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.

Reduction of ketones cannot be carried out with which of the following reagents ?

Answer»

Sodium borohydride or Lithium aluminium hydride
Zinc AMALGAM and CONCENTRATED HCl
Hydrazine and KOH in ethylene glycol
Hydrogen in presence of PALLADIUM in BARIUM sulphate and quinoline

Answer :D
2.

Reduction of hexose A (molecular formula C_(6)H_(12)O_(6)) with sodium borohydride gives compound B and C. Compound B is optically inactive, whereas compound C is optically active. Which of the following is compound A?

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D - fructose
D - glucose
D - mannose
D - galactose

Answer :A
3.

Reduction of ester with Na//C​_(2)​H_(5)​OH is called as :

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Birch reduction
Bouvealt - BLANC reduction
STEPHENS reduction
Clemmesen reduction

Solution :`R-oversetOoverset(||)C-OR OVERSET(Na+CHOH)to R-CH_2OH +ROH`
4.

Reduction of glucose by HI suggest that

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presence of OH groups
presence of -CHO group
cyclic STRUCTURE of glucose
six carbon ATOMS are ARRANGED in straight chain

Answer :D
5.

Reduction of CH_(3)CH_(2)NO_(2) with LiAlH_(4) gives:

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`CH_(3)CH_(2)NH_(2)`
`CH_(3)NH_(2)`
`CH_(3)CH_(2)OH`
`CH_(3)CH_(3)`.

Solution :`CH_(3)CH_(2)NO_(2)OVERSET(LiAlH_(4))rarr CH_(3)CH_(2)NH_(2)`
6.

Reduction of chlorobenzene by LiAlH_(4) gives

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toluene
benzene
diphenyl
phenyl

Answer :B
7.

Reduction of glucose with NaBH_(4)gives

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SORBITOL
mannitol
n-Hexane
both (a) and (b)

SOLUTION :`NaBH_(4)`- reduces - `CHO`group to give PHOSPHATE.
8.

Reduction of benzenediazonium chloride with SnCl_(2) + HCl

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Aniline
Phenylhydrazine
Azobenzene
Hydrazobenzene

Solution :SEE PROPERTIES of DIAZONIUM salt (B section)
9.

Reduction of carbonyl compounds to alkanes with NH_2-NH_2 and NaOH is called:

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CLEMMENSEN REDUCTION
Wolfe-Kishner reduction
WURTZ's reaction
none

Answer :B
10.

Reduction of benzenediazonium chloride with Zn/HCl gives

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aniline
phenylhydrazine
azobenzene
hydrazobenzene

Solution :`C_(6)H_(5)N_(2)CL OVERSET(Zn//HCl)to C_(6)H_(5)NH_(2)+NH_(3)`
11.

Reduction of can be carried out with

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CATALYTIC REDUCTION
`Na//C_2H_5OH`
Wolff -Kishner reduction
`LiAIH_4`

ANSWER :C
12.

Reduction of carbonyl compound to alkanes using ZN/Hg + HCI is called:

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CLEMMENSEN REDUCTION
Wolfe-Kishner reduction
WURTZ reaction
None of the above

Answer :A
13.

Reduction of benzene diazonium chloride with Zn/HCl gives

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PHENYL hydrazine
hydrazine hydrate
aniline
azo benzene

Answer :A::C
14.

Reduction of bensaldehyde with zinc and hydrochloric and gives

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SOLUTION :
15.

Reduction of benzenediazonium chloride with ethyl alcohol gives phenol.

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ANSWER :F
16.

Reduction of aromatic nitro compounds using Fe and HCl gives:

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AROMATIC oxime
aromatic hydrocatbon
aromatic PRIMARY amine
aromatic amide

Answer :C
17.

Reduction of aromatic nitro compounds using Fe and HCl gives____

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AROMATIC oxime
aromatic hydrocarbon
aromatic PRIMARY amine
aromatic amide

Answer :C
18.

Reduction of aromatic nitro compounds using Fe and HCI gives

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AROMATIC oxime
aromatic hydrocarbon
aromatic primary amine
aromatic amide.

