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.

Additionof water to 3-methylbut-1-ene in the presence of dil. H_(2)SO_(4) gives mainly

Answer»

3-Methylbutan-1-ol
3-Methylbutan-2-ol
2-Methylbutan-1-ol
2-Methylbutan-2-ol

Solution :`UNDERSET("3-Methylbut-1-ene")(CH_(3)-underset(CH_(3))underset(|)(CH)-CH=CH_(2))overset(H^(+))RARR`
`CH_(3)-underset((2^(@)))(underset(CH_(3))underset(|)(CH))-overset(+)(CH)=CH_(3)overset("Hydride SHIFT")rarr`
`CH_(3)-underset(3^(@))underset(CH_(3))underset(|)overset(+)(C)-CH_(2)-CH_(3)overset(OH^(-))rarrunderset("2-Methylbutan-2-ol")(CH_(3)-underset(CH_(3))underset(|)overset(OH)overset(|)(C)-CH_(2)-CH_(3))`
Discussion
2.

Addition of surface to water leads .

Answer»

An increase in ANGLE of contact
An increase in surface tension
A decrease in angle of contact and increase in surface tension .
A decrease in angle of contact and increase in surface tension .

Solution :Addition of SOAP redues surface tension of system.
Discussion
3.

Addition of sulphuric acid in alkene and addition of sulphuric acid in alkene with addition reaction.

Answer»

SOLUTION :Dissociation of sulphuric acid `(H_(2)SO_(4) hArr H^(DELTA+)O^(delta-)SO_(3)H)` fragement are `H^(delta+)` and `HSO_(4)^(-)` or `""^(-)OSO_(3)H` positive and negative radicals on `pi`-bond as per Markovnikov rule electrophilic ADDITION reaction.
(i) `H_(2)SO_(4) rarr overset(+)(H)+bar(O)SO_(2)OH`

(ii) Propene is reacted with`H_(2)SO_(4)` where `overset(+)(H)` is electrophilic species. More stable carbocation `CH_(3)overset(+)(C)HCH_(3)` is formed first, negative part hydrogen sulfate, is added as addition product and formed ISOPROPYL hydrogen sulphare.

This reaction occurs at normal temperature and ractants are added in ratio 1:1 and formation of alkyl hydrogen sulpare.
Discussion
4.

Addition of sodium hydroxide solution to a weak acid (HA) results in a buffer of pH 6 . If ionisation constant of HA is 10^(-5) , the ratio of salt to acid concentration in the buffer solution will be

Answer»

`10 : 1`
`4 : 5`
`1 : 10`
`5 : 4`

ANSWER :A
Discussion
5.

Addition of phposphate and nitrate/feticlizers in to water leads to

Answer»

Increased GROWTH of decomposers
REDUCED algal growth
Increased algal growth
Niutrient enrichment (eutrophication)

Solution :addition of phosphate and nitratefertilizers into water LEADS to NUTRIENT enrichment or eutrophication
Discussion
6.

Addition of ozone to acetylene gives a product which is

Answer»

monocyclic
bicyclic
TRICYCLIC
tetracyclic

Answer :B
Discussion
7.

Addition of phosphate fertilizers into water leads to

Answer»

Increased growth of DECOMPOSERS
REDUCED ALGAL growth
Increased algal growth
Nutrient ENRICHMENT (eutrophication)

ANSWER :D
Discussion
8.

Addition of mineral acid to an aqueous solution of borax, the following compound is formed :

Answer»

boron hydride
pyroboric acid
metaboric acid
orthoboric acid

Solution :Orthoboric acid is formed.
`Na_(2)B_(4)O_(7) +H_(2)SO_(4) rarr underset(" Orthoboric acid ") (4H_(3)BO_(3)) + Na_(2)SO_(4)`
Discussion
9.

Addition of HX on alkene proceed through formation of carbocation . This reaction is follows Markovnikov's rule. According to markovnikov's rule addition of electrophile occurs on that carbon of alkene which have more number of -H atom Ph-CH = CH_2 underset(HX)to underset("(major)")(Ph-overset(x)overset(|)CH-CH_3) + underset("(minor)")(Ph-CH_2-CH_2-X) Which of the following alkenes will give markovnikov's reaction ?

