Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

Predict whether the reaction CO_((g)) + H_2O_((g)) to CO_(2(g)) + H_(2(g)) is spontaneousor not. The standard free energies of formation of CO_((g)), H_2O_((g)) and CO_(2(g)) are – 137.27, –228.6 and –394.38 kJ mole"^(-1) respectivley.

Answer»

SOLUTION :`Delta_r^@ = - 28.51 KJ`
Discussion
2.

Predict whether the entropy change of the system in each of the following processes is positive or negative. (a) CaCO_(3)(s) to CaO(s)+CO_(2)(g) (b) N_(2)(g)+3H_(2)(g) to 2NH_(3)(g) (c ) N_(2)(g)+O_(2)(g) to 2NO(g) (d) HCl(g)+NH_(3)(g) to NH_(4)Cl(s) (e) 2SO_(2)(g)+O_(2)(g) to 2SO_(3)(g) (f) Cooling of N_(2)(g) from 20^(@)C to -50^(@)C

Answer»

Solution :GASEOUS substances generally POSSESS more ENTROPY than solids. So whenever the products CONTAIN more moles of a gas than the reactants, the entropy change is probably positive. And hence, DeltaS is
(a) positive
(B) negative
(c ) small, the sign of DeltaS is impossible to predict
(d) negative
(e ) negative
(f) negative
Discussion
3.

Predict whether the following reactions occur under standard state conditions : (a) Oxidationof Ag_((S)) by Cl_(2(g)).E_("Ag")^(0) = 0.8V, E_("Cell")^(0) = 1.36V (b) Reduction of Fe^(3+) to Fe^(2+) by Au_((s)) E_(Fe^(3+),Fe^(2+))^(0) = 0.77 V, E_(("Au"))^(2) = 1.4V

Answer»

Solution :(a) SINCE THESTANDARD reduction potentialof `Cl_(2(g))` ishigher`(E_(cl^(2+)//cl^(-0))^(0) - = 1.36V) `than thatof `Ag,(E_(Ag^(2+)//Ag)^(0) = 0.8V)`,chlorine will oxidise `Ag_((S))` to `Ag^(+)` .
The redox reactionwill be , `2Ag_((s)) + Cl_((2)) to2Ag^(+) + 2Cl_((aq))^(-)`
(B) Sincethestandardreductionpotential of Auis higher`(E_(Cl^(2+)//Cl^(-))^(0) = - 1.4V)` thanthat of `Fe^(3+)`, Fe^(2+)`couple`(E_(Fe^(3+),Fe^(2+))^(0) = 0.77V)` , Auwillnot reduce `Fe^(3+)`, to `Fe^(2+)` but Au can oxidise `Fe^(2+)` to `Fe^(3+)` sinceAu has moretendency to accpet electrons.
Discussion
4.

Predict whether the following reaction (s) is (are) feasible or not (i) Fe+Zn^(2+)toFe^(2+)+Zn,E_(Zn)^(@)=-0.76,E_(Fe)^(@)=-0.44V (ii) Zn+2Ag^(+)toZn^(2+)+2Ag,E_(Zn)^(@)=-0.76V,E_(Ag)^(@)=-0.80V

Answer»

SOLUTION :(i) `E_(CELL)^(@)=E_(Fe//Fe^(2+))^(@)+E_(Zn^(2+)//Zn)^(@)=+0.44+(-0.76)=-0.32V` LTBRGT (ii) `E_(cell)^(@)=E_(Zn//Zn^(2+))^(@)+E_(Ag^(+)//Ag)^(@)=+0.76+0.80=1.56V`.
Discussion
5.

Predict whether the reaction 2Ag_((s))+Zn_((aq))^(2+)rarr 2Ag_((aq))^(+)+Zn_((s)) is feasible or not.

