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.

Hyperconjugation describes the orbital interactions between the p-systems and the adjacent s-bond of the substituent group(s) in organic compounds. Hyperconjugation is also called as Baker and Nathen effect. The necessary and sufficient condition for the hyperconjugation are : i) Compound should have at least on sp2 hybrid carbon of either alkene, carbocation or alkyl free radical. ii) A-carbon with respect to sp2 hybrid carbon should have at least one hydrogen. Hyperconjugation are of three types: (i) s(C-H), p-conjugation. (iii) s(C-H), positive charge conjugation iv) s(C-H), odd electron conjugation The hyperconjugation may be represented as Number of resonating structures due to hyperconjugation = (n + 1) where n is the number of a-hydrogen. Greateris the number of such forms, more is the stability of the species under considersation. Which of the following carbocations will show highest number of Hyperconjugation forms?

Answer»

`CH_(3)-overset(+)(C)H_(2)`
`H_(3)C-underset(underset(CH_(3))(|))(overset(+)(C)H)`
`H_(3)C-underset(underset(CH_(3))(|))overset(overset(CH_(3))(|))(C^(+))`
`H_(3)C-underset(underset(CH_(3))(|))overset(overset(CH_(3))(|))(CH_(2)-C^(+)`

Solution :C has 9-hyper conjugative STRUCTURE
For a-3, HC, For b-6, Hc, For d-8, Hyper conjugative structures
Discussion
2.

Hyperconjugation describes the orbital interactions between the p-systems and the adjacent s-bond of the substituent group(s) in organic compounds. Hyperconjugation is also called as Baker and Nathen effect. The necessary and sufficient condition for the hyperconjugation are : i) Compound should have at least on sp2 hybrid carbon of either alkene, carbocation or alkyl free radical. ii) A-carbon with respect to sp2 hybrid carbon should have at least one hydrogen. Hyperconjugation are of three types: (i) s(C-H), p-conjugation. (iii) s(C-H), positive charge conjugation iv) s(C-H), odd electron conjugation The hyperconjugation may be represented as Number of resonating structures due to hyperconjugation = (n + 1) where n is the number of a-hydrogen. Greateris the number of such forms, more is the stability of the species under considersation. Hyperconjugation is possible in which of the following species ?

Answer»

`CH_(3)-BAR(C)H-CH_(3)`
`C_(6)H_(5)-CH_(3)`
`H_(2)C=CH_(2)`
`H_(3)C-UNDERSET(underset(CH_(3))(|))OVERSET(overset(CH_(3))(|))(C)-CH=CH_(2)`

Solution :
No other structure has the REQUIRED condition.
Discussion
3.

Hyperconjugation is also known as

Answer»


ANSWER :no-bond
Discussion
4.

Hyper conjugation phenomenon is possible in:

Answer»

`H_(2)C=CH_(2)`
`C_(6)H_(5)CH=CH_(2)`
`CH_(3)CH_(2)-CH=CH_(2)`
`(CH_(3))_(3)C-CH=CH_(2)`

Solution :`alpha` HYDROGENS must PRESENT
Discussion
5.

Hyper conjugation is most useful for stabilizing which of the following carbocations ? (a)neo-Pentyl (b) tert-Butyl (c) iso-Propyl (d)Ethyl (e) Methyl

Answer»


ANSWER :B
Discussion
6.

Hyper Conjugation is also known as

Answer»

no bond resonance
Baker - NATHAN effect
both (a) and(B)
none of these

Answer :C
Discussion
7.

Hyper conjugation involves overlap of the following orbitals

Answer»

`sigma-sigma`
`sigma-p`
`p-p`
`pi-pi`

SOLUTION :`sigma-p`
Discussion
8.

Hyper conjugation is most useful for stabilising which of the following carbocations ?

Answer»

Neo-pentyl
Tert-BUTYL
Iso-propyl
Ethyl

Solution :Hyper conjugation is useful for STABILISING tertiary butyl carbocation with nine HYDROGEN atoms at the `alpha`-positions.
`{:(""CH_(3)),("|"),(H_(2)C-C o+),("|"),(""CH_(3)):}`
Discussion
9.

Hynud oxygen is an example of ……………

Answer»

SOLUTION :GASEOUS solution
Discussion
10.

Hydroxylamine reduces iron III according to the equaction, 4Fe^(3+) + 2NH_(2)OH rarr N_(2) O + H_(2)O + 4 Fe^(2+) + 4H^(+). Orpm II thus produced is estimatedf by tiration with standard KMnO_(4)solution. The reaction is MnO_(4)^(-) + 5Fe^(2+)+ 8H^(+) rarr Mn^(2+) + 5Fe^(3+) + 4H_(2)O. A 10 mL of hydroxyamine solution was diluted to one litere. 50 mL of this diluted soltuion was boiledwith an excessof Fe^(3+) solution. The resulting solution required 12mL of 0.02M KMnO_(4) solutionfor complete oxidationof Fe^(2+). Calculatethe weight of NH_(2)OH in one litre of orignal solution.