Answer»

`F_3 C - CH =CH_2 OVERSET(HCL)to`
`H_3N^(+)-CH = CH_2 overset(HCl)to `
`O_2N - CH = CH_2 overset(HBr)to`

Solution :A) Because `F_3C- overset(OPLUS)(C)H-CH_3` is unstable due to powerful - I effect of `CF_3` GROUP
B) `H_3^(oplus)N - CH - CH_2 overset(HCl)to H_3overset(oplus)N -CH_2 - Cl`.
Because `NH_3 -CH-CH_3` unstable due to powerful - I effect of `NH_3` group
C) `NO_2 - CH - CH_2 NO_2 - CH_2 - Br `
Because `NO_2 - overset(oplus)CH_2 - CH_3` is unstabledue to -I effect to `NO_2` group
D)
is more STABLE I effect `-CH_3` groups and .9. hypen conjugative structures
Discussion
10.

Addition of HX on alkene proceed through formation of carbocation . This reaction is follows Markovnikov's rule. According to markovnikov's rule addition of electrophile occurs on that carbon of alkene which have more number of -H atom Ph-CH = CH_2 underset(HX)to underset("(major)")(Ph-overset(x)overset(|)CH-CH_3) + underset("(minor)")(Ph-CH_2-CH_2-X) Which of the following alkenes can produce diastereomers

Answer»

`CH_3-CH=CH-CH_3`


`CH_3-CH=CH_2`

Solution :Optically inactive ALKENE does not PRODUCES DIASTEREOMERS on addition of HX Optically active ALKENES produce diastereomers if a chiral carbon in generated uring addition
A) Is not optically active
B) is not optically active
C) Is optically active D) Is not optically active
Discussion
11.

Addition of HOCl to allyl alcohol gives

Answer»

2-Chloropropane-1, 3-diol
3-Chloropropane-1, 2-diol
2, 3-Dichloropropanol
1, 2,3-Trichloropropane

Solution :
Discussion
12.

Addition of HBr to propene yields 2-bromopropane, while inpresence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism.

Answer»

SOLUTION :Addition of HBr to propene is an ionic electrophilic addition REACTION in which the electrophile, i.e., `H^+` first ADDS to give a more stable `2^@` carbocation. In the 2nd second, the carbocation is rapidly attacked by the nucleophile `Br^-` ION to give 2-bromopropane

In presence of benzoyl peroxide , the reaction is still electrophilic but the electrophile here is a `oversetdot(Br)` free radical which is obtained by the action of benzoyl peroxide on HBr

In the first step, `oversetdot(Br)` radical adds to propene in such a way as to generate a more stable `2^@` radical . In the second step, the free radical thus obtained rapidly abstracts a hydrogen atom from HBr to give 1-bromopropane.

it is evident that although both reactions (i.e., in presence or absence of benzoyl peroxide ) are electrophilic addition but it is due to different order or sequence of addition of H and Br atoms which gives different PRODUCTS.
Discussion
13.

Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism.

Answer»

SOLUTION :ADDITION to UNSYMMETRICAL ALKENES.
Discussion
14.

Addition of HBr to propene yields 2-bromo-propene, while in the presence of benzoyl peroxide, the same reaction yields 1-bromo-propane. Explain and give mechanism.

Answer»

SOLUTION :(a) Addition to HBR to propene, yields 2-bromo-propane and it is electrophilic addition reaction.
(B) Addition to Hbr to propene in PRESENCE of peroxide, forms 1-bromopropane, it is free-radical addition reaction.
Discussion
15.

Addition of HBr to propene gives 2 - bromo propane, this reaction is intiated by

Answer»

`H^(o+)`<BR>`Br^(-)`
`Br`
`Br^(o+)`

ANSWER :A
Discussion
16.

Addition of HBr to 2-butene gives two products while that of HBr to 1-buane gives three products . Explain why ?

Answer»

SOLUTION :Addition of HBr to 2-butene gives 2-bromobutane which being a chiral molecule EXISTS as two STEREOISOMERS ( I and II) .<BR>
In contrast, 1-butene is an unsymmetrical molecule, therefore , it gives two products, i.e., 2-bromobutane (major product ) and 1-bromobutane (minor product). Since 2-bromobutane is a chiral molecule, it exists as two stereoisomers (I+II) as shown below above . While 1-bromobutane being a achiral molecule exists as such (III).
`underset"1-Butene"(CH_3-CH_2-CH=CH_2) overset"HBr"to underset"I+II"(CH_3-CH_2-undersetunderset(Br)|overset**CH-CH_3)+underset"III"(CH_3CH_2CH_2CH_2-Br)`
Thus, in all three products (I, II and III ) are FORMED.
Discussion
17.