Answer»

Solution :The cell half REACTIONS are
`{:("Anode",:,2Ag_((s))rarr2Ag_((AQ))^(+)+2e^(-),E^(@)=0.80V),("Cathode",:,Zn_((aq))^(+)+2e^(-)rarrZn_((s)),E^(@)=-0.763V):}`
`E_("cell")^(@)=E_("cathode")^(@)-E_("anode")^(@)`
`E_("cell")^(@)=-0.763V-0.80V`
`=-.1563V`
Since `E_("cell")^(@)`is NEGATIVE, the given reaction is not feasible.
Discussion
6.

Predict whether a spontaneous reaction will occur:

Answer»

When a piece of silver wire is DIPPED in a `ZnSO_4` SOLUTION
When lodine is added to a NABR solution
When a piece of Zn metal is dipped in a `NiSO_4` solution
When a chlorine gas is bubbled through a KI solutiön

Answer :C::D
Discussion
7.

Predict whether Ag^(+) can oxidize Pb to Pb^(2+) under standard state conditions. E_(Ag)^(@)=0.799V and E_(Pb)^(@)=-0.126V.

Answer»

SOLUTION :(a) (i) Since the standard reduction potential of Ag under standard CONDITIONS is higher `(E_(Ag^(+)//Ag)=0.799V)` than that of `Pb(E_(Pb^(2+)//Pb)^(@)=-0.126V), Ag^(+)` have higher TENDENCY to gain electrons than Pb.
(ii) Hence `Ag^(+)` acts as a STRONGER oxidising AGENT and oxidises Pb to `Pb^(2+)` while `Ag^(+)` gets reduced to Ag.
The redox reaction will be, `Pb_((s))+2Ag_((aq))^(+) rarr Pb_((aq))^(2+)+2Ag_((s))`
Discussion
8.

Predict titration curve for titration of Na(2)CO_(3) & HCl

Answer»

Solution :`Na_(2)CO_(3)+HClrarrNaHCO_(3)+NaCl`
`pH_(1)="Half equivalence point with respect to" 1^(st)` neutralisation =`pK_(a_(2))`(:.solution will be a buffer consisting of equal amount of `CO_(3)^(2-)&HCO_(3)^(-))`
`pH_(1^(st eqpt))=(pK_(a_(1))+pK_(a_(2)))/(2)( :."AMPHIPROTIC ANION")`
`NaHCO_(3)+NaClrarr H_(2)CO_(3)+NaCl`
`pH_(2)`=half equivalence point with respect to `2^(nd)`neutralisation =`pK_(a_(2))` ( :.solution will be a buffer)
consisting of equal amount of `HCO_(3)^(-)&H_(2)CO_(3))`
`pH_(2^(ndaqpt))=1//2(pK_(a_(1))-LOGC)(":.weak acid")`
Discussion
9.

Predict the type of radioactive decay process that is likely for each of the following nuclides (a) ""_(92)^(228)U (b) ""_(5)^(8)B (c ) ""_(29)^(68)Cu

Answer»

SOLUTION :(a) `alpha` (B) `BETA^(+)` or K-capture (c ) `beta` EMISSION
Discussion
10.

Predict the temperature for the carbon reduction of Cu_(2)O from the Ellingham diagram.

Answer»

SOLUTION :500 - 600 K temperature is ideal for the reduction of `Cu_(2)O`.
Discussion
11.

Predict the tendency of undergoing possible mechanism for the athyl, isopropyl and t-butyl halides. (ii) Complete the following equations showing major product with reaction mechanism (S_(N)1,S_(N)2,E1,E2).

Answer»

Solution :When `S_(N)1` mechanism favoured it means that the activation ENERGY for this path is less than that for the `S_(N)2`. Practically, more stable the lon, the lower is the activation energy and more favoured will be the mechanism. As we KNOW, the order of stability of carbonium ion is `1^(@)LT2^(@)lt3^(@)`and hence the TENDENCY for the `S_(N)1` mechanism to operate will be `t-BuXgt" ios PROPYL "XgtEtX`. The tendency willbe reverse for the `S_(N)2` mechanism,
Discussion
12.

Predict the structure of MgO crystal and coordination number of its cation in which the cation and anion radii are equal to 65 pm and 140 pm respectively.

