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.

The reaction of neopentyl alcohol with concentrated HCl gives

Answer»

neopentyl CHLORIDE
2-chloro-2-methylbutane
2-methyl-2-butene
a mixture of neopentyl chloride and 2-methyl-2-butene

Solution :`CH_(3)-underset(CH_(2)OH)underset(|)OVERSET(CH_(3))overset(|)(C)-CH_(3)-overset(H^(+))rarr CH_(3)-underset(CH_(2)^(+))underset(|)overset(CH_(3))overset(|)(C)-CH_(3) overset("Alkyl shift")rarr CH_(3)-underset(+)overset(CH_(3))overset(|)(C)-CH_(2)CH_(3) overset(Cl^(-))rarr CH_(3)-underset(Cl)underset(|)overset(CH_(3))overset(|)(Cl)-CH_(2)CH_(3)`
Discussion
2.

The reaction of NaCl and K_(2)Cr_(2)O_(7) mixture with conc. H_(2)SO_(4) result in the formation of:

Answer»

`CrO_(2)Cl_(2)`
`CrOCl_(2)`
`CrO_(2)Cl`
`(CrOCl_(2))_(2)`

SOLUTION :Red VAPOURS of chromyl `(CrO_(2)Cl_(2))` are formed
Discussion
3.

The reaction of n-hexane in presence of catalyst at 773 K and 10-20 bar pressure to form benzene is known as ………reaction.

Answer»

Hydration
Hydrogenation
Dehydration
DEHYDROGENATION

ANSWER :D
Discussion
4.

The reaction of methyl trichloroacetate (C Cl_(3)CO_(2)Me) with sodium methoxide (NaOMe) generates

Answer»

carbocation
CARBENE
carbanion
carbon radical

Solution :A carbene, i.e., DICHLOROCARBENE isgenerated ACCORDING to the FOLLOWING MECHANISM.
Discussion
5.

The reaction of methyl magnesium bromide with ethyl alcohol gives

Answer»

METHYL alcohol
methane
isopropyl alcohol
ethane

SOLUTION :`underset("Ethy ALCHOL")(C_(2)H_(5)OH) + underset("bromide")underset("Methyl magnesium")(CH_(3)MgBr)rarr underset("Methane")(CH_(4))+C_(2)H_(5)OMgBr`
Discussion
6.

The reaction of H_2O_2 with X does not liberate gaseous product. Which of the following is X ?

Answer»

`PbO_2`
`KMnO_4//H^+`
`PBS`
`Cl_2`

ANSWER :C
Discussion
7.

The reaction of HBr with ethene is an example of reaction

Answer»

Polymerisation
Substitution
Addition
Condensation

Answer :C
Discussion
8.

The reaction of Ethanol with H_(2)SO_(4) does not give

Answer»

Ethylene
Diethyl ether
Acetylene
Ethyl hydrogen sulphate

Solution :Acetylene cannpt be produced from ethanol and `H_(2)SO_(4)`.
Discussion
9.

The reaction of cynamide, NH_(2)CN(s), with dioxygen was carriedout in a bomb calorimeter, and DeltaU was found to be - 742.7 k J mol^(-1) at 298 K. Calculate the enthalpy change of the reaction at 298 K NH_(2)CN(s) + (3)/(2)O_(2)(g)rarrN_(2)(g)+CO_(2)(g) +H_(2)O(l)

Answer»

<P>

Solution :`Deltan_(g)=(n_(p) -n_(p))_(g)= 2-(3)/(2) = +(1)/(2)` mol
`DeltaH =DeltaU +Deltan_(g) RT= - 742.7 KJ mol^(-1) +(+(1)/(2)mol)(8.314xx10^(-3)kJ K^(-1) mol^(-1))(298K)`
`= - 742.7 +1.2kJ mol^(-1) = - 741.5 kJ mol^(-1)`
Discussion
10.

