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89801.

(A) Fehiling's reagent is a test for all aliphatic aldehydes. (R) Aliphatic aldehydes can be easily oxidesed even with mild oxidising agents.

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If both (A) and (R) are CORRECT and (R) is the CORECT explanation of (A)
If both (A) and (R) are correct but (R) is not correct explanation of (A)
If (A) is correct but (R) is INCORRECT.
If (A) is incorrect but (R) is correct

Answer :A
Discussion
89802.

(A): Fe_(3)O_(4) is ferrimagnetic at room temperature but becomes paramagnetic at 850 K (R): The magnetic moments in Fe_(3)O_(4) ale aligned equally in parallel and antiparallel directions which on heating randomise

Answer»

Both (A) and (R) are TRUE and (R) is the correct EXPLANATION of (A)
Both (A) and (R) are true and (R) is not the correct explanation of (A)
(A) is true but (R) is false
Both (A) and (R) are false

Answer :C
Discussion
89803.

(A): Fe^(+3) is more stable than that of Fe^(+2). (R) : Fe^(+3) ion has half filled 3d orbital whereasFe^(+2) does not.

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Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A)
Both (A) and (R) are true and (R) is not the correct explanation of (A)
(A) is true but (R) is FALSE
Both (A) and (R) are false

Solution :Both (A) and (R) are true and (R) is the correct explanation of (A)
Discussion
89804.

A fcc element (atomic mass =60) has a cell edge of 400pm. Its density is :

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`6.23 G cm^-3`
`6.43 g cm^-3`
`6.53 g cm^-3`
`6.63 g cm^-3`

ANSWER :A
Discussion
89805.

A face-centred cubic element (atomic mass 60) has a cell edge of 400 pm. What is its density?

Answer»


ANSWER :6.23 g/cc
Discussion
89806.

A face centered cubic solid of an element ( atomic mass 60) has a cube edge of 4Å. Calculate its density.

Answer»

SOLUTION :Edge length`= 4 Å = 4 xx 10^(-8) cm`
VOLUME of unit cell = ` ( 4 xx 10^(-8) cm)^(3)`
`64 xx 10^(-24) cm^(3)`

Mass of the unitcell = Number of atoms in the unit cell ` xx ` mass of each atom
Number of atomsin the fcc unitcell
` 8 xx 1/8 + 6 xx 1/2= 4`
mass of oneatom =` (" AtomicMass")/( " Avogadro number") `
` = 60/( 6.023 xx 10^(23))`
Mass of the unitcell= ` ( 4 xx 60)/( 6.023 xx 10^(23))`
Densityof UINT cell = ` ("Mass of unit cell")/( " Volume of unit cell") `
` Rightarrow( 4 xx 60)/( 6.023 xx 10^(23)) xx 1/(64 xx 10^(-24))`
` Rightarrow6.2 g cm^(-3)`
Discussion
89807.

A face centered cubic solid of an element (atomic mass 60) has a cube edge of 4 Å. Calculate its density.

Answer»

Solution :For FCC UNIT cell n = 4
`"Edge length (a)"=4Å=4xx10^(-8)cm`
`"MASS (M)"="60 g MOL"^(-1)`
`"Density "(rho)=?`
`rho=(nM)/(a^(3)N_(A))`
`=(4xx"60 g mol"^(-1))/((4xx10^(-8)cm)^(3)xx(6.023xx10^(23)"mol"^(-1)))`
`=("240 g mol"^(-1))/(6.4xx10^(-23)cm^(3)xx6.023xx10^(23)"mol"^(-1))`
`=(240)/(38.54)GCM^(-3)`
`rho=6.227gcm6(-3)`
Discussion
89808.

A face-centred cubic solid of an element A has a largest sized guest atom B at the body centre octahedral hole if insertion of B doesn't affect the original unit cell dimension, determine the packing fraction of the solid.

Answer»

Solution : In the given solid, there is one B and four A per unit cell.

Also, under the condition of largest possible size of B, it will be in contact of A present at the face CENTRES only and the FOLLOWING relatinship will exist :`4r_(A)sqrt(2)` and`2(r_(A)+r_(B))=a`
solving `(r_(B))/(r_(A))=0.414`
Now,packing fraction
`(phi)=(4pi)/(3)(4r_(A)^(3)+r_(B)^(3))xx(1)/(a^(3))=(4pi)/(3)(4r_(A)^(3)+r_(B)^(3))xx(1)/(1sqrt(2r_(A)^(3)))`
`(PI)/(12SQRT(2))[4+((r_(B))/(r_(A)))^(3)] =(pi)/(12(sqrt(2)))[4+(0.414)^(3)]=0.7536`
Discussion
89809.

(A) F-F bond is stronger than Cl -Cl bond (R ) Atomic size of F is larger than that of Cl.

Answer»

Both (A) and (R ) are TRUE and (R ) is the CORRECT EXPLANATION of (A)
Both (A) and (R ) are true and (R ) is not the correct explanation of (A)
(A) is true but (R ) is FALSE
Both (A) and (R ) are false

Answer :D
Discussion
89810.

