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WORRY IS A MISUSE OF IMAGINATION.
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"People with integrity do what they say they are going to do. Others have excuses." Rudyard Kipling, the celebrated English author and poet, once said, "We have forty million reasons for failure, but not a single excuse." Yet today we are literally inundated with a tidal wave of excuses from every direction. In fact, it seems everyone has a reason, explanation or justification for not doing what they were supposed to do. Why do so many of us crank out one excuse after another for virtually everything we fail to do? Well, for starters, excuses are easy. In fact, they're way too easy. After all, making excuses doesn't require any effort or commitment on our part. All we have to do is toss out excuse after excuse and we feel we're off the hook, since the best excuses always absolve us of any personal responsibility whatsoever. While getting in the habit of making of excuses is easy, excuse making doesn't get any of us anywhere close to where we want to go in life. Sooner or later all of our years of excuses eventually catch up with us. Before we realize it, the best of life has slipped away in a lazy, hazy, crazy blur of excuses. Ninety-nine percent of the failures come from people who have the habit of making excuses. Hold yourself responsible for a higher standard than anybody else expects of you, never excuse yourself. The person who really wants to do something invariably finds a way to get it done. And for those who don't want to do something; well, one excuse is just as good as another I suppose. What it all boils down to is simply this: what kind of person do you really want to be? Do you want to make excuses - or make something happen instead? It's time to turn all of your excuses loose once and for all. Each and every time you fall short, pick yourself up, learn from your mistakes and immediately get going again. No complaining, no explaining and absolutely no excuses allowed. You will find that the minute you stop making excuses and start finding a way to get the job done, you'll start making your life everything it could be and should be... and so much more. Get rid of the excuses and you can get anywhere you've ever dreamed of going.
Presenting forever positive ideas to your success. Yours truly
Pramod Maheshwari, B.Tech., IIT Delhi
Editor : Pramod Maheshwari XtraEdge for IIT-JEE
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AUGUST 2010
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Volume-6 Issue-2 August, 2010 (Monthly Magazine) NEXT MONTHS ATTRACTIONS
CONTENTS INDEX
Regulars ..........
Much more IIT-JEE News. Know IIT-JEE With 15 Best Questions of IIT-JEE Challenging Problems in Physics,, Chemistry & Maths Key Concepts & Problem Solving strategy for IIT-JEE. Xtra Edge Test Series for JEE- 2011 & 2012
PAGE
NEWS ARTICLE
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IITian ON THE PATH OF SUCCESS
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KNOW IIT-JEE
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22-year-old becomes youngest IIT teacher Delhi power firm, IIT tie-up to reduce power loss Prof. Mohit Renderia & Dr. Rajiv Laroia Previous IIT-JEE Question
Study Time........ DYNAMIC PHYSICS S Success Tips for the Months • " Always bear in mind that your own resolution to succeed is more important than any other thing." • "God gave us two ends. One to sit on and one to think with. Success depends on which one you use; head you win -- tails, you lose." • "The ladder of success is best climbed by stepping on the rungs of opportunity." • "Success is getting what you want. Happiness is wanting what you get." • "The secret of success in life is for a man to be ready for his opportunity when it comes." • "I don't know the key to success, but the key to failure is trying to please everybody." • "The secret of success is to be in harmony with existence, to be always calm… to let each wave of life wash us a little farther up the shore."
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8-Challenging Problems [Set# 4] Students’ Forum Physics Fundamentals Capacitor - 2 Work power energy & Conservation Law
CATALYSE CHEMISTRY
30
Key Concept Reaction Mechanism Solid State Understanding : Physical Chemistry
DICEY MATHS
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Mathematical Challenges Students’ Forum Key Concept Vector Permutation & Combination
Test Time .......... XTRAEDGE TEST SERIES
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Class XII – IIT-JEE 2011 Paper Class XI – IIT-JEE 2012 Paper
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AUGUST 2010
22-year-old becomes youngest IIT teacher
MUMBAI: IITians often liken the generation gap between themselves and their teachers to that between MS-DOS and Windows. This semester, however, the students on the Powai campus can look forward to someone much closer to their age: a physics teacher who has just entered his 20s. At 22, Tathagat Avatar Tulsi, who has never studied in a classroom, plans to ask his students how they would want to be taught. "I have never taught in a class. But I believe I can come down to the level of a student and help them understand the subject," he said. Having completed high school when he was nine, his graduation in science at 10, an MSc in Physics at 12, and his PhD in Quantum Computing from the Indian Institute of Science (IISc), Bangalore, at 21, Tulsi says he is going to write to the Limca Book of Records to include him as the youngest faculty member in the country. Having achieved a lot pretty early in life, Tulsi may seem like a young man in hurry, but he has set a huge task for himself— to come up with an important scientific discovery, which will probably lead him to his ultimate dream: to own that shining piece of gold with Alfred Nobel on the obverse.