The reaction of cyanamlde, NH_(2) CN_((s)), with dioxygen was carried out in a bomb calorimeter, and DeltaU was found to be -742.7 "kJ mol"^(-1) at 298 K. Calculate enthalpy change for the reaction at 298 K. NH_(2) CN_((g)) + (3)/(2) O_(2(g)) to N_(2(g)) + CO_(2(g)) + H_(2) O_((l))

Answer»

SOLUTION :`Deltan_((G)) = n_(p) - n_(R)`
`=2-(3)/(2) = +(1)/(2) "mol"`
`Deltan_((g)) = Delta U + Deltan_((g)) "RT"`
`= -742.7 + (+(1)/(2) ) (8.314 xx 10^(-3) ) (298)`
`= -742.7 + 1.2`
`= -741.5 "kJ mol"`
Discussion
11.

The reaction of cyanamide, NH_2CN(s) with O_2 was carried out in a bomb calorimeter and Delta U was found to be -742.7 kJmol^-1 at 298 K. Calculate enthalpy change for the reaction at 298K.NH_2CN(s)+3/2O_2(g)rarrN_2(g)+CO_2(g)+H_2O(l)

Answer»

SOLUTION :`Delatan(G)=(1+1)-3/2=+1/2`
`DeltaU=-742.7 KJ mol^-1 R=8.314xx10^-3 kJ mol^-1k^-1DeltaH=DeltaU+DeltanRT=-742.7+1/2xx8.314xx10^-3xx298=-741.5 kJ`
Discussion
12.

The reaction of cyanamide (s), with dioxygen was carried out in a bomb calorimeter and DeltaU was found to be -742.7 kJ mol^(-1) at 298 K. Calculate the enthalpy change for the reaction at 298 K. NH_(2)CN_((g))+(3)/(2)O_(2)(g)toN_(2(g))+CO_(2(g))+H_(2)O_(l)

Answer»

SOLUTION :`-743.94 KJ "MOLE" ^(-1)`
Discussion
13.

The reaction of Cl_(2) with X gives bleaching powder . X is

Answer»

`CAO`
`Ca(OH)_(2)`
`Ca(O Cl)_(2)`
`Ca(C lO_(3))_(2)`

SOLUTION :`Caunderset(X)((OH)_(2)) + Cl_(2) to Ca(O Cl) Cl + H_(2)O` .
Discussion
14.

The reaction of (CH_(3))_(2)C(OH)CH_(2)CH_(3) with copper at 300^(@)C gives

Answer»

`CH_(3)CO(CH_(3))_(2)` and `H_(2)`
`CH_(3)-UNDERSET(CH_(3))underset(|)(C)=CH-CH_(3)` and `H_(2)O`
`(CH_(3))_(2)C = CHCH_(3)` and `H_(2)O`
`CH_(3)CH = CHCH_(2)CH_(3)` and `H_(2)`

Solution :The tertiary alcohols undergoes DEHYDRATION at `300^(@)C` to form an alkene with maximum number of `-R` groups and water.
`CH_(3)-underset(CH_(3))underset(|)overset(OH)overset(|)(C)-CH_(2)CH_(3) underset(300^(@)C)overset(Cu)rarr CH_(3)-underset(CH_(3))underset(|)(C)=CHCH_(3)+H_(2)O`
Discussion
15.

The reaction of C_6H_5CH=CHCH_3 with HBr produces

Answer»

`C_6H_5CH_2CH_2CH_2Br`<BR>
`C_6H_5undersetunderset"Br"|CHCH_2CH_3`
`C_6H_5CH_2undersetunderset"Br"|CH_2CH_3`

Solution :ADDITION of `H^+` preferentially occurs on `C_2` rather than `C_1` since it GENERATES a carbocation which is stabilized by RESONANCE over the BENZENE ring.
`C_6H_5-overset1CH=overset2CHCH_3overset(H^+)to underset"(Stabilized by resonance over the benzene ring)"(C_6H_5-overset+CH-CH_2CH_3) overset(Br^-)to underset"1-Bromo-1-phenylpropane"(C_6H_5-undersetunderset(Br)|CH-CH_2CH_3)`
Discussion
16.