Solution :`C_6 H_5 NO_2 +6[H] OVERSET(Fe//HCl)to underset("Aromatic "1^@ "amine")(C_6 H_5 NH_2 +2H_2 O)`
19.

Reduction of aromatic nitro compounds using Fe and HCl gives ______________ .

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AROMATIC oxime
aromatic hydrocarbon
aromatic PRIMARY amine
aromatic amide.

Solution :`C_6 H_5 NO_2 +6[H] OVERSET(Fe//HCl)to underset("Aromatic "1^@ "amine")(C_6 H_5 NH_2 +2H_2 O)`
20.

Reduction of an alkyne to the double bond stage can yield a cis alkene or trans alkene (except in cases where triple bond is at the end of the chain ) Reduction of alknes with sodium or lithium in liquid NH_(3) yields predominantly trans alkene. Hydrogenation of alkynes with Lindlar's catalyst or a nickel boride called P_(2) catalyst yeilds is alkene (as high as 98%) If Pd/C/H_(2) is used in the absence of a catalyst poison, two equivlanet of H_(2) are added forming alkanes. Select the correct statement(s)

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Trans form of vinylic anion is more stable
cis form of vinylic anion is more stable
RATE determining step OCCURS prior to the step in which vinylic anion reacts
Rate determing step ocurs afte vinylic anoins has reacted

ANSWER :A
21.

Reduction of an oxide ore with carbon at high temparature is called

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ANSWER :A
22.

Reduction of an alkyne to the double bond stage can yield a cis alkene or trans alkene (except in cases where triple bond is at the end of the chain ) Reduction of alknes with sodium or lithium in liquid NH_(3) yields predominantly trans alkene. Hydrogenation of alkynes with Lindlar's catalyst or a nickel boride called P_(2) catalyst yeilds is alkene (as high as 98%) If Pd/C/H_(2) is used in the absence of a catalyst poison, two equivlanet of H_(2) are added forming alkanes.Reduction of alkynes to trans alkene by Li//NH_(3) is carried out in the presence of dry ice. This is because of

Answer»

radical anion is formed at the temperature of dry ice
intermediate if formed with dry ice
`NH_(3)` REMAINS inliquidstate at the temperature of dry ice
Li remains in liquid STATE at the temperature of dry ice

ANSWER :C
23.

Reduction of an alkyne to the double bond stage can yield a cis alkene or trans alkene (except in cases where triple bond is at the end of the chain ) Reduction of alknes with sodium or lithium in liquid NH_(3) yields predominantly trans alkene. Hydrogenation of alkynes with Lindlar's catalyst or a nickel boride called P_(2) catalyst yeilds is alkene (as high as 98%) If Pd/C/H_(2) is used in the absence of a catalyst poison, two equivlanet of H_(2) are added forming alkanes.Lindlar's catalyst can be prepared in precipitating palladium calcium carbonate and treating it with lead acetate or quinoline. This treatement is

Answer»

partically deactivities the catalyst WHICHCAUSES REDUCTION of ALKYNE to alkene
increases the rate of hydogention
selectively gives only cis-isomer
selectively gives trans-isomer

Answer :C
24.

Reduction of an ester with lithium aluminium hydride gives

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TWO acids
two ALDEHYDES
ONE molecule of ALCOHOL and another of CARBOXYLIC acid
two alcohols

Answer :D
25.

Reduction of an alkyne to the double bond stage can yield a cis alkene or trans alkene (except in cases where triple bond is at the end of the chain ) Reduction of alknes with sodium or lithium in liquid NH_(3) yields predominantly trans alkene. Hydrogenation of alkynes with Lindlar's catalyst or a nickel boride called P_(2) catalyst yeilds is alkene (as high as 98%) If Pd/C/H_(2) is used in the absence of a catalyst poison, two equivlanet of H_(2) are added forming alkanes.Shown below is the first step in hte synthesis of hte important perfume constituent, cis-Jasmone which reagents you choose to carry out this last step?

Answer»

`Li//NH_(3)`
Lindlar's catalyst
`Na//C_(5)H_(5)OH`
`Pt//H_(2)`

ANSWER :B
26.

