94420 + Interview Questions in Chemistry in Class 12

1.	Compounds with high heat of formation are less stable because
Answer» HIGH TEMPERATURE is required to synthesise them molecules of such compounds are distorted it is difficult to synthesise them energy rich state leads to instability Solution :Heat of formation is the heat evolved or absorbed when one mole of a matter is formed from its constituent elements. POSITIVE value of heat of formation indicates that the compound is less stable than its elements. Negative value of heat of formation indicates that the compound is associated with less energy as compared to its constituents and the compound is more stable. A high energy system is unstable. it tends to undergo SPONTANEOUS change in order to acquire a state of MINIMUM energy and maximum stability.

Discussion

2.	Compounds with antiseptic properties are...
Answer» `CHCl_3` `CHI_3` BORIC ACID 0.3 PPM AQUEOUS solution of `Cl_2` Answer :B::C

Discussion

3.	Compounds with antiseptic properties are ........
Answer» `CHCI_(3)` `CHI_(3)` Boric acid 0.3 ppm AQUEOUS SOLUTION of `Cl_(2)` Answer :B::C

Discussion

4.	Compoundswith antiseptic properties are "……….."
Answer» `CHCl_(3)` `CHI_(3)` boric acid `0.3 "ppm"` aqueous solution of `Cl_(2)` Solution :Antiseptics are the chemical substances which either kill or prevent the growth of microorganisms. (a) `CHCl_(3)` (CHLOROFORM) was used as an anaesthesia in surgery but not its is used in the production of the freon refrigerant `R-22`. (b) lodoform `(CHI)_(3)` produces iodine on coming in contact with skin, so it is used as an antiseptic for wounds. (c ) Boric acid `(H_(2)BO_(3))` in dilute aqueous solution is a weak antiseptic for EYES. THUS, chemical messenger gives messagle to the all without entereing the cell. (d) Chloric in the concentration of `0.2`, to `0.4 "ppm"` in aqueous solution and sulphur dioxide in very LOW concentrations, are disinfectants.

Discussion

5.	Compounds which rotate plane of plane polarised light in clockwise direction are knownas
Answer» DEXTRO ROTATORY LEAVE rotatory recemic MIXTURE OPTICAL in active Answer :A

Discussion

6.	Compounds which do not give coupling reaction with benzenediazonium chloride in acidic medium is/are
Answer» PHENOL Aniline Nitrobenzene Chlorobenzene Solution :

Discussion

7.	Compounds which contain two lone pairs and four bonds pairs are
Answer» `XeF_2` `XeF_4` `XeOF_2` `XeO_3` SOLUTION :`2l.p + 4b.p "" XeF_4.XeOF_4`

Discussion

8.	Compounds which contain one lone pair and six bond pairs are
Answer» `XeO_2 F_2 ` `XeO_4` `XeF_6` `XeOF_4` SOLUTION : He has low BOILING POINT due to weak intermolecular ATTRACTIVE force.

Discussion

9.	Compounds which can give periodic cleavage are :
Answer» SOLUTION :N//A

Discussion

10.	Compounds used in Holme's signal are ……………… .
Answer» `"PHOSPHINE"+"ACETYLENE"` `H_(3)PO_(3)+H_(3)PO_(4)` `"Calcium carbide"+"calcium phosphide"` `"Calcium carbonate"+"calcium phosphate"` SOLUTION :`"Calcium carbide"+"calcium phosphide"`

Discussion

11.	Compounds thatwill not exhibit geometrical isomerism-
Answer» `CH_(3)CH=CH_(2)` ANSWER :A::B::C

Discussion

12.	Compound(s) that on hydrogenation produce(s) optically inactive compound(s) is (are)
Answer» SOLUTION :Compounds (b) and (d) being enantiomers upon hydrogenationn given the same reduction product (I) Compound (I) being symmetrical does not contain a chiral carbon and hence is optically inactive. Thus, OPTION (b) and (d) are CORRECT. please find out yourself that OPTIONS (a) and (c) give optically active products.

