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.

Although chlorine is an electron withdrawing group, yet it is ortho-para directing in electrophilic aromatic substitution reactions, why ?

Answer»

SOLUTION :THISIS explainedon thebasisof resonance
Dueto resonanceelectrondensityis moreat -o- andp-positionsthereforeelectrophilicsubsitutionoccursat o- and -p-positions
Discussion
2.

Although chlorine is an electron withdrawing group, yet it is ortho-, para- directing in electrophilic aromatic substitution reactions. Why?

Answer»

Solution :Chlorine WITHDRAWS electrons through inductive effect and releases electrons through resonance. Through inductive effect, chlorine destabilises the intermediate carbocation formed during the ELECTROPHILIC substitution.

Through resonance, halogen tends to stabilise the carbocation and the effect is more pronounced at ortho- and para- positions. The inductive effect is STRONGER than resonance and causes net electron withdrawal and thus causes net deactivation. The resonance effect tends to oppose the inductive effect for the attack at ortho- and PARAPOSITIONS and hence makes the deactivation less for ortho- and paraattack. Reactivity is thus CONTROLLED by the stronger inductive effect and orientation is controlled by resonance effect.
Discussion
3.

Although carbon and hydrogen are better reducing agents, by they are not used to reduce metallic oxides at high temperature.Why?

Answer»

Solution :This is because CARBON and HYDROGEN REACT with METALS to form carbides and HYDRIDES respectively at high temperature.
Discussion
4.

Although carbon and hydrogen are better reducing agents but they are not used to reduce metallic oxides at high temperatures. Why?

Answer»

Solution :This is due to formation of metal carbides and metal HYDRIDES at high temperatures.
`CaO + 3C OVERSET(2273K)(rarr) + CaC_2 + CO`
`Ca + H_2 to CaH_2`
Discussion
5.

Althoughcarbonand hydrogenare betterreducingagentsbuttheyare notused toreducemetallicoxidesat high temperatures.Why ?

Answer»

SOLUTION :Thereasonbeingthat ahightemperatures,metalsreactwith carbonandhydrogento formtheircarbidesandhydridesrespectively. Forexample,
` CaO(s)+3 C (s) overset ( 2273 K) toCa C _2(s)+CO(g) `
`CA(s)+H_2(g) overset (" Heat")toCaH _2(s) `
Discussion
6.

Although both polymers are prepared by free radical processes, poly (vinyl chloride) is amorphous and poly (vinylidene chloride) (saran) is highly crystalline. How do you account for the different? (vinylidene chloride is 1,1-dichloroethene).

Answer»

Solution :
As poly (vinyl chloride) is ABLE to show stereoisomerism and further it is formed by a free radical PROCESS, it is atactic (chlorine atoms DISTRIBUTED randomly), the molecules fit together poorly.
Poly (vinylidene chloride) has two identical substituents on each and the chains fit together well.
Discussion
7.

Although both allyl alccohol and 1-propanol are primary alcohols, they can still be distinguished by Lucas reagent. Explain how?

Answer»

Solution :The ally carbocation is resonannce stabilzied and hence is as stable as a `3^(@)` carbocation.

Therefore, on TREATEMENT with Lucas reagent, allyl alcohol gives white turbidity in LESS than 1 min. 1-Propanol, on the other hand, does not react with Lucas reagent.
`underset("Allyl alcohol")(CH_(2)=CH-CH_(2)OH) underset(("Lucas reagent"))overset(HCl+ZnCl_(2))to underset("Turbidity in " lt " 1min".)(CH_(2)=CH-CH_(2)CL)underset("1-Propanol")(CH_(3)CH_(2)CH_(2)OH) underset(("Lucas reagent"))overset(HCl+ZnCl_(2))to`No action.
Discussion
8.

Although amino group is o- and p-directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of mnitroaniline. Give reason.

Answer»

Solution :Thisis because in thestronglyacidicconditionsof NITRATION, MOSTOF theanilineof converted intoanilniumion and `""^(+)NH_3` WHICHIS m - DIRECTING
Discussion
9.

Although amino group is o-and p-directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline . Give reason.

Answer»

Solution :This is because in the STRONGLY ACIDIC conditions of NITRATION, most of the aniline is converted into ANILINIUM ion and `""^(+)NH_3` which is m-directing.
Discussion
10.

