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.	Diborane hydrolyses to give :
Answer» `B_(2)O_(3)` `H_(3)BO_(3)` `HBO_(2)` does not hydrolyse. Solution :`B_(2)H_(6)+6H_(2)Orarr UNDERSET(Borica ACID")(2H_(3)BO_(3))+6H_(2)`

2.	Diborane has ……………….. B-H bonds.
Answer» ANSWER :EIGHT

3.	Diborane combines with ammonia at 120^(@)C to give
Answer» `B_(2)H_(6).NH_(3)` `B_(2)H_(6).2NH_(3)` `B_(2)H_(6).3NH_(3)` `B_(2)H_(6).4NH_(3)` ANSWER :B

4.	Diazotizationof aniline gives
Answer» NITROBENZENE DINITROBENZENE benzaldehyde benzene diazoniumchloride Answer :D

5.	Diazotisation reaction is used to prepare
Answer» alcohol phenol aldehyde ketone Answer :B

6.	Diazotisation of aniline followed by hydrolysis gives
Answer» nitrophenol phenol chlorobenzene benzoic acid Answer :B

7.	Diazotisation can be carried out by the action ofNaNO_2 and dilute HClat ice cold temperature on :
Answer» Aromatic secondary amine Aromatic PRIMARY amine Aromatic NITRO COMPOUND ALIPHATIC amine Answer :B

8.	Diazonium salts are stable only at :
Answer» Very HIGH temperature Very low temperature High temperature NONE ANSWER :B

9.	Diazonium salt formation and coupling reaction: When a reaction mixture of phenyl amine and nitrous acid is kept below 10^(@)C, a diazonium salt is formed. This reaction is called diazotization reaction. The diazonium ion, -N_(2)^(+), is rather unstable and decomposes readily to nitrogen. However, delocalization of the diazonium from pi-bond electron over a benzene ring phenyl diazonium sufficiently for it to form at low temperature. The phenyl diazonium ion behaves as an electrophile, and will attack another arene molecule such as phenol. Electrophilic subsitiution takes at the 4 position, producing 4-hydroxy phenyl azobenzene. The reaction is known as coupling reaction. The compound formed is an energetically stable, yellow azo dye (the azo group is -N=N-) The stability is due to extensive delocalisation of electrons via the nitrogen- nitrogen double bonds. The product which is a red azodye obtained on reacting benzene diazonium chloride with one of the following compounds :
Answer» ANSWER :B

10.	Diazonium salt formation and coupling reaction: When a reaction mixture of phenyl amine and nitrous acid is kept below 10^(@)C, a diazonium salt is formed. This reaction is called diazotization reaction. The diazonium ion, -N_(2)^(+), is rather unstable and decomposes readily to nitrogen. However, delocalization of the diazonium from pi-bond electron over a benzene ring phenyl diazonium sufficiently for it to form at low temperature. The phenyl diazonium ion behaves as an electrophile, and will attack another arene molecule such as phenol. Electrophilic subsitiution takes at the 4 position, producing 4-hydroxy phenyl azobenzene. The reaction is known as coupling reaction. The compound formed is an energetically stable, yellow azo dye (the azo group is -N=N-) The stability is due to extensive delocalisation of electrons via the nitrogen- nitrogen double bonds. The azo dye obtained on reacting 4-aminophenol with nitrous acid (in dilute hydrochloric acid) below 10^(@)C and coupling the resulting diazonium salt with phenol is :
Answer» ANSWER :C

11.	Diazonium salt decomposes as C_(6)H_(5)N_(2)^(+)Cl^(-) to C_(6)H_(5)Cl+N_(2) At 0^(@)C, the evolution of N_(2) becomes two faster when the initial concentration of the salt is doubled. Therefore, it is
Answer» a FIRST order REACTION a second order reaction independent of the initial concentration of the salt a zero order reaction Solution :As doubling the initial CONC. Doubles the RATE of reaction, order=1