Discussion

13.	Compounds P(C_6H_10) does not have any geometrical isomer.On ozonolysis, two products R(C_3H_4O) and Q(C_3H_6O) are formed.R gives negative iodoform test while Q responds positively towards I_2//NaOH solution.S, another isomer of P is an unsymetrical alkene and on ozonolysis produces T(C_6H_(10_7)) which also gives a yellow precipitate with l_2//NaOH solution and also positive test with Tollen's reagent Which of the following does not represent any of the molecules amongst P,Q,R,S & T.
Answer» SOLUTION :

Discussion

14.	Compounds present product mixture :
Answer» `Me-OVERSET(O)overset(\|\|)(C )-OH` SOLUTION :

Discussion

15.	Compounds P and R upon ozonolysis produce Q and S, respectively. The molecular formula of Q and S is C_(8)H_(8)O. Q undergoes cannizzaro reactions but not haloform reaction. Whereas S undergoes haloform reactions but not cannizzaro reactions. (i) P overset((i) O_(3)//CH_(2)Cl_(2))underset((ii)Zn//H_(2)O)tounderset((C_(8)H_(8)O))(Q) "" (ii) R overset((i) O_(3)//CH_(2)Cl_(2))underset((ii) Zn//H_2O)tounderset((C_(8)H_(8)O))(S) The option (s) with suitable combination of P and R , respectively is (are)
Answer» ANSWER :C::D

Discussion

16.	Compounds of nitrogen, phosphorus and sulphur such as ammonia, phosphoric acid and sulphuric acid are used in fertilizer industry. Write the chemical equation for the preparation of phosphoric acid (H_3PO_4) from H_3PO_3
Answer» SOLUTION :`4H_3PO_3overset"HEAT"to3H_3PO_4+PH_3`

Discussion

17.	Compounds I and II may be grouped as
Answer» diastereomers epimers `C-2` epimers all of the three Solution :When STRUCTURE `I` and `II` are `C-2` epimers. It IMPLIES that these are epimers and diastereomers too.

Discussion

18.	Compounds I and II can be distinguished by using reagent. (I) -4-Amino-2-methylbut-3-en-ol (II) -4-Amino-2,2dimethylbut-3-yn-1-ol
Answer» `NaNO_(2)//HCl`<BR>`Br_(2)//H_(2)O` `HCl//ZnCl_(2)` (ANHYDROUS) `Cu_(2)Cl_(2)//NH_(4)OH` Solution :

Discussion

19.	Compounds I and II can be distinguished by using reagent. {:(""I,""II),("4-Amino-2methylbut-3-en-2ol","4-Amino-2,2-dimethylbut-3-yn-1-ol"):}
Answer» `NaNO_2//HCl` `Br_2//H_2O` `HCl//ZnCl_2` (ANHYDROUS ) `Cu_2Cl_2 // NH_4 OH` Solution : (I) gives immediately TURBIDITY by LUCAS reagent and (II) does not GIVE turbidity appriciably.

Discussion

20.	Compounds having the same molecular formula but different strucutres are cllassified as:
Answer» METAMERISM optical isomerism structural ISOMERS functioal GROUP isomers. ANSWER :C

Discussion

21.	Compounds having same number and kind of atoms but different arrangement of atoms in their molecules are called:
Answer» allotropes isotopes isomers polymers Answer :C

Discussion

22.	compounds formed when the noble gases get entrapped in the cavities of crystal lattices of certain organic compounds are known as:
Answer» INTERSTITIAL Compounds Clathrates Hydrates Picrates Answer :B

Discussion

23.	Compounds formed by the union of two different halogens are called.......
Answer» SOLUTION :INTERHALOGEN COMPOUNDS

Discussion

24.	Compounds containing both amino and COOH groups are known as
Answer» Diamines UNKNOWN Amino acids ENZYMES ANSWER :C

Discussion

25.	Compounds containing both -NH_2 and -COOH groups are called
Answer» Proteins Dicarboxylicacids Amino ACIDS `ALPHA`-HYDROXY acids Answer :C

Discussion

26.	Compounds A(C_9 H_8 O_3) on reduction with Zn-Hg&HCI gave a carboxylic acid B(C_9 H_(10) O_3). B on further oxidation gave bezoic acid. A on strong heating gave a ketone C(C_8H_8 O). C reacts with I_2 -, NaOH to give yellow solid along with a salt. The salt on treatment with acid gave benzoic acid.Study the paragraph carefully and answer the following questions give belowthe structureof'C' is
Answer» SOLUTION :

