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.	Explain the terms with example : Homopolymer
Answer» Solution :A POLYMER made from IDENTICAL repeating UNIT of one monomer is called homopolymer. Examples : POLYTHENE, PVC, Nylon - 6.

2.	Homolytic fission of C-C bond in ethane gives an intermediate in which carbon is….hybridised:
Answer» `SP^3` `sp^2` sp `sp^2d` ANSWER :A

3.	Homolytic fission of C-C bond in ethane gives an intermediate in which carbon is
Answer» `sp^(3)` hybridised `sp^(2)` hybridised sp hybridised `sp^(3)` d hybridized Solution :`CH_(3)-CH_(3)UNDERSET("BOND fission")OVERSET("Homolytic")rarr underset("Methyl free radicals")(CH_(3)^()+CH_(3)^())` FRE radicals is formed which is `sp^(2)`-hybridized.

4.	Homologous compounds have:
Answer» Same CHEMICAL properties Same molecular weight Same PHYSICAL properties Same MELTING point and BOILING point Answer :A

5.	Homolytic and heterolytic bond dissociation energies are
Answer» Same Equal Different None of these Answer :C

6.	H.O.M.O(Highest Occupied Molecular Orbital ) of CO molecular is :
Answer» Non-bonding M.O. with SLIGHT antiboding character Non-bonding M.O. with slight boding character Pure non-bonding M.O. None of above Solution :ACCORDING to coulsion MODEL, H.O.M.O. (Highest OCCUPIED MOLECULAR Orbital ) of CO molecule is non-bonding M.O. with slight antibonding character.

7.	Homogeneous mixture of camphor in N_(2) is an example of
Answer» GAS solute - Solid SOLVENT Solid solute - LIQUID solvent Solid solute - Gas Solvent Liquid solute - Solid solvent Answer :C

8.	Homogeneous mixing and compressibility both result from the fact that the molecules are far apart in gases. Mixing occurs because individual gaseous molecules have little interaction with their neighbours and, assuming that no reaction takes place, the chemical identities of those neighbours are irrelevant. Compressibility is possible in gases because only about 0.1% of the volume of a typical gas is taken up by the molecules themselves under normal circumstances, the remaining 99.9% is empty space. Assume molecules are spherical of radius 1A . Volume occupied by molecules in one mole of a gas at NTP is :
Answer» `2.52 m^(3)` `2.52 XX 10^(-4) m^(3)` `2.52 xx 10^(-6) m^(3)` `2.52 xx 10^(-2) m^(3)` Answer :C

9.	Hofmann rearrangement In the Hofmann rearrangement an unsubstituted amide is treated with sodium hydroxide and bromide to give a primary amine that has one carbon lesser than starting amide. In the migrating group is chiral then is configuration is retained. Electron releasing effects in the migrating group increases reactivity of Hofmann rearrangement. Which of the following compounds (s) cannot give Hofmann rearrangement :
Answer» `CH_(3)-CH_(2)-overset(O)overset(\|\|)(C)-NH-Ph` `Ph-overset(O)overset(\|\|)(C)-NH_(2)` ANSWER :B

10.	Hofmann's bromamide reaction is to convert
Answer» acid to alcohol alcohol to acid amide to amine amine to amide. Solution :Hofmann's bromamide reaction - AMIDES when HEATED with bromine and caustic SODA or caustic potash solution, yield primary amine containing one carbon atom LESS than the amide.