12.	Diazonium salt formation and coupling reaction: When a reaction mixture of phenyl amine and nitrous acid is kept below 10^(@)C, a diazonium salt is formed. This reaction is called diazotization reaction. The diazonium ion, -N_(2)^(+), is rather unstable and decomposes readily to nitrogen. However, delocalization of the diazonium from pi-bond electron over a benzene ring phenyl diazonium sufficiently for it to form at low temperature. The phenyl diazonium ion behaves as an electrophile, and will attack another arene molecule such as phenol. Electrophilic subsitiution takes at the 4 position, producing 4-hydroxy phenyl azobenzene. The reaction is known as coupling reaction. The compound formed is an energetically stable, yellow azo dye (the azo group is -N=N-) The stability is due to extensive delocalisation of electrons via the nitrogen- nitrogen double bonds. Benzene diazonium chloride on reaction with phenol in weakly basic medium gives :
Answer» DIPHENYL ETHER p-hydroxy AZOBENZENE CHLOROBENZENE benzene Answer :B

13.	Diazomethene (CH_2N_2) on decomposition forms singlet methylene (":"CH_2) which gets attached to different non-equivalent C-H bonds of alkanes formed when pentane (CH_3CH_2CH_2CH_2CH_3) reacts with singlet methylene. Assuming methylene to be highly reactive and less selective, calculate the probable amounts the formed alkanes.
Answer» SOLUTION :Three alkanes are formed when pentane reacts with singlet methylene because there are three non-equivalent `C-H` bonds in pentane molecule. So, the alkanes formed are: `underset("hexane")(CH_(3)CH_(2)CH_(2)CH_(2)CH_(2)CH_(3))` `underset("2-methylpentane")(CH_(3)overset(overset(CH_(3))(\|))(C)HCH_(2)CH_(2)CH_(3)) " " underset("hexane")(CH_(3)CH_(2) overset(overset(CH_(3))(\|))(C)HCH_(2)CH_(3))` As methylene is HIGHLY reactive and less selective, its insertion occurs in a random fashion. So, the amounts of the formed compounds are CALCULATED on the basis of probablility factor and number of equivalent `C-H` bonds. For example, PERCENTAGE of hexane. `(CH_(3)CH_(2)CH_(2)CH_(2)CH_(2)CH_(3)) = ((6)/(12)) xx 100= 50` Percentage of 2-methylpentane `CH_(3) - overset(overset(CH_(3))(\|))(C)H - CH_(2)CH_(2)CH_(3) = ((4)/(12)) xx 100 = 33.3` Percentage of 3-methylpentane `= ((2)/(12)) xx 100 = 16.7`

14.	Diazo-coupling is useful to prepare some
Answer» PESTICIDES proteins Dyes vitamins Solution :

15.	Diazo coupling is useful to prepare
Answer» Pesticides DYES Proteins Vitamins Solution :Dyes.

16.	Diazomethane reacts with carboxylic acids to produce :
Answer» Ester Alcohol Amine Lmines Answer :A

17.	Diazo- coupling is useful to prepare :
Answer» PESTICIDES Dyes Proteins Vitamins ANSWER :B

18.	Diaspore is:
Answer» `Al_2O_3.H_2O` `Al_2O_3.2H_2O` `Al_2O_3` `Al_2O_3.3H_2O` ANSWER :A

19.	Diane diol worked for several days to prepare compounds shown. She labelled and slept. When she got up all lables had slipped. Her friend Glycol said that They could be easily distuingished by two experiments. Which of the m were optically active. How many products were obtained when each is treated with periodic acid. after experimentation Diane found the following results 1. Compounds A, E, F were optically active and B, C and D were optically inactive. 2. one product was obtained from the reaction of A, B and D with periodic acid. 3. Two products were obtained from the reaction of F with periodic acid. 4. C and E didnt react with periodic acid. The structure of the compounds were Answer the following based on the above observations, On the basis of the observation which compounds cannot be distinguished ?
Answer» COMPOUNDS C and F compounds B and D compounds B and C compounds D and E Answer :B