Answer»

Solution :Radius of cation (`Mg^(2+)`) = 65 pm.
Radius of anion (`O^(2-)`) = 140 pm
Radius ratio, `(r_(+))/(r_(-))=(r_(Mg^(2+)))/(r_(O^(2-)))=(65)/(140) =0.464`
The radius ratio lies between 0.414-0.732. HENCE the EXPECTED structure is octahedral with coordination number of SIX.
Discussion
13.

Predict the structure of propane -1,2 diol

Answer»

`CH_(2)(OH) -CH_(2)CH_(2)OH`
`HOCH_(2)-CH_(2)OH`
`CH_(3)CH(OH)CH_(2)OH`
None of these

ANSWER :C
Discussion
14.

Predict the state of the solute in the solution in the following situations : When 'I' is found to be more than one.

Answer»

SOLUTION :When .i. is FOUND to be more than ONE.
The SOLUTE is DISSOCIATED in the solution.
Discussion
15.

Predict the state of the solute in the solution in the following situations : When 'I' is found to be less than one.

Answer»

SOLUTION :When .i. is FOUND to be LESS than ONE.
The SOLUTE is associated in the solution.
Discussion
16.

Predict the species produced during heterolytic fission of the following:

Answer»


`(CH_(3))_(3)CBR`
`CH_(3)CH_(2)Li`
`(CH_(3))_(2)C-UNDERSET(H)underset(|)O^(+)-CH_(3)`

Solution :`(CH_(3))_(2)CHOH`
Discussion
17.

Predict the sign/value of DeltaS^(@) for the following reaction2H_(2)S_((g))+3O_(2(g))rarr2H_(2)O_((g))+2SO_(2(g))

Answer»

`+ve`
`-ve`
Zero
Can not be predicted

SOLUTION :LESS VOLUME on the product SIDE.
Discussion
18.

Predict the sign of DeltaS in the following processes. Give reasons for your answer. (a) N_(2(g))+3H_(2(g))rarr2NH_(3(g)) (b) CO_(2(g))rarrCO_(2(s))

Answer»

Solution :`N_(2(G))+3H_(2(g))rarr2NH_(3(g))`
In the reaction, 4 moles of gaseous reactants form 2 moles of gaseous products `(DELTA n LT 0)`. Therefore disorder decreases and hence entropy decreases, `DeltaS lt 0.`
Discussion
19.

Predict the sign of DeltaS in the following processes. Give reasons for your answer : N_(2_((g)))+3H_(2_((g))) to 2NH_(3_((g)))

Answer»

Solution :`N_(2_((g)))+3H_(2_((g))) to 2NH_(3_((g)))`
In the reaction, 4 moles of GASEOUS reactants form 2 molesgaseous products `(Deltan lt 0)`. Therefore disorder decreases and HENCE entropy decreases, `DeltaS lt 0.`
Discussion
20.

Predict the sign of DeltaS in the following processes. Give reasons for your answer : Fe_(2)O_(3_((s)))+3H_(2_((g))) to 2Fe_((s))+3H_(2)O_((g))

Answer»

SOLUTION :`Fe_(2)O_(3_((s)))+3H_(2_((g)))to 2Fe_((s))+3H_(2)O_((g))`
In the REACTION number of moles of gaseous reactants and PRODUCTS are same, hence `DeltaS=0`
Discussion
21.

Predict the sign of Delta S forthe following process: H_(2)O_((g)) rarr H_(2)O_((l))

Answer»

SOLUTION :NEGATIVE or - ve.
Discussion
22.

Predict the shape of ClF_3 on the basis of VSPER theory.

Answer»

Solution :`ClF_3` has T- shapestructure with 3 BOND PAIRS and 2 lone pairs.
Discussion
23.

Predict the shape and the asked angle (90^(@)" or more or less") in the following case SO_(3)^(2-) and the angle O-S-O.

Answer»

SOLUTION :PYRAMIDAL :

ANGLE : `O-S-O` more than `90^(@)`
Discussion
24.