11.	Homogeneous system among the following is
Answer» milk sand in WATER urea in water BENZENE in water Answer :B

12.	Hofmann's hypobromite reaction affords a method of :
Answer» Preparing a TERTIARY AMINE Preparing a mixture of AMINES Stepping down a SERIES Stepping up a series ANSWER :C

13.	Hofmann rearrangement In the Hoffmann rearrangement an unsubstituted amide is treated with sodium hydroxide and bromine to give a primary amine that has one carbon lesser than starting amide. R-oversetoverset(O)(\|\|)C-NH_2+NaOH+Br_2toR-N=underset("isocyanate")(C=O)overset("hydrolysis")toR-NH_2 Mech : If the migrating group is chiral then its configuration is retained . Electron releasing effects in the migrating group increases reactivity of Hofmann rearrangement. H_3C-H_2C-undersetundersetunderset((R))(CH_3)(\|)overset(**)CH-oversetoverset(O)(\|\|)C-NH_2+CH_3-CH_2-undersetundersetunderset((S))(CH_3)(\|)oversetoverset(D)(\|)C-oversetoverset(O)(\|\|)C-overset(15)(NH_2)overset(Br_2//KOH)toProducts
Answer» B & C both are CORRECT Solution :In Hofmann rearrangement CONFIGURATION of central atom REMAIN uncharged as the rearragement is through cross over.

14.	Hofmann rearrangement In the Hoffmann rearrangement an unsubstituted amide is treated with sodium hydroxide and bromine to give a primary amine that has one carbon lesser than starting amide. R-oversetoverset(O)(\|\|)C-NH_2+NaOH+Br_2toR-N=underset("isocyanate")(C=O)overset("hydrolysis")toR-NH_2 Mech : If the migrating group is chiral then its configuration is retained . Electron releasing effects in the migrating group increases reactivity of Hofmann rearrangement. Which of the following compound cannot give Hofmann rearrangement :
Answer» `CH_3-CH_2-oversetoverset(O)(\|\|)C-NH-Ph` `Ph-oversetoverset(O)(\|\|)C-NH_2` Solution :Only PRIMARY AMIDE gives Hofmann REARRANGEMENT

15.	Hofmann rearrangement In the Hoffmann rearrangement an unsubstituted amide is treated with sodium hydroxide and bromine to give a primary amine that has one carbon lesser than starting amide. R-oversetoverset(O)(\|\|)C-NH_2+NaOH+Br_2toR-N=underset("isocyanate")(C=O)overset("hydrolysis")toR-NH_2 Mech : If the migrating group is chiral then its configuration is retained . Electron releasing effects in the migrating group increases reactivity of Hofmann rearrangement. Arrange the following amides according to their relative reactivity when react with Br_2 in excess of strong base
Answer» IVgtIgtIIgtIII IIgtIgtIIIgtIV IIgtIVgtIIIgtI IIgtIgtIVgtIII Solution :Electron donating GROUPS, ACCELERATES the process of REARRANGEMENT & electron with DRAWING groups retard it.

16.	Hofmann rearrangement during the conversion of an amide to amme is
Answer» intenn olecular intramolecular both (a) and (B) NONE of these. Answer :B

17.	Hofmann rearrangement during the conversion of an amide to amine involves……..rearrangement :
Answer» Intermolecular Intramolecular Both (a) and (B) None Answer :B

18.	Hofmann bromide degradation reaction is shown by
Answer» `ArNH_2` `ArCONH_2` `ArNO_2` `ArCH_2 NH_2` SOLUTION :`ArCONH_2`

19.	Hofmann Bromide Degradation reaction is shown by____
Answer» `ArNH_(2)` `ArCONH_(2)` `ArNO_(2)` `ArCH_(2)NH_(2)` ANSWER :B

20.	Hofmann bromamide reaction involves the conversion of amides into amines. Which of the following undergo at fastest rate under identical conditions
Answer» SOLUTION :EDG `uarr` RATE of `"Rea"^(N)uarr`

21.	Hoffmann's bromamide reaction is used to convert
Answer» an amine to AMIDE an amide to amine an alcohol to acid an acid to alcohol. Solution :`underset("Amide")(RCONH_(2))+Br+4KOH OVERSET("Heat")(rarr) underset("ANILINE")(R-NH_(2))+2KBr+K_(2)CO_(3)+2H_(2)O`