20.	Diaspore is _________.
Answer» `Al_(2)O_(3)` `Na_(3)AlF_(6)` `Al_(2)O_(3).H_(2)O` `Al_(2)O_(3).2H_(2)O` ANSWER :C

21.	Diane diol worked for several days to prepare compounds shown. She labelled and slept. When she got up all lables had slipped. Her friend Glycol said that They could be easily distuingished by two experiments. Which of the m were optically active. How many products were obtained when each is treated with periodic acid. after experimentation Diane found the following results 1. Compounds A, E, F were optically active and B, C and D were optically inactive. 2. one product was obtained from the reaction of A, B and D with periodic acid. 3. Two products were obtained from the reaction of F with periodic acid. 4. C and E didnt react with periodic acid. The structure of the compounds were Answer the following based on the above observations, Which structure could be suggested for A
Answer» II III IV VI Answer :B

22.	Diane diol worked for several days to prepare compounds shown. She labelled and slept. When she got up all lables had slipped. Her friend Glycol said that They could be easily distuingished by two experiments. Which of the m were optically active. How many products were obtained when each is treated with periodic acid. after experimentation Diane found the following results 1. Compounds A, E, F were optically active and B, C and D were optically inactive. 2. one product was obtained from the reaction of A, B and D with periodic acid. 3. Two products were obtained from the reaction of F with periodic acid. 4. C and E didnt react with periodic acid. The structure of the compounds were Answer the following based on the above observations, Which structure could be compound F
Answer» II III IV VI Answer :C

23.	Diamond structure can be considered as ZnS (Zinc blend) structure in which each Zn^(2+) in alternate tetrahedral void and S^(2-) in cubic close pack arrangement is replaced by one carbon atom.If C C covalent bond length in diamond is 1 .SÅ,what is the edge length of diamond unit cell (2 =8).
Answer» 3.46Å 6.92 Å 1.73 Å 3 Å Answer :A

24.	Diamond structure can be considered as ZnS (Zinc blend) structure in which each Zn^(2+) in alternate tetrahedral void and S2– in cubic close pack arrangement is replaced by one carbon atom. If C – C covalent bond length in diamond is 1.5 Å, what is the edge length of diamond unit cell (z = 8).
Answer» ANSWER :3.46 Å

25.	Diamonds are formed from graphite under high pressure in coal mines. Calculate the equilibrium pressure (in atm) at which graphite is converted to diamonds at 25^(@)C (assumed constant) given densities of rho_("graphite")=2g//c c & rho_("diamond")=3g//c c (DeltaG_(f)^(@)) for diamonds is "3 k J m"^(-1) from graphite
Answer» ANSWER :15001

26.	Diamond is harder than graphite because
Answer» Graphite is planar DIAMOND has FREE electron Graphite is `sp^(3)` hybridised None of these Solution :In the crystalline lattice of diamond, each carbon atom is linked to FOUR other carbon atoms tetrahedrally by using `sp^(3)`-hybrid orbitals giving rise to a rigid three dimensional network of carbon atoms. For this reason, diamond is extremely hard. Graphite has a HEXAGONAL layer structure in which each carbon atom is linked to to three other carbon atoms by using `sp^(2)` hybrid orbitals. the hexagonal layers in graphite are held together by weak vander WAAL's force. For this reason, graphite is soft.

27.	Diamond is a non-conductor of electricity but a good conductor of heat-why?
Answer» SOLUTION :Due to absence of free electrons, it is a non-conductor of electricity. It has the highest known THERMAL CONDUCTIVITY because thermal motion is distributed in its 3D -structure very effectively.