Addition of HBr to 2-pentene gives

Answer»

2-bromopentane only
3-bromopentane only
2-bromopentane and 3-bromopentane
1-bromopentane and 3-bromopentane

Solution :Addition of a proton to 2-pentene can give two CARBOCATIONS (I and II)

which are almost equally stable since CARBOCATION (I) is stabilized by five (3+2) HYPERCONJUGATION structures and carbocation (II) is stabilized by FOUR (2+2) hyperconjugation structures . Therefore , nucleophilic ATTACK on these gives the coresponding bromoalkanes in almost equal amounts . Of course , the yield of 2-bromopentane will be slightly more than that of 3-bromopentane.
Discussion
18.

Addition of H_(2)O into alkene.

Answer»

SOLUTION :`H_(2)O` means `overset(+delta)(H) overset(-delta)(OH)`. Alkene reacts with `H_(2)SO_(4)`, an addition of `H_(2)O, -OH` (alcohol) is formed.
This reaction is addition of `H_(2)O` or dydrolysis of alkene or FORMATION of alocohol from alkene. This reaction is electrophilic addition reaction. markonikov rule is applied.
`underset("Ethene")(CH_(2)=CH_(2))+H_(2)O underset((H_(2)SO_(4)))overset(100^(@)C H^(+), Delta)rarr underset("ETHANOL")(CH_(3)-CH_(2)OH)`
`underset("PROPENE")(CH_(3)CH=CH_(2))+H_(2)O underset(H_(2)SO_(4))overset(373 K H^(+))rarr underset("Propane-2-ol")(CH_(3)-underset(OH)underset(|)(CH)-CH_(3))`
`underset("2-Methyl prepene")(CH_(3)-underset(CH_(3))underset(|)(C)=CH_(2)+H_(2)O)underset(H_(2)SO_(4))overset(H^(+))rarr underset("2-Methyl prepane-2-ol")(CH_(3)-underset(CH_(3))underset(|)overset(OH)overset(|)(C)-CH_(3))`
Discussion
19.

Addition of H_2 to 2-butyne in presence of Lindlar's catalyst . Gives cis-2-butene while with Na/in liq. NH_3 gives trans-2-butene.Why so ? Explain.

Answer»

Solution :In presence of Lindlar's catalyst (Pd deposited over `BaSO_4` and partially POISONED by addition of S or quinoline), `H_2` gets first absorbed over the catalyst as H atoms and then transferred to the same face of 2-butyne giving cis-2-butene.
HOWEVER, in CASE of Na in liquid `NH_3`, (Birch REDUCTION), the reduction OCCURS through two electron transfers and two proton transfer giving trans-2-butene .
Discussion
20.

Addition of catalyst in a system

Answer»

INCREASES equilibrium CONCENTRATIONS
No EFFECT on equilibrium concentrations
Decreases equilibrium concentrations
Increases rate of forward REACTION and decreases rate of BACKWARD reaction.

Solution :Catalyst does not effect on equilibrium concentrations.
Discussion
21.

Addition of bromine to cis-3 hexene gives

Answer»

RACEMIC dibromide
a MIXTURE of DIASTEREOMERIC dibromides
optically ACTIVE dibromide
meso dibromide

Answer :A
Discussion
22.

Addition of bromine to buta-1,3- diene gives

Answer»

1,2- ADDITION product only
THREE carbon atoms
1,4-addition product only
SIX carbon atoms

SOLUTION :
Discussion
23.

Addition of BrC CI_(3) to propene in the presence of peroxides gives 3-bromo-1, 1, 1-trichloro-2-methylpropane.

Answer»

SOLUTION :False: `CH_(3)CH=CH_(2) underset(ROOR)overset(BrC Cl_(3)) underset("2-Bromo-1, 1, 1-trichlorobutane")(CH_(3)underset(BR)underset(|)CH-CH_(2)C Cl_(3))`
Discussion
24.

Addition of Br, takes place redily with

Answer»

`CH_(2)=CH_(2)`
`CH_(3)-CH=CH_2`
`CH_(3)-CH=CH-CH_(3)`
`CH_(3)-underset(CH_3)underset(|)(C) = CH -CH_3`

ANSWER :D
Discussion
25.

Addition of I mole of N_(2) is made to an equilibrium mixture of PCl_(5) in a piston fittcd cylinder. Which of the following is correct.

Answer»

No effect on equilibrium constant `K_c` or `K_p`
The degree of dissociation of `PCI_5` decreases
The VOLUME of CONTAINER does not change
No effect on equilibrium concentration

Solution :INERT gas does not AFFECT `K_(p)` or `K_(C)` volume changes equilibrium shifts
Discussion
26.

Addition of 1 equivalent of HCL to CH_(2)=C=CH_(2) gives

Answer»




ANSWER :B
Discussion
27.