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The "wonder boy", who suffered humiliation in August 2001 when a delegation of scientists taken by the department of science & technology to Lindau in Germany for an interaction with Nobel laureates, suggested that he was not a thinker, but a "fake prodigy" who had "mugged up" theories. Putting that behind, the Patna boy will stay on the Powai campus in the faculty quarters and work towards achieving that dream. That "not-so-distant" goal is probably why Tulsi chose teaching over a vocation. "I want to pursue my research and at IIT-B, I will have the leisure to continue my research and one day set up a lab focused on quantum computation in our country." Going to foreign shores is currently not on Tulsi’s plans. He chose the Powai college over Waterloo University, Canada, and the Indian Institute of Science Education & Research (IISER), Bhopal, both of which had also offered him teaching jobs.
Delhi power firm, IIT tieup to reduce power loss New Delhi: The BSES Yamuna Power Limited (BYPL) and the Indian Institute of Technology (IIT)-Delhi have come together to improve the quality and reliability of power supply by reducing transmission and distribution losses. A memorandum of understanding (MoU) was signed Monday between BYPL CEO Ramesh Narayanan and Anil Wali, managing director, Foundation for Innovation and Technology Transfer (FITT) - a society established by IIT-Delhi, to focus on how to bring the next-level or 6
"SMART" technology to the power distribution business and to keep pace with technological innovations taking place in the transmission business. Both BYPL and IIT-Delhi will appoint one principal project investigator each. The cost of the projects will be borne by BYPL. IANS.
Forests ministry teams with IITs for Ganga management plan
New Delhi: The Indian Institutes of Technology (IIT), the premier higher technical educational institutions of the Ministry of Human Resource Development (HRD) have committed themselves to the responsibility of development of a management plan of the Ganga river basin called the National Ganga River Basin Management Plan Project (NGBRM). Seven IITs have come together for this purpose. The IITs have accepted this societal challenge as part of their response to the present-day challenges of the Indian society. "It is required to ensure that the flow of the river Ganga must be continuous (Aviral Dhara), the river must have longitudinal and lateral connectivity, the river must have adequate space for its various AUGUST 2010
functions and the river must not be seen as a carrier of waste loads (Nirmal Dhara)," stated an official press release. The management plan will outline the strategy and the actions that need to be undertaken for the maintenance and restoration of the Ganga basin. The management plan should take into account the constraints of population, urbanization, industrialization and agriculture activities. The IITs will form several thematic groups and each group will develop a detailed outline for the improvement of ecological health of the basin system. Besides the thematic groups, the IITs will also integrate in a holistic manner, all the issues into a comprehensive management plan. In order to develop this plan, discussions will be held with local, state and other agencies who have to deal with the maintenance of the basin system. The management plan will also take into account the experience of earlier attempts of Ganga Action Plans. The HRD ministry and the Ministry of Environment & Forests are coming together to support the initiatives of the IITs. The work is estimated to be carried out in a period of 18 months. The funding for this project is estimated to be about Rs.15 crores. An agreement has been signed between the Directors of seven IITs and the Ministry of Environment & Forests in the presence of Minister of State for HRD, Kapil Sibal and Mister of State for Environments and Forest, Jairam Ramesh. This initiative will involve not only faculty and students of seven IITs but will also take help from experts from other institutes and universities also.
IIT seeks robotic solution in conflict KOLKATA: A group of students from IIT-Kharagpur is working on a prototype of an Autonomous
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Ground Vehicle (AGV) which, if it proves successful, may be developed by the Defence Research and Development Organisation (DRDO) for use by security agencies for Low-Intensity Conflict (LIC).
Gupta, Sarbartha Banerjee, Subhagato Dutta, Rahul Das, Anindita Bhattacharya and A Srinivas Reddy submitted an excellent design. Officials are waiting to find out how the prototype performs.
