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.

Aniline is basic in nature. Justify this statement.

Answer»

Solution :The lone PAIR of electrons on nitrogen atom in ANILINE makes it base. Aniline reacts with mineral acids to FORM salt.
`underset("Aniline")(C_(6)H_(5)-overset(..)(N)H_(2))+HClhArr underset("Anilinium chloride")(C_(6)H_(5)overset(+)(N)H_(3)CL^(-))`
Discussion
2.

Aniline is a weaker base than cyclohexylamine. Explain.

Answer»


ANSWER :The lone pair of ELECTRONS on nitrogen is delocalised due to resonance with BENZENE ring. There is no such delocalisation in cyclohexylamine. THEREFORE ANILINE is a weaker base than cyclohexylamine.
Discussion
3.

Aniline is a stronger base than benzyl amine.

Answer»

SOLUTION :ANILINE is a WEAKER BASE than BENZYLAMINE.
Discussion
4.

Aniline is a weaker base than benzyl amine.

Answer»

SOLUTION :ANILINE is a WEAKER BASE than BENZYLAMINE.
Discussion
5.

Aniline is a:

Answer»

PRIMARY BASE
Secondary base
Tertiary base
NEUTRAL COMPOUND

ANSWER :A
Discussion
6.

Aniline in a set of the following reactions yielded a coloured compound Y:

Answer»




SOLUTION :
Discussion
7.

Aniline is soluble in

Answer»

more basic than ammonia
more basic than p- AMINO phenol
more basic than p-nitro aniline
as basic as METHYL AMINE

Answer :C
Discussion
8.

Aniline in a set of reactions yielded a product

Answer»

`C_(6)H_(5)CH_(2)NH_(2)`
`C_(6)H_(5)NHCH_(2)CH_(3)`
`C_(6)H_(5)NHOH`
`C_(6)H_(5)CH_(2)OH`

SOLUTION :
Discussion
9.

Aniline hydrogen sulphate on heating with sulphuric acid at 453-473 K produces

Answer»

BENZENE SULPHURIC acid
anthranilic acid
aniline
Sulphanilic acid.

Answer :D
Discussion
10.

Aniline in a set o reactions yield a product D The structure of the product D would be

Answer»

`C_(6)H_(5)CH_(2)OH`
`C_(6)H_(5)CH_(2)NH_(2)`
`C_(6)H_(5)NHOH`
`C_(6)H_(5)NHCH_(2)CH_(3)`

ANSWER :A
Discussion
11.

Anilinehave higher B.Pthan correspendingalkane or ethers due to

Answer»

RESONACE stabilizationof benzenering
resonancestabilizationof aniliumion
more HYDROPHOBIC natureof `C_(6)H_(5)` groupthan `C_(2)H_(5)`GROUP
morehydrophobicnature`C_(6)H_(5)` group than`C_(2)H_(5)` group

Solution :`C_(6) H_(5)`- groupis morewaterhatingthan `C_(2)H_(5) ` group.
Discussion
12.

Aniline gives meta derivative as major product with

Answer»

`CH_3COCl//"PYRIDINE"`
` HNO_3 +H_2SO_4`
` Br_2//"WATER"`
` CH_3Cl//"pyridine"`

ANSWER :B
Discussion
13.

Aniline gets coloured on standing in air for a long time. Why? Or Why does aniline turn blackish brown in on open air.

Answer»

Solution :Due to strong electron-donating effect (+R-effect) of the `NH_(2)` group, the electron density ono the benzene ring increases. As a result, aniline is easily oxidised on standing in AIR for a long time to FORM coloured blackish BROWN products.
Discussion
14.

Aniline gets coloured on standing in air for a long time. Why?

Answer»

Solution :Due to strong electron DONATING EFFECT(+R effect) Of `NH_(2)` group, the electron density on the BENZENE RING increases. As a result ANILINE is easily oxidised on standing in air for a long time to form coloured products.
Discussion
15.

