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Get Ready for IIT-JEE 1
Advanced Problems
in
Organic Chemistry
for
IIT-JEE
By : M.S. Chouhan
1. Several factors (steric, electronic, orbital interactions etc.) can affect the inversion barrier of
an amine. In the given pair which data is correctly placed ?
i – PrMe
||
vsNN vs N Ni – PrMe Me Me(a) (b) Me Me
+ DG = 20.5 kcal/molDG = 7.0 kcal/molDG = 0.2 kcal/molDG = 7.9 kcal/mol
Me Cl
| |
vsN N
Me Me(c) Cl Cl (d*) All of these
+ ++ +
DG = 7.9 kcal/molDG = 22.9 kcal/mol
2. Circle represent most acidic hydrogens in these molecules. Which of the following is correct
representation ?
O
OH MeO C2
OH HO(a) (b) CO H2CF3
S
(c) (d*) All of these
F
3. The pH at which maximum hydrate is present in an solution of oxaloacetic acid, is :
O O O
|| || ||
H-O-C-C-CH-C-O-H2
pK=2.2 pK=3.98a a
(a*) pH = 0 (b) pH = 12 (c) pH = 4 (d) pH = 6 2 Get Ready for IIT-JEE
4. Which of the following isomeric hydrocarbons is most acidic ?
(a) (b*) (c) (d)
5. Compound A and B, both were treated with NaOH, producing a single compound C.
O O
CH CH3 3
-+ HO¾fi C. Compound (C) is :
heat
OH OH
(A) (B)
O
OC — CH OO 3
(a*) (b) (c) (d)
O
14
C label
OTs RCO H2
6. ¾fi ; Product of above reaction is :
no label
OCOROCOR
(a) (b)
(c*) both a and b (d) None of these
CH O OCH3 3
CN
150°C H C=CH — CN27. ¾fi(A)+(B)¾¾fi
CH O CCH3 3
(C)
Compound (B) reacts acrylonitrile to give (C), structure of compound (A) is :
CH2 OCH3
(a) (b*) (c) (d)
CH2 Get Ready for IIT-JEE 3
8. Increasing order of rate of reaction with HNO H SO is :3 2 4
O O O
O O OO O
O
(i) (ii) (iii)
(a) iii < ii< i (b) ii < iii < i
(c) i < iii < ii (d*) i < ii < iii
9. Match the columns I, II and III (Matrix).
Col umn-I Col umn- II Column- IIII
Re ac tion Nature of product formed Num ber of chiral center pres -
ent in product. (Consider
only one isomer in case of
racemic mixture or
Diastereomer)
Br2¾fi
CCl4(a) (p) (w)Racemic mix ture 0H
CH3
Br2 ¾fi
CCl4(b) (q) (x)Meso 1H
CH3
Br(c) 2 (r) (y) ¾fi Diastereomer 2
CCl4CH3
CH H3 Br2
C = =C (d) (s) (z)CCl Vic i nal dihalide 34
H CH3
O O
O 4 Get Ready for IIT-JEE
OH HH C3
H
10.
HH
HO
Ethynylestradiol (1)
The synthetic steroid ethynylestradiol (1) is a compound used in the birth control pill.
3A. How many sp hybridised carbon atoms are present in compound (1)?
(a) 8 (b) 9 (c) 10 (d) 11 (e) 12
2B. How many sp hybridised carbon atoms are present in compound (1) ?
(a) 4 (b) 5 (c) 6 (d) 7 (e) 8
C. How many sp hybridised
(a) 2 (b) 4 (c) 6 (d) 8 (e) 10
D. Which of the following functional group is contained in compound (1) ?
(a) A ketone (b) An alcohol (c) A carboxylic acid (d) An ester
E. How many asymmetric (stereogenic) centres are present in compound (1) ?
(a) 2 (b) 3 (c) 4 (d) 5
1. In the prep aration of iron from haematite (Fe O ) by the reaction with carbon2 3
Fe O + C¾fiFe+ CO2 3 2
How much 80% pure iron could be produced from 120 kg of 90% pure Fe O ?2 3
(a*) 94.5 kg (b) 60.48 kg
(c) 116.66 kg (d) 120 kg
-12. van der Waals constant b of helium is 24 mL mol . Find molecular diameter of helium.