The six-member IIT team has bagged the first rank in Phase-I in the Student Robot Competition, 2010, organised by DRDO. As many as 240 colleges and institutes from across the country participated in the first phase of the contest, where third- and fourth-year students were asked to design an Autonomous Ground Vehicle for Low Intensity Conflict'. Only 14 teams were shortlisted for Phase-II. Among the entries from the east, the Indian School of Mines team from Dhanbad was ranked fourth. The National Institute of Technology, Rourkela, ranked 14th.
"The idea of the competition is to harness the innovative ideas of our student community to the National Robotics Program of India. The demands made from the participants are enormous. The robot will have to complete a closed loop obstacle course of 500 metres within an anticipated time of 20 minutes, using autonomous navigation. For the first 350 metres, the robot would have to navigate with the help of lanefollowing' by colour detection. While doing this, it would have to avoid static positive obstacles, cross over slopes, staircases and corrugations. It would have to navigate the remaining 150 metres with the help of GPS waypoints. Maximum width of the robot would have to be 1 metre, maximum speed of 10 km per hour and minimum turning radius of 5 metres. It would have to carry an additional payload of 10 kg. The robot would have to be selfpowered in all respects," the official said. In Phase-I, teams submitted designs with system configuration details. In the next phase, the selected teams would have to build a prototype and make it perform before the judges. Initially, 10 teams were supposed to be shortlisted for Phase-II. Given the nature of the papers submitted, it was finally decided to shortlist 14. Each team received a cash award of Rs 1,00,000.
"Participants were told that the AGV will be a combat vehicle of the future and assigned tasks that a conventional manned vehicle cannot perform. These are basically autonomous robots to be used by security agencies engaged in LIC in urban and unstructured environments. These robots would be used in LIC and Explosive Ordnance Disposal (EOD) programmes in undesirable, hazardous and potentially lifethreatening environments," said a senior DRDO official. The AGV would have to be armed with sensors, software and other equipment to help it negotiate harsh terrain, identify and designate targets, engage and neutralise them. The vehicle would also have to detect minefields and neutralise them. In short, the AGV would be an autonomous off-road robotic platform that would navigate rough terrain and avoid natural and man-made obstacles in the shortest possible time. According to the team of experts who judged the entries, the IITKharagpur team, comprising Nalin 7
The teams are now busy building their prototypes. The final competition will be held at the Combat Vehicles Research and Development Establishment (CVRDE), Chennai, between September 27-29, where the prototypes would be tried out.
AUGUST 2010
Success Story This articles contains stories of persons who have succeed after graduation from different IIT's
Dr. Rajiv Laroia B.Tech., IIT – Delhi
Prof. Mohit Randeria received his Bachelor’s Degree in Electrical Engineering from IIT Delhi in 1980. He obtained M.S. Degree from the California Institute of Technology, USA in 1982 and Ph.D. Degree from Cornell University, USA in 1987. Prof. Mohit Randeria is presently Professor of Physics at the Ohio State University in Columbus, Ohio. Prof. Randeria started his academic career as an Assistant Professor of Physics at SUNY Stony Brook in 1989, after a few years of post-doctoral work at Cornell and Illinois. He then joined the Argonne National Lab in Illinois as a Staff Scientist in the Materials Science Division, and after four years there, resumed his academic career at the Tata Institute of Fundamental Research (TIFR), Mumbai in 1995. He spent nine years in the Theoretical Physics Department at TIFR as Reader, Associate Professor, and Professor. Since 2004, he has been a Professor of Physics at Ohio State University. Prof. Randeria is an acknowledged expert in the area of Theoretical Condensed Matter Physics. His research interests are focused, at present, on high temperature superconductivity, strongly correlated systems, and ultra-cold atomic gases. Prof. Randeria is the author of over hundred research papers in Condensed Matter Physics. He is the recipient of a prize in Condensed Matter Physics, awarded in honour of Nobel Laureate Phillip Anderson by the International Center of Theoretical Physics, Trieste, Italy in 2002. He has also been awarded the Swarnajayanti Fellowship in 1998, the B.M. Birla Science prize in 1999, the S.S. Bhatnagar Award in 2002, and the George A. Miller Visiting Professorship at University of Illinois, Urbana-Champaign in 2002-2003. In honouring Prof. Mohit Randeria, IIT Delhi recognizes the outstanding contributions made by him as a Researcher and Scientist. Through his achievements, Prof. Mohit Randeria has brought glory to the name of the Institute.