28.	Diamond has fcc crystal structure in which each carbon atom is attached with four other carbon atoms, then the number of carbon atoms per unit cell in diamond are.
Answer» 12 6 8 4 ANSWER :C

29.	Diamond has each of the following properties except :
Answer» HIGH MELTING point ability to CONDUCT ELECTRICITY inertness to chemicals extreme hardness. Solution :DIAMOND does not conduct electricity.

30.	Diamond belongs to the crystalsystem :
Answer» cubic triclinic tetragonal HEXAGONAL SOLUTION : DIAMOND BELONGS to cubic SYSTEM.

31.	Diamond - bad conductor of electricity.
Answer» Solution :In DIAMOND, all four valance electrons of carbon are involved in BONDING there is no FREE electrons for CONDUCTIVITY.

32.	Diamond and silicon carbide are the examples of………………….. solids.
Answer» SOLUTION :COVALENT

33.	Diamond and graphite are shown to be allotropic forms of carbon by the fact that
Answer» diamond is hard but GRAPHITE is soft diamond is transparent while graphics is opaque they have DIFFERENT crystal structures both from `CO_(2)` when burnt. Solution :Diamond and graphite are the allotropes of carbon because both on burning produce `CO_(2)` and thus show similar CHEMICAL PROPERTIES.

34.	Diamond and graphite are:
Answer» Isomers Isotopes Allotropes Polymers Answer :C

35.	Dialysis is a method of purification of sols. But prolonged dialysis of the sol makes it unstable. Why?
Answer» Solution :Traces of electrolytes in the SOL, impart CHARGE to the dispersed phase particles making it stable. PROLONGED dialysis REMOVES all the electrolytes, thus making the sol unstable.

36.	Diamagnetic oxide of chlorine is:
Answer» `ClO_3` `Cl_2O_6` `ClO_2` None Answer :B

37.	Dialysis can separate
Answer» GLUCOSE and FRUCTOSE Glucose and SUCROSE Glucose and NaCl glucose and proteins Answer :D

38.	Dialkyl ethers react with very few reagents other acids. The only reactive sites that molecules of a dialkyl ether has to another reactive substance are the C H bonds of the alkyl groups and the O group of the ether linkage. Heating dialkyl ethers with very strong acids. (HI, HBr, and H_(2)SO_(4)) causes them to undergo reactions in which the carbon-oxygen bond breaks. When mixed ethers are used, the alcohol and alkyl iodide that form depend on the nature of the alkyl groups. Mechanism is by S_(N)^(2) reaction or S_(N)^(1). B gives positive lucas test in a few seconds. Which is 'B'.
Answer» ANSWER :B

39.	Dialkyl ethers react with very few reagents other acids. The only reactive sites that molecules of a dialkyl ether has to another reactive substance are the C H bonds of the alkyl groups and the O group of the ether linkage. Heating dialkyl ethers with very strong acids. (HI, HBr, and H_(2)SO_(4)) causes them to undergo reactions in which the carbon-oxygen bond breaks. When mixed ethers are used, the alcohol and alkyl iodide that form depend on the nature of the alkyl groups. Mechanism is by S_(N)^(2) reaction or S_(N)^(1).
Answer» ANSWER :B

40.	Dialkyl ethers react with very few reagents other acids. The only reactive sites that molecules of a dialkyl ether has to another reactive substance are the C H bonds of the alkyl groups and the O group of the ether linkage. Heating dialkyl ethers with very strong acids. (HI, HBr, and H_(2)SO_(4)) causes them to undergo reactions in which the carbon-oxygen bond breaks. When mixed ethers are used, the alcohol and alkyl iodide that form depend on the nature of the alkyl groups. Mechanism is by S_(N)^(2) reaction or S_(N)^(1). What is the coorect order of reactivity towards conc. HI assuming S_(N)^(2) type cleavage?
Answer» `I GT IV gt III gt I` `I gt III gt IV gt II` `II gt I gt IV gt III` `IV gt II gt III gt I` Answer :B