(A) F- centre brings about colour (R) Vacancy with a trapped cation is called F- centre

Answer»

Both (A) and (R) are true and (R) is the CORRECT explanation of (A)
Both (A) and (R) are true and (R) is not the correct explanation of (A)
(A) is true but (R) is false
Both (A) and (R) are false

Answer :C
Discussion
89811.

A : Extraction of iron metal from iron oxide ore is carried out by heating with coke. R: The reaction Fe_(2)O_(3(s)) to Fe_((s)) + 3/2 O_(2(g))is a spontaneous process.

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Both assertion and reason are true and reason is the correct explanation of assertion. 
Both assertion and reason are true but reason is not the correct explanation of assertion. 
Assertion is true but reason is FALSE. 
Assertion is false but reason is true. 

Solution :Extraction of iron metal from iron oxide ore is carried out by heating with coke and flux (calcium carbonate). Flux is a slag forming substance. It converts infusible impurities into fusible slag.
The reaction : `Fe_(2)O_(3(s)) to Fe_((s)) + 3/2 O_(2(g))`is not a SPONTANEOUS PROCESS. `FeO` is converted to `FeO` at about `400^@C FeO` is converted to Fe at about `800^@C – 1000^@C`
Discussion
89812.

(a) Explain why on addition of 1 mole glucose to 1 litre water the boling point of water increases .5 (b) Henry's law constant for CO_(2) in water is 1.67cc10^(8)Pa at 298K.Calculate the number of moles of CO_(2) in 500 ml of soda water when packed under 2.53xx10^(5) Pa at same temperature.

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Solution :(a) Boiling point of solution having 1 mol of glucose is higher than that of pure solvent because the VAPOUR pressure of solution is lower than that of pure solvent and vapour pressure increases with increase in TEMPERATURE. Hence, the solution has to be heated more to make the vapour pressure equal to atmospheric pressure.
(b) `K_(H)=1.67xx10^(8)Pa`
Henry's law, `P_(CO_(2))=2.53xx10^(5)Pa`
`P_(CO_(2))=K_(H)xxx_(CO_(2))`
`x_(CO_(2))=(P_(CO2))/(K_(H))=(2.53xx10^(5))/(1.67xx10^(8))`
`n_(CO2)/(n_(H2O))=1.525xx10^(-3)`
`nH_(2)O=(500)/(18)=27.78[500ml~~500g]`
`n_(CO2)=1.515xx10^(-3)xx27.78`
`n_(CO2)=42.01xx10^(-3)"moles"`
Discussion
89813.

(a) Explain why electrolysis of aqueous solution of NaCl gives H_2 at cathode and Cl_2 at anode. Write overall reaction. (b) The resistancee of a conductivity cell containing 0.000M KCl solution at 298K is 1500 Omega. Calculate the cell constant if conductivity of 0.001M KCl solution at 298K is 0.146 times 10^-3 S cm^-1.

Answer»

Solution :(a) Criteria: The Criteria is as FOLLOWS: The substance which possess higher standard reduction potential is reduced first at the cathode. Whereas the substance which possesses LOWER reduction potential than water will get oxidised at the anode.
`NaClleftrightarrowNa^(+) (AQ)+Cl^(-) (aq)`
`2H_2O (l) to O_2(g)+4H^++4e^-`
AT cathode:
`2H_2O(l)+2e^(-) to H_2+2OH^-) :E^@=-0.83V`
`Na^(+) (aq) +e^(-) toNa(s), E^@=-2.71V`
Atanode: `1/2 O_2 (g)+H_(2) (g) to H_2O, E^@=1.23V`
`2CL^(-) (aq)+2e^(-) to Cl_2(g),E^@=-1.36V`
(b) `R=1500 OmegaMolarity =0.001 KCl`
Cell const.=?
K (conductivity)=`0.146 times 10^-3 S cm^-1`
Cell const.= Conductivity `times` OBS. resistance
`=0.146 times 10^-3 times 1500`
`=219 times 10^-3 cm^-1`
Discussion
89814.

(a) Explain why a nonsymmetrical ether is not usually prepared by heating a mixture of ROH and R'OH in acid. (b) Why is it possible to prepare t-butyl ethyl ether by heating a mixture of t-butanol and ethanol? (c ) Would you get any di-t-butyl ether from this reaction? Explain. (d) Can t-butyl ethyl ether be made by heating H_(2)C=CH(CH_(3))_(2) and ethanol?

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Solution :(a) A mixture of three ethers, i.e. R-O-R, R-O-R and R-O-R is obtained.
(b) When alcohol is `3^(@)`, its OXONIUM ion easily loses water to form a carbocation, which is solvated by the other `2^(@)` or `1^(@)` alcohol to give the mixed ether preferentially.
This is an example of an `S_(N)1` mechanism.
`Me_(3)COH_(2)^(+)overset(-H_(2)O)rarrMe_(3)C^(+)underset(-H^(+))overset(HOCH_(2)CH_(3))rarrMe_(3)COCH_(2)CH_(3)`
(c ) No. t-Butanol does not solvate the `3^(@)` carbocation readily because of STERIC hindrance.
(d) YES, The addition of `H^(+)` to the alkene gives the same `Me_(3)C^(+)` INTERMEDIATE.
Discussion
89815.