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Dr. Rajiv Laroia received his Bachelor’s Degree in Electrical Engineering from IIT Delhi in 1985. He obtained his Ph.D. in Electrical Engineering from University of Maryland, College Park in 1992. Dr. Laroia is presently Senior Vice President of Technology at Qualcomm, USA. Dr. Laroia joined Siemens Research Labs in Munich, Germany after graduating from IIT Delhi. In 1992, he joined AT&T Bell Labs after obtaining his Ph.D. In 2000, Dr. Laroia founded Flarion Technologies, a venture backed company to develop and commercialize a novel all-IP mobile wireless broadband technology. He served as the CTO of Flarion and in that role provided the vision and led the development of technology and products for the company. In 2006, Flarion was acquired by the wireless technology giant Qualcomm, where he currently serves as Senior Vice President (Technology). Dr. Laroia is one of the world’s leading researchers and innovators in the field of communication. He holds more than 50 patents and has more than 100 others pending. His early research focused on wire-line communication. He has significant technology contributions to V.34 and V.90 International Standards for sending data over telephone lines. The technology he invented is used in virtually all dial-up modems in the world. Since 1997, Dr. Laroia has been working in the field of mobile wireless communication. The technology he and his team developed at Flarion is now incorporated in all three major next generation international wireless standards UMTS LTE, UMB and Wimax. Dr. Laroia has won numerous industry awards. In 2006, he was inducted to the Innovation Hall of Fame at the University of Maryland. He is a fellow of the IEEE. In honouring Dr. Rajiv Laroia, IIT Delhi recognizes the outstanding contributions made by him as an Entrepreneur and Technologist. Through his achievements, Dr. Rajiv Laroia has brought glory to the name of this Institute. 8
AUGUST 2010
KNOW IIT-JEE By Previous Exam Questions
2.
PHYSICS 1.
A small body attached to one end of a vertically hanging spring is performing SHM about it's mean position with angular frequency ω and amplitude a. If at a height y* from the mean position, the body gets detached from the spring calculate the value of y* so that the height H attained by the mass is maximum. The body does not interact with the spring during it's subsequent motion after detachment. (aω2 > g) [IIT-2005]
P0
m Sol. The total energy of the spring-mass system at any position of mass above the mean position is the sum of the follows. (a) Gravitation potential energy of mass (b) Kinetic energy of mass (c) Elastic potential of spring. The mass will reach the highest point when its mechanical energy [Sum of (a) and (b)] is maximum. This is possible when elastic potential energy of system is zero. ⇒ The mass should detach when the spring is at its natural length. Let L = Natural length of spring when mass m is hanging at equilibrium the L
L Kl
l
mg = kl ; ⇒y= ⇒ y=
where
mg mg l= k
mg k g
ω2 g ω2
Mean Position of oscillation
C
(c) As ∆U = 0 in cyclic process, hence ∆Q = ∆W ∆QAB + ∆QCA + ∆QBC = ∆W, PV PV ∆QBC = P0V0 – 0 0 = 0 0 2 2 As net heat is absorbed by the gas during path BC, temp. will reach maximum between B and C.
[Q K = mω2] < a (given)
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A
2V0 V0 (a) the work done by the gas. (b) the heat rejected by the gas in the path CA and the heat absorbed by the gas in the path AB; (c) the net heat absorbed by the gas in the path BC; (d) the maximum temperature attained by the gas during the cycle. Sol. n = 1 = no. of moles, For monoatomic gas : 5R 3R Cp = , Cv = 2 2 Cyclic process A → B ⇒ Isochoric process C → A ⇒ Isobaric compression (a) Work done = Area of closed curve ABCA during cyclic process. i.e. ∆ABC 1 1 ∆W = × base × height = V0 × 2P0 = P0V0 2 2 (b) Heat rejected by the gas in the path CA during Isobaric compression process ∆QCA = nCp∆T = 1 × (5R/2)(TA – TC) 2P0 V0 PV TC = , TA = 0 0 , I×R I×R 5 5R P0 V0 2P0 V0 = – P0V0 ∆QCA = − 2 2 R R Heat absorbed by the gas on the path AB during Isochoric process ∆QAB = nCv∆T = 1 × (3R/2) (TB – TA) 3R 3P0 V0 P0 V0 − = = 3P0V0 2 1× R 1× R
y0
K
One mole of an ideal monatomic gas is taken round the cyclic process ABCA as shown in figure. Calculate. [IIT-1998] P B 3P0
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(d) Equation for line BC 2P P = – 0 V + 5P0, As PV = RT hence, V0 RT P= [For one mole] [as y = mx + c] V 2P ...(1) ∴ RT = – 0 V2 + 5P0V V0
For maximum;
⇒ ⇒
2P0 dT = 0, – × 2V + 5P0 = 0; dV V0
5V0 4 Hence from equation (1) and (2)
∴ V=
...(2)
Substituting the value of q1' from (2) in (1) q1 ' + q2' = σ × 20 πR2 2 3q 2 ' = σ × 20 πR2 2 q2 ' 5 σ = × 2 3 2 4π(2R )
⇒
New charge density on bigger sphere q2 ' 5σ = = 2 6 4π(2R )
4.