-10 -8(a) 1.335·10 cm (b) 1.335·10 cm
-8 -8(c*) 2.67 10 cm (d) 4.34·10 cm Get Ready for IIT-JEE 5
2+ 2-3. There exist an equi librium between solid SrSO and Sr and SO ion in aque ous medium.4 4
The possible equilibrium states are shown in figure as thick line. Now, if equilibrium is
dist urbed b y a dd it ion of (a) Sr(NO ) and (b) K SO and dotted line represent approach3 2 2 4
of system toward’s equilibrium. Match the columns given below :
2+ 2+ 2+ 2+Sr Sr Sr Sr
(iii) (iv) (v) (vi)
2– 2– 2– 2–SO SO SO SO4 4 4 4
(I) addition of Sr(NO ) (II) addition of K SO3 2 2 4
(a) (I) (iii), (II) (iv) (b*) (I)(iv),(II) (v) (c) (I) (vi), (II) (v) (d) (I) (iv), (II) (v)
4. Determine the potential of the following cell :
+ –3 –Pt|H (g, 0.1 bar)|H (aq, 10 M )||MnO (aq, 0.1 M ), 2 4
2+ +Mn (aq, 0.01 M ), H (aq, 0.01 M )|Pt
Given : E° =1.51 V
– 2+MnO |Mn4
(a) 1.54 V (b*) 1.48 V (c) 1.84 V (d) none of these
5. 0.1 M KI and 0.2 M AgNO are mixed in 3 : 1 volume ratio. The depression of freezing point3
-1of the resulting solution will be [K (H O)=1.86 K kg mol ] :f 2
(a) 3.72 K (b) 1.86 K (c) 0.93 K (d*) 0.279 K
6. In an atomic bcc, what fraction of edge is not covered by atoms?
(a) 0.32 (b) 0.16 (c*) 0.134 (d) 0.268
7. Select the correct statement(s) :
(a*) A solution is prepared by addition of excess of AgNO solution in KI solution. The3
charge likely to develop on colloidal particle is positive.
(b*)The effects of pressure on physical adsorption is high if temperature is low
(c) Ultracentrifugation process is used for preparation of lyophobic colloids.
(d) Gold number is the index for extent of gold plating done
8. For a first order homogeneous gaseous reaction, A¾fi2B+C
If the total pressure after time t was P and after long time (t¥) was P then k in terms of t ¥
P , P and t is :t¥
2.303 P 2.303 2Pæ ö æ ö¥ ¥(a) k= log (b) k= logç÷ ç÷
t ŁP-Pł t ŁP-Pł¥ t ¥ t
2P2.303 æ ö¥(c*) k = log (d) none of theseç ÷
t Ł3(P - P )ł¥ t
9. Fixed mass of an ideal gas contained in a 24.63 L sealed rigid vessel at 1 atm is heated from
-73°C to 27°C. Calculate change in Gibb’s en ergy if entropy of gas is a function of
-2tem per a ture as S=2+10 T (J/K): (Use 1 atm L=0.1 kJ)
(a) 1231.5 J (b) 1281.5 J (c*) 781.5 J (d) 0 6 Get Ready for IIT-JEE
10. Column-I Col umn-II
h(A) Or bital an gu lar mo men tum of an (P) s(s+1)
elec tron 2p
(B) (Q)An gu lar mo men tum of an elec tron n(n+2)
in an orbit
nh(C) Spin an gu lar mo men tum of an (R)
elec tron 2p
h(D) Magnetic moment of atom (S) l(l+1)
2p
* Answer : A – S, B – R, C – P, D – Q
11. Which is the correct order of ionization energies?
– – – –(a) F > F> Cl > Cl (b*) F > Cl > Cl > F
– – – – –(c) F > Cl > Cl > F (d) F > Cl > F > Cl
12. Ionization energy of an element is :
(a*) Equal in magnitude but opposite in sign to the electron gain enthalpy of
the cation of the element
(b) Same as electron affinity of the element
(c*) Energy required to remove one valence electron from an isolated
gaseous atom in its ground state
(d) Equal in magnitude but opposite in sign to the electron gain enthalpy of the anion of the
element
13. O F is an unstable yellow orange solid and H O is a colourless liquid, both have O—O2 2 2 2
bond and O—O bond length in H O and O F respectively is :2 2 2 2
(a) 1.22Å, 1.48Å (b*) 1.48 Å, 1.22 Å
(c) 1.22Å, 1.22Å (d) 1.48Å, 1.48Å
14. Which of the following equilibria would have highest and lowest value of K at a commonp
temperature.