(a) Explain the preparation of Nylon-6, 6 with equation. (b) What are thermoplastic polymers? Give an example (c )Write the structure of isoprene (2-methyl-1,3-butadiene).

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Solution :(a) It is obtained by the condensation polymerization of hexamethylenediamine with adipic ACID under hig! pressure and at high temperature.
`underset("Adipic acid")(nHOOC - (CH_2)_(4)-COOH) + underset("Hexamethylenediamine")(nH_2N-(CH_2)_(6) -NH_(22) underset("High Pressure")overset(533K)(rarr)[overset(H)overset(|)(N)-(CH_2)_(6)-overset(H)overset(|)(N)-overset(O)overset(||)(C )-(CH_2)_(4)-overset(O)overset(||)(C )]_(4)`
(b) These are the linear or slightly branched long chain molecules capable of repeatedly softening on heating and hardening on cooling,
Example:Polythene or POLYSTYRENE or Polyvinyl chloride
(c) `underset(("Isoprene"))underset("Cis-2-methylbute 1,3-diene")(CH_2=underset(CH_3)underset(|)(C )-CH=CH_2)` .
Discussion
89816.

(a) Explain the preparation of Buna-N with equation. Name the monomer present in the following polymer (i) Poly vinyl chloride (ii) Natural rubber. ( c ) Give an example of bio-degradable polymer.

Answer»

Solution :By Co-polymerisation of 1,3 Butadiene and Acryld nitrile in the PRESENCE of peroxide or SODIUM catalyst.
(b) (i) VINYL chloride or Chloro ethene (ii) isoperene (OR) 2-Methyl 1,3 Butadiene.
( C) PHBV.
Discussion
89817.

Name the monomer present in the following polymer i) Poly vinyl chloride. Ii) Natural rubber.

Answer»

Solution :By co-polymerization of 1,3 -butadiene and ACRYLO nitrile in the presence of peroxide or sodium catalyst.

(b) (i) Vinyl chloride OR CHLOROETHANE.
(ii) ISOPRENE OR 2-methyl -1,3 - butadiene .
(c) Poly `beta`- hydroxybutyrate - co `beta`-hydroxy valerate (PHBV) OR Nylon 2-Nylon 6 OR polyglycodic acid (PGA) OR polylactic acid (PLA) OR polycaprolactone (PCL) OR Any natural polymer.
Discussion
89818.

Answer any FOUR of the following questions. b. How do you prepare methoxy ethane by Williamson's ether synthesis?

Answer»

SOLUTION :(a) Step 1 : FORMATION of proptonated alcohol.

Step 2 : Formation of carboncation :
`H-UNDERSET(H)underset(|)overset(H)overset(|)(C )-underset(H)underset(|)overset(H)overset(|)(C )-overset(H)overset(|)(underset(ddot)O)-Hoverset("Slow")( Step 3: Elimination of proton -

(b) When sodium sodium ethoxide is heated with methyl iodide (in alcoholic medium), methoxyethane is formed
`underset("Sodium ethoxide")(CH_3CH_2)-ONa) + underset("Methoxy iodine")(I-CH_3) underset(Delta)overset("Alcohol")(rarr)underset("Methoxy methane")(CH_3CH_2-O-CH_3+NaI)`
DETAILED ANSWER:
(b) In a Williamson synthesis, an ether is formed by reacting an alcohol and an alkyl halide in the presence of a base.
When `CH_3-I` is mixed with the base, like `NaOH`, the alcohol is deprotonated, leaving a negatively charged oxygen. This acts as a nucleophile and attacks the carbon bonded to the halogen(iodide). The halogen, a good leaving group, is released, leaving behind methoxy ethane.
Discussion
89819.

(a) Explain the mechanism of acid catalysed dehydration of ethanol to ethene.(b) How do you prepare methoxy ethane. by Williamoson's ether synthesis?

Answer»

Solution :(b)METHOXY ethane can be PREPARED by heating methyl CHLORIDE with SODIUM ethoxide.
`CH_(3) - CH_(2) – ONA + CH_(3) – Cl overset(Delta)to CH_(3) – CH_(2) – O –CH_(3) + NaCl`
Sodiumethoxide Methylchloride Methoxyethane
Discussion
89820.

(a) Explain the mechanism of acid catalysed dehydration of ethanol to (b) Write general equation for preparation of ether by Williamsonsynthesis.(c) Among alcohols and phenols which one is more acidic?