Predict the products of the following reactions : (i) CH_(3)-CH_(2)-CH_(2)-O-CH_(3)+HBr rarr (ii) (iv) (CH_(3))_(3)C-OC_(2)H_(4)overset(HI)rarr

Answer»

Solution :(i) Both the alkyl groups ATTACHED to the oxygen atom are primary, therefore, attack of `Br^(-)` ion occurs on the smaller methyl GROUP giving propan-1-ol and bromomethane as the products.
`underset("N-Propyl methyl ether")(CH_(3)-CH_(2)-CH_(2)-O-CH_(3))+HBr overset(373K)rarr underset("Propan-1-ol")(CH_(3)-CH_(2)-CH_(2)-OH)+underset("Bromomethane")(CH_(3)-Br)`
(ii) `C_(6)H_(5)-O` bond has some double bond character DUE to resonance and hence is STRONGER than `O-C_(2)H_(5)` bond. Therefore, the cleavage takes place at the weaker `O-C_(2)H_(5)` bond to yield phenol and bromoethane as the products.

(iii) `-OC_(2)H_(5)` group is activating as well as o-, p - directing. Therefore, nitration of ethoxybenzene will give a mixture of 2- and 4- ethoxynitrobenzene in which 4 - nitroethoxybenzene dominates due to steric hindrance at 2 - position.

(iv) The reaction follows `S_(N)1` MECHANISM leading to the formation of tert - butyl iodide and ethanol. tery - butyl carbocation is much more stable than ethyl carbocation.
Step 1 :
Discussion
25.

Predict the products of the following reactions : (i) CH_(3) - underset(CH_(3))underset(|)C=O overset((i) H_(2)N-NH_(3))underset((ii)KOH//"Glycol",Delta) to ? (ii) C_(6)H_(5)-CO-CH_(3) overset(NaOH//I_(2)) to ?+ ? (iii) CH_(3)COONa overset(NaOH//CaO)underset(Delta) to ?

Answer»

Solution :(i) `CH_(3)-underset(CH_(3))underset(|)C = O overset((i) H_(2)N, NH_(2))underset((ii) KOH//"GLYCOL",Delta) to underset("Propane")(CH_(3)CH_(2)CH_(3))`
(ii) `C_(6)H_(5) - CO - CH_(3) overset(NaOH//I_(2))to underset("Sod.benzoate")(C_(6)H_(5)COONa) + underset("Iodoform")(CHI_(3))`
(III) `CH_(3)COONa overset(NaOH//CaO)underset(Delta)to underset("METHANE")(CH_(4))`
Discussion
26.

Predict the products of the following reactions : (iii) R-CH=CH-CHO+NH_2-overset(O)overset(||)(C )-NH-NH_2overset(H^+)(rarr) (iv)

Answer»

Solution :
(ii) `NH_2` attached to `NH` is more nucleophilic than `NH_2`, attached to `C=O`group, therefore, reaction OCCURS through `NH_2` attached to NH to GIVE the CORRESPONDING semicarbozone, i.e., `R-CH=CH-CH= O + H_2 NNHCONH_2 RARR R -CH=CH-CH= NNHCONII_2 + H_2O`
(IV)
Discussion
27.

Predict the products of the following reactions CH_3-CH_2-CH_2-O-CH_3+HBr rarr

Answer»

SOLUTION :`CH_3-CH_2-CH_2-OH` and `CH_3Br`
Discussion
28.

Predict the products of the following reactions.

Answer»


ANSWER :`(##ANE_PKE_CHE_XII_C12_E01_017_S01##)`
Discussion
29.

Predict the products of the following reactions.

Answer»


ANSWER :`(##ANE_PKE_CHE_XII_C12_E01_015_S01##)`
Discussion
30.

Predict the products of the following reactions

Answer»

SOLUTION :
Discussion
31.

Predict the products of the following reactions.

Answer»


ANSWER :`(##ANE_PKE_CHE_XII_C12_E01_014_S01##)`
Discussion
32.

