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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Arrange the following compounds in increasing order of their acid strength: Propane-1-ol, 2, 4, 6-trinitrophenol, 3-nitrophenol, 3,5-dinitrophenol , phenol, 4-methylphenol. |
| Answer» Propan-1-ol, 4-methylphenol, phenol, 3-nitrophenol, 3, 5-dinitrophenol, 2,4,6-trinitrophenol. | |
| 2. |
State the equations for the preparation of the following compounds . (Equations need not to be balance). (i) Chlorobenzene from aniline (in two steps ) (ii) N-Propyl amine from ethyl chloride (in two steps ) |
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Answer» ii. `underset ("Ethyl chloride" ) (CH_3CH_2Cl) + Alc.KCN rarr CH_3 CH_2 CN` ` underset ((Reduction)) overset (H_2//Ni)rarr underset ("n-Propylamine") (CH_3CH_2CH_2NH_2)`. |
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| 3. |
The ratio of molecular mass to empirical formula mass for glucose isA. 2B. 4C. 6D. 8 |
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Answer» Correct Answer - C Molecular mass of glucose `(C_6H_(12)O_6)`=180 Empiriacal formula `(CH_(2)O)` mass =30 `(Mol.Mass)/(Empirical formula mass )=(180)/(30)=6` |
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| 4. |
Write the structures for the Ethoxybutane compounds. |
| Answer» Ethoxybutane `to CH_(3)-CH_(2)-O-CH_(2)-CH_(2)-CH_(2)-CH_(3)` | |
| 5. |
Write the structures for the Ethoxyethane compounds. |
| Answer» Ethoxyethane `to CH_(3)-CH_(2)-O-CH_(2)-CH_(3)` | |
| 6. |
Give the major products that are formed by heating of the `CH_(3)-CH_(2)-CH_(2)-O-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-CH_(2)-CH_(3)` ethers with HI. |
| Answer» `CH_(3)CH_(2)CH_(2)OH + CH_(3)CH_(2)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-I` | |
| 7. |
Identify the reactant needed to form t-butylalcohol from acetone. |
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Answer» When acetone reacts with methyl magnesium bromide followed by the hydrolysis forms t-butyl alcohol. `underset("Acetone")(CH_(3)-COCH_(3))+underset("Methyl magnesium bromide")(CH_(3)MgBr) to CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-OMgBr overset(H_(2)O)to underset("t-butyl alcohol")(CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-OH)` |
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| 8. |
Give the major products that are formed by heating of the `CH_(3)-CH_(2)-overset(CH_(3))overset(|)CH-CH_(2)-O-CH_(2)-CH_(3)` ethers with HI. |
| Answer» `CH_(3)-CH_(2)-underset(CH_(3))underset(|)CH-CH_(2)OH+CH_(3)CH_(2)I` | |
| 9. |
Suggest a suitable oxidising reagent for the following conversions: A. a.`MnO_(2)` in `(A)` and `CrO_(3)` (in glacial acetic acid) in `(B)`.B. b.`CrO_(3)` in (A) and `MnO_(2)` in `(B)`.C. c.Both are correctD. d.Both are incorrect |
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Answer» Correct Answer - A In `(A)` the oxidation of allylic `1^(@)` alcohol to aldehyde is carried out with `MnO_(2)`, while in `(B)` the oxidation of `2^(@)` alcohol to ketone is carried out with `CrO_(3) + CH_(3)COOH`. So the answer is `(a)`. |
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| 10. |
The following is not an appropriate reaction for the preparation of t-butyl ethyl ether. `C_(2)H_(5)O"Na"+CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-Cl to CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-OC_(2)H_(5)` Write a suitable reaction for the preparation of t-butylethyl ether. |
| Answer» `CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-underset(* *)overset(bar(* *))O overset(+)Na+CH_(3)CH_(2)Cl to CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-OC_(2)H_(5)` | |
| 11. |
Benzoic acid and carbolic acid can be distiguished by:A. a.Aqueous `NaHCO_(3)`B. b.Netural `FeCl_(3)`C. c.Aqueous `NaOH`D. d.Aqueous `NH_(3)` |
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Answer» Correct Answer - A::B a. Phenol is also called carbolic acid. Benzoic acid is more acidic than phenol. So the acid reacts with `NaHCO_(3)` to give `CO_(2)(g)`, while phenol does not react. b. Acids give buff-coloured precipitate with netural `FeCl_(3)` solution, e.g., `3PhCOOH+FeCl_(3) rarr underset("(But coloured ppt.)")underset("(Ferric benzoate)")((PhCOO)_(3)Fe)+3HCl` `3PhOH+FeCl_(3) rarr underset("(Violet colouration)")underset("(Ferric phenoxide)")((PhO)_(3)Fe+3HCl)` c. Both acid and phenol react with aqueous `NaOH` to give their corresponding salts. `PhCOOH+NaOH rarr PhCOONa` `PhOH+NaOH rarr PhONa` Both acid and phenol react with `NH_(3)`. `PhCOOH+NH_(3) rarr PhCONH_(2) "Benzamide") +H_(2)O` `PhOH+NH_(30 rarr PhNH_(2) ("Aniline")+H_(2)O` So the answers are `(a)` and `(b)`. |
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| 12. |
Which of the following compounds will not give haloform reaction?A. a.alpha-Phenyl ethanolB. b.AcetophenoneC. c.Ethyl bromideD. d.`(MeCO)_(2)O` |