Answer»

Solution :(b)When alkyl halide reacts with sodium alkoxide ETHERS are obtained.
`R-ONa + R^(1) X to underset(ETHER)(R-O-R^(1) + NaX)`
`C_(2)H_(5)ONa + C_(2)H_(5)Br to C_(2)H_(5)OC_(2)H_(5) + NaBr`.
(c)Phenols are more acidic than alcohols. This is because the PHENOXIDE ION formed from the ionisation of phenol is stabilised by resonance.
Discussion
89821.

(a) Explain the mechanism of a nucleophilic attack on the carbonyl group of an aldehyde or a ketone. (b) An organic commpound (A) (molecular formula C_(8)H_(16)O_(2)) was hydrlysed with dilute sulphuric acid to give a carboxylic acid(B) and an alcohol (C). Oxidatin of (C) with chromic acid also produced (B). On dehydratin (C) gives but-1-ene. Write the equations for the reactions involved.

Answer»

Solution :(a) Mechanism of nucleophilic addition reactions :
attack from the top face ` (##SB_CHM_XII_DB_2010_E01_034_S01.png" width="80%">
A nucleaphile attacks the electrophilic carbon atom of the polar carbonyl group from a direction perpendicular to the plane of `sp^(2)` hybridised orbitals of carbonyl carbon. The hybridisation of carbon changes from `Sp^(2) to sp^(3)`in this process and a tetrahedral alkoxide intermediate is produced.
This intermediate CAPTURES a proton from the REACTION MEDIUM to give the electrically neutral product. The net result is additon of `Nu^(-)and H^(+)`across the carbon oxygen double BOND as shown in Fig. (b) Q.30 (or) (b) Outside Delhi, Set-I., 2009. .
Discussion
89822.

(a) Explain the mechanism involved in the conversion of ethanol into ethene.(b) An organic compound with molecular formula C_(6)H_(6)O gives white precipitatewith bromine water. Identify the functional group in the organic compound and write the chemical equation for the reaction.

Answer»

Solution :(a) (1) FORMATION of protonated alcohol `CH_(3)-CH_(2) -OH +overset(+)(H)overset(“fast”)hArrCH_(3) -CH_(2) -overset(+)OH_(2)`
Formation of carbocation
`CH_(3) -CH_(2) -overset(+)OH_(2) overset(“SLOW”) hArrCH_(3) -overset(+)CH_(2) +H_(2) O`
(3) Elimination of PROTON to form ethene
`CH_(3) -overset(+)(CH_(2))hArr CH_(2) =CH_(2) + H^(+)` (b) The organic compound is phenol
Discussion
89823.

(a) Explain the magnetic behaviour of transition metals. (b) Write the electronic configuration of Ni(28), Zn(30).

Answer»

Solution : Most of the d-block elements and their COMPOUNDS are attracted by a magnet. This MEANS that these are paramagnetic in nature. This property is due to the unpaired electrons present in their (n-1) d-orbitals. The more the number of unpaired electrons, the more will be magnetic character. If there are no unpaired electrons, the substance will be diamagnetic e.g. Zn, Cd, Hg etc.
(B) `Ni_(28): 1s^(2)2s^(2) 2p^(6)3s^(2)3p^(6)4s^(2)3d^(8)`
`Zn_(30): 1s^(2)2s^(2) 2p^(6)3s^(2)3p^(6)4s^(2)3d^(10)`
Discussion
89824.

How does anisole react with methyl chloride?

Answer»

SOLUTION :(a) Mechanism INVOLVES THREE steps
Steps 1. FORMATION of protonated ALCOHOL :
Alcohols react with sulphuric acid to form protonated alcohol.

Step - 2 :Formation of Carbocation :
The protonated alcohol loses a molecule of water to form ethyl carbocation.

Step 3 : Formation of ethene by elimination of a proton :
The carbocation loses a proton to form ethene.

(b) Anisole undergoes Friedal crafts reactions with methyl chloride in presence of anhydrous aluminium chloride. Methyl groups are introduced at ortho and para positions.
Discussion
89825.

Write the mechanism of aicd catalysed dehydration of ethanol to ethane.

Answer»

Solution :(a) Mechanism involves three steps
Steps 1. Formation of protonated alcohol :
Alcohols REACT with sulphuric acid to FORM protonated alcohol.

Step - 2 :Formation of Carbocation :
The protonated alcohol loses a molecule of water to form ethyl carbocation.

Step 3 : Formation of ethene by ELIMINATION of a PROTON :
The carbocation loses a proton to form ethene.

(B) Anisole undergoes Friedal crafts reactions with methyl chloride in presence of anhydrous aluminium chloride. Methyl groups are introduced at ortho and para positions.
Discussion
89826.

a. Explain the laboratory method of preparation fo p-bromoacetanilide from acetailide. b. Mention a general test forthe following: (i) Carboydrates (ii) Oils and fats.

Answer»

SOLUTION :
2 gm of acetanilide +5ml of glacial acetic acid +5ml of BROMINE in acetic acid is mixed in a conical flask. The mixture is KEPT aside for 5 MINUTES and contents are poured into cold WATER. A white precipitate of p-bromo acetanilide is obtained.
(Quantity of the resctants not a value point).
b. (i) Molisch.s test
(ii) Acrolein test
Discussion
89827.