Discussion

27.	Compounds (C ) and (D) respectively are :
Answer» `AgNO_3 and Hg(NO_3)_2` `AgNO_3 and Cu(NO_3)_2` `Pb(NO_3)_2 and Bi(NO_3)_3` `AgNO_3 and Bi(NO_3)_3` Solution :`Pb^(2+)+2I^(-)toPbI_2 darr ` (YELLOW), INSOLUBLE in `NH_3` solution. `PbI_2 + 2I^(-)to [PbI_4]^(2-)`(soluble complex) `PbI_2` is moderately soluble in boiling WATER giving a colourless solution. `Bi^(2+)+3I^(-)to BiI_3 darr` (black) , `BiI_3 darr+I^(-) to [BiI_4]^(-)` (soluble complex)

Discussion

28.	Compounds A(C_9 H_8 O_3) on reduction with Zn-Hg&HCI gave a carboxylic acid B(C_9 H_(10) O_3). B on further oxidation gave bezoic acid. A on strong heating gave a ketone C(C_8H_8 O). C reacts with I_2 -, NaOH to give yellow solid along with a salt. The salt on treatment with acid gave benzoic acid.Study the paragraph carefully and answer the following questions give belowThe structureof 'A'is
Answer» SOLUTION :

Discussion

29.	Compounds A(C_9 H_8 O_3) on reduction with Zn-Hg&HCI gave a carboxylic acid B(C_9 H_(10) O_3). B on further oxidation gave bezoic acid. A on strong heating gave a ketone C(C_8H_8 O). C reacts with I_2 -, NaOH to give yellow solid along with a salt. The salt on treatment with acid gave benzoic acid.Study the paragraph carefully and answer the following questions give belowThestructureof 'B' is
Answer» SOLUTION :

Discussion

30.	Compounds A(C_9 H_(16) ) decolourises Br, in C CI_4 A reacts alkaline KMnO_4 to give a diol. A on ozonolysis gave B(C_7 H_(12)O) and acetaldehyde. B reacts in presence of sodium hydroxide to give two products C(C_7 H_(12)O_2) and sodium salt which on nentralisation gave D(C2H,20). Con oxidation also gave 'D'. On the basis of the reactions give above answer the following questions. The structure of 'D' is likely to be
Answer» any of the above SOLUTION :

Discussion

31.	Compounds A(C_9 H_(16) ) decolourises Br, in C CI_4 A reacts alkaline KMnO_4 to give a diol. A on ozonolysis gave B(C_7 H_(12)O) and acetaldehyde. B reacts in presence of sodium hydroxide to give two products C(C_7 H_(12)O_2) and sodium salt which on nentralisation gave D(C2H,20). Con oxidation also gave 'D'. On the basis of the reactions give above answer the following questions.The structure of 'B' is likely to be
Answer» SOLUTION :

Discussion

32.	Compounds A(C_(10) H_(10)) decolourises Br, in C CI_4 Ozonolysis of A gave B(C_(10) H_(10),O_2). B reacts hydrazine and base followed by heating in ethylene glycol gave a hydrocarbon C(C_(10) H_(14)) C on oxidation gave a dicarboxylic acid which form anhydride easily.thestructureof compound 'A' is
Answer» Any of the above Solution :Dicarboxylicacideasilyformsanhydride=

Discussion

33.	Compounds A(C_(10) H_(10)) decolourises Br, in C CI_4 Ozonolysis of A gave B(C_(10) H_(10),O_2). B reacts hydrazine and base followed by heating in ethylene glycol gave a hydrocarbon C(C_(10) H_(14)) C on oxidation gave a dicarboxylic acid which form anhydride easily.Conversion of B to C is called
Answer» Wolff-Kishner REDUCTION CLEM MENSEN’s reduction ROSENMUND's reduction Sabatier-Senderen's reduction SOLUTION :Wolff -Kishnerreducation