Reduction of an aldehyde produces

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PRIMARY ALCOHOL
Monocarboxylic acid
Secondary alcohol
Tertiary alcohol

Solution :`UNDERSET("Aldehyde")(R-CHO) overset("Reduction")rarr underset(1^(@)"Alcohol")(R-CH_(2)-OH)`
27.

Reductionof alkyl nitrilewith sodiumand ethanalis called

Answer»

CARBYLAMINE
Mendius REACTION
Catalyticreduction
Clemmenson's REDUCTION

ANSWER :B
28.

Reduction of alkyl produces:

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SECONDARY amine
primary amine
tertiary amine
amide

Answer :B
29.

Reduction of aldehydes with Lia1H_4 or NaBH_4 GIVES 1^@ alcohol

Answer»


ANSWER :T
30.

Reduction of aldehydes and ketones into hydrocarbons using zinc amalgam and conc. HCl is called :

Answer»

Wolff-Kishner Reduction
CLEMMENSEN reduction
Cope Reduction
Dow Reduction

Solution :R - `overset(O)overset(||)(C ) - R. overset(ZN - Hg//HCl)(rarr) R - CH_(2) - R.`
Clemmensen Reduction.
31.

Reduction of an aldehyde with H_2 /Pd gives a 1^@ alcohol.

Answer»


ANSWER :T
32.

Reduction of Aldehydes and Ketones to hydrocarbon take place in the presence of

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ZN AMALGAM and HCl acid
`Pd//BaSO_(4)`
Anhydrous `AlCl_(3)`
`Ni//Pt`

Answer :A
33.

Reduction of Al_(2)O_(3) is occurred at low potential and high current through electrolysis. If 4.0xx10^(4) ampere current is pass through molten Al_(2)O_(3) solution for 6 hours then how much aluminium is obtained ? (atomic mass of aluminium is 27 gm/mol at 100% efficiency)

Answer»

`8.1xx10^(4)GM`
`2.4xx10^(5)gm`
`1.3xx10^(4)gm`
`9.0xx10^(3)gm`

Solution :Having `Q=ixxt`
`Q=4.0xx10^(4)xx6xx60xx60C`
`=8.64xx10^(8)C`
Now, 96500 C liberate 9 gm AL, So, `8.64xx10^(8)C`
So, `(9)/(96500)xx8.64xx10^(8)` gram Al is liberated
`=8.05xx10^(4)` gram Al.
34.

Reduction of aldehydes and ketones produces

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`1^@` ALCOHOLS
`2^@` alcohols
`3^@` alcohols
`1^@` or `2^@` alcohols

Answer :D
35.

Reduction of aldehyde with HCl gives:

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PRIMARY alcohol
Secondary alcohol
Alkane
Tertiary alcohol

Answer :C
36.

Reduction of acetone in the presence of sodium borohydride gives

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1-propanol
2-propanol
Propene
n-propane

Solution :`CH_(3)-OVERSET(O)overset(||)(C )-CH_(3) overset(NaBH_(4))underset(("Reduction"))RARR underset(2-"propanol")(CH_(3)-overset(OH)overset(|)(CH)-CH_(3))`
37.

Reduction of acetyl chloride with H_(2) in presence of Pd gives

Answer»

`CH_(3)COCH_(3)`
`C_(2)H_(5)OH`
`CH_(3)COOH`
`CH_(3)CHO`

Solution :`CH_(3)COCl+H_(2)UNDERSET("REDUCTION")OVERSET(Pd)toCH_(3)CHO+HCl`
38.

Reduction of Ag^+ ions, provided by Tollen's reagent to silver metal is known as _____.

Answer»

SOLUTION :SILVER MIRROR
39.

Reduction of a ketone with H_2 /Pd gives a 3^@ alcohol.

Answer»


ANSWER :F
40.

Reductionof a ketogroup to a methylene group is converted by using

Answer»

`ZnHg + CONC. HCL`
`NaHg + ` WATER
`Sn + conc. HCl`
`Zn + CH+_(3) COOH`

Answer :A
41.

Reduction in presence of amalgamated zinc and conc. HCl is known as

Answer»

Mendius REDUCTION
WURTZ REACTION
Hoffmanns reaction
Clemmenson's reduction

ANSWER :D
42.