(a*) BeCO ƒ BeO+CO (b) CaCO ƒ CaO+CO3 2 3 2
(c) SrCO ƒS rO+CO (d*) BaCO ƒBaOCO3 2 3 2
15. In the structure of H CSF , which of the following statement is/are correct? 2 4
(a*) Two C—H bonds are in the same plane of axial S—F bonds
(b) Two C—H bonds are in the same plane of equatorial S—F bonds
(c*) Total six atoms are in the same plane
(d) Equatorial S—F plane is perpendicular to plane of p-bond
16. In the isoelectronic series of metal carbonyl, the CO bond strength is expected to increase in
the order :
+ - - +(a) [Mn(CO) ]<[Cr(CO) ]<[V(CO) ] (b*)[V(CO) ]<[Cr(CO) ]<[Mn(CO) ]6 6 6 6 6 6
- + + -(c) [V(CO) ] <[Mn(CO) ]<[Cr(CO) ] (d) [Cr(CO) ]<[Mn(CO) ]<[V(CO) ]6 6 6 6 6 6 Get Ready for IIT-JEE 7
2+ 4–17. [Fe(H O) ] and [Fe(CN) ] differ in :2 6 6
(a) geometry, magnetic moment (b) geometry, hybridization
(c*) magnetic moment, colour (d) hybridization, number of d-electrons
18. Which of the following statements is true ?
2+(a) In [PtCl (NH ) ] the cis form is optically inactive while trans form is optically active2 3 2
3–(b*)In [Fe(C O ) ] , geometrical isomerism does not exist while optical2 4 3
isomerism exists
(c) In Mabcd, square planar complexes show both optical as well as geometrical isomerism
(d) In tetrahedral complex, optical isomerism cannot be observed
4– –1 2–19. The CFSE for [(CoCl) ] complex is 18000cm . The D for [CoCl ] will be :6 4
–1 –1 –1 –1(a) 18000cm (b) 16000 cm (c*) 8000 cm (d) 2000 cm
X Y Z20. A B A C
Pale yellow Yellow Regenerated Deep blueDeep blue
solution solution solution
The sequential unknown reagents is/are:
+ 2+(a) H O /neutral medium, H O /H , Cu salt solution2 2 2 2
+ -3+(b*) H O / H , H O / OH , Fe2 2 2 2
- +2+(c) H O /OH , H O /H , Co salt solution2 2 2 2
- 2+(d) H O /neutral medium, H O /OH , Fe salt solution2 2 2 2
21. Se lect cor rect state ment(s):
(I) When excess FeCl solution is added to K [Fe(CN) ] solution, in addition to 3 4 6
III II - II III -Fe [Fe (CN) ] , Fe [Fe (CN) ] is also formed due to side redox reaction6 6
II III -(II) When FeCl is added to K [Fe(CN) ] solution, in addition to Fe [Fe (CN) ] , 2 3 6 6
III II - Fe [Fe (CN) ]6
III II - II III -(III) Fe [Fe (CN) ] is paramagnetic while Fe [Fe (CN) ] is diamagnetic6 6
III II - II III -(IV) Fe [Fe (CN) ] is diamagnetic while Fe [Fe (CN) ] is paramagnetic6 6
(a*) I, II (b) III, IV (c) both (a) and (b) (d) None of these
22. Col umn-I Col umn-II
(Pair of com plexes) (Property which is similar in given pair)
3– 2+(A) (P) Mag netic mo ment[Fe(CN) ] and [Co(NH ) ]6 3 6
2+ 4–(B) (Q) Ge om e try[Fe(H O) ] and [Fe(CN) ]2 6 6
4–(C) (R) Hy bridi sa tion[Ni(CN) ] and [Ni(CO) ]4 4
2+ 2–(D) (S)[Ni(H O) ] and [NiCl ] Num ber of d-electrons2 6 4
* Answer : A – P, Q, R; B – Q, S; C – P, Q, R, S; D – P, S 8 Get Ready for IIT-JEE
1P A S S A G EP A S S A G E
+H O CO SO Na S/I Ag /salt2 2 2 2 2
Na¾fiA¾fiB¾fiC¾fiD¾fiE (complex)
D
a. The compound B and C are :
(a) Na CO , Na SO (b) NaHCO , Na SO2 3 2 4 3 2 4
(c*) Na CO , Na SO (d) None of these2 3 2 3
b. The compound D is :
(a) Na SO (b) Na S O (c) Na S O (d*) Na S O2 4 2 4 6 2 2 5 2 2 3
c. Oxidation number of each ‘S’ atom in compound D :
(a) +2, +2 (b) +4, 0 (c*) +6, –2 (d) +5, –1
2P A S S A G EP A S S A G E
The molecule in which an atom is associated with more than 8 electrons is known as
hypervalent molecule and less than 8 electrons is known as hypovalent molecule. All molecules must have dp- bonding but the molecules having back bonding
need not to have always dp--bonding.