Answer»


ANSWER :`39.6 G//"LITRE"`
Discussion
11.

Hydroxyl ion is considered as a strong base. Give reasons.

Answer»

SOLUTION :Hydroxyl ION is the conjugate base of water, `H_2 O HARR H^(+) +OH^(-)` Water is a very weak acid. Hence its conjugate base, hydroxyl ion is a strong base.
Discussion
12.

Hydroxyl amine reduces iron (III) according to following equation NH_(2)OH+ Fe_(2) (SO_(4))_3 rarr N_(2) (g) +H_2O+FeSO_4 +H_(2) SO_4 Which statement is correct

Answer»

n-factor for Hydroxyl AMINE is 12
EQUIVALENT WEIGHT of `Fe_(2) (SO_4)_(3)` , is M/2
6 meq of `Fe_(2)(SO_(4))_3` , is contained in 3 millimoles of ferric SULPHATE
all of these

Solution :`NH_(2)OH rarr N_(2) , ` n - factor =2
Discussion
13.

Hydroxides of beryllium are _________ in nature

Answer»

SOLUTION :AMPHOTERIC
Discussion
14.

Hydroxide of following ion is highly water soluble

Answer»

`Ni^(2+)`
`AL^(3+)`
`K^(+)`
`AG^(+)`

ANSWER :C
Discussion
15.

Hydrolytic reaction of fats by caustic soda is known as

Answer»

Acetylation
Carboxylation
Esterification
SAPONIFICATION

Solution :ALKALINE HYDROLYSIS of FATS is CALLED Saponification.
Discussion
16.

Hydrolytic conversion of sucrose into glucose and fructose is known as

Answer»

Induction
INVERSION
Insertion
Inhibition

Solution :Inversion
Discussion
17.

Hydrolysis of trichloromethane with aqueous KOH gives

Answer»

Potassium FORMATE
Acetylene
Chloral
Methanol

Solution :`CHCl_(3) + 3KOH RARR HCOOH + 3KCl + H_(2)O`
`HCOOH + KOH rarr underset("formate")underset("Potassium")(HCOOK + H_(2)O)`
In excess of KOH potassium formate is formed
Discussion
18.

Hydrolysis of sucrose givesSucrose + H_(2)O hArr Glucose + Fructose Equilibrium constant K_(c) for the reaction is 2 xx 10^(13) at 300 K. Calculate Delta G^(@) at 300 K.

Answer»

SOLUTION :`DELTAG^(@) = - 2.303 RT log K_(c) =- 2.303 xx ( 8.314JK^(-1) mol^(-1)) xx 300 K xx log ( 2 xx 10^(13))`
`= - 2.303 xx 8.314 xx 300 xx 13.301 J mol ^(-1) = - 76402 J mol^(-1) = -764 xx 10^(4) J mol^(-1)`
Discussion
19.

Hydrolysis of sucrose gives, Sucrose + H_2O hArr Glucose + Fructose Equilibrium constant K_c for the reaction is 2xx10^13 at 300 K. Calculate DeltaG^ө at 300 K.

Answer»

Solution :`DeltaG^ө`=-RT LN `K_c`
=-2.330 RT `log_10 K_c`
`therefore DeltaG^ө =-2.303 (8.314 "J mol"^(-1)K^(-1)) (300 K) log (2xx10^13)`
=-2.303(8.314)(300)(13.3010)
=-76336.4
`=-7.634 XX 10^4 "J mol"^(-1)`
Discussion
20.

Hydrolysis of sodium carbonate produce .... solution.

Answer»

Acidic
Basic
Amphoteric
Neutral

Solution :`CO_(3)^(-2) +H_(2)O to HCO_(3)^(-) +OH^(-)`
Discussion
21.

Hydrolysis of SiCl_(4) gives

Answer»

`Si(OH)_(4)`
`SiOCI_(2)`
`SiO_(2)`
`H_(2)SiO_(4)`

Solution :`SiCl_(4(l))+2H_(2)O_((l))to SiO_(2(s))+4HCl_((AQ))`
Discussion
22.

Hydrolysis of phophodiester groups is the back bone of DNA, has DeltaG^circ=-5.5kcal//mol at 27^circC. Approximate equilibrium costant for the hydorysis reaction is :

Answer»

`10^9`
`10^4`
`10`
`10^10`

ANSWER :B
Discussion
23.