41.	Dialkyl sulphides are known as :
Answer» Sulphonal Mercaptan Thioethers Thioesters Answer :C

42.	……………….diagrams normally consist of plots of triangle_(f)G^(0) vs T for the formation of oxides of common metals.
Answer» SOLUTION :ELLINGHAM

43.	Dialkyl cadmium reacts with a compound to from a ketone. The compound is:
Answer» Acit Acid chloride Ester CO Answer :B

44.	Diacetone alcohol is obtained when
Answer» 2 MOLECULES of ACETONE condense in presence of barium hydroxide 3 molecules of acetone condense in presence of barium hydroxide 3 molecules of acetone polymerise in presence of CONC `H_(2)SO_(4)` 3 molecules of acetone condese in presence of conc `H_(2)SO_(4)` Answer :A

45.	Diacetone alcohol is obtained by the reaction of :
Answer» Acetone and ethanol Acetone and CONC. `H_2SO_4` Acetone and `BA(OH)_2` Acetone and `Al(OH)_3` ANSWER :C

46.	Diabetes is detected using ......by testing urine of patients
Answer» Fehling's solution Tollen's reagent Benedict's solution Baeyer's reagent Solution :Benedict solution contains `CuSO_(4),Na_(2)CO_(3)` and sodium citrate. This permits FORMATION of a complex , which lowers the concentration of Cu (II) ions to such asn extent that it does not permit the precipitation of insoluble `Cu(OH)_(2)`. Benedict solution is more stable than Fehling's solution and is not AFFECTED by substance like creatine and uric acid present in urine. hence it is preferred to detect glucose in urine.

47.	Di-tert-glycols rearrange in the presence of acid to give alpha-trtiary ketones. The trivial name of the simplest glycol of this type is pinacol, and this type of reaction therefore is named pinacol rearrangement (in this specific case, the reaction is called a pinacol-pinacolone rearrangement). The rearrangement involves 4 steps. one of the hydroxyl group is protonated in the first step. A molecule of water is eliminated in the second step and a tertiary carbocation is formed. the carbocation rearranges in the third step into a more stable carboxonium ion via a [1, 2] rearrangement. In the last step, the carboxonium ion is deprotonated and the product ketone is obtained. Product (A) is :
Answer» `CH_(3)-overset(O)overset(\|\|)(C)-underset(CH_(3))underset(\|)overset(CH_(3))overset(\|)(C)-CH_(3)` None of these Solution :

48.	Diabetes is detected by testing urine of the patient with
Answer» TOLLEN's REAGENT NESSLER's reagent Benedict's reagent BRADY's reagent SOLUTION :Benedict.s reagent

49.	Di-tert-glycols rearrange in the presence of acid to give alpha-trtiary ketones. The trivial name of the simplest glycol of this type is pinacol, and this type of reaction therefore is named pinacol rearrangement (in this specific case, the reaction is called a pinacol-pinacolone rearrangement). The rearrangement involves 4 steps. one of the hydroxyl group is protonated in the first step. A molecule of water is eliminated in the second step and a tertiary carbocation is formed. the carbocation rearranges in the third step into a more stable carboxonium ion via a [1, 2] rearrangement. In the last step, the carboxonium ion is deprotonated and the product ketone is obtained. How many products obtained in above reaction ?
Answer» 1 2 3 4 Solution :(B) (b) 2 PRODUCTS are FORMED during REACTION

50.	Di-tert-glycols rearrange in the presence of acid to give alpha-trtiary ketones. The trivial name of the simplest glycol of this type is pinacol, and this type of reaction therefore is named pinacol rearrangement (in this specific case, the reaction is called a pinacol-pinacolone rearrangement). The rearrangement involves 4 steps. one of the hydroxyl group is protonated in the first step. A molecule of water is eliminated in the second step and a tertiary carbocation is formed. the carbocation rearranges in the third step into a more stable carboxonium ion via a [1, 2] rearrangement. In the last step, the carboxonium ion is deprotonated and the product ketone is obtained. Product 'P' is :
Answer» SOLUTION :