Although Aluminium is above hydrogen in the electrochemical series, it is stable in air and water. Explain.

Answer»

Solution :DUE to the FORMATION of PROTECTIVE OXIDE layer on its surface `(Al_(2)O_(3))`.
Discussion
11.

Although aldehydes are easily oxidisable to the corresponding carboxylic acids yet propanal can conveniently be prepared by the oxidation of propanol by acid potassium dichromate.

Answer»

Solution :Aldehydes can be obtained in quantitative YIELD by the oxidation of primary alcohols provided these are REMOVED from the reaction mixture as soon as they are formed to prevent their further oxidation to carboxylic acids.
Since alehydes having boilng points less than 373K (b.p. of `H_(2)O`)can be easily removed by distillation, THEREFORE, propanal (b.p. 323K) can be easily prepared from propanonl-1 by distilling it from the alcohol-acid dichromate solution.
Discussion
12.

Although AgCl is insoluble in water, it readily dissolves upon the addition of ammoniaAgCl(s) + 2NH_3(aq) iff Ag(NH_3)_2^(-) (aq)+ Cl^(-)(aq)(a) What is the equilibrium constant for this dissolving process?(b) Ammonia is added to a solution containing excess AgCl (s). The final volume is 1 litre and the resulting equilibriuin concentration of NH, is 0.80 M. Calculate the number of moles of AgCl dissolved, the molar concentration of Ag(NH_3)_2and the number of moles of NH, added to the original solution. K_(sp) (AgCl) = 1.8 xx 10^(-10) and K_f [Ag(NH_3)_2^+]= 1.7 xx 10^7 .

Answer»

Solution :`(a) 3.1 XX 10^(-3) ` (b) 0.045 mole, 0.045 mole, 0.89 mole
Discussion
13.

Although aldehydes are easily oxidisable, propanal can conveniently be prepared by oxidation of propanol by acidified potassium dichromate. Why ?

Answer»

SOLUTION :The aldehydes can beconveniently oxidised to their RESPECTIVE acids only if they are not removed from the reaction MIXTURE during preparation . As the boiling point of propanal is low (323K) , it is conveniently distilled out during the reation . This AVOIDS its further oxidation.
Discussion
14.

Although (+3) oxidation states is the characteristic oxidation state of lanthanoids but cerium shows (+4) oxidation state also. Why?

Answer»

Solution :In (+4) OXIDATION state, a noble gas configuration `[Xe] 4f^(0) 5d^(0) 6S^(0)` is attained
Discussion
15.

Although +3 oxidation state is the characteristic oxidation state of lanthanoids but cerium shows+4 oxidation state also. Why ?

Answer»

Solution :`._(58) Ce= [Xe]^(54) 4f^(2) 5d^(0) 6s^(2) . Ce^(3+) = [Xe]^(54) 4f^(1) `. Further lossof ONE electron GIVES stable `4f^(0)` configuration. Hence, it SHOWS `+4` oxidation state ALSO.
Discussion
16.

Although +3 oxidation state is the characteristic oxidation state of lanthanoids but cerium shows +4 oxidation state also. Why?

Answer»

SOLUTION :The electronic CONFIGURATION of Ce is `-4f^(1)5d^(1)6s^(2)`. Usually `5d^(1)` and `6s^(2)` electrons are lost by the lanthanoids in their REACTIONS i.e., they exhibit +3 oxidation STATES. But Ce exhibit +4 oxidation state also because it gains extra stability by losing `4f^(1)` electron because it will give rise to completely filled ORBITALS.
Discussion
17.

Alternating copolymers are commonest copolymers. Comment

Answer»

Solution :Alternating copolymers are usually produced when TWO monomers which differ in their reactive groups undergo CONDENSATION. A dicarboxylic acid reacts with a diol and the CHAIN is propagated to give a polyester like polyethylene terephthalate (PET).
PET also called dacron or terylene is an alternating copolymer. Nylon 6,6 is ANOTHER FAMILIAR condensation alternating copolymer. Buna type rubbers, on the other hand, are addition alternating copolymers.
Discussion
18.

Although +3 is the characteristic oxidation state for langthanoids but certain also shows +4 oxidation state because "…...................."

Answer»


Solution :`._(58) Ce=[Xe]^(54) 4f^(2) 5d^(0)6S^(2) `. Hence, `Ce^(4+) = [Xe]^(54)4f^(0) `. Hence, (B) and (C ) are TRUE.
Discussion
19.

Although (+3) is the characteristic oxidation state for lanthanoids but cerium also shows (+4) oxidation state because…

Answer»

It has variable ionisation enthalpy
It has a tendency to ATTAIN NOBLE gas configuration
It has a tendency to attain `f^(0)` configuration
It resembles `Pb^(4+)`

Solution :Ce shows (+4) oxidation STATE because in (+4) oxidation state, Ce attains `f^(0)` configuration i.e., noble gas configuration
Discussion
20.

Name of the protein which is responsible for gout is

Answer»

RIBOSE unit
nitrogenous base
phosphate
none of the above

Answer :D
Discussion
21.

Also explain the behaviour of these ions in magnetic field.

Answer»

Solution :`Ti^(+4)` and `Ag^(+)` are DIAMAGNETIC in NATURE. So they will be REPELLED by the magnetic field. `Co^(2+)` being PARAMAGNETIC in nature is attracted by the magnetic field.
Discussion
22.

alpha-particles can be detected using

Answer»

THIN aluminum
Barium sulphate
Zinc sulphide screen
Gold foil

Solution :Rutherford FIRST of all used zinc sulphate (ZnS) as phosphor in the DETECTION of `ALPHA`-particles
Discussion
23.

Alpha particles are... Times heavier (approximately) than neutrons

Answer»

2
4
3
`2 (1)/(2)`

Solution :`alpha`-particles are 4 times HEAVIER than NEUTRONS
Discussion
24.

Alpha rays consist of a stream of

Answer»

`H^(+)`
`He^(+2)`
Only electrons
Only neutrons

Solution :`alpha`-rays consist of a STREAM of `He^(2+)`
Discussion
25.

alpha - particles can be detected using

Answer»

Thin ALUMINUM SHEET
Barium sulphate
ZINC sulphide screen
GOLD foil

ANSWER :C
Discussion
26.

alpha -nitroalkene is converted into nitroalkane by

Answer»

OXIDATION
TREATING it with `KNO_(2)` and followed by hydrolysis
acid hydrolysis
alkaline hydrolysis

Answer :C
Discussion
27.

Alpha particle is….times heavier than neutron:

Answer»

2
4
3
25

Answer :B
Discussion
28.

alpha-maltose can be hydrolysed to glucose according to the following reaction: a - C_(12)H_(22)O_(11)(aq) + H_(2)O(l)to 2C_(6)H_(12)O_(6)(aq). Given standard enthalpy of formation of H_(2)O(l), C_(6)H_(12)O_(6)(aq) C_(12)H_(12)O_(11) are-285, -1263 and - 2238 kJ/mol respectively. Which of the following statements is true?

Answer»

The hydrolysis reaction is endothemic
Heat LIBERATED on combustion of 1.0 mol `alpha`=- MALTOSE is greater than the heat liberated on combustion of 2.0 mol glucose.
Increasing temperature will INCREASE the degree of hydrolysis of `alpha`-maltose
Enthalpy of reaction will remain same even it SOLID `alpha`-maltose is TAKEN in the reaction.

Answer :B
Discussion
29.

alpha-maltose consists of

Answer»

ONE `alpha`-D-GLUCOPYRANOSE unit and one
`beta`-D- glucopyranose unit with 1-2 glycosidic linkage
two `alpha`-D-glucopyranose units with 1-4 glycosidic linkage
two `beta`-D-glucopyranose units with 1-4 glycosidic linkage

ANSWER :D
Discussion
30.

alpha-hydroxypropanoic acid can be prepared from ethanal by following the steps given in the sequence.

Answer»

Treat withHCN followed by acidic hydrolysis.
Treat with `NaHSO_(3)` followed by REACTION with `Na_(2)CO_(3)`.
Treat with `H_(2)SO_(4)` followed by hydrolysis.
Treat with `K_(2)Cr_(2)O_(7)` in presence of suphuric acid.