Discussion

34.	Compounds A undergoes hydrolysis to produce a colourless gas with rotten fish smell. The gas gives a vortex ring. The gas is
Answer» `PH_3` `P_(2)O_3` `P_(2)O_5` `P_2S_3` SOLUTION :The colour of `PH_3` is ROTTEN fish smell. The gas GIVES a VORTEX RINGS is `PH_3`

Discussion

35.	CompoundsA (molecularformulaC_(3) H_(8) O) is treatedwith acidifiedpotassiumdichromate toforma product B( molecularformulaC_(3) H_(6) O) B.Formaa shiningsilvermirroron warmingwith ammonicalsilver nitrateB whentreatedwith aqueoussolutionof H_(2) NCONHNH_(2) HCI and sodiumacetategivesa product C. Identifythe structureof C.
Answer» `CH_(3) CH_(2) CH- NNHCONH_(2)` `CH_(3)- underset(CH_(3))underset(\|)(C ) = NNHCONH_(2)` `CH_(3) - underset(CH_(3)underset(\|)(C ) =NCONHNH_(2)` `CH_(3) CH_(2) CH= NCONHNH_(2)` Solution : Reactionof Bindicatesthat BIS analdehydethus Bshouldbe `C_(2) H_(5) CHO` or `CH_(3) CH_(2)CHO`and thereforeC shouldbe `CH_(3) CH_(2) CH=NNHCONH_(2)` `CH_(3) CH_(2) CH_(2) OHoverset([O])(to) CH_(3) CH_(2)CHO` `overset( "Silver mirror test ") (to)CH_(3) CH_(2) COOH+ 2 AG ` `CH_(3) CH_(2) CHO + H_(2) NNHCONH_(2)` SEMICARBAZIDE `to CH_(3) CH_(2) CH=NNHCONH_(2)` semicarbazone [C]

Discussion

36.	Compounds A, B and C are isomeric alcohols with formula C_(5)H_(12)O. A on oxidation given ketone, B gives acid while C is not oxidised, A gives test with I_(2)//NaOH. The three isomerice alcohols react with HBr with decreasing rates C gt A gt B. Identify A and B.
Answer» ANSWER :A,C

Discussion

37.	Compounds A and C in the following reactions are CH_(3)CHO underset((ii)H_(2)O)overset((i)CH_(3)MgBr)to (A) overset(H_(2)SO_(4),Delta)to(B) overset((i)gt" Hydroboration, (ii) Oxidation")to(C)
Answer» Identical Positional isomers Functional isomers Optical isomers Solution :`underset("ETHANAL")(CH_(3)CHO) underset((II).H_(2)O)overset((i)CH_(3)MgBr)to underset("Propan-2-ol (A)")(CH_(3)-overset(OH)overset(\|)(C)H-CH_(3)) underset(("Dehydration"))overset(H_(2)SO_(4),Delta)to underset("PROPENE (B)")(CH_(3)-CH=CH_(2)) overset("Hydroboration-oxidation")to` Thus, `CH_(3)CHOHCH_(3)(A)` and `CH_(3)CH_(2)CH_(2)OH(C)` are position isomers, i.e., option (b) is correct.

Discussion

38.	Compounds (A) and (C ) in the following reactions are CH_(3)CHO overset((i)CH_(3)MgBr)underset((ii)H_(2)O)to(A) overset(H_(2)SO_(4),Delta)to(B)overset("Hydroboration oxidation")to (C)
Answer» identical POSITIONAL isomers FUNCTIONAL isomers OPTICAL isomers. SOLUTION :

Discussion

39.	Compounds (A) and (B) respectively are :
Answer» `Cu(NO_3)_2 and pb(NO_3)_2` `Cu(NO_3)_2 and Hg(NO_3)_2` `Cu(NO_3)_2 and AgNO_3` `Hg(NO_3)_2 and AgNO_3` Solution :`2CU^(2+)+5I^(-)toCu_2I_2darr`(white)+`I_3^(-)` (brown solution DUE to dissolution of `I_2`) `Hg^(2+)+2I^(-)toHgI_2dar (red) , HgI_2 darr +2I^(-)to [HgI_4]^(2-)` (soluble complex)

Discussion

40.	Compounds A and B react with a common reagent according to first order kinetics in both cases. If 99% of A must react before 1% of B has reacted. What is the minimum ratio for their respective rate constants? ("Given":(2)/(2-log^(99)) = 458)
Answer» ANSWER :458