Anilinedonotreactswith

Answer»

`Br_(2) ` + water
CONC.`H_(2)SO_(4)`
conc. `H_(2)SO_(4)`
`CH_(3) - Cl`thepresenceof `AlCl_(3)`

Solution :Anilinedoes notundergoes FriedelCraftreactionbecause`AlCl_(3)`bondedwith`- NH_(2) ` toform`- NH_(2)^(+)Cl`
Discussion
16.

Aniline doesn't react with

Answer»

`"DIL. "HCl`
`"dil. "NAOH`
`CH_(3)CHO`
`Br_(2)" WATER"`

Answer :B
Discussion
17.

Aniline does not undergo friedel-crafts reaction. Why?

Answer»

Solution :ANILINE being BASIC combines with the Lewis acid catalyst `AlCl_3` used in Friedel-Crafts REACTION and forms salt. So Friedel-Crafts reaction does not OCCUR.
Discussion
18.

Why aniline does not undergo Friedel - Crafts reaction ?

Answer»

SOLUTION :Aniline forms salts with Friedel - Crafts catalyst `AlCl_(3)`. In this salt .N. ATOM acquires POSITIVE charge and acts as strong deactivating GROUP. Hence, the reactivity is decreased.
Discussion
19.

Aniline differs from ethyl amine in :

Answer»

BASIC NATURE
ACETYLATION
CARBYLAMINE REACTION
Reaction with aldehyde

Answer :A
Discussion
20.

Aniline can be separated from phenol using

Answer»

`Na_(2)CO_(3)`
`NaNO_(2)+ HCl` at `0^(@) C`
`NaCl`
Acidified `KMnO_(4)`

Solution :Aniline and phenol can be separated by adding `NaNO_(2) + HCl` as aniline forms DIAZONIUM salt anilinium chloride, while phenol does not. Hence, choice (b) is correct. Choice (a) is INCORRECT because `Na_(2)CO_(3)` NEITHER reacts with aniline nor with phenol. Choice (c ) is incorrect because `NaCl` does not REACT withphenol. Choice (d) is incorrectbecause `KMnO_(4)` oxidies both aniline and phenol.
Discussion
21.

Aniline cannot be prepared by Gabriel phthalimide synthesis.

Answer»

Solution :In Gabriel phthalimide synthesis of aniline, POTASSIUM phthalimide REQUIRES the treatment with CHLOROBENZENE or bromo-benzene. Sincearyl halides do not undergo NUCLEOPHILIC substitution reaction, chlorobenzene or BROMOBENZENE does not react with potassium phthalimide to give N-phenylphthalimide and hence aniline cannot be prepared by Gabriel phthalimide synthesis.
Discussion
22.

" Aniline + benzoylchloride "overset(NaOH)(to) C_6H_5-NH-COC_6H_5 this reaction is known as

Answer»

Friedal - CRAFT's REACTION
HVS reaction
Schootten- Baumann reaction
Cannizaro reaction

SOLUTION :SCHOTTEN - Baumann reaction
Discussion
23.

Aniline + benzoylchloride overset(NaOH)(to)C_(6)H_(5)-NH-COC_(6)H_(5) this reaction is known as………………….

Answer»

Friedel-crafts reaction
HVZ reaction
Schotten-Bauman reaction
none of these

Answer :C
Discussion
24.

Aniline can be converted into Benzene by

Answer»

DIAZOTIZATION
diazotization followed by TREATING with `H_(3)PO_(2)`
treating with `H_(3)PO_(2)`
diazotization followed by treating with steam

Answer :B
Discussion
25.

Aniline + Benzene diazonium chloride to X. Identify X.

Answer»

ORANGE DYE
YELLOW dye
malachite GREEN dye
madder dye

Solution :yellow dye
Discussion
26.

Aniline at low temperature.

Answer»

Solution :Aniline is dissolved in dilute hydrochloric acid, well COOLED solution of sodium nitrite in water is added at `0-5^(@)`. Benzene diazonium chloride is OBTAINED. This is called diazotisation REACTION.
`NaNO_2+HCl toHNO_2+NaCl`
`C_6H_5N.H_2+O.N.OH+H.Cloverset(0-5^(@)C)toC_6H_5-N=N-Cl+2H_2O`
Discussion
27.

