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.

Propanal-1 and pentan-3-one are the ozonolysis products of an alkene. What is th structural formula of alkene?

Answer»

`CH_(3)CH_(2)-underset(CH_(2)CH_(3))underset(|)(C)=CH-CH_(2)CH_(3)`
`CH_(3)CH_(2)-CH=CH-overset(CH_(2)CH_(3))overset(|)(C)H-CH_(3)`
`CH_(3)-underset(CH_(2)CH_(3))underset(|)(C)=overset(CH_(2)CH_(3))overset(|)(C)-CH_(3)`
`CH_(3)CH_(2)CH_(2)CH_(2)-CH=CH-CH_(2)CH_(3)`

SOLUTION :`CH_(3)CH_(2)-underset(CH_(3)-CH_(2))underset(|)(C)=O+O=underset(H)underset(|)(C)-CH_(2)-CH_(3) to CH_(3)-CH_(2)-underset(CH_(2)-CH_(3))underset(|)(C)=CH-CH_(2)-CH_(3)`
Discussion
2.

Propanal and pentan-3-one are the ozonolysis products of an alkene ? What is the structural formula of the alkene ?

Answer»

Solution :(i)WRITE the structures of PROPANAL and pentan-3-one with their oxygen atoms facing each other. Remove oxygen atoms and join the TWO fragments by a DOUBLE bond, the structure of the alkene is
Discussion
3.

Propanal and pentan-3-one are the ozonolysis products of an alkene? What is the structural formula of the alkene?

Answer»

SOLUTION :The PRODUCTS of OZONOLYSIS are
Discussion
4.

Propan-1-ol can be prepared from propene by

Answer»

`H_(2)O //H_(2)SO_(4)`
`Hg(OAc)_(2)//H_(2)O` followed by `NaBH_(4)`
`B_(2)H_(6)` followed by `H_(2)O`
`CH_(3)CO_(2)H//H_(2)SO_(4)`

Solution :`6CH_(3)CH=CH_(2)+B_(2)H_(6) rarr 2(CH_(3)CH_(2)CH_(2))_(3)B UNDERSET(OH^(-))OVERSET(H_(2)O_(2))rarr 6CH_(3)CH_(2)CH_(2)OH+B(OH)_(3)`.
Discussion
5.

Propan-1-ol and ethanol can be distinguished by

Answer»

LUCAS test
Iodoform test
Victor Meyer's test
All of these

Solution :Propan-1-ol does not RESPOND to iodoform test on reaction with `I_(2)` and `NAOH`.
Discussion
6.

Propadiene overset(HCl(1eq))to A overset(O_2// LiAlH_4)to B + C (B and C are organic products ). If molecular mass of B is 32 then approximate molecular mass of C is

Answer»

98
68
58
46

Solution :A is 2-chloro PROPENE. B and C are
Discussion
7.

Proof spirit contains about

Answer»

`40%` alcohol by weight
`50%` alcohol by weight
`25%` alcohol by weight
`10%` alcohol by weight

Solution :`49.3%` by weight.
Discussion
8.

Proof spirit (by volume) is a mixture of

Answer»

`67.1%` ETHYL Alcohol `+32.9%` WATER
`95.87%` Ethyl Alcohol `+4.13%` Water
`57.1%` Ethyl Alcohol `+42.9%` Water
None of the above

Answer :C
Discussion
9.

Prolonged oxidation of Ethane-1,2,-diol using periodic acid gives :

Answer»

ACETIC acid
Formic acid
Oxalic acid
Glyoxal

Solution :`{:(CH_(2)-CH_(2)),("|""|" overset(HIO_(4))rarr2HCHO),(OH""OH):}`
`UNDERSET("Further")(HCHO overset([O])RARR HCOOH)`
Discussion
10.

Prof. Jonh gave a task to synthesize to three of his stundent ali, divakar and micheal. Difference routes were adopted by them. Ali's method Which of the following statements is correct?

Answer»

Ali and DIVAKAR will GET DESIRED product but YIELD will be better in divakar method
Ali and Micheal will get desired product but yield will be better in micheal method
Only micheal will get desired product
All of them will get desired product.

Answer :B
Discussion
11.

Products. The total number of pi bonds in the products is

Answer»


SOLUTION :NUMBER of `pi` BONDS is 7 on PRODUCT 'X''
Discussion
12.