(a) Explain the following terms: (i) Rate of a reaction.(ii) Activation energy of a reaction. (b)The decomposition of phosphine, PH_(3) proceeds accordingto the following equation: Rate =k[PH_(3)] The half-life of PH_(3) is 37.9 s at 120^(@)C. (i) How much time is requiredfor 3//4th of PH_(3) to decompose? (ii) What fraction of the original sample of PH_(3) remains behind after 1 minute?

Answer»


ANSWER :(i) 75.83s, (II) 0.3342
Discussion
89828.

(a) Explain thefollowing terms : (i) Rate of a reaction (ii) Activation energy of a reaction (b) Thedecompositon of phosphine , PH3, proceeds according to the following equation : 4 PH_(3) (g) to P_(4)(g) + 6 H_(2) (g) It is found that the reaction follows the following rate equation :Rate = k [PH_(3)]. The half-life ofPH_(3)" is "37.9" s at "120^(@)C. (i) How much time is required for 3//4^("th") "of " PH_(3) todecompose ? (ii) What fraction of the original sample of PH_(3) remains behind after 1 minute ?

Answer»

Solution :(a) (i) Rate of a reaction : Therate of a reaction can be defined as thechange in concentration of a reactant or proudct in unit time.
(II) Activation energy of a reaction : The MINIMUM extra amount of energy absorbed by the reactant molecules so that their energy becomes equal to threshold VALUE is called activation energy. It is denoted by EA.
(b) (i)`t_(1//2) = 37.9 ` s
Letinitial concentration = a
` :.x = 3/4a`
Using the formula
`t=(2.303)/klog.a/(a-x)`
` t=((2.303)/(0.693))/(t_(1//2))log. a/(a-3/4a)""[:'k=(0.693)/t_(1//2)]`
`t = 2.303 xxt_(1//2)/(0.693)log.q/(1/4a)`
` t = (2.303xxt_(1//2))/(0.693) xx log 4 =(2.303 xxt_(1//2)xx2log 2)/(0.693)`
`t = (2.303 xx37.9 xx0.3010)/(0.693)""[:. t_(1//2) = 37.9s]`
` = (52.544)/(0.693) = 75. 82 sec`.
(ii) Heret = 1 min = 60s
` andt_(1//2) = 37.9s`
Using the formula
` t = (2.303)/ klog. ([A]_(0))/([A]_(t))`
` and k = ( 0.693)/(t_(1//2))`
` :. "" t = (2.303)/((0.693)/t_(1//2))log.[A]_(0)/[A]_(t)`
`t=2.303 xx t_(1//2)/(0.693) xx log.[A]_(0)/[A]_(t)`
` 60 = (2.303 xx37.9)/(0.693) log . [A]_(0)/[A]_(t) `
`:. ""log.[A]_(0)/[A]_(t)=(60 xx 0.693)/(2.303xx37.9) = 0.4768`
`[A]_(0)/[A]_(t) = anti-log(0.4768) = 2.997`
` [A]_(t)/[A]_(0) = 1/(2.997) = 0.3337 = 33.37 %`
Discussion
89829.

(a) Explain the following terms : (i) Rate of a reaction. (ii) Activation energy of a reaction. (b) The decomposition of phosphine, PH_(3), proceeds according to the following equation : 4" PH"_(3)(g)to P_(4)(g)+6" H"_(2)(g) It is found that the reaction follows the following rate equation : Rate =k[PH_(3)] The half-life of PH_(3) is 37.9 s at 120^(@)C. (i) How much time is required for 3//4th of PH_(3) to decompose. (ii) What fractional of the original sample of PH_(3) remains behind after 1 minute ?

Answer»

SOLUTION :`"[""(b) (i) "t=(2.303)/(0.693//t_(1//2))LOG""(a)/(a-(3)/(4)a)=75.82" s"`
(ii) `log""([A]_(0)) /([A])=(kt)/(2.303)=((0.693//37.9)(60))/(2.303)=0.4763`
`[A]_(0)//[A]="Antilog "0.4763=1.2994" or "[A]//[A]_(0)=1//1.2994=0.77=77%"]"`
Discussion
89830.

(a) Explain the following : (i) Lyophilic coloid is more stable than lyophobic colloid (ii) Coagulation takes place when sodium chloride solution is added to colloidal solution of ferric chloride (iii) Sky appears to be blue.

Answer»
Discussion
89831.

Explain the Crystal field splitting is an octahedral field.

Answer»

Solution :(a) When ligand APPROACH the central metal ion the electrons in the d-ORBITALS of central metal ion will be repelledby the lone paris of the ligand. Because of there interactions, the degeneracy of the d-orbitals of the metal ion is lost and it splits into two sets of orbitals having different energies. The splitting of d-orbital is as SHOWN in the figure.
Discussion
89832.

a) Explain the buffer action in a basic buffer containing equimolar ammonium hydroxide and ammonium chloride.