In the given circuit R1 = 2R4 = 6 ohms E1 = 3E2 = 2E3 = 6 volts ; ; C = 5µF R3 = 2R2 = 4 ohms Find the current in R3 and the energy stored in the [IIT-1988] capacitor. E1 R1
2
2P0 5V 5V × 0 + 5P0 0 V0 4 4 25P0 V0 25 25 = –2P0V0 × + = P0V0 16 8 4 25 P0 V0 ∴ Tmax = 8 R
RTmax = –
C
Sol.
R1 = 6Ω
σ
q'1
For sphere of radius R
For sphere of radius 2R
R2 = 2Ω i2 i2+i1 F R3 = 4Ω 2V = E
D
σ=
σ=
A i1 B i2
4πR 2 q2 4π(2R ) 2
⇒
q1' =
.
E3 = 3V
R4 = 4Ω
C
Applying Kirchoff's law in ABFGA 6 – (i1 + i2) 4 = 0 Applying Kichoff's law in BCDEFB I2 × 3 – 3 – 2 + 2i2 + (i2 + i1) 4 = 0 Putting the value of 4 (i1 + i2) = 6 in (2) 3i2 – 5 + 2i2 + 6 = 0 1 ∴ i2 = – A 5 Sybstituting this value in (i) we get 1 i1 = 1.5 – – = 1.7 A 5 Therefore current in R3 = i1 + i2 = 1.7 – 0.2 = 1.5 A To find the p.d. across the capacitor VE – 2 – 0.2 × 2 = VG ∴ VE – VG = 2.4 V 1 CV2 ∴ Energy stored in capacitor = 2
q1
q2 = σ × 16πR2 When the two spheres are connected then the potential on the two spheres will be same. There will be a rearrangement of charge for this to happen. Let q1' and q2' be the new charges on the two spheres. Since the total charge remains the same ...(1) q'1 + q'2 = q1 + q2 = σ × 20 πR2 Also Since V1 = V2 1 q1 ' 1 q2 ' = 4πε 0 R 2πε 0 2R
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E1 = 6V
i2
q'2
⇒
q2 ' 2
G
2
q1 = σ × 4πR2
∴
R4
5µF = C
V
Connecting Wire
R3
E3
E
V
R2
E2
Two isolated metallic solid spheres of radii R and 2R are charged such that both of these have same charge density σ. The spheres ares are located far away from each other, and connected by a thin conducting wire. Find the new charge density on the bigger sphere. [IIT-1996] σ σ q2 Sol. q1 2R R 3.
..(2) 10
…(1) …(2)
AUGUST 2010
→
1 × 5 × 10–6 × (2.4)2 2 = 1.44 × 10–5 J
=
5.