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Answer» Correct Answer - D In `(a), (Me-underset(Ph)underset(|)overset(alpha)(CH)-OH)` contains `(Meunderset(|)(C)H-OH)` group , so it will give iodoform test. In `(b), (PhCOMe)` sontains `(MeCO-)` group, so it will give iodoform test. In `(c ) C_(2)H_(5)Br` under basic condition in iodoform test first is converted into `(C_(2)H_(5)OH)`, which contains `(Meunderset(|)(C )HOH)` group. So it will give iodoform test. In `(d), (MeCO)_(2)O` is an anhydride and does not give Iodoform test. So the answer is `(d)`. |
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| 13. |
The following is not an appropriate reaction for the preparation of t-butyl ethyl ether. `C_(2)H_(5)O"Na"+CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-Cl to CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)C-OC_(2)H_(5)` What would be the major product of this reaction ? |
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Answer» The major product of the given reaction is 2-methylprop-1-ene. It is because sodium ethoxide is a strong nucleophile as well as strong base. Thus elimination reation predominates over substitution. |
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| 14. |
When ` C_6 H_5N_2Cl` is reduced with ` Na _2 SnOO_2`, the product is :B. C. |
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Answer» Correct Answer - B Diazo group can be replaced by (H) on reeduction with `Na_2 SnO_2` or `H_3 PO _2` warm with `C_2 H_5 OH`. |
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| 15. |
The product `(A)` is:A. a.B. b.C. c.D. d. |
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Answer» Correct Answer - B The reduction of `ArNO_(2)` group under neutral conditions with `(Zn dust + NH_(4)Cl)` or `(Al-Hg+H_(2)O)` gives hydroxylamine compound. So the answer is `(b)`. |
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| 16. |
Which of the following compounds do(es) not react with `H_(2)+Pd+C`?A. a.`Me- -= -Me`B. b.C. c.`MeCOMe`D. d.`Me-N_(3)` |
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Answer» Correct Answer - B::C `H_(2)+Pd+C (charcoal)` reduces `(C-=C)` to `(C=C), (-NO_(2))` benzyl alcohol `(PhCh_(2)OH)` to toluene `(PhCH_(3))`, and azides `(RN_(3) to RNH_(2))`. So, it does not react with ketone and `(C=C)` bond. Hence, the answers are `(a)` and `(c )`. |
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| 17. |
Methanamide is reduced to methanamine with:A. a.`LAH`B. b.`NaBH_(4)`C. c.`H_(2)+Ni`D. d.`B_(2)H_(6)` |
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Answer» Correct Answer - A::C::D `underset("Methanamide")(H-overset(O)overset(||)(C)-NH_(2))overset("Reagent in ")underset({:((a) or (c) ),(or (d)):})tounderset("Methyle amine")(H-CH_(2)-NH_(2))` |
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| 18. |
The product `(A)` and `(B)` are:A. a.B. b.C. c.D. d. |
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Answer» Correct Answer - A `NH_(4)HS` selectively reduces `(-NO_(2))` group para to the `(Me) (EDG)` group, whereas `(SnCl_(2)+HCl)` selectively reduces `(-NO_(2))` group ortho to the `(ME) (EDG)` group. So the answer is `(a)`. |
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| 19. |
Which of the following compounds reacts with `NaCNBH_(3)`?A. a.`Me- CH=NH`B. b.`Me_(2)C=N-Me`C. c.D. d.`PhNO_(2)` |
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Answer» Correct Answer - A::B `NaCNBH_(3)` selectively reduces imines to give `1^(@)` or `2^(@)` amines. So the answers are `(a)` and `(b)`. |
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| 20. |
`overset(NaBH_(4))(leftarrow) PhCH=CH-CHOunderset(2. H_(3)O^(+))overset(1. LAH, ether)(rarr)(A)` The products `(A)` and `(B)` are:A. a.B. b.C. c.D. d. |
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Answer» Correct Answer - D `LAH` reduces `(CHO)` group to `(CH_(2)OH)` and also reduces double bond when `Ph` group is attached to `beta`-position of double bond, whereas `NaBH_(4)` reduces only `(-CHO)` group to `(CH_(2)OH)` group. So the answer is `(d)`. |
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| 21. |
Nitrobenzene is subjected to reduction with zinc dust and ammonium chloride. The main product formed will be:A. BenzenamineB. AnilineC. N-PhenylhydroxylamineD. None of these |
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Answer» Correct Answer - C `PhNO_2overset(Zn_2+NH_4Cl)to underset(N-"Phenyl hydroxylamine")(PhNHOH)` |
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| 22. |
Which of the following is an appropraite set of reactants for the preparation of 1- methoxy-4- nitrobenzene ? |
| Answer» Correct Answer - (ii) | |
| 23. |
Nitrobenzene on eelctrolutic reduction gies :.A. AzobebzeneB. HydrazobebzeneC. AminophenolD. Aniline |
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Answer» Correct Answer - D `PhNO_2 underset ("Weakly acidic medium" )overset ("Electrolytic reduction")rarr PhNH_2` |
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| 24. |
In order to distinguigh between ` C_2H_5NH_2` and ` C_6H_5NH_2` which of the following reagenis is useful ?A. Hinsberg reagentB. p-NaphatheolC. Benzene diazomium chlorideD. None of the above |