Add : (i) 93.4 + 4.03 (ii) 77.86, 14.12 and 5.7

Answer»


Solution :(i) Add : `{:(93.4),(4.03),(bar(97.43)):}`
The CORRECT ANSWER is 97.4
(Digit 3 is to be DELETED)
(ii) Add : `{:(77.86),(14.12),(" 5.7"),(bar(97.68)):}`
The correct answer is 97.7
(Digit 6 has been ROUNDED off to 7)
Discussion
28.

Add 6.65xx10^(4) and 8.95xx10^(3).

Answer»

SOLUTION :Exponent is MADE same for the both NUMBERS
`6.65 XX 10^(4) +0.895 xx 10^(4)=7.545 xx 10^(4)`
Discussion
29.

Add 18.11, 12.0 and 2.021 and report the result in significant figures.

Answer»

SOLUTION :18.11+12.0+2.021 = 32.131
Here, 12.0 has ONE digit after the decimal POINT and the result should be REPORTED only up to one digit after decimal point. Hence round off the result to 32.1
NUMBER of significant figures is 3
Discussion
30.

Adam's catalyst is

Answer»

Platinum
Iron
Molybdenum
Nickel

Solution :GENERALLY TRANSITION elements acts as CATALYSTS. Adam's catalyst is ANOTHER name of platinum.
Discussion
31.

Adamantive is the crystalline form of

Answer»

Aluminium
BORON
Thallium
Beryllium

Solution :ADAMANTIVE is the CRYSTALLINE FORM of boron.
Discussion
32.

Acyl halide is formed by reacting PCl_(5) with

Answer»

Alcohol
Ester
Amide
Both carboxylic ACIDS as well as esters

Solution :The answer are based upon the REACTIONS GIVEN in Comprehensve REVIEW.
Discussion
33.

Active species involved in the process C_6H_6 rarr C_6 H_5SO_3H is

Answer»

`SO_3`
`HSO_4^(-)`
`SO_3^(+)`
`SO_2`

ANSWER :A
Discussion
34.

Active mass of 5.6 lit N_2 at STP

Answer»

22.4M
0.25M
`1/(22.4)M`
4M

Solution :ACTIVE mass = `(N)/(V//"(LIT)")=((5.6)/(22.4))/(5.6)=(1)/(22.4)M`
Discussion
35.

Active mass of 0.64g SO_(2) in 10 lit vessel is

Answer»

`10^(-2)` M
`10^(-3)` M
`10^(-1)` M
0.64g

Solution :Active mass = `(n)/(V//"(LIT)")`
Discussion
36.

Action of water or dilute mineral acids on metals can give.....

Answer»

MONO hydrogen
Tritium
Di hydrogen
Trihydrogen

Answer :C
Discussion
37.

Action of nitrous acid on ethylamine gives

Answer»

`C_(2)H_(6)`
`C_(2)H_(5)OH`
`C_(2)H_(5)OH` and `C_(2)H_(4)`
`C_(2)H_(5)OH` and `NH_(3)`

Solution :`C_(2)H_(5)-NH_(2)OVERSET(HNO_(2))rarrC_(2)H_(5)OH+N_(2)+H_(2)O`
Discussion
38.

Action of caustic acid soda on aluminium hydroxide gives a compound having formula

Answer»

`Al_(2)(OH)_(4)`
`Na_(2)Al(OH)_(4)`
`NaAlO_(2)`
`Na_(3)AlO_(3)`

SOLUTION :`underset("CAUSTIC soda")(NAOH) + Al(OH)_(3) rarr underset("Sod. Metaaluminate")(NaAlO_(2)+2H_(2)O)`
Discussion
39.

Action of acetylene on dilute H_(2)SO_(4)gives

Answer»

ACETIC acid
Acetone
ACETALDEHYDE
ETHYL alcohol

Answer :C
Discussion
40.

Action of a heterogeneous catalyst depends upon

Answer»

mass
solubility
PARTICLE SIZE
none.

Solution :ACTION of heterogeneous CATALYST DEPENDS upon particle size.
Discussion
41.

Across theperiod , Ionisation energy _________

Answer»

increases
decreases
does not very
First decreases and then increases

Solution :ACROSS the PERIOD , IONISATION ENERGY is increases and down the group , Ionisation energy is and down the group ,Ionisation energy is decreases.
Discussion
42.