The reaction of C_(6) H_(5) CH= CHCH_(3) with HBr porduces :

Answer»

`C_(6) H_(5) CH UNDERSET(Br)(CH_(2)) CH_(3)`
`C_(6)H_(5)CH_(2) underset(Br)(CHCH_(3))`
`C_(6)H_(5)CH_(2)CH_(2)CH_(2) Br`

SOLUTION :`C_(6) H_(5) CH underset(Br)(CH_(2)) CH_(3)`
Discussion
17.

The reaction of biphenyl with HOCl in the presence of a strong acid gives

Answer»




SOLUTION :`HOCltoHO^(-)+CL^(+)` ELECTROPHILIC SUBSTITUTION.
Discussion
18.

The reaction of aromatic acyl chloride and phenol in the presence of a base NaOH or pyridine is called

Answer»

Kolbe's reaction
Perkin's reaction
Sandmeyer's reaction
Schotten Baumann reaction

Solution :`C_(6)H_(5)COCl+C_(6)H_(5)OH underset("or pyridine")overset(NAOH)rarr C_(6)H_(5)COOC_(6)H_(5)+HCl`.
Discussion
19.

The reaction of aqueous KMnO_(4) with H_(2)O_(2) in acidic conditions gives

Answer»

`MN^(4+) and O_(2)`
`Mn^(2+) and O_(2)`
`Mn^(2+) and O_(3)`
`Mn^(4+) and MnO_(2)`

ANSWER :B
Discussion
20.

The reaction of aqueous KMnO_4 with H_2O_2 in acidic conditions gives :

Answer»

`MN^(4+)` and `O_2`
`Mn^(2+) ` and `O_2`
`Mn^(2+)` and `O_3`
`Mn^(4+)` and `MnO_2`

SOLUTION :`3H_2SO_4 + 2KMnO_4 + 5H_2O_2 to 5O_2 + 2MnSO_4 + 8H_2O + K_2SO_4`
Discussion
21.

The reaction of acetyleneand propyne with HgSO_4 in presence of H_2SO_4 produces respectively

Answer»

acetone and acetladehyde
ACETALDEHYDE and acetone
PROPIONALDEHYDE and acetone
acetone and propionaldehyde

SOLUTION :ACETYLENE GIVES acetaldehyde while propyne gives acetone
Discussion
22.

The reaction of acetaldehyde with HCN followed by hydroysis gives a product which exhibits

Answer»

Metamerism
Tautomerism
Enantiomerism
Geometrical isomerism

Solution :LACTIC ACID is FORMED
Discussion
23.

The reaction of

Answer»




Solution :
The intermediate carbocation I FORMED is resonance STABILISED DUE to the electron donating effect of `-OH` group at p-position. The addition of `Br^(-)` of this carbocation gives the addition product (B).
Discussion
24.

The reaction of 1-phenylpropane with limited amount of chlorine in the presence of light gives mainly .

Answer»

4-chloropropylbenzene
1-chloro-1-phenylpropane
3-chloro-1-phenylpropane
2-chloro-1-phenylpropane

Answer :B
Discussion
25.

The reaction occurs between H_2 and I at 731 K temperature equilibrium concentration of H_2, I_2 and HI are as under in Table Calculate "[HI]"/([H_2][I_2]) and [HI]^2/([H_2][I_2]) and discuss of obtainedresults and derive chemical equilibrium rule.

Answer»

Solution :
The result of A is different but B is same, so expression of B explain the law of chemical EQUILIBRIUM. At definite temperature, the multiple of active mass of product and multiple of active mass of reactant is constant.
OR
The ratio of multiple of concentration of product with PROPER POWER of MOLES of the multiple of concentration of reactant with proper power of moles is constant.
Discussion
26.

The reactionNa_(2)S+I_(2)+Na_(2)SO_(3)rarrNa_(2)S_(2)O_(3)+2NaIis called

Answer»

SPRINGS REACTION
Friedal CRAFT reaction
Blanc reduction
Solvay reaction.