22.	Hoffmann bromanide degradation is
Answer» CONDENSATION REACTION ADDITION reaction Molecular rearrangement reaction Coupling reaction Solution :Definition

23.	Hoffmann bromamidereaction involves conversion of a carboxylic acid amide into an amine with a loss of a carbon atom on treatment with aqueous sodium hypobromite. Thus Hoffmann result in shortening of a carbon chain. R=overset(O)overset(\|\|)C-NH_(2)underset(NaOH)overset(Br_(2))rarrR-NH_(2)+NaBr+Na_(2)CO_(3) Mechanism of the reaction is :
Answer» NONE of these Solution :

24.	Hoffmann degradation of m-bromo benzamide gives:
Answer» Aniline m - Bromoaniline Bromobenzene m - Bromoethyl benzene. Solution :

25.	Hoffmann bromamidereaction involves conversion of a carboxylic acid amide into an amine with a loss of a carbon atom on treatment with aqueous sodium hypobromite. Thus Hoffmann result in shortening of a carbon chain. R=overset(O)overset(\|\|)C-NH_(2)underset(NaOH)overset(Br_(2))rarrR-NH_(2)+NaBr+Na_(2)CO_(3) Mechanism of the reaction is : Product (A) :
Answer» `Ph-NH_(2)`<BR>`Ph-CH_(2)-NH_(2)` `Ph-NH-CH_(2)` Solution :`PhCONH_(2)underset(KOH)overset(Br_(2))rarrPhNH_(2)`

26.	Hoffmann bromamidereaction involves conversion of a carboxylic acid amide into an amine with a loss of a carbon atom on treatment with aqueous sodium hypobromite. Thus Hoffmann result in shortening of a carbon chain. R=overset(O)overset(\|\|)C-NH_(2)underset(NaOH)overset(Br_(2))rarrR-NH_(2)+NaBr+Na_(2)CO_(3) Mechanism of the reaction is : Which of the following will not giveHoffmann bromamide reaction.
Answer» `CH_(3)-OVERSET(O)overset(\|\|)C-NH_(2)` Solution :`2^(@)` AMIDE do not GIVE HOFFMANN.

27.	Hoffmann bromamidereaction involves conversion of a carboxylic acid amide into an amine with a loss of a carbon atom on treatment with aqueous sodium hypobromite. Thus Hoffmann result in shortening of a carbon chain. R=overset(O)overset(\|\|)C-NH_(2)underset(NaOH)overset(Br_(2))rarrR-NH_(2)+NaBr+Na_(2)CO_(3) Mechanism of the reaction is : Number of moles of NaOh consumed in abovereaction.
Answer» 1<BR>2 3 4 Solution :`RCONH_(2)+Br_(2)+4NaOHrarrRNH_(2)`

28.	Hoffmann bromamide reaction can be used to reduce the length of carbon chain by one carbon atom.
Answer» ANSWER :1

29.	Hoffmann bromamide or hypobromite reactions is given by
Answer» amines esters alcohols amides ANSWER :D

30.	What is Hofmann bromamide degradation reaction ?
Answer» `Ar-NH_2` `Ar-CONH_2` `Ar-NO_2` NONE of these. Answer :B

31.	Hoffmann bromamide degradation reaction is shown by:
Answer» `ArNH_(2)` `ArCONH_(2)` `ArNO_(2)` `ArCH_(2)NH_(2)` Answer :B

32.	Hoffman Bromamide degradation reaction involves :
Answer» `Ar-NH_2` `Ar-CONH_2` `Ar-NO_2` NONE of these Answer :B

33.	Hoe many alkenes on catalytic hydrogenation give isopentane as a product (consider only structural isomers) ?
Answer» `2` `3` `4` `5` SOLUTION :N//A