Answer»

SOLUTION :(a) Dissociation of BUFFER components
`NH_4 OH_((aq)) hArr NH_(4(aq))^(+)`
`NH_4 CI to NH_(4)^(+)+CI^+`
Addition of `OH^(-)`
The added `H^+` ions are neutralized by `NH_4 OH` and there is no appreciable decrease in pH.
`NH_4 OH_((aq))+H^+to NH_(4(aq))^(+) +H_2 O(I)`
Addition of
`NH_(4(aq))^(+)+OH_((aq))^(-) to NH_4 OH_((aq))`
The added `OH^(-)` ions react with `NH_(4)^(+)` to produce unionized `NH_(4) OH`. Since `NH_4 OH` is a weak base, there is no appreciable INCREASEIN pH.
Discussion
89833.

(a) Explain the action of Cone. HCl on KMnO_4 crystals(b) Write the structure of perchloric acid.

Answer»

Solution :(a) When `KMnO_(4)`is treated with cone. HC1 chlorine is liberated.
`2KMnO_(4) +16HCI —rarr 2KCl + 2MnCI_(2) + 8H_(2)O + 5Cl_(2)`
(b)
Discussion
89834.

Explain Stephen's reduction with an example.

Answer»

Solution :Stephan.s reduction: Nitriles on reduction with stannous chloride and hydrochloria acid gives imine HYDRO chloride which on hydrolysis gives corresponding aldehyde. This reaction is called Stephan.s reduction.
`SnCl_(2)+2HCl rarr SnCl_(2)+2[H]`
`underset("(NITRATE)")(R-C-=N+2[H])overset(SnCl_(2)+"Conc. HCl")underset("Ether")rarr underset("(imine)")(R-CH=NH+HCl overset(H_(2), "boil")rarr underset("(aldehyde)")(R-overset(O)overset(||)C-H+NH_(4)Cl)`
Example
`underset("(Ethane Nitrate)")(CH_(3)-C-=N+2[H])overset(SnCl_(2)+"Conc. HCl")underset(290K)rarr underset("(Acetaldimine hydrochloride)")(CH_(3)CH=NH+HCl)overset(H_(3)O" boil")rarr underset("(ACETALDEHYDE)")(CH_(3)-overset(O)overset(||)C-H+NH_(4)Cl)`
Discussion
89835.

(a) Explain SN^(2) mechanism with an example.(b) Name the product formed when chloromethane reacts with (i) aqueous KOH & (ii) alcoholic AgCN. (c) Give an example of polyhalogen compound.

Answer»

SOLUTION :(a)
(b) (i) When CHLOROMETHANE reacts with aq. KOH gives methanol.
(ii) When chloromethane reacts with alcoholic ARCN gives methyl ISOCYANIDE.
(c) Chloroform `[CHCI_(3)]` (or) carbon tetrachloride `(C CI_(4))`
Discussion
89836.

(a) Explain S_(N)1 mechanism for the conversion of tertiary butyl bromide to tertiary butyl alcohol. (b) Complete the following reactions : (i) CH_(3) - CH = CH_(2) + HI rarr (ii) (iii) CH_(3)CH_(2)Br underset("Aq Ethanol")overset(AgCN)rarr

Answer»

Solution :In this step, polarised C Br bond UNDERGOES cleavage to produce a CARBOCATION and a bromide ion.

In this step, carbocation is attacked by nucleophile to form the product (formation of C-OH bond).

(b) (i) `CH_(3) - CHI - CH_(3)`
(ii)
(iii) `CH_(3)CH_(2)NC`
(a) Conversion of tertiary butyl bromide to tertiary butyl alcohol using `SN^(1)` mechanism occurs as FOLLOWS :
Step - 1 : Formation of carbocation :
`underset("(tertiary butyl bromide)")(CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C) - Br) overset("slow step")rarr underset("(carbocation)")(CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C^(+)))+Br^(-)`
Step - 2: Fast step and formation of racemic mixture of product :
`CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C^(+)) + underset("(Nucleohile)")(OH^(-)) overset("fast")rarr underset("(tertiary butyl alcohol)")(CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-OH)`
(b) (i)This is an example of unimolecular nuclephile substitution reaction i.e. rate `prop` [substrate]
(b) (i) `CH_(3)-CH=CH_(2) + HI rarr CH_(3)-underset(I)underset(|)(CH) - CH_(3)`
(ii)
(iii) `CH_(3).CH_(2).Br underset("aq.thanol")overset(AgCN)rarr underset("(ETYL isocynaide)")(CH_(3).CH_(2).NC)+AgBr`
Discussion
89837.

How does propanone (CH_(3)COCH_(3)) reacts with hydrazine?Give equation.

Answer»

Solution :(a) Benzoyl CHLORIDE is hydrogenated over catalyst, palladium on barium sulphate.
This REACTION is CALLED Resenmund reduction.

(B) When hydrazie is TREATED with acetone in the presence of acid as catalyst, propanone (acetone), acetone hydrazone formed.

(c) Chromyl chloride.
Discussion
89838.