Predict the products of the following reactions.

Answer»


ANSWER :`R-CH=CH-CH=N-NH-CO-NH_2`
Discussion
33.

Predict the products of the following reactions

Answer»

SOLUTION :
Discussion
34.

Predict the products of electrolysis of the following: (i) An aqueous solution of AgNO_(3) with silver electrodes (ii) An aqueous solution of AgNO_(3) with platinum electrodes. (iii) A dilute aqueous solution of H_(2)SO_(4) with platinum electrodes. (iv) An aqueous solution of CuCl_(2) with platinum electrodes. (Given E_(Ag^(+)//Ag)^(@)=+0.80V, E_(Cu^(2+)//Cu)^(@)=+0.34V)

Answer»

Solution :(i) Electrolysis of aqueous solution of `AgNO_(3)` with silver electrodes.
`AgNO_(3)(s)+aqtoAg^(+)(aq)+NO_(3)^(-)(aq)`
`H_(2)OhArrH^(+)+OH^(-)`
At cathode: `Ag^(+)` ions have lower discharge potential than `H^(+)` ions. Hence, `Ag^(+)` ions will be deposited as Ag in preference to `H^(+)` ions.
Alternatively, we have stanard reduction potential potentials as
`Ag^(+)(aq)+e^(-)toAg(s),E^(@)=+0.80V`
`H^(+)(aq)+e^(-)to(1)/(2)H_(2)(g),E^(@)=0.00V`
As `Ag^(+)` ions hae HIGHER standard reduction potential than that of `H^(+)` ions, hence `Ag^(+)` ions will be reduced more easily and deposited as Ag.
At anode: As Ag anode is attached by `NO_(3)^(-)` ions, Ag of the anode will dissolve to form `Ag^(+)` ions in the solution.
`AgtoAg^(+)+e^(-)`
Alternatively, out of the three possible oxidation reactions occurring at the anode, i.e.,
`AgtoAg^(+)+e^(-),2OH^(-)toH_(2)O+(1)/(2)O_(2)+e^(-) and NO_(3)^(-)toNO_(3)+e^(-)`,
Ag has highest oxidation potential. hence, Ag of anode is oxidized to `Ag^(+)` ions which pass into the solution.
(ii) Electrolysis of aqueous solution of `AgNO_(3)` using platinum electrodes.
At cathode: Same as above. ltBrgt At anode: As anode is not attackable, out of `OH^(-) and NO_(3)^(-)` ions, have lower discharge potential. hence, `OH^(-)` ions will be DISCHARGED in preference to `NO_(3)^(-)` ions, which then decompose to give out `O_(2)`.
`OH^(-)(aq)toOH+e^(-),4OHto2H_(2)O(L)+O_(2)(g)`
(iii) Electrolysis of dilute `H_(2)SO_(4)` with platinum electrodes.
`H_(2)SO_(4)(aq)to2H^(+)(aq)+SO_(4)^(2-)(aq)`
`H_(2)OhArrH^(+)+OH^(-)`
At cathode: `H^(+)+e^(-)toH,H+HtoH_(2)(g)`
At anode: `OH^(-)toOH,e,4OHto2H_(2)O+O_(2)(g)`
Thus, `H_(2)` is LIBERATED at the cathode and `O_(2)` at the anode.
(iv) Electrolysis of aqueous solution of `CuCl_(2)` with platinum electrodes.
`CuCl_(2)(s)+aqtoCu^(2+)(aq)+2Cl^(-)(aq)`
`H_(2)OhArrH^(+)+OH^(-)`
At cathode: `Cu^(2+)` ios will be reduced in preference to `H^(+)` ions
`Cu^(2+)+2e^(-)toCu`
At anode: `Cl^(-)` ions will be oxidized in preference to `OH^(-)` ions
`Cl^(-)toCl+e^(-),Cl+CltoCl_(2)(g)`
Thus, Cu will be deposited on the cathode and `Cl_(2)` will be liberated at the anode.
Discussion
35.