Solution :Springs reaction.
Discussion
27.

The reaction M2A(g) + B (g) hArr 3 C (g) + D (g)is begun with the concentrationsof A and B both at intial value of 1.00 M. Whenequilibrium is reached , the concentrations of D is measured and is reached , the concentration of D is measured and found to be 0*25 M . The value for the equilibriumconstant for this reaction is given by the expresion

Answer»

`[(0*75)^(3)(0*25)] div [(1.00)^(2)(1.00)]`
`[(0*75)^(3) (0*25)]div [(0*50)^(2) (0*75)]`
`[(0*75)^(3) (0*25)]div [(0*50)^(2) (0*75)]`
`[(0*75)^(3) (0*25)] div [(0*75)^(2) (0*25)]`

Solution :` 2 A (G) + B (g) hArr 3 C (g) + D (g)`
`{:(,2A(g),+,B(g),hArr,3 C(g),+,D(g)),("Intial",1.0 M,,1.0 M,,0,,0),("conc.",,,,,,,),("Eqm.",1-(2xx0.25),,1-0.25,,3xx0.25,,0.25),("conc".,0.5 M,,0.75 M,,0.75 M,,):}`
` K=([C]^(3) [D])/([A]^(2) [B])= ((0.75)^(3) (0.25))/((0.5)^(2) (0.75))`
Discussion
28.

The reaction is oxidation - reduction type and takes place in presence of 50% aqueous orethanolic alkali One molecule of the aldehyde (missing of a- hydrogen atom )oxidizes, to corresponding carboxlic acid and the other is being reduced to the corresponding alcohol. Mechanism : the mechanism involves two distinct steps Nucleophilic attack and hydride ion transfer. A. The A is

Answer»

LACTIC acid
Mandelic acid
Salicyic acid
Malonic acid

Solution :
Discussion
29.

The reaction is oxidation - reduction type and takes place in presence of 50% aqueous orethanolic alkali One molecule of the aldehyde (missing of a- hydrogen atom )oxidizes, to corresponding carboxlic acid and the other is being reduced to the corresponding alcohol.Mechanism : the mechanism involves two distinct steps Nucleophilic attack and hydride ion transfer.

Answer»




SOLUTION :
Discussion
30.

The reaction is oxidation - reduction type and takes place in presence of 50% aqueous orethanolic alkali One molecule of the aldehyde (missing of a- hydrogen atom )oxidizes, to corresponding carboxlic acid and the other is being reduced to the corresponding alcohol. Mechanism : the mechanism involves two distinct steps Nucleophilic attack and hydride ion transfer.

Answer»




All of these

Solution :
Discussion
31.

The reaction is fastest when X is

Answer»

`Cl`
`NH_(2)`
`OC_(2)H_(5)`
`OCOR`

SOLUTION :Refer to COMPREHENSIVE review
Discussion
32.

The reaction is Decomposition but it's not redox reaction

Answer»

`2HgO to 2H_(2)+O_(2)`
`2H_(2)O to 2H_(2)+O_(2)`
`MgCO_(3) to MGO +CO_(2)`
`KClO_(3) to 2KCl +3O_(2)`

ANSWER :C
Discussion
33.

The reaction is CH_(2) = CH- CH_(3) + HBr rarr CH_(3) - CHBr-CH_(3)

Answer»

NUCLEOPHILIC addition
Electrophilic substitution
Electrophilic addition
Free RADICAL addition

Answer :C
Discussion
34.

The reaction is balanced if, 5H_(2)O_(2) + XClO_(2) +2OH^(-) rarr XCl^(-) YO_(2)+6H_(2)O

Answer»

`X = 5, Y = 2`
`X = 2, Y = 5`
`X = 4, Y = 10`
`X = 5, Y = 5`

Solution :`2xx{{:(ClO_(2),rarr,Cl^(-)),(+2OH^(-),,2H_(2)O),(,,+1E^(-)):}:|{:(5H_(2)O_(2),rarr,5O_(2)),(10OH^(-),,10H_(2)O),(,,10e^(-)):}}xx5`
Discussion
35.