a. Which of the molecule is not hypovalent but completes its octet :
(a) AlCl (b) AlBr (c*) AlF (d) BF3 2 3 3
b. Which of the following molecule is having complete octet :
(a) BeCl (dimer) (b) BeH (dimer) (c) BeH (s) (d*) BeCl (s)2 2 2 2
c. Which of the following molecule is not having dp--bonding :
(a) SO (b) P O (c) PF (d*) B N H2 4 10 3 3 3 6
3P A S S A G EP A S S A G E
An unknown mixture contains one or two of the following : CaCO , BaCl , AgNO ,3 2 3
Na SO , ZnSO and NaOH. The mixture is completely soluble in water and solution2 4 4
gives pink colour with phenolphthalein. When dilute hydrochloric acid is gradually
added to the solution, a precipitate is formed which dissolves with further addition of the
acid.
a. Which of the following combination of compounds is soluble in water?
(a) BaCl and AgNO (b) AgNO and NaOH2 3 3
(c) BaCl and Na SO (d*) ZnSO and excess NaOH2 2 4 4
b. The aqueous solution of mixture gives white precipitate with dil. HCl which dissolves in
excess of dil. HCl. It confirms:
(a) BaCl+NaOH (b) Na SO+NaOH2 2 4
(c*) ZnSO+NaOH (d) AgNONaOH4 3
c. The white pre cip i tate is:
(a) ZnSO (b) Na ZnO4 2 2
(c*) Zn(OH) (d) ZnCl2 2 Get Ready for IIT-JEE 9
Mechanics
for
IIT-JEE
Vol. 1
By : Er. Anurag Mishra
1. Initially blocks A and B are given impulse in opposite directions as shown in figure. Now we
have to calculate the :
(i) Maximum stretch in spring B A
2uu(ii) Maximum velocity of block A and B in ground 0 0
m 2mframe.
(iii)Minimum velocity of block A and B in ground
(a)frame.
Fig. 4.21
(2m)(2u)-mu0 0Sol. u= =uCM 03m
At initial instant
fififi $ v A CM=v A g-v CM g=ui0
fififi $ v =v -v =-2uiB CM CM g 0B g
At maximum stretchInitial state
u = 0 u = 02u B/CM A/CM 0 u0m 2m u u0 0
l + x0 m
CM frame CM frame
(b)
Fig. 4.21
From work energy equation in CM frame
We get w =DKEspring system CM
1 1 (2m)(m)2 2 -kX=0- (3u)m 02 2 (2m+m)
1 1 22 æö 2 kX=ç m÷(3u)m 0Łł2 2 3
6m
or X=um 0k 10 Get Ready for IIT-JEE
Blocks return to
relaxed state
u = uu = 2u A/CM 0B/CM 0
m 2m u0ìBlock
What appears in
diagrams
CM frame
representing
situation when
block returns
to relaxedí
u = 3uu = 0A/g 0 B/g state
m 2mî
What appears in
ground frame
(c)
Fig. 4.21
When spring again regains its natural length in CM frame.
Most Important Concept
fififi
ground CM ground v =v +vA A CMframe frame frame
fififi
ground CM groundsimilarly v =v +vB B CMframe frame frame
Note that velocity of any block in ground frame is superposition of two velocity vectors, velocity of block
in CM frame and velocity of CM with respect to ground.
fi fi
v ground is maximum when v CM has maximum magnitude and is in same direction as vector A Aframe frame

v ground.CM frame
fi fifi
ground CM groundSimilarly v is minimum when v and v vectors are in opposite direction.A A CMframe frame frame
fififiCM ground | v |= | v |+| v |=2u uA fi fiCM 0 0A max frame frame |v | = u |v | = B/CM 0 A/CM 2
fififi uCM ground 0Similarly v |= | v | +| v |=3u B AB max B CM 0frame frame
Minimum velocity of A is attained when block is at equilibrium CM frame
fifi
Fig. 4.22position (spring is relaxed) and v and v areCM gA CM
opposite to each other.

Thus | v |=0Amin

Minimum velocity of B is attained at the instant B is moving toward left (opposite v ) and velocityCM
magnitude is u (see figure)0

Thus | v |=0Bmin