation of methylene blue. Treatment of the aqueous solution of Y with the reagent potassium hexacyanoferrate(II) leads to the formation of an intense blue precipitate. The precipitate dissolves on excess addition of the reagent. Similarly, treatment of the solution of Y with the solution of potassium hexacyanoferrate (III) leads to a brown coloration due to the formation fo Z. 16.
The compound X is A) NaNO3 C) Na2 SO4 Sol: D
17.
The compound Y is A) MgCl2
PAGE B) NaCl
D) Na2 S
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C) FeCl3
D) ZnCl2
Sol: C
FeCl3 + K 4 [ Fe(CN )6 ] → Fe4 [ Fe(CN )6 ] Prussian blue
18.
The compound Z is A) Mg 2 [ Fe(CN )6 ]
B) Fe[ Fe(CN ) 6 ]
C) Fe4 [ Fe(CN )6 ]3
D) K 2 Zn3 [ Fe(CN )6 ]2
Sol: B
FeCl3 + K3 [ Fe(CN )6 ] → Fe [ Fe(CN )6 ] Brown
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SECTION - IV Matrix - Match Type This section contains 2 questions. Each question contains statements given in two columns, which have to be matches. The statements in COLUMN-I are labelled A, B, C and D, while the statements in COLUMNII are labelled p, q, r, s and t. Any given statement(s) in COLUMN-II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following examples:
19
Match each of the compounds in Column I with its characteristic reaction(s) in Column II. Column I Column II A) CH 3CH 2CH 2CN B) CH 3CH 2OCOCH 3
PAGE
p) Reduction with Pd − C / H 2 q) Reduction with SnCl2 / HCl
C) CH 3 − CH = CH − CH 2 OH
r) Development of foul smell on treatment with chloroform and alcoholic KOH
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D) CH 3CH 2CH 2CH 2 NH 2
(DIBAL-H) t) Alkaline hydrolysis Sol: A → p, q, s, t ; B → p, s, t; C → p; D → r
20.
Match each of the diatomic molecules in Column I with its property/properties in Column II. Column I Column II A) B2
p) Paramagnetic
B) N 2
q) Undergoes oxidation
C) O2−
r) Undergoes reduction
D) O2
PAGE
s) Bond order ≥ 2 t) Mixing of ‘s’ and ‘p’ orbitals.
Sol: A → p, q, r; B → q, r , s; C → p, q, r , s; D → p, q, r
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