Hydrolysis of ozonide of but-1-ene gives

Answer»

ETHYLENE only
Acetaldehyde and FORMALDEHYDE
PROPIONALDEHYDE and formaldehyde
Acetaldehyde only

Answer :C
Discussion
24.

Hydrolysis of one mole of peroxodisulphuric acid produces.....

Answer»

TWO moles of sulphuric acid.
two moles of PEROXOMONOSULPHURIC acid.
one MOLE of sulphuric acid and one mole of peroxomonosulphuric acid.
one mole of sulphuric acid, one mole of peroxomonosulphuric acid and one mole of hydrogen peroxide.

Answer :C
Discussion
25.

Hydrolysis of ethyl acetate is catalysed by aqueous:-

Answer»

`Na_(2)SO_(4)`
`K_(2)SO_(4)`
`H_(2)SO_(4)`
`BaSO_(4)`

Solution :`CH_(3)C O OC_(2)H_(5)+HOHunderset("Catalyst")overset(Conc." "H_(2)SO_(4))toCH_(2)CO OH+C_(2)H_(5)OH`
Discussion
26.

Hydrolysis of ..... Carbide , we get methane

Answer»

CALCIUM
BERYLLIUM
ALUMINIUM
Both Be and AL

ANSWER :D
Discussion
27.

Hydrolysis of an ester gives acid A and alcohol B. The acid reduces Fehling's solution. Oxidation of alchohol B gives acid A. The ester is :

Answer»

METHYL formate
ethyl formate
methyl acetate
ethyl acetate

SOLUTION :
`HCOOH` REDUCES FEHLING solution
Discussion
28.

Hydrolysis of an ester gives a carboxylic acid which on Kolbe's electrolysis yields ethane. The ester is

Answer»

ETHYL methonoate
methyl ethanoate
propylamine
Ethylamine

Solution :`underset("Methyl ethanoate")(CH_(3)COOCH_(3)) underset(-CH_(3)OH)overset(NAOH)rarr underset("Sodium acetate")(CH_(3) COONa) underset(-CO_(2), - NaOH, -H_(2))overset("Kolbe's electrolysis")rarr underset("Ethane")(CH_(3) - CH_(3))`
Discussion
29.

Hydrolysis of alkyl halide is an example for ..............

Answer»

SOLUTION :NUCLEOPHILIC SUBSTITUTION
Discussion
30.

Hydrolysis of alkyl halides is an example for which type of reaction? Explain

Answer»

SOLUTION :Hydrolysis of alkyl halides is an example for nucleophilic substitution reactions.
`CH_(3)Br overset("AQUEOUS OH^(ө))to CH_3OH + Br^(ө)`
Hence , `OH^ө` is the incoming nucleophile or attacking species and `Br^ө` is the LEAVING GROUP.
Discussion
31.

Hydrolysis of 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
32.

Hydrolysis of a compound C_(9)H_(10)ClBr (P) yields C_(9)H_(10)O(Q) (Q) gives positive haloform test ? Strong oxidation of (Q) yields a dibasic acid which gives only two mono-nitro derivative. What is the structure of (P) ?

Answer»




SOLUTION :
Discussion
33.

Hydrolysis constant for NH_4 Cl is 4xx 10^(-10). What is the dissociation constant of ammonium hydroxide ?

Answer»

SOLUTION :` 2.5 XX 10^(-5)`
Discussion
34.

Hydrolysis constant of salt derived from strong acid and weak base is 2xx 10 ^(-5).The dissociation constant of the weak base is

Answer»

` 5xx10 ^(-8) `
` 5xx10 ^(-9) `
` 5xx10 ^(-10) `
` 2xx10 ^(-19) `

Solution :` K_h= (K_w)/(K_b)rArr K_b =(1xx 10 ^(-14))/(2xx 10 ^(-5)) = 5 XX 10 ^(-10) `
Discussion
35.

Hydrolith, a source of Hydrogen is

Answer»

NaH
`CaH_(2)`
LiH
`BaH_(2)`

ANSWER :B
Discussion
36.

Hydrogne peroxide was discovered by ____________

Answer»

Chadwick
J.J Thomson
Urey
J.L. Thenard

Answer :D
Discussion
37.

The hydrogenation of vegetable oil in the presence of nickel catalyst forms vanaspati ghee. Give reason.

Answer»


ANSWER :1
Discussion
38.

Hydrogenation of the adjoining compound in the presence of poisoned palladium catalyst gives.

Answer»

an OPTICALLY active compound
an optically INACTIVE compound
a racemic mixture
a DIASTEREOMERIC mixture

Answer :B
Discussion
39.