The reaction involves in the oil of winter green test is Salicylic acid underset(conc. H_(2)SO_(4))overset(Delta)rarr Product The product is treated with Na_(2)CO_(3) solutions. The missing reagent in the above reaction is

Answer»

PHENOL
`NAOH`
Ethanol
Methanol

Solution :
Discussion
36.

The reaction involved during the removal of temporary hardness of water is

Answer»

`Ca(HCO_(3))_(2)overset(Delta)toCaCO_(3)+H_(2)O+CO_(2)uarr`
`CaCl_(2)+Na_(2)CO_(3)toCaCO_(3)darr+2NaCl`
`MgSO_(4)+Na_(2)Al_(2)Si_(2)O_(8)xH_(2)O to MgAl_(2)Si_(2)O_(8)xH_(2)O+Na_(2)SO_(4)`
`2CaCl_(2)+(NaPO_(3))_(6)toNa_(2)[Ca_(2)(PO_(3))_(6)]+4NaCl`

Answer :A
Discussion
37.

The reaction in which sodium alkoxide react with haloalkane to form ether in called …………………… .

Answer»

SOLUTION :Williamson.s SYNTHESIS
Discussion
38.

The reaction in which hydrogen poeroxide is a reducing agent is:

Answer»

`Pbs+4H_(2)O_(2)rarrPbSO_(4)+4H_(2)O`
`2KI+H_(2)O_(2)rarr2KOH+I_(2)`
`H_(2)SO_(3)+H_(2)O_(2)rarrH_(2)sO_(4)+H_(2)O`
`Ag_(2)O+H_(2)O_(2)rarr2Ag+H_(2)O+O_(2)`

SOLUTION :`H_(2)O_(2)` has reduced `Ag_(2)O` is silver (Ag)
Discussion
39.

The reaction in which hydrogen peroxide acts as a reducting agent is .

Answer»

`PBS+ 4H_(2)O_(2) rarr PbSO_(4) + 4H_(2)O`
`2KI + H_(2)O_(2) rarr 2KOH + I_(2)`
`2FeSO_(4) + H_(2)SO_(4) + H_(2)O_(2) rarr Fe_(2)(SO_(4)) _(3) + 2H_(2)O`
`Ag_(2)O + H_(2)O_(2) rarr 2Ag+ H_(2)O +O_(2)`

Answer :D
Discussion
40.

The reaction in which electrons are transferred from one reactant to another is called a ..........

Answer»

SOLUTION :REDOX REACTION
Discussion
41.

The reaction in terms of intermediates and type of reaction is given below. Mark the incorrect option.

Answer»

`CH_(3)-overset(CH_(3))overset(|)(C)=CH_(2)+HB R to` Carbocation intermediate
`CH_(3)-overset(CH_(3))overset(|)(C)=CH_(2)+HB r overset("peroxide")to`FREE radical intermediate

Solution :
of electrophilic addition reaction.
Discussion
42.

The reaction H_3PO_2+D_2OtoH_2DPO_2+HDO indicates that hypo-phosphorus acid is

Answer»

TRIBASIC acid
dibasic acid
mono BASIC acid
none of these

Answer :C
Discussion
43.

The reaction H_2O_2toH_2O+O_2 represents,

Answer»

OXIDISING of `H_2O_2`
REDUCING of `H_2O`
dispropornation of oxygen
acidic NATURE of `H_2O_2`

ANSWER :C
Discussion
44.

The reaction H_(2(g))+I_(2(g)) harr 2HI_((g)) is carried out in a 1 litre flask.If the same reaction is earned out in a 2 litre flask at the same temperature, the equilibrium constant will be

Answer»

same
doubled
halved
decreased

Solution :Equlibrium CONSTANT (`K_(eq)`) is INDEPENDENT on VOLUME of vessel
Discussion
45.