Predict the products of the following reactions

Answer»

SOLUTION :
Discussion
36.

Predict the products of electrolysis obtained at the electrodes in each case when the electrodes used are of platinum : (i) An aqueous solution of AgNO_3. (ii) An aqueous solution of H_2SO_4.

Answer»

Solution :(i) `AgNO_(3)(aq) to Ag^(+) (aq) + NO_(3)^(-) (aq)`
`H_(2)O (l) to H^(+) (aq) + OH^(-) (aq)`
At cathode: `Ag^(+) (aq) + E^(-) to O_(2) + 2H^(+) (aq) +4e^(-)`
`H_(2)O(l) to H^(+) (aq) + OH^(-) (aq)`
At cathode: `2H^(+) (aq) + 2e^(-) to O_(2) + 2H^(+) (aq) + 4e^(-)`
or `2H_(2)O (l) to O_(2) + 4H^(+) (aq) + 4e^(-)`
Hydrogen gas is produced at the cathode and oxygen at the anode.
Discussion
37.

Predict the products of the following reactions.

Answer»


ANSWER :`(##RES_CHM_CCAK_E01_026_A01##)`
Discussion
38.

Predict the products of electrolysis of an aqueous solution of CuBr_(2) .

Answer»

Solution :The probable reactions at CATHODE are :
`(i) CU^(2+) (aq) + 2e^(-) to Cu (s) , "" E^(Theta) = 0.34 V`
(ii) `2 H_(2) O (l) + 2e^(-)to H_(2) (g) + 2 OH^(-) (aq) , E^(Theta) = -0.83` V
Since the reduction potential of `Cu^(2+)` ions is higher than that of water , copper will be reduced preferably at cathode ,
The probable reactions at anode are :
`2 Br^(-) (aq) to Br_(2) (g) + 2e^(-) "" E^(Theta) = 1.08 V`
`H_(2) O(l) to (1)/(2) O_(2) (g) + 2 H^(+) + 2e^(-) E^(Theta) = 1.23` V
Since the reduction potential of `Br^(-)` is LESS than water , it will be readily oxidised to `Br_2` at anode. Therefore , the reactions are :
At cathode : `Cu^(2+) (aq) + 2 e^(-) to Cu (s)`
At anode : `2 Br^(-) (aq) to Br_(2) (g) + 2 e^(-)`
Discussion
39.

Predict the products of electrolysis obtained at the electrodes in each case by using platinum electrodes : (i) An aqueous solution of AgNO_(3) using platinum electrodes (ii) An aqueous solution of CuSo_(4) using attackable electrodes.

Answer»

SOLUTION :(i)Ag is DEPOSITED at cathode.
`O_(2)` is evolved at anode.
(II)`Cu^(2+)` IONS are deposited at cathode. Equivalent amount of Cu from anode goes into the solution.
Discussion
40.

Predict the products of electrolysis in each of the following: (i) An aqueous solution of AgNO_3 with silver electrodes. (ii) An aqueous solution of AgNO_3 with platinum electrodes. (iii) A dilute solution of H_2SO_4 with platinum electrodes. (iv) An aqueous solution of CuCl_(2) , with platinum electrodes.

Answer»