Hydrogenation of C_(6)H_(5)CHOH - CHOH over Rh - Al_(2)O_(3) catalyst in methanol gives

Answer»

`C_(6)H_(5)CH_(2)COOH`
`C_(6)H_(11)CH_(2)COOH`
`C_(6)H_(5)CHOHCH_(2)OH`
`C_(6)H_(11)CHOH - COOH`

Solution :When the animal CHARCOAL is MIXED with acetic acid, it absorbed the impurity used in the acetic acid, therefore concentration of acetic acid solution decreases
Discussion
40.

Hydrogenation of benzoyl chloride in the presence of Pd//BaSO_(4) gives

Answer»

BENZYL alcohol
benzaldehyde
benzoic acid
phenol

Answer :B
Discussion
41.

Hydrogenation of benzoyl chloride in the presence of Pd and BaSO_(4) gives

Answer»

BENZYL alcohol
benzoic acid
bezaldehyde
toluene

Answer :C
Discussion
42.

Hydrogenation of benzoyl chloride in the presence of Pb on BaSO_(4) gives

Answer»

Benzyl ALCOHOL
Benzaldehyde
Benzoic ACID
Phenol

Solution :See PROPERTIES of acid CHLORIDE
Discussion
43.

Hydrogenation of alkyne is

Answer»

ADDITION reaction
ELIMINATION reaction
REARRANGEMENT reaction
Substitution reaction

Solution :Alkynes have TRIPLE bond.
Discussion
44.

Hydrogenation of alkenes and alkynes takes place in presence of certain catalysts. In sabatier-Senderen's reaction, the addition of hydrongen taken place in presence of nickel catalyst. Controlled hydrogenation of alkyne in presence of Lindlar's catalyst gives cis-alkene. Non terminal alkynes are reduced by Na or Li metal dissolved in liquid NH_(3). In this reaction trans alkenes are formed. Q. The product of the following reaction is

Answer»




ANSWER :B
Discussion
45.

Hydrogenation of alkenes and alkynes takes place in presence of certain catalysts. In sabatier-Senderen's reaction, the addition of hydrongen taken place in presence of nickel catalyst. Controlled hydrogenation of alkyne in presence of Lindlar's catalyst gives cis-alkene. Non terminal alkynes are reduced by Na or Li metal dissolved in liquid NH_(3). In this reaction trans alkenes are formed. Q. In which of the following the reaction is most exothermic

Answer»




ANSWER :B
Discussion
46.

Hydrogenation is done of hydrogen A of unsaturated hydrocarbon on addition of H_(2). Ozonolysis of A gives acetone, acetaldehyde and propane-1, 3-diol. Give the structure and name of X ?

Answer»

Solution :(i) On hydrogenation of A, 2 moler of `H_(2)` is added. Therefore in A, there are `2pi`-bond therefore in A there is 1 double bond and 1 triple bond.
(ii) OZONOLYSIS of A gives three product.
`underset("Acetone")overset((a))(CH_(3)COCH_(3))+underset("Acetaldehyde")overset((b))(CH_(3)CHO)+ overset((c))(CHOCH_(2)CHO)`
No. of carbon `"3+2+3=8"`
(III) From above PRODUCTS, A is diene and (a), (b) and (c) has double bonds reacts with X, STRUCTURE is as follows 2-methyl hepta-2, 5-diene.
Formed product :
`(H_(3)C-overset(CH_(3))overset(|)(C))/((a))=(CH-CH_(2)-CH)/((c))=(CH-CH_(3))/((b))`
Hydrocarbon (A)
Discussion
47.

Hydrogenation of adjoining compound in the presence of poisoned palladium catalyst gives

Answer»

An OPTICALLY active COMPOUND
An optically inactive compound
A racemic inactive compound
A diastereomeric mixture

Solution :Hydrogenation with poisoned PALLADIUM brings about C is hydrogenation of ALKYNE and does not affect DOUBLE bond.
Discussion
48.

Hydrogen will not reduce heated

Answer»

CUPRIC OXIDE
FERRIC oxide
stannic oxide
ALUMINIUM oxide

Answer :D
Discussion
49.

Hydrogen will not reduce....

Answer»

heated CUPRIC oxide.
heated FERRIC oxide.
heated STANNIC oxide.
heated ALUMINIUM oxide.

Answer :D
Discussion
50.

Hydrogen will not reduce

Answer»

HEATED CUPRIC oxide
heated FERRIC oxide
heated stannic oxide
heated aluminium oxide.

Solution :Fluorspar (`CaF_(2)` ) does not contain aluminium.
Discussion