Reduction of 2-Butyne with Na in liquidNH_3 givespredominantly:

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n-butane
Trans-2-butene
No reaction
Cis-2-butene

Answer :B
43.

Reduction half reaction for the reaction :Fe^(2+)+MnO_(4)^(-)toFe^(3+)+Mn^(2+)is

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1. `MnO_(4)^(-)+8H^(+)+5e^(-)toMn^(2+)+4H_(2)O`
2. `Fe^(2+)toFe^(3+)`
3. `MnO_(4)^(-)+Fe^(2+)toFe^(3+)+Mn^(2+)`
4. `MnO_(4)^(-)+5e^(-)toMn^(2+)+4(O)`.

ANSWER :A
44.

Reducing sugar can reduce :

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ALDEHYDES to alcohols
FERRIC salts to ferrous salts
chlorates to chlorides
Fehling SOLUTION to cuprous oxide.

ANSWER :D
45.

Reducing property of monosaccharide is due to the presence of

Answer»

`-OH` group
Keto group
Acetal group
Anomeric HYDROXYL group

Answer :D
46.

Reducing power of A,B,C and D are w,x,y and z respectively such that z gt y gt x gt z. The one which will displace all others is

Answer»

A
B
C
D

Solution :APPLICATION of EMF SERIES.
47.

Reducing property of SO_(2) is represented by the following reaction

Answer»

`2H_(2)S+SO_(2) rarr 3S+2H_(2)O`
`I_(2)+SO_(2)+H_(2)O rarr SO_(4)^(2-) +2I^(-)+4H^(+)`
`3Fe+SO_(2) rarr 3FeO+FeS`
`4NA+3SO_(2) rarr Na_(2)SO_(3) +Na_(2)S_(2)O_(3)`

ANSWER :B
48.

Reducing property of SO_(2) is shown in A) 2H_(2)S+SO_(2)rarr 3S+2H_(2)O B) I_(2)+SO_(2)+2H_(2)Orarr SO_(4)^(-2)+2I^(-)+4H^(+) C) 3Fe+SO_(2)rarr 2Fe+FeS

Answer»

A
B
A, B
A, C

ANSWER :B
49.

Reducing agent used to convert benzene diazonium salt to benzene is

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`Zn+NH_(4)Cl`
`Na+C_(2)H_(5)-OH`
`C_(2)H_(5)-OH`
`LiAlH_(4)`

Answer :C
50.

Redox reactions play a pivotyal role in chemistry and biology. The values of standard redox potential (E^(@)) of two half-cell reactions decidewhich way the reaciton is expected to proced. A simple exampler is a Daniel cell in which zinc goes into solution and copper gerts deposited. Given below are a set of half-cell reactions (acidic medium) along with theirE^(@) (V with respect to normal hydrofgen electrode) values. Using this data : I_(2) + 2e^(-) rarr 2I^(-)"" E^(@) = 0.54, CI_(2) + 2e^(-) rarr 2CI^(-) ""E^(@) = 1.36, .Mn^(3+) + e^(-) rarr Mn^(2+) ""E^(@) = 1.50, Fe^(3+) + e^(-) rarr Fe^(2+)"" E^(@) = 0.77, O_(2) + 4H^(+) + 4e^(-) rarr 2H_(2)O ""E^(@) = 1.23, Soldium fusion extract, obtained from aniline, on treatment with iron (II) sulphatge and H_(2)SO_(4) in presence of airgives a Prussion buleprecipitate. The blue colour is due to the formation of :

Answer»

`Fe_(4)[Fe(CN)_(6)]_(3)`
`Fe_(3)[Fe(CN)_(6)]_(2)`
`Fe_(4)[Fe(CN)_(6)]_(2)`
`Fe_(3)[Fe(CN)_(6)]_(3)`

Solution :`Na + C + N rarr NACN`
`Fe^(2+) + 6CN^(-) rarr [Fe(CN)_(6)]^(4-)`
`4Fe^(3+) + 3[Fe(CN)_(6)]^(4-) rarr underset("Prussion blue")(Fe_(4)[Fe(CN)_(6)]_(3darr)`