Aniline and ethylamine resembles in:

Answer»

Solubility
ACTION with `HNO_2`
Action of GRIGNARD REAGENT
Coupling REACTION

Answer :C
Discussion
28.

Anilin is heatedwith conc H_(2)SO_(4)gives

Answer»

o-amino benzenesulphonic ACID
p-amino BENZENE sulphonic acid
m-amino benzenesulphonicacid
isophthalic aicd

Answer :B
Discussion
29.

Anilie is treated with a mixture of NaNO_(2) and H_(3)PO_(2), the product formed is

Answer»

Aniline diazonium hypophosphite
BENZENE
Anilinium hypophosphite
Aniline diazonium hypophosphite.

Solution :When aniline is treated with a MIXTURE of sodium NITRITE and hypophosphorous acid `(H_(3)PO_(2))`, the product formed is benzene as shown below
`C_(6)H_(5)NH_(2) overset(NaNO_(2)//H_(3)PO_(2))rarr C_(6)H_(5) N_(2)^(+) H_(2) PO_(2)^(-) UNDERSET(-H_(3)PO_(3))overset(H_(2)O //H_(3)PO_(2))rarr C_(6)H_(6) + N_(2)`
Discussion
30.

Anhydrous MgCl_(2) is obtained by heating the hydrated salt with …..

Answer»


ANSWER :DRY HCL
Discussion
31.

Anhydrous MgCl_(2) can be prepared by heating MgCl_(2)*6H_(2)O :

Answer»

1. in a CURRENT of DRY HCl
2. with carbon
3. with lime
4. until it fuses

Answer :A
Discussion
32.

Anhydrous mercurous chloride can be prepared by:

Answer»

the reduction of `HgCl_(2)` with `SnCl_(2)` solution
the REACTION of `HgCl_(2)` with Hg
the reaction of Hg with CONCENTRATED HCl

Answer :B
Discussion
33.

Anhydrous magnesium chloride can be prepared by heating MgCl_(2).6H_(2)O

Answer»

with CONCENTRATED HCL
in an atmosphere of NITROGEN
in a CURRENT of dry HCl gas
in an atmosphere of HYDROGEN

Answer :C
Discussion
34.

Anhydrous magnesium chloride can be prepared by heating MgCl_2.6H_2O

Answer»

In a CURRENT of DRY HCL gas
with carbon
Until it fuses
With lime

Answer :A
Discussion
35.

Anhydrous FeSO_4 is:

Answer»

WHITE
Black
Green
Brown

Answer :A
Discussion
36.

Anhydrous CuCl_2 and CuBr_2 exist as :

Answer»

MONOMER
Dimer
Trimer
POLYMER

ANSWER :D
Discussion
37.

Anhydrous copper sulphate is colourless, but hydrated copper sulphate is blue. Explain ?

Answer»

Solution :In ANHYDROUS copper sulphate, the splitting is very LESS. It absorbs energyin infrared REGION.It is colourless. In hydrated copper sulphate, water molecules are present in the COORDINATION sphere. The splitting of d-orbitals is reasonably increased. If it absorbs energy in visible region of light.Hence it is exhibit COLOUR.
Discussion
38.

Anhydrous calcium chloride is often used as a dessicant. In the presence of excess of CaCl_(2), the amount of the water taken up is governed by K_(p) = 6.4 x× 10^(85) for the following reaction at room temperature, CaCl_(2)(s) + 6H_(2)O(g)hArrCaCl_(2).6H_(2)O(s). What is the equilibrium vapour pressure of water in a closed vessel that contains CaCl_(2)(s) ?

Answer»


ANSWER :0.821 ATM
Discussion
39.

Anhydrous CaCl_(2) is not recommended as a drying agent for alcohols and amines.