(a) Explain ionozation isomerism with an example .

Answer»

Solution : TWO or more compound having the same molcecular formula but furnish different ions in soultion are called ionosation isomers . Athe phenomemon is called ionisation isomerism
Ex: `[C0(NH_3)_5Br]`So_4 and `[(Co(NH_3)_5 SO _4 ]` Br
Homoleptic Complex in which a metal is bound to only ONE kind of DONOR group are called homolepticcomplex
e.g. `[Co(NH_3)_6 ]^3+`
Discussion
89839.

Write the mechanism of acid catalysed dehydration of ethanol to ethene.

Answer»

Solution :(a) when sodium phenate is heated with CARBON dioxide to `140^(@)C` under 6-7 ATM pressure, sodium SALICYLATE is OBTAINED which on acidification gives SALICYLIC acid.
Discussion
89840.

(a) Explain : (i) Kolbe's reaction (ii) Reimer - Tiemann's reaction. (b) Complete the following : (CH_(3))_(2)CH-Ohoverset((O))rarr ?overset((O))rarr

Answer»

SOLUTION :
(b) `(CH_(3))_(2)CHOH overset((O))underset(K_(2)Cr_(2)O_(7)//H^(+))RARR (CH_(3))_(2)C=Ooverset(50%HNO_(3))underset((O))rarr CH_(3)COOH`
Discussion
89841.

(a) Explain Etards reaction.(b) Name the products A and B in the following reaction.2CH_(3)CHO overset("dil NaOH")hArr "A"overset("Delta")underset(-H_(2)O)hArr B (c ) Namé the reagent used in the decarboxylation of carboxylic acid.

Answer»

Solution : (a) When TOLUENE treated with `CrO_(2)CI_(2)` & `CS_(2)` followed by HYDROLYSIS benzaldehyde is formed.

(B) A=3-hydroxybutanal or Aldol
(c ) SODALIME
Discussion
89842.

Explain Hoffmann bromamide degradation reaction and write the general equation for the reaction involved.

Answer»

Solution :a) Amides on heating with BROMINE & aq NaOH gives `1^(@)` amines. This REACTION is called Haff man.s bromamide reaction.
`UNDERSET("Amide")(R-CO-NH_(2)+4NaOH) RARR underset("Amine")(R-NH_(2)+2NaBr+Na_(2)CO_(3)+2H_(2)O)`
Discussion
89843.

(a) Explain esterification reaction between acetic acid and ethyl alcohol as example.(b) Boiling point of alcohol is greater than the boiling point of hydrocarbons of coparable molar masses, Why?(c ) What is the effect of -NO_(2)group on the acidic strength of phenol?Give reason.

Answer»

Solution :(a) When acetic acid is HEATED with ethyl alcohol in the presence of conc.`H_(2)SO_(4)`, ethyl acetate is formed.
`CH_(3)COOH + C_(2)H_(5)OH OVERSET("conc. "H_(2)SO_(4))to CH_(3)COOC_(2)H_(5) + H_(2)O`
(b) Hydrogen bonding
(c ) Acidic strength of phenol increases. Because -`NO_(2)` is electron withdrawing group.
Discussion
89844.

(a) Explain carbylamine reaction by taking methyl amine as an example. (b) Why do primary amine have higher boiling point than tertiary amines? (c) Give an example for a coupling reaction of diazonium salt and give its chemical equation.

Answer»

Solution :(a) When a primary amine is heated with CHLOROFORM and alcoholic KOH gives ISOCYANIDES. ,
`R-NH_(2) +CHCI_(3) +3KOHoverset(“heat”) to R- NC+3KCI+ 3H_(2) O`
Eg. `R=CH_(3)`
` CH_(3) - NH_(2) +CHCI_(3) +3KOH overset(”heat”)to CH_(3) - NC +3KCI+3H_(2)O`
(b) Due to hydrogen bonding (OR) INTERMOLECULAR assciation
Benzene diazonium chloride (BDC)reacts wit5hy phenol to give azo dye.
Discussion
89845.

(a)Explain briefly the collision theory of bimolecular reactions . [Or] [b]Discuss about the hydrolysis of salf of weak acid and week base and derived pH value for the solution.

Answer»