Discussion

41.	Compounds 'A' and 'B' react according to the following chemical equation A(g) + 2B(g) to 2C(g) Concentration of either 'A' or 'B' were changed keeping the concentrations of one of the reactants constant and rates were measured as a function of initial concentration. Following results were obtained. Choose the correct option for the rate of equations for this reaction.
Answer» RATE of = `K[A]^(2)[B]` Rate = `k[A][B]^(2)` Rate = `k[A][B]` Rate = `k[A]^(2)[B]^(-)` ANSWER :B

Discussion

42.	Compounds A and B react according to the following chemical equation : A(g) + 2B(g) rarr2C(g) Concentration of either A or B were changed keeping the concentrations of one of the reactantconstants and rates were measured as a function of initial concentration, Following results were obtained. Choose the correct option for the rate equations for this reaction.
Answer» Rate = `k[A]^(2)[B]` Rate = `k[A][B]^(2)` Rate = `k[A][B]` Rate = `k[A]^(2)[B]^(0)` Solution :Let Rate law, `r=k[A]^(x)[B]^(y)""....(i)` On putting values, we get `0.10 = K [0.30]^(x)[0.30]^(y) ""....(ii)` `0.40 = K[0.30]^(x)[0.60]^(y)""....(iii)` `0.20=K[0.60]^(x)[0.30]^(y)""....(iv)` From EQS. (ii) and (iii), we get `(0.1)/(0.4)=[(0.30)/(0.30)]^(x)[(0.30)/(0.60)]^(y)` `(1)/(4)=[(1)/(2)]^(y)implies y = 2` From Eqs.(iii) and (iv), we get `(0.40)/(0.20)=((0.30)/(0.60))^(x)((0.60)/(0.30))^(y)` `2=((1)/(2))^(x)(2)^(y)` On putting the values of y, we get `2=((1)/(2))^(x)(2)^(2)` `implies ""x=1` On putting the values of x and y in Eq. (i), we get `r=k[A]^(1)[B]^(2)=k[A][B]^(2)`

Discussion

43.	Compounds A & B on reaction with HCl released gases P & Q saperately ''Q'' turns di chromate solution green. Hence ''B' can be
Answer» `Na_(2)SO_(3)` `Na_(2)CO_(3)` `Na_(2)HCO_(3)` `NaCl` SOLUTION :Conceptual

Discussion

44.	Compounds A and B are the structural isomers of each other with molecular formula C_(4)H_(9)Br. Compound 'A' is a linear chain and optically inactive while 'B' is optically active. When compound 'A' is treated with aq. KOH solution, it gives compound 'C' whereascompound 'B' when treated with aq.KOH solution, it gives compound 'D'. Identify compound 'C' and 'D'.
Answer» `CH_(3)-CH_(2)-CH_(2)-CH_(2)-OH , CH_(3)-UNDERSET(CH_(3))underset(\|)OVERSET(CH_(3))overset(\|)C-OH` `CH_(3)-overset(CH_(3))overset(\|)CH-CH_(2)-OH,H_(3)C-underset(CH_(3))underset(\|)overset(CH_(3))overset(\|)C-OH` `{:(""CH_(2)-CH_(3)),(""\|),(CH_(3)-CH_(2)-CH_(2)-CH_(2)-OH","H_(3)C-CH-OH):}` `{:(""CH_(3)""CH_(2)-CH_(3)),(""\|""\|),(CH_(3)-CH-CH_(2)-OH"," CH_(3)-CH-OH):}` Answer :C

Discussion

45.	Compounds (A) and (B) are:
Answer» `NaClO_(3), NACLO` `NaOCl_(2), NaOCl` `NaClO_(4),NaClO_(3)` `NaOCl,NaClO_(3)` ANSWER :D