I
120º
0.11µ 0 IQv ˆ 0.11 3µ 0 IQv ˆ j– i 2a 2a ∴ Instantaneous acceleration
O
0.11µ 0 IQv ˆ F a= m = ( j − 3 ˆi ) 2am m (b) In uniform magnetic field, force on a current loop is zero. Further, magnetic dipole moment of the loop
x
P
→
will be, M =(IA) kˆ Here, A is the area of the loop. 1 1 A = (πa2) – [2 × a sin 60º] [a cos 60º] 3 2
N (a) If a particle with charge +Q and mass m is placed →
at the centre P and given a velocity V along NP (see figure), find its instantaneous acceleration. (b) If an external uniform magnetic induction field
=
∴ M =(0.61 Ia2) kˆ
B = B ˆi is applied, find the force and the torque acting on the loop due to this field. Sol. (a) Magnetic field at the centre P due to arc of circle, Subtending an angle of 120º at centre would be : y M a a
N
→
y
→
B = B ˆi
→
CHEMISTRY A sample of hard water contains 96 ppm of SO42– and 183 ppm of HCO3– with 60 ppm of Ca2+ as the only cation. How many moles of CaO will be required to remove HCO32– from 100 kg of this water ? If 1000 kg of this water is treated with the amount of CaO calculated above, what will be the concentration (in ppm) of residual Ca2+ ions ? (Assume CaCO3 to be completely insoluble in water). If the Ca2+ ions in one litre of the treated water are completely exchanged with hydrogen ions, what will be its pH ? (one ppm means one part of the substance in one million part of [IIT-1997] water, mass/mass) Sol. In 106 g(= 1000 kg) of the given hard water, we have amount of SO42– ions = 96 g amount of HCO3– ions = 183 g 96 g So amount of SO42– ions = = 1 mol 96 g mol −1 6.
v 60º
x
1 1 µ I (field dut to circle) = 0 3 3 2a 0.16µ 0 I µ I (outwards) = 0 (outwards) = a 6a
B1 =
→
0.16µ 0 I ˆ k a Magnetic field due to straight wire NM at P : µ I B2 = 0 (sin 60º + sin 60º) 4π r Here, r = a cos 60º µ I ∴ B2 = 0 (2 sin 60º) 4π a cos 60º µ I 0.27µ 0 I (inwards) or B2 = 0 tan 60º = 2π a a → 0.27µ 0 I ˆ or B 2 = – k a → → → 0.11µ 0 I ˆ ∴ B net = B1 + B 2 = – k a Now, velocity of particle can be written as, → 3v ˆ v ˆ j v = v cos 60º ˆi + v sin 60º ˆj = i+ 2 2 Magnetic force
or B1 =
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→
Given,
∴ τ = M × B = (0.61 Ia2B) ˆj
x
+Q P
πa 2 a2 – sin 120º = 0.61 a2 3 2
→
→
60º r 60º
→
→
a
I
→
=
A wire loop carrying a current I is placed in the x-y plane as shown in figure. [IIT-1991] y v M +Q
→
Fm = Q( v × B )
and amount of HCO3– ions =
183 g 61 g mol −1
= 3 mol
These ions are present as CaSO4 and Ca(HCO3)2. 3 Hence, amount of Ca2+ ions = 1 + = 2.5 mol 2 The addition of CaO causes the following reactions: CaO + Ca(HCO3)2 → 2CaCO3 + H2O 1.5 mol of CaO will be required for the removal of 1.5 mol of Ca(HCO3)2 in form of CaCO3. In the treated water, only CaSO4 is present now. Thus, 1 mol of Ca2+ ions will be present in 106 g of water. Hence, its concentration will be 40 ppm. 11
AUGUST 2010
Molarity of Ca2+ ions in the treated water will be 10–3 mol l–1. If the Ca2+ ions are exchanged by H+ ions then, Molartiy of H+ in the treated water = 2 × 10–3 M Thus, pH = – log(2 × 10–3) = 2.7
2 CH3 – CH – CH – CH3 Y and Z (C6H12) H→
Ni
CH3 CH3 2,3-dimethyl butane
The above alkane can be prepared from two alkenes CH3 – C = C – CH3 and CH3 – CH – C = CH2
The vapour pressure of ethanol and methanol are 44.5 mm and 88.7 mm Hg respectively. An ideal solution is formed at the same temperature by mixing 60 g of ethanol and 40 g of methanol. Calculate the total vapour pressure of the solution and the mole fraction [IIT-1986] of methanol in the vapour. Sol. Given that, Pe0 = 44.5 mm Hg For ethanol (C2H5OH), M(C2H5OH) = 2 × 12 + 5 × 1 + 1 × 16 + 1 × 1 = 46 m(C2H5OH) = 60 g 60 m ∴ Moles of ethanol, ne = = = 1.3 46 M 7.