Products of the following reaction Me_2C=CHCH_3 underset((ii)(CH_3)_2S)overset((i)O_3)to ? Are

Answer»

`CH_3CHO+CH_3COOH`
`Me_2CO+ CH_3CHO`
`Me_2CO+ CH_3COOH`
`2 Me_2CO`

Solution :`(CH_3)_2S` is USED for DECOMPOSITION of OZONIDES to give almost pure products .
Discussion
13.

Products obtained in above Wurtz reaction is/are .

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SOLUTION :
Discussion
14.

Products is

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SOLUTION :(d)CANNIZARO REACTION
Discussion
15.

products formed are ?

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SOLUTION :
is ATTACKED at POSITION ANDC is FORMED
Discussion
16.

products formed are

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SOLUTION :
Discussion
17.

Products A and B formed in the following reactions are respectively:

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SOLUTION :N//A
Discussion
18.

Products = ?

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SOLUTION :ALDOL CONDENSATION.
Discussion
19.

Product will be

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Both A and B

Answer :C
Discussion
20.

Product/s. The products formed are

Answer»

`CH_3 - OVERSET(O)overset(||)C-COOH`
`HOOC - COOH`
`CH_3-overset(O)overset(||)C-overset(O)overset(||)C-COOH`
Both (1) and (2)

Solution :`CH_3 C - overset(O)overset(||)C-COOH and HOOC - COOH`
Discussion
21.

Product of the reaction is

Answer»

RACEMIC
Diastereomers
MESO
PURE enantiomers

SOLUTION :Racemic
Discussion
22.

Product of the given reaction CH_(3)-CH=CH-Choverset(O_(3)//CH_(2)Cl_(2))underset(-78^(@)C)rarr will be :

Answer»

`CH_(3)-CHO`
`CH_(3)-COOH`
`CH_(3)-underset(OH)underset("|")"CH"-underset(OH)underset("|")"CH"-CH_(3)`

SOLUTION :
Discussion
23.

Product of the reaction is

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`(CH_(3))_(2)C=CH_(2)`

Solution :NITRATION mixture gives `NO_(2)^(+)` as a ELECTROPHILE.
Tert-Butyl group was replaced by `NO_(2)`, then it makes benzene ring as deactivating group.
Discussion
24.

Product of the Clemmensen reduction is ,

Answer»




ANSWER :3
Discussion
25.

Product of addition of HBr on the alkene C_(6)H_(5) - CH = CH - CH_3 is

Answer»

`C_6H_5-UNDERSET(BR)underset(|)(CH)-CH_2-CH_3`
`C_6H_5-CH_2-underset(Br)underset(|)(CH)-CH_3`
`C_6H_5-CH_2-CH_2-CH_2-Br`
`Br-C_6H_4-CH_2-CH_2-CH_3`

ANSWER :A
Discussion
26.

Product of above reaction is :

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ANSWER :B
Discussion
27.

Product obtained in above reaction is :

Answer»

`3CH_(3)CHO`
`3HCHO`
`3HCOOH`
`3CH_(3)OH`

ANSWER :B
Discussion
28.

Product obtained in above reaction are :

Answer»




ANSWER :D
Discussion
29.

…………product obtained from the chlorination of benzene sulphonic acid in presence of FeCl_(3).

Answer»

SOLUTION :m CHLOROBENZENE sulphonic ACID.
Discussion
30.

product not dormed is

Answer»




Solution :After ozonolysis base attacksmore acidic HYDROGEN forming CARBANION. NUCLEOPHILE created ATTACKS carbonyl carbon.
Discussion
31.

Product M cannot respond with :

Answer»

2,4 -DNP
AMMONICAL SILVER nitrate
Sodium hypoiodite
Sodium bicarbonate

Answer :D
Discussion
32.

product is

Answer»




SOLUTION :
Discussion
33.

Product is:

Answer»


ANSWER :`(##RES_CHM_ROHR_E01_019_A01##)`
Discussion
34.

Product formed is

Answer»




SOLUTION :
Discussion
35.