Solution :Electrolysis of AQUEOUS solution of `AgNO_(3)` with silver electrodes:
The ions produced by the dissociation are as under:
`AgNO_(3) (s) + aq to Ag^(+) + NO_(3)^(-3) (aq)`
`H_(2)O At cathode : `Ag^+` ions have lower DISCHARGE potential than H+ ions. Hence, `Ag^+` ions will be deposited in preference to `H^+` ions.
At anode : As Ag anode is attacked by `NO_(3)^(-)`ions, Ag of the anode will dissolve to form `Ag^+` ions in the solution.
`Agto Ag^(+) + E^(-)`
(ii) Electrolysis of aqueous solution of `AgNO_3` using platinum electrodes : At cathode : Same as in (i).
At anode : Out of `OH^(-)`and `NO_3^(-)`ions, `OH^(-)`ions have lower discharge potential. Hence, `OH^(-)` ions will be discharged in preference to `NO_3^(-)` ions. OH ions decompose to give out `O_(2)`as under:
`OH^(-)(aq) to OH + e^(-), 4OH to 2H_(2)O (l) +O_(2)(g)`
(III) Electrolysis of dilute `H_(2)SO_(4)` with platinum electrodes:
`H_(2)SO_(4)` and `H_(2)O` ionise as under:
`H_(2)SO_(4)(aq) to 2H^(+) (aq) + SO_(4)^(2-) (aq)`
`H_(2)O At cathode: `H^(+) + e^(-) to H, H + H to H_(2)(g)`
At cathode: `OH^(-) to OH +e^(-), 4OH to 2H_(2)O + O_(2)(g)` (because of lower discharge potential of `OH^(-)`)
(iv) Electrolysis of aqueous solution of `CuCl_2` with platinum electrodes :
`CuCl_(2)(s) + aq to Cu^(2+) (aq) + 2Cl^(-)(aq)`
`H_(2) O At cathode: `Cu^(2+)` ions will be reduced in preference to `H^(+)` ions
`Cu^(2+) + 2e^(-) to Cu`
At anode: `Cl^(-)` ions will be oxidised in preference to `OH^(-)` ions.
`Cl^(-) to Cl + e^(-), Cl + Cl to Cl_(2)(g)`
THUS, Cu will be deposited on the cathode and `Cl_2` will be liberated at the anode.
Discussion
41.

Predict the products of electrolysis in each of the following: (i) An aqueous solution of AgNO_(3) with silver electrodes. (ii) An aqueous solution of AgNO_(3) with platinum electrodes.

Answer»

Solution :(i) Electrolysis of aqueous solution of `AgNO_(3)` using silver electrodes :
`AgNO_(3)(s)overset("water")toAg^(+)(aq)+NO_(3)^(-)(aq)`
`H_(2)OhArrH^(+)OH^(-)`
cathode. `Ag^(+)` ions have lower discharge potential than `H^(+)` ions. Hence `Ag^(+)` ions will be deposited as silver (in preference to `H^(+)` ions)
At ANODE. Since silver electrode is attacked by `NO_(3)^(-)` ions, Ag anode will DISSOLVE to form `Ag^(+)` ions in the solution.
`Ag""toAg^(+)+e^(-)`
(ii) Electrolysis of aqueous solution of `AgNO_(3)` using platinum electrodes :
At cathode. Same as above.
At anode. Since silver is not attacked, out of `OH^(-)` and `NO_(3)^(-)` ions, `OH^(-)` ions have lower discharge potential and hence `OH^(-)` ions will be discharged in preference to `NO_(3)^(-)`. The `OH^(-)` will decompose to give `O_(2)`.
`OH^(-)(aq)toOH+e^(-)`
`4OHto2H_(2)O(l)+O_(2)(g)`
Discussion
42.

Predict the product(s) of (A) Friedel-Crafts acylation of anisole (methoxybenzene) with acetyl chloride in the presence of one equivalent of AlCl_(3) follwed by H_(2)O. (B) Friedel-Crafts alkylation of a large excess of ethylbenzene with chloromethane in the presence of AlCl_(3).

Answer»


ANSWER :`(##MOT_CON_JEE_CHE_C05_E03_006_A01##)`
Discussion
43.

Predict the products in the given reaction 2HCHO overset(conc NaOH)(to)

Answer»

`HCOOH + CH_3O^(-) NA^(+)`
`CH_3CH_2OH+HCOO^(-)Na^(+)`
`CH_3OH+HCOO^(-)Na^+`
`CH_3OH+HCOOH`

SOLUTION :CANNIZZARO reaction .
`2HCHO underset(Delta)overset(conc NAOH)(to) CH_3OH+HCOO^(-) Na^(+)`
Discussion
44.