Ni
CH3 CH3
(Y) H2 Ni
CH3 – CH – C = CH2 CH3 CH3
CH3 – CH – CH – CH3 CH3 CH3
(Z)
Both, Y and Z can be obtained from following alkyl halide : Cl CH3 – C – CH – CH3
K-t-butoxide
CH3 CH3
∆; –HCl
2-chloro-2,3-dimethyl butane (X)
nm 1.25 1.25 = = 1.3 + 1.25 2.55 ne + nm According to Raoult's law, 44.5 × 1.3 Pe = Pe0 xe = = 22.69 mm Hg 2.55 88.7 × 1.25 = 43.48 mm Hg and Pm = Pm0 xm = 2.55 Hence, total vapour pressure of the solution, PT = Pe + Pm = 22.69 + 43.48 = 66.17 mm Hg According to Dalton's law, (in vapour form) Pm = PTx´m Hence, mole fraction of methanol in vapour form, P 43.48 x´m = m = = 0.66 66.17 PT
CH2 = C — CH – CH3 + CH3 – C = C – CH3 CH3 CH3 (Z) 20%
CH3 CH3 (Y) 80%
Cl
Hence,
X, CH3 – C – CH – CH3 CH3 CH3 Y, CH3 – C = C – CH3 CH3 CH3
Z, CH3 – CH – C = CH2 CH3 CH3
An organic compound (X), C5H8O, does not react appreciably with Lucas reagent at room temperatures but gives a precipitate with ammonical AgNO3 solution. With excess CH3MgBr; 0.42 g of (X) gives 224 ml of CH4 at STP. Treatment of (X) with H2 in the presence of Pt catalyst followed by boiling with excess HI gives n-pentane. Suggest structure of (X) [IIT-1992] and write the equations involved. Sol. Lucas test sensitive test for the distinction of p, s, and t-alcohol. A t-alcohol gives cloudiness immediately, while s-alcohol within 5 minutes. A p-alcohol does not react with the reagent at room temperature. Thus, the present compound (X) does not react with this reagent, hence it is a p-alcohol.
9.
An alkyl halide X, of formula C6H13Cl on treatment with potassium t-butoxide gives two isomeric alkenes Y and Z(C6H12). Both alkenes on hydrogenation give 2, 3-dimethyl butane. Predict the structures of X, Y [IIT-1996] and Z. Sol. The alkyl halide X, on dehydrohalogenation gives two isomeric alkenes. 8.
K − t − butoxide C 6 H13Cl → Y + Z C 6 H12
Both, Y and Z have the same molecular formula C6H12(CnH2n). Since, both Y and Z absorb one mol of H2 to give same alkane 2, 3-dimethyl butane, hence they should have the skeleton of this alkane.
XtraEdge for IIT-JEE
(Z)
CH3 CH3
xm =
∆ ; – HCl
2,3-dimethyl
The hydrogenation of Y and Z is shown below : H2 CH3 – C = C – CH3 CH3 – CH – CH – CH3
For methanol (CH3OH), Pm0 = 88.7 mm Hg M(CH3OH) = 1 × 12 + 3 × 1 + 1 × 16 + 1 × 1 = 32 m(CH3OH) = 40 g 40 m ∴ Moles of methanol, nm = = = 1.25 32 M ne 1.3 1.3 ∴ xe = = = 1.3 + 1.25 2.55 ne + nm
X
CH3 CH3 butene-1
CH3 CH3 2,3-dimethyl butene-2 (Y)
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AUGUST 2010
(X) = C4H6.CH2OH(p-alcohol) Since the compound gives a ppt. with ammonical AgNO3, hence it is an alkyne containing one –C≡ CH, thus (X) may be written as : HC≡C –C2H4 – CH2OH (X) It is given that 0.42 g of the compound (which is 0.005 mol) produces 22.4 ml of CH4 at STP (which is 0.01 mol) with excess of CH3MgBr. This shows that the compound (X) contains two active H atoms (H atom attached to O, S, N and –C≡CH is called active). Of these, one is due to the p-alcoholic group (–CH2OH) and the other is due to the –C≡CH bond, since both these groups are present in (X), hence it evolves two moles of CH4 on reaction with CH3MgBr. H – C≡C. C 2 H 4 – CH2OH + 2CH3MgBr →
10. Compound (X) on reduction with LiAlH4 gives a hydride (Y) containing 21.72% hydrogen along with other products. The compound (Y) reacts with air explosively resulting in boron trioxide. Identify (X) and (Y). Give balanced reactions involved in the formation of (Y) and its reaction with air. Draw the structure of (Y). Sol. Since B2O3 is formed by reaction of (Y) with air, (Y) therefore should be B2H6 in which % of hydrogen is 21.72. The compound (X) on reduction with LiAlH4 gives B2H6. Thus it is boron trihalide. The reactions are shown as: 4BX 3 + 3LiAlH4 → 2B 2 H 6 + 3LiX + 3AlX3 (X)
(X = Cl or Br) B 2 H 6 + 3O2 → B2O3 + 3H2O + heat (Y)
(X)
Structure of B2H6 is as follows: Hb
BrMgC≡C–C2H4 – CH2OMgBr + 2CH4 Moreover, the treatment of (X) with H2/Pt followed by boiling with excess of HI gives n-pentane (remember that 2HI are required to convert one –CH2OH into CH3). This shows that the compound (X) contains a straight chain of five carbon atoms.