Discussion

46.	Compound (X)C_(9)H_(10)O in inert to Br_(2)//"CC"l_(4). Vigorous oxidation with hot alkaline KMnO_(4)//ObarH yields C_(6)H_(5)COOH . (X) gives precipitate with 2,4,-dinitrophenyl hydrazine. How casn these isomers be distinguished by the usual chemical tests? Following are possible isomers of X: (I) C_(6)H(5)-CH_(2)-CH_(2)-CHO CH_(6)H_(5)-underset(CH_(3))underset(\|)(CH)-CHO (III) C_(6)H_(5)-CH_(2)-overset(O)overset(\|\|)(C)-CH_(3) (IV) CH_(6)H_(5)-overset(O)overset(\|\|)(C)-CH_(2)-CH_(3)
Answer» I give RED ppt with Fehling SOLUTION and II & III canbe distinguished by iodoform test I & II canbe distinguished by simple CHEMCIAL method I & II give red ppt. with Fehlilng solution and III & IV can be distinguished by iodoform test II give red ppt. Fehling solution and I & IV can be distinguished by iodoform test. Solution :By the condition of the question.I & IV both compounds do not GIE iodoform test ad I & II both are GIVES similar test

Discussion

47.	Compound Y,C_7H_8O is insoluble in water, dil HCl and aqueous NaHCO_3 it dissolves in dilute NaOH.When Y is treated with bromine water it is converted rapidly into a compound of formula C_7H_6OBr_3.Identify the structure of Y
Answer» SOLUTION :

Discussion

48.	Compound (X)C_9H_10O is inert to Br_2//C Cl_4.Vigorous oxidation with hot alkaline KMnO_4//OH yield C_6H_5COOH (X) gives precipitate with 2,4-Dintrophenyl hydrazine. How can these isomers be distinguished by the usual chemical tests ? Following are possible isomers of X : (i)C_6H_5-CH_2-CH_2-CHO , (ii)C_6H_5-undersetunderset(CH_3)(\|)CH-CHO (iii)C_6H_5-CH_2-oversetoverset(O)(\|\|)C-CH_3 , (iv)C_6H_5-oversetoverset(O)(\|\|)C-CH_2-CH_3
Answer» I gives red ppt. with FEHLING solution and II & III can be distinguished by iodoform TEST I & II can be distinguished by simple chemical method I & II GIVE red ppt. with Fehling solution and III & IV can be distinguished by iodoform test II gives red ppt. with Fehling solution and I & IV can be distinguished byiodoform test Solution :I & IV both compounds do not give iodoform test and I & II both given similar test

Discussion

49.	Compound (X) with molecular formula C_(3)H_(8)O is treated with acidifiedpotassium dichromate to forma product(Y) with molecularformula C_(3)H_(6)O. (Y) does not form a shining silver mirror on warming with ammoniacal AgNO_(3). (Y) when treated with an aqueous solution ofNH_(2)CONHNH_(2).HCland sodium acetate, gives a product (Z). The structure of (Z) is
Answer» `CH_(3)CH_(2)CH=N NHCONH_(2)` `(CH_(3))_(2)C=N NHCONH_(2)` `(CH_(3))_(2)C=NCONHNH_(2)` `CH_(3)CH_(2)CH=NCONHNH_(2)` Solution :`{:(CH_(3)-overset(H)overset(\|)underset((X))underset(CH_(3))underset(\|)C-OH overset(K_(2)Cr_(2)O_(7))underset(H^(+))to CH_(3)-underset((Y))underset(CH_(3))underset(\|)C=O),(""underset(CH_(3)COONa)underset(+)underset(NH_(2)CONHNH_(2)*HCl)(" ")darr),(""(CH_(3))_(2)Cunderset((Z))(=N N)HCONH_(2)):}`

Discussion

50.	Compound X with a molecular formula C_(4)H_(11)N did not react with Hinsberg's reagent, but reacted with one mole of CH_(3)I to form a salt. What is the structure of X ?
Answer» Solution :The structure of X is `UNDERSET("N, N-Dimethylethanamine")(H_(3)C-overset(C_(2)H_(5))overset("\|")underset(CH_(3))underset("\|")("N": ))` Since the compound does not react with `HNO_(2)`, it must be a TERTIARY amine. Tertiary amine reacts with one mole of `CH_(3)I` to from a quaternary SALT i.e. `[(CH_(3))_(3)N^(+)-C_(2)H_(5)]I^(-)`.

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

Compounds with high heat of formation are less stable because

Compounds with antiseptic properties are...

Compounds with antiseptic properties are ........