Product formed on heating Pb(NO_(3))_(2) are

Answer»

`PBO,N_(2),O_(2)`
`Pb(NO_(2))_(2),O_(2)`
`PbO,NO_(2),O_(2)`
`Pb,N_(2),O_(2)`

Solution :`Pb(NO_(3))_(2)toPbO+NO_(2)+O_(2)`
Discussion
36.

Product can be

Answer»




Solution :(A),(B),(C ),(D)
Due to Aldol CONDENSATION as WELL as dispropornation REACTION.
Discussion
37.

Product ( C ) of the reaction is

Answer»




SOLUTION :
(SALT of `BETA`-KETOACID)
Discussion
38.

Product (C) is

Answer»




ANSWER :B
Discussion
39.

Product C is :

Answer»




ANSWER :D
Discussion
40.

Product (B) in this reaction is ,

Answer»




SOLUTION :
CYCLOPENTANONE
Discussion
41.

Product (A) undergoes how many stroctoral isomers ?

Answer»


Solution :
(2) those are (i) positinal isomers.(II) FUNCTIONAL GROUP isomerism.
Discussion
42.

Product (A) is not ?

Answer»




SOLUTION : is OBTAINED
Discussion
43.

Product (A) is:

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Solution :Once a SMALL amount of `OsO_(4)` is used up the `Os(VI)` by product is oxidized with is the reaction MIXTURE by the amine oxide acts as the ultimate OXIDANT.
Discussion
44.

product A is :

Answer»




SOLUTION :
(L.D.A = LITHIUM di Isopropyl Amide)
L.D.A is sterically hindred BASE, so removes proton from less SUBSTITUTED `alpha-c` of ketone
Discussion
45.

Product (A) is ,

Answer»




Solution :In basic medium halide UNDERGO elimination reaction(`E_(2)`ELIMANATION BIMOLECULAR)
Discussion
46.

product (A) and (B) in above reactionis :

Answer»

`O^(_)-overset(O)overset(||)S-O-H,O^(-)-overset(O)overset(||)S-O-CH_(3)`
`O^(_)-overset(O)overset(||)S-O-H,O^(-)-""_(o+)underset(O^(-))underset(|)overset(O)overset(||)S-O-CH_(3)`
`O^(_)-overset(O)overset(||)S-O-CH_(3),O^(THETA)-""^(o+)underset(CH^(3))underset(|)overset(O)overset(||)S-H`
`O^(-)-""^(o+)underset(H)underset(|)overset(O)overset(||)S-O,O^(-)-underset(CH_(3))underset(|)overset(O)overset(||)S-O^(-)`

Solution :
( sulphur is better nucleophoile than OXYGEN )(but oxygen is better base than sulhur)
(A) Is FORMED by ACID -base reaction therefore oxygen will REACT.
(B) oid formed by electrophioe and nueophile reaction therefore sulphur wuill react.
Discussion
47.

product

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SOLUTION :Clemmensons REDUCTION
Discussion
48.

Which of the following method of separation can be applied to the mixture of liquids having different boiling points ?

Answer»

SIMPLE DISTILLATION
FRACTIONAL distillation
CRYSTALLIZATION
Fractional crystallization

ANSWER :A
Discussion
49.

Among the reactions, F_(2(g)) + 2e^(-) to 2F_((g))^(-) and Cl_(2(g)) + 2e^(-) to 2Cl_((g))^(-) which is more feasible ? Give the reason.

Answer»

Solution :`F_(2(g)) + 2E^(-) to 2F_((g))^(-)` is easy.
Though electron gain ENTHALPY of `Cl_((g))` to give `Cl_((g))^(-)` is more than that of `F_((g))` to give the bond DISSOCIATION of `F_(2(g))` is very less than that of `Cl_(2(g))`.
Discussion
50.

Process (A) : F_(2(g)) + 2e^(-) rarr 2F^(-)_((g)) , Process (B) : Cl_(2(g)) +2e^(-) rarr 2Cl_(g)^(-) which of these process is easy ? Why ?

Answer»

SOLUTION :`F_(2(g)) + 2e^(-) rarr 2F_((g))^(-)` is EASY.
Though electron GAIN enthalpy of `Cl_((g))` to give `Cl_((g))^(-)` is more thab that of `F_((g))`to give `F_((g))^(-)` , the bond dissociation of `F_((2(g)))` is very less than that of `Cl_(2(g))`
Discussion