Acidity or alkalinity of a solution depend upon the concentration of hydrogen ion relative to that of hydroxyl ions. The product of hydrogen ion & hydroxyl ion concentration is given by K_(w) = [H^(+)][OH^(-)] the valueof which depends only on the temperature & not on the individual ionic concentration. If the concentratoin of hydrogen ions exceeds that of the hydroxyl ions, the solution is said to be acidic, whereas, if concentrations of hydroxyl ion exceeds that of the hydrogen ions, the solution is said to be alkaline. The pH corresponding to the acidic and alkaline solutions at 25^(@)C will be less than and greter than seven, respectively. To confirm the above facts 0.5 M CH_(3)COOH is taken fro the experiments. [Given : K_(a) of acetic acid = 1.8 xx 10^(-5)] To what volume at 25^(@)C must 1 dm^(3) of this solution be diluted in order to double the pH

Answer»

`3.37 XX 10^(4) DM^(3)`
`1 xx 10^(3) dm^(3)`
`1.68 xx 10^(4) dm^(3)`
`3.18 xx 10^(3) dm^(3)`

Solution :`M_(1)V_(1) = M_(2)V_(2)`
Amount of water to be added `= V_(2) - V_(1)`
Discussion
43.

Acidity or alkalinity of a solution depend upon the concentration of hydrogen ion relative to that of hydroxyl ions. The product of hydrogen ion & hydroxyl ion concentration is given by K_(w) = [H^(+)][OH^(-)] the valueof which depends only on the temperature & not on the individual ionic concentration. If the concentratoin of hydrogen ions exceeds that of the hydroxyl ions, the solution is said to be acidic, whereas, if concentrations of hydroxyl ion exceeds that of the hydrogen ions, the solution is said to be alkaline. The pH corresponding to the acidic and alkaline solutions at 25^(@)C will be less than and greter than seven, respectively. To confirm the above facts 0.5 M CH_(3)COOH is taken fro the experiments. [Given : K_(a) of acetic acid = 1.8 xx 10^(-5)] pH of the solution will be:-

Answer»

`2.52`
`2.22`
`5`
`3.92`

Solution :`[H^(+)] = SQRT((k_(a)c))`
`k_(a) = 1.8 XX 10^(-5)`
`c = 5 xx 10^(-1) M`
`:. [H^(+)] = sqrt(1.8 xx 10^(-5) xx 5 xx 10^(-2)) = sqrt(9 xx 10^(-6)) = 3 xx 10^(-3)`
Discussion
44.

Acidity of phenol was explained by .....

Answer»

I EFFECT
E-effect
R-effect
HYPER CONJUGATING effect

SOLUTION :R-effect
Discussion
45.

Acidity of phenol was explained by .............

Answer»

SOLUTION :R -EFFECT
Discussion
46.

Acidity of BF_3 can be explained on the basis of which of the following concepts ?

Answer»

Arrhenius concept
Bronsted Lowry concept
Lewis concept
Bronsted Lowry as well as Lewis concept.

Solution :Lewis in 1923 DEFINED an acid as is the substance which accepts an electron PAIR and BASE which donates an electron pair. As `BF_3` is an electron DEFICIENT COMPOUND, hence, it is a Lewis acid.
Discussion
47.

Acidity of BF_(3) can be explained on the basis of which of the following concepts?

Answer»

ARRHENIUS concept
Bronsted Lowry concept
Lewis concept
Bronsted Lowry as WELL as Lewis concept.

Solution :`BF_(3)` is electron deficient compound and HENCE is a Lawis ACID.
Discussion
48.

Acidity of acetylene, water and ammonia decreases in the order _______

Answer»


ANSWER :WATER , ACETYLENE , AMMONIA
Discussion
49.

Acidified solution of chromic acid on treatment with H_2O_2 yields.

Answer»

`CrO_3+H_2O+O_2`
`Cr_2O_3+H_2O+O_2`
`CrO_5+H_2O`
`H_2O_7+H_2O+O_2`

ANSWER :C
Discussion
50.

Acidified KMnO_(4) oxidizes acid to CO_(2). What is the volume ( in litre) of 10^(-4) M KMnO_(4) required to completely oxidize 0.5 litre of 10^(-2) M oxalic acid in acid medium ?

Answer»

125
1250
200
20

Solution :`2KMnO_(4)+3H_(2)SO_(4)+5{:(COOH,),(|"",),(COOH,):}toK_(2)SO_(4)+2MnSO_(4)+10CO_(2)+8H_(2)O`
`(M_(1)V_(1))/(n_(1))(KMnO_(4))=(M_(2)V_(2))/(n_(2)){:((COOH,),(|"",),(COOH,)):}`
`(10^(-4)xxV_(1))/(2)=(10^(-2)xx0.5)/(5)`
`V_(1)=20 L`
Discussion