B
2
1
ZnCl + HCl
(X)
Ag – C≡C – CH2CH2CH2OH + NH4NO3
Br MgC≡C.CH2CH2CH2OMgBr + 2CH4
Pentanol-1 CH3CH2CH2CH2CH3 n-pentane
The production of 2 moles of CH4 is confirmed as the reactions give 224 ml of CH4. Q 84 g(X) gives = 2 × 22.4 litre CH4 2 × 22.4 × 0.42 ∴ 0.42 g (X) gives = 84 = 224 ml of CH4
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121.5º
11. Find the values of a and b so that the function x + a 2 sin x, 0 ≤ x ≤ π/ 4 f(x) = 2 x cot x + b, π/ 4 ≤ x ≤ π/2 a cos 2 x − b sin x , π / 2 < x ≤ π [IIT-1989] is continuous for 0 ≤ x ≤ π Sol. As, f(x) is continuous for 0 ≤ x ≤ π π π ∴ R.H.L. at x = = L.H.L. at x = 4 4
CH3CH2CH2CH2CH2OH
∆, –H2O; –I2
B
MATHEMATICS
White ppt.
2 HI
1.33Å
Hb Ht 1.77Å Thus, the diborane molecule has four two-centre-two electron bonds (2c-2e– bonds) also called usual bonds and two three-centre-two-electron bonds (3c-2e– bonds) also called banana bonds. Hydrogen attached to usual and banana bonds are called Ht (terminal H) and Hb (bridged H) respectively
2 Room → No reaction H − C ≡ C − CH 2 CH 2 CH 2 OH temp.
2H2/Pt
97º
Ht
Ht 1.19Å
4-pentyne-1-ol
2CH3MgBr
Hb B
The different equations of (X) are :
NH3
Hb Ht
HC≡C.CH2 CH2 – CH2OH (X)
AgNO3
Ht
or
CH3CH2CH2CH2CH3 + H2O + I2
4 3
B
Ht
n-pentane On the basis of abvoe analytical facts (X) has the structure : 5
Ht
Ht
H 2 / Pt → H – C≡C–C2H4 – CH2OH 2 CH3CH2.C2H4 – CH2OH HI 2 → ∆
(Y)
π π π π ⇒ 2. cot + b = + a 2 . sin 4 4 4 4 π π ⇒ +b= +a 2 4
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AUGUST 2010
{neglecting 30º, as not possible} ⇒ 2A + 60º = 150º ⇒ A = 45º again from (1), sin (60º + 2c) = –1/2 ⇒ 60º + 2C = 210º, 330º ⇒ C = 75º or 135º Also from (1) sin (C – A) = ½ C – A = 30º, 150º, 195º for A = 45º, C = 75º and C = 195º (not possible) ∴ C = 75º Hence, A = 45º, B = 60º, C = 75º
π ....(i) 4 π π also, R.H.L at x = = L.H.L at x = 2 2
⇒ a–b=
π π 2π π − b sin = 2. . cot + b ⇒ a cos 2 2 2 2 ⇒ –a–b=b ⇒ a + 2b = 0 ...(ii) 3π −3π From (i) and (ii), a = and b = 2 4 dy at x = –1, when 12. Find dx (sin y)
π sin x 2
+
14. If exp {(sin2x + sin4x + sin6x + ...... ∞). ln 2} satisfies the equation x2 – 9x + 8 = 0, find the value of cos x π ,0