Solution :(a) Collision theory on the kinetic theory is based on the kinetic theory of GASES. According to this theory.chemicalreaction occur as a RESULT of collisions between the reacting molecules Let us understand this If a consider the following reaction
`A_(2(g))+B_(2(g))rarr2AB_((g))`
If we consider that , the reaction between `A_2 and B_2`molecules proceeds through collision between them , then the would be proportional to the number of collisions per second . Rate `prop`Number of molecules colliding per litre per second (or) Rate of `prop` collision rate.
The number of collisions is directly proportional to the concentration of both `A_2 and B_2` . Collision rate `prop [A_2][B_2]`Collision rate `= Z[A_2][B_2]`where , Z is a constant .
The Collision rate in gas can be CALCULATED from kinetic theory of gases. For a gas at room temperature (298K) and 1 atm pressure, each molecule undergoes approximately `10^9`collisions per second i.e I collision in `10 ^(-9)`second This if every collision resulted in reaction, the reaction would be complete in `10^(-9)`second In actual practice does not happen . It implies that all collisions are not effective to lead to the reaction. In order to react, the Colliding molecules must posses a minimum energy calledactivation energy . The molecules that collide with less energy that activation energy will remain intact and no reaction occurs.
Fraction of effective collisions (f) is given by the expression , `f -=e^((-E_a)/(RT))` .
Fraction of collisions is further reduced dut to orientation factor i.e., even if the reactant collide with sufficient energy , then will not unless the orientation of the reactant molecules is suitable for the formation of the transition state .
The fraction of effective collisions (f) having proper orientation is given by the steric factor P.
Rate `=pxxfxx` collision rate
Rate `Pxxe^.(-E_a)/(RT)xxZ[A_2][B_2]""......(1)`
As per the rate law , Rate `=k[A_2][B_2]""...(2)`
Where k is the rate constant .
On comparing equation (1) and (2) , the rate constant k is ,
`k=PZe^(((-E_a)/(RT))`
[OR]
(b) i. Consider the hydrolysis of ammonium acetate
`CH_3COONH_(4(aq))rarrCH_3COO_((aq))^(-)+NH_(4(aq))^+`
ii. In this case both the cation `(NH_4^+) and (CH_3COO^-)` anion have the tendency to react with water .
`CH_3COO^(-)+H_2^OhArrCH_3COOH+OH^(-)`
`NH_4^(+)H_2OhArrNH_4OH+H^+`
iii. The nature of the solution depends on the strength of acid (or) base i.e. , if `K_a gt K_b` , then the solution is acidic and `pH lt 7` if `K_a lt K_b` then hte solution is basic and `pH gt 7`.
IV. The relation between the DISSOCIATION constant `K_a , K_b` and hydrolysis constant is given by the following expression.
`K_a. K_b.K_h=K_w`
v. pH of the solution
`pH = 7 + 1//2 pK_a - 1//2 pK_b`
Discussion
89846.

(a) Explain carbyl amine reaction with equation. (b) How does nitrobenzene is reduced to aniline? Give equation. (c) Write the IUPAC name of underset(CH_3)underset(|)(C_6H_6-N-CH_3)

Answer»

Solution :(a) Aliphatic and aromatic primary amines on HEATING with chloroform and ethanolic potassium hydroxide form isocyanides or carbylamines. This REACTION is known as carbylamines reaction.
`UNDERSET(1^(st)"amine")(R-NH_2) + underset("Chloroform")(CHCl_3) + 3KOH_((alc)) underset(Delta)(rarr) underset("Carbylamine")(R-NC) + 3KCl + 3H_2O`
(b)NItrobenzene is reduced to aniline by passing hydrogen GAS in the presence of finely divided nickel, palladium or platinum.

OR
Nitrobenzene is reduced to anline with metals in acidic MEDIUM.

(c) N, N - dimethylaniline.
Discussion
89847.

(a) Explain Cannizaro's reaction with an example. (b) Name the product obtained by the reaction of acetyl chloride with dimethylcadmium. (c) Explain the reaction between carboxylic acid and PCI5

Answer»

Solution :(a) TWO molecules of aldehydes do not contain a-hydrogen atoms are reacted with strong solution of NaoH to produce alcohol and salt of carboxylic acid.

(b) Acetone (c ) when carboxylic acid reaction with PCI_(5) to form coressponding acid chloride.
` RCOOH+PCI_(5) TO RCOCI+POCI_(3)+HC`I
eg.` CH_(3) COOH+PCFI_(5) to underset(ACETYL chride)CH_(3) COCI+POCI_(3)+HCI`
Discussion
89848.

(a) Explain about Daniell cell.

Answer»

Solution : Each cell consisting of two electrodes.
At one electrode oxidation TAKES places and at ANOTHER electrode REDUCTION takes place.
When an electrode is dipped in it.s solution half cell is FORMED.
When these two half cells combined together with salt bridge is called Daniel cell and without salt bridge is called galvanic cell.
Discussion
89849.

(A) Exhaust systems of supersonic jet aeroplanes slowly depleting the concen-tration of ozone layer. (R ) NO _((g)) + O _(3 (g)) to NO _(2(g)) + O _(2 (g))

Answer»

Both (A) and (R ) are true and (R ) is the correct explanation of (A)
Both (A) and (R ) are true and (R ) is not the correct explanation of (A)
(A) is true but (R ) is FALSE
Both (A) and (R )are false

ANSWER :A
Discussion
89850.

(A) Every substance is super conducting at room temperature. (R) A super conducting substance offers resistance to the flow of electricity

Answer»

Both (A) and (R) are TRUE and (R) is the correct EXPLANATION of (A)
Both (A) and (R) are true and (R) is not the correct explanation of (A)
(A) is true but (R) is false
Both (A) and (R) are false

Answer :D
Discussion