34.	HO - CH_(2) CH_(2)CH_(2) - overset(O)overset(\|\|)(C) - OHon reaction with H^(+)forms .
Answer» ANSWER :A

35.	HNO_(3) + P_(2)O_(3) to A + B A is an oxiacid of phosphorous and B is a oxide of Nitrogen. What will be A & B ?
Answer» `H_(3)PO_(4), N_(2)O_(3)` `HPO_(3), N_(2)O_(3)` `HPO_(3), N_(2)O_(5)` `H_(3)PO_(3), N_(2)O_(5)` ANSWER :C

36.	HNO_3 oxidises
Answer» `H_2O_2` `H_2S` `SO_2` `All` ANSWER :D

37.	HNO_(3) is more volatile than water. If an aqueoussolution of HNO_(3) is taken then which of the following graph correctly represents variation of vapour pressure/boiling point with composition?
Answer» ANSWER :A

38.	HNO_(3) is manufactured by:
Answer» BirkelandEyde's PROCESS HALBER's process CONTACT's process Fischer-Tropsch's process. ANSWER :A

39.	HNO_3 is manufactured by:
Answer» BIRKELAND and Eyde.s process Haber.s process Contact.s process Fischer-Tropsch.s process Answer :A

40.	HNO_(3) in aqueous solution yields
Answer» `NO_(3)^(-)` and `H^(+)` `NO_(3)^(-) ` and `H_(3)O^(+)` `NO_(2)^(-)`and `OH^(-)` `N_(2)O_(5)` and `H_(2)O` ANSWER :B

41.	HNO_3 has the following properties:
Answer» OXIDIZING and bleaching Acidic and oxidizing Basic and reducing Reducing and bleaching Answer :B

42.	HNO(3) can be in Al vessel bu NaOH cant't be, because
Answer» `Al` developes a protective layer of oxide with `HNO_(3)` `Al` reacts with `HNO_(3)` to DISSOLVE it `Al` reacts with `NaOH` FORMING `NaAlO_(2)` `Al` develops a protective layer of oxide with `NaOH`. Solution :Al turns passive due to formation of protective oxide layer, HENCE reaction ceases. But Al dissolves in `NaOH` forming `NaAlO_2` hence it cannot be stored in Al vessels. `2AL + 2NAOH + 2H_2O rarr 2NaAlO_2 + 3H_2`.

43.	HNO_(2) acts as an oxidant with which one of the following reagent :-
Answer» `KMnO_(4)` `H_(2)S` `K_(2)Cr_(2)O_(7)` `Br_(2)` Answer :B

44.	HNO, cannot be used as an acid medium along with KMnO_(4) Why?
Answer» Solution :`HNO_(3)` cannot be USED SINCE it is a good oxidising agent and it REACTS with reducing agents in the REACTION.

45.	Hl when heated with conc. H_2SO_3, it forms
Answer» `HIO_(3)` `KIO_(3)` `I_(2)` All of these Solution :`2HI+H_(2)SO_(4) to I_(2)+SO_(2)+2H_(2)O`

46.	Hl cannot be prepared by the action of conc. H_(2)SO_(4) on Kl because
Answer» HI is a stronger acid than `H_(2)SO_(4)` HL is more VOLATILE than `H_(2)SO_(4)` conc. `H_(2)SO_(4)` is an oxidant and HI is a reductant. conc. `H_(2)SO_(4)` is a dehydrating agent. ANSWER :C

47.	HI was heated in a sealed tube at 400^(@)C till the equilibrium was reached. HI was found to be 22% decomposed. The equilibrium constant for dissociation is
Answer» 0.282 0.0796 0.0199 1.99 Answer :C

48.	Histamine is responsible for nasal congestion associated with ________________.
Answer» SOLUTION :COMMON COLD

49.	HIO_3 on heating gives
Answer» `I_2` `O_2` `I_2O_5` HI Answer :C

50.	Histidine can be converted into histamine by
Answer» DECARBOXYLATION of histidine Dehydration of histidine Decarboxylation of histamine Dehydrohalogenation of histamine Answer :A