The reaction, H_(2(g)) + 1/2O_(2(g)) to H_2O_((l)), is spontaneous. The DeltaS^(@) = - 163.1 J mol^(-1) K^(-1). The absolute entropies of H_(2(g)) and O_(2(g)) are 130.6 JK^(-1)mol^(-1) and 205 JK^(-1) mol^(-1) respectively. Calculate the absolute entropy of water.

Answer»

Solution :CHANGE in STANDARD entropy,
`DeltaS^(@) = S_(H_2O_(l))^(@) - [S_(H_2)^(@) + 1/2S_(O_2)^(@)]`
Absolute entropy of WATER = `69.9 JK-mol^(-1).`
Discussion
46.

The reaction for which K_(C) lt K_(P) at a given temperature ( gt 50 k)is/are:

Answer»

`PCI_5(G)hArrPCI_3(g)+CI_2(g)`
`CO_2(g)+C(s)hArr2CO(g)`
`CAO(s)+CO_2(g)hArrCaCO_3(s)`
`I_2(g)hArr2I(g)`

ANSWER :A::B::D
Discussion
47.

The reaction condtions leading to best yields of C_2H_5CI are

Answer»

`C_6H_6(EXCESS)+CI_2overset(UV" "light)to`
`C_6H_6+CI_2underset("ROOM temp.")overset("DARK")to`
`C_2H_6+CI_2("excess")overset("UV light")to`
`C_2H_6+CI_2overset("UV light")to`

ANSWER :A
Discussion
48.

The reaction conditions used for converting 1,2-dibromoethane to ethylene are

Answer»

Zn, ALCOHOL, `Delta`
KOH, alcohol, `Delta`
KOH, WATER, `Delta`
NA, alcohol, `Delta`

Answer :A
Discussion
49.

The reaction, CO_((g)) + 3H_(2(g)) hArr CH_(4(g)) + H_2O_((g))is at equilibrium at 1300K in a 1L flask. It also contain 0.30 mol of CO, 0.10 mol of H_2 and 0.02 mol of H_2O and an unknown amount of CH_4 in the flask. Determine the concentration of CH_4 in the mixture. The equilibrium constant, K_c for the reaction at the given temperature is 3.90.

Answer»

Solution :`{:("EQUILIBRIUM reaction :",CO_((g))+,3H_(2(g)) HARR , CH_(4(g))+ , H_2O_((g))),("Mol at equilibrium :",0.30,0.10,x ,0.02),("Volume = 1L So, mol L"^(-1):,0.30,0.10,x,0.02):}`
`K_c`=3.90, where, x=Concentration of `CH_4` equilibrium
According to chemical equilibrium constant,
`K_c=([CH_4][H_2O])/([CO][H_2]^3)`
`therefore 3.90=((x)(0.02))/((0.30)(0.10)^3)`
`therefore x=((3.9)xx(0.30)(1xx10^(-3)))/0.02`
`=58.5xx10^(-3)`
`=5.85xx10^(-2) "mol L"^(-1)`
Discussion
50.

The reaction , CO(g) + 3 H_(2) (g) hArr CH_(4) (g) + H_(2)O (g),is at equilibrium at 1300 K in a 1 L flask. It also contains 0*30 mol of CO, 0*10 mol of H_(2) and 0*02 " mol of " H_(2)O and an unknown amount of CH_(4) " in the flask. Determine the concentration of " CH_(4)in the mixture. The equilibrium constant, K_(c), for the reaction at the given temperature is 3*90.

Answer»

Solution :` K_(c) = ([ CH_(4)][H_(2)O])/([CO][H_(2)]^(3))`
` :. 3* 90 = ([CH_(4)][H_(2)O])/([CO][H_(2)]^(3)) "" ("Molar conc = No. of moles because volume of flask = 1 L")`
`[CH_(4)] = 0*0585 M = 5*85 xx 10^(-2) M`
Discussion