Compoundswith antiseptic properties are "……….."

Compounds which rotate plane of plane polarised light in clockwise direction are knownas

Compounds which do not give coupling reaction with benzenediazonium chloride in acidic medium is/are

Compounds which contain two lone pairs and four bonds pairs are

Compounds which contain one lone pair and six bond pairs are

Compounds which can give periodic cleavage are :

Compounds used in Holme's signal are ……………… .

Compounds thatwill not exhibit geometrical isomerism-

Compound(s) that on hydrogenation produce(s) optically inactive compound(s) is (are)

Compounds present product mixture :

Compounds of nitrogen, phosphorus and sulphur such as ammonia, phosphoric acid and sulphuric acid are used in fertilizer industry. Write the chemical equation for the preparation of phosphoric acid (H_3PO_4) from H_3PO_3

Compounds I and II may be grouped as

Compounds I and II can be distinguished by using reagent. (I) -4-Amino-2-methylbut-3-en-ol (II) -4-Amino-2,2dimethylbut-3-yn-1-ol

Compounds I and II can be distinguished by using reagent. {:(""I,""II),("4-Amino-2methylbut-3-en-2ol","4-Amino-2,2-dimethylbut-3-yn-1-ol"):}

Compounds having the same molecular formula but different strucutres are cllassified as:

Compounds having same number and kind of atoms but different arrangement of atoms in their molecules are called:

compounds formed when the noble gases get entrapped in the cavities of crystal lattices of certain organic compounds are known as:

Compounds formed by the union of two different halogens are called.......

Compounds containing both amino and COOH groups are known as

Compounds containing both -NH_2 and -COOH groups are called

Compounds (C ) and (D) respectively are :

Compounds A(C_(10) H_(10)) decolourises Br, in C CI_4 Ozonolysis of A gave B(C_(10) H_(10),O_2). B reacts hydrazine and base followed by heating in ethylene glycol gave a hydrocarbon C(C_(10) H_(14)) C on oxidation gave a dicarboxylic acid which form anhydride easily.thestructureof compound 'A' is

Compounds A(C_(10) H_(10)) decolourises Br, in C CI_4 Ozonolysis of A gave B(C_(10) H_(10),O_2). B reacts hydrazine and base followed by heating in ethylene glycol gave a hydrocarbon C(C_(10) H_(14)) C on oxidation gave a dicarboxylic acid which form anhydride easily.Conversion of B to C is called

Compounds A undergoes hydrolysis to produce a colourless gas with rotten fish smell. The gas gives a vortex ring. The gas is

Compounds A, B and C are isomeric alcohols with formula C_(5)H_(12)O. A on oxidation given ketone, B gives acid while C is not oxidised, A gives test with I_(2)//NaOH. The three isomerice alcohols react with HBr with decreasing rates C gt A gt B. Identify A and B.

Compounds A and C in the following reactions are CH_(3)CHO underset((ii)H_(2)O)overset((i)CH_(3)MgBr)to (A) overset(H_(2)SO_(4),Delta)to(B) overset((i)gt" Hydroboration, (ii) Oxidation")to(C)

Compounds (A) and (C ) in the following reactions are CH_(3)CHO overset((i)CH_(3)MgBr)underset((ii)H_(2)O)to(A) overset(H_(2)SO_(4),Delta)to(B)overset("Hydroboration oxidation")to (C)

Compounds (A) and (B) respectively are :

Compounds A and B react with a common reagent according to first order kinetics in both cases. If 99% of A must react before 1% of B has reacted. What is the minimum ratio for their respective rate constants? ("Given":(2)/(2-log^(99)) = 458)

Compounds A & B on reaction with HCl released gases P & Q saperately ''Q'' turns di chromate solution green. Hence ''B' can be

Compounds (A) and (B) are:

Compound Y,C_7H_8O is insoluble in water, dil HCl and aqueous NaHCO_3 it dissolves in dilute NaOH.When Y is treated with bromine water it is converted rapidly into a compound of formula C_7H_6OBr_3.Identify the structure of Y

Compound X with a molecular formula C_(4)H_(11)N did not react with Hinsberg's reagent, but reacted with one mole of CH_(3)I to form a salt. What is the structure of X ?