Predict the products: (i) CH_(3)CH_(2)OH overset(K_(2)Cr_(2)O_(7))underset(H_(2)SO_(4)) to A overset(SOCl_(2)) to B overset(NH_(2)) to C overset(NaOBr) to D (ii) CH_(3)CH_(2)Cl overset(KCN) to A overset(OH^(-))underset(H_(2)O_(2)) to B overset(P_(2)O_(5)) to C (iii) CH_(3)COOHoverset(C_(2)H_(5)OH)underset("conc"H_(2)SO_(4)) to A overset(CH_(3)MgBr)underset(H^(+)//H_(2)O) to B overset(CH_(3)MgBr)underset(H_(2)O//H^(+)) to C

Answer»

Solution :The PRODUCTS A, B, C, D are identified with the help of the following reactions :
(i)`CH_(3)CH_(2)OH overset(K_(2)Cr_(2)O_(7))UNDERSET(H(2)SO_(4)) to CH_(3)COOH overset(SOCl_(2))to CH_(3)COCl overset(NH_(3))to CH_(3)CONH_(2) overset(NaOBr)to CH_(3)NH_(2)`
(ii) `CH_(3)CH_(2)CL overset(KCN) to underset(A)(CH_3)CH_(2)C-= N overset(OH^(-))underset(H_(2)O_(2)) to CH_(3)CH_(2) - underset(B)overset(O)overset(|\)C-NH_(2) overset(P_(2)O_(3))underset("Dehydration")to underset( C)(CH_(3)CH_(2)C -= N + H_(2)O)`
(iii) `CH_(3)COOH overset(C_(2)H_(5)OH)underset("conc "H_(2)SO_(4)) to underset(A)(CH_(3)COOC_(2)H_(5)) overset(CH_(3)MgBr)underset(H_(2)O//H^(+)) to CH_(3) -underset(B)overset(O)overset(||)C-CH_(3) overset(CH_(3)MgBr)underset(H_(2)O//H^(+)) to CH_(3) - underset(CH_(3))underset(|)overset(OH)overset(|)C-CH_(3)`
Discussion
45.

Predict the products in the following sequence of reactions.

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SOLUTION :
Discussion
46.

Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents : (i) PhMgBr and then H_3O^(+) "" (ii) Tollens' reagent (iii) Semicarbazide and weak acid "" (iv) Excess ethanol and acid (v) Zinc amalgam and dilute hydrochloric acid.

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F

Solution :
Discussion
47.

Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents: (i) PhMgBr and then H_(3)O^(+) (ii) Tollens' reagent (iii) Semicarbazide and weak acid (iv) Excess ethanol and acid (v) Zinc amalgam and dilute hydrochloric acid.

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SOLUTION :.
Discussion
48.

Predict the products formed when cyclohexanecarbaldehyde reacts with following reagents: (a) PhMgBr and then H_(3)O^(+). (b) Tollens' reagent.

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SOLUTION :
Discussion
49.

Predict the products expexted of the following reactions:

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SOLUTION :
Discussion
50.

Predict the products formed in the following cases: (i) (A) reacts with PhMgBr and then hydrolysed. (ii) (A) reacts with hydrazine and is then heated with KOH and ethylene glycol.

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SOLUTION :(i) `underset((A))underset(CH_(3))underset(|)underset(CH_(2))underset(|)(CH_(3)CH_(2)C)=O+PhMgBr to CH_(3)CH_(2)-underset(CH_(3))underset(|)underset(CH_(2))underset(|)overset(Ph)overset(|)C-OMgBr overset(H_(2)O)to underset("3-Phenylpentane-3-ol")(CH_(3)CH_(2)-underset(CH_(3))underset(|)underset(CH_(2))underset(|)overset(Ph)overset(|)C-OH)`
(ii) `underset((A))(CH_(3)CH_(2)COCH_(2)CH_(3)) underset({:(KOH//"Ethylene glycol"),(" (Wolff-Kishner)"),("reduction"):})overset(NH_(2)NH_(2))to underset(N-"PENTANE")(CH_(3)CH_(2)CH_(2)CH_(2)CH_(3))`
Discussion