Solution :
Discussion
31.

alpha-hydroxy acetic acid is also known as :

Answer»

Lactic acid
OXALIC acid
GLYCOLIC acid
None

Answer :C
Discussion
32.

alpha- Helix is found in

Answer»

DNA
RNA
LIPID
PROTEIN

ANSWER :D
Discussion
33.

alpha-helix is a secondary structure of proteins formed by twisting of polypeptide chain into right handed screw like structures. Which type of interactions are responsible for making the alpha-helix structure stable ?

Answer»

SOLUTION :Hydrogen BOND between `-NH-` group and `gt C=O` group of the peptide bond stabilises `alpha`-helix STRUCTURE of PROTEINS.
Discussion
34.

alpha-Helix is a secondary structure of proteins formed by twisting of polypeptide chain into right handed screw like structures. Which type of interactions are responsible for making the alpha- helix structure stabel ?

Answer»

Solution :The `alpha`-helix STRUCTURE of polypeptide chain is stabilized by intramolecular H-bonding in which `gt C=O` of one AMINO acid residue in one TURN FORMS a H-bond with `-NH-` group of the fourth amino acid in the adjacent turn.
Discussion
35.

alpha - Helix is a secondary structure of protein formed by twistingof polypeptidechain into righthanded screw likestructure . Whichtype of interactions areresponsiblefor makingthe alpha - helix structurestable ?

Answer»

Solution :In a `ALPHA -` helix , a polypeptidechain is stabilised by the formationof hydrogenbondsbetween `- NH`-GROUP of amino acids in oneturn with the groups of amino acidsbelongingto adjacent turn.
Discussion
36.

alpha-helix and beta- pleated structures of proteins are classified as :

Answer»

PRIMARY structure
SECONDARY structure
TERTIARY structure
QUATERNARY structure

Answer :B
Discussion
37.

alpha-halogenation of carboxylic acid is called

Answer»

GATTERMANN reaction
Riemer-Tiemer reaction
Sandmeyer'sreaction
HVZ reaction

Answer :A::C::D
Discussion
38.

alpha - H atomsof nitroalkane is

Answer»

ACIDIC
BASIC
NEUTRAL
can'tbe PREDICTED

ANSWER :A
Discussion
39.

alpha-glucose and beta-glucose are

Answer»

Isomers
Anomers
Epimers
Tautomers

Answer :B
Discussion
40.

alpha and beta -glucose differ in theorientation of OH goup around

Answer»

NUMBER of OH groups
CONFIGURATION
CONFORMATION
size of hemiacetal RING .

Answer :B
Discussion
41.

alpha-D- Glucose and beta - D glucose differ from each other due to difference in one carbon with respect to its?

Answer»

Size of hemiacetal RING
Number of OH groups
Configuration
CONFORMATION

Answer :C
Discussion
42.

alpha-D-Glucose and beta-D-Glucose are:

Answer»

Epimers
ANOMERS
ENANTIOMERS
ACETALS

ANSWER :B
Discussion
43.

alpha-D(+)-glucose and beta-D(+)- glucose are

Answer»

ENANTIOMERS
CONFORMERS
EPIMERS
ANOMERS

ANSWER :D
Discussion
44.

alpha -D(+) glucose and beta-D(+) glucose are :

Answer»

enantiomers
geometrical isomers
epimers
anomers.

Answer :D
Discussion
45.

alpha-D glucose and beta-D glucose are :

Answer»

ANOMERS
2-Epimers
3-Epimers
Enantiomers

Answer :A
Discussion
46.

alpha-D-Glucopyranose and beta-D-Glucopyranose are

Answer»

Anomers
Epimers
Diastereomers
Enantiomers

Answer :A::C
Discussion
47.

alpha-D glucose and beta-D glucose are :-

Answer»

EPIMER
ANOMER
Anomer
Anomer

ANSWER :B
Discussion
48.

alpha-D-glucopyranose and beta-D-glucopyranose are :

Answer»

stereosiomers
diastereoisomers
enentiomers
anomers

Solution :N//A
Discussion
49.

alpha-D-galactose and beta-D-galactose are

Answer»

epimers
metamers
anomers
tautomers

Answer :C
Discussion
50.

alpha-(D)-(-)fructose and beta-(D)-(-) fructose are

Answer»

anomers
epimers
diastereoisomers
tautomers

Answer :A
Discussion