Answer»

Solution :Alcohols and amines combine with anhyd. `CaCl_(2)` to form complexes. For example, with `C_(2)H_(5)OH`, it GIVES a complex of MOLECULAR formula. `CaCl_(2).3C_(2)H_(5)OH`
Discussion
40.

Anhydrous CaCl_(2) is used as drying agent because it

Answer»

ADSORB and ABSORBS WATER molecules
Absorbs water molecules
Adsorb water molecules
None

Answer :B
Discussion
41.

Anhydrous CaCl_2 can be used for drying

Answer»

ALCOHOLS
AMMONIA
`H_2S`
methane

Solution :Anhydrous `CaCl_2` cannot be used for DRYING alcohols, ammonia and `H_2S` (SEE A-level information).
Discussion
42.

Anhydrous aluminium chloride is a……………. Substance.

Answer»


ANSWER :HYGROSCOPIC
Discussion
43.

Anhydrous aluminium chloride fumes in air because of :

Answer»

HYDRATION
HYDROLYSIS
OXIDATION
decomposition

Answer :B
Discussion
44.

Anhydrous aluminium chloride exists as:

Answer»

monomer
dimer
trimer
polymer

Answer :B
Discussion
45.

Anhydrous AlCl_(3) is used as a catalyst in the friedel-crafts reaction because it is

Answer»

Electron rich
Soluble in ETHER
Insoluble to chloride and aluminium ions
Electron deficient

SOLUTION :`AlCl_3` is an electron deficient COMPOUND. It generates electrophile in the REACTION
`CH_3Cl+AlCl_3 to underset"Electrophile"(CH_3^(+))+AlCl_4^-`
Discussion
46.

The catalyst used in Friedel-Crafts reaction is :

Answer»

ELECTRON rich
soluble in ether
insoluble
electron deficient

Answer :D
Discussion
47.

Anhydrous AlCl_(3) is used in Firedel-Craft reaction because it is

Answer»

ELECTRON RICH
electron DEFICIENT
SOLUBLE in ether
ionise to Al and Cl ions

Answer :B
Discussion
48.

Anhydrous AlCl_3is covalent. From the data given below, predict whether it would remain covalent or become ionic in aqueous solution. (Ionization energy for Al = 5137 kJ "mole"^(-1), Delta H_("hydration")for Al^(3+) = -4665 kJ "mole"^(-1) , Delta H_("hydration") for Cl^(-) = -381 kJ "mole"^(-1)).

Answer»

Solution : TOTAL ENERGY evolved due to hydration = -4665 - 36-381) = -5808 kJ/mole. As this RELEASED energy is greater than ionization energy (5137 kJ/mole) of AL, `AlCl_3`can be ionic in aqueous solution.
Discussion
49.

Anhydrous AlCl_(3) is covalent . From the data given below predict whether it would remain as a molecule or converts into ions in aqueous solution . [L.E. for Al Cl_(3) = 5173 kJ/mol] DeltaH hydration for Al^(3+) = -4665 kJ/mol, Delta H_("hydra") for Cl^(-) = -381 kJ/mol .

Answer»


ANSWER :IONIC
Discussion
50.

Anhydrous AlCl_(3) is covalent. From the data given below : Lattice energy = 5137 kJ/mol DeltaH hydration for Al^(+3) = –4665 kJ/mol DeltaH hydration for Cl^(-) = -381 kJ/mol Correct statement is :-

Answer»

The solution will consist of `AL^(+3) & CL^(-)`
The solution will consist of hydrated `Al^(+3) & Cl^(-)`
It will remain covalent in aqueous solution
None

Solution :`DELTAH`hydration for `Al^(+3)= -4665 K^(-1)`MOL
`DeltaH`hydration for `Cl^(-) = -381` kJ/mol
In `AlCl_(3) to 3Cl^(-)`
so, `DeltaH` hydrationfor `3Cl^(-) = -381 xx 3 = -843`
Total hydration enthalpy
`=-4665+ -843 = -5508`
Lattice energy = 5137
Hydration energy is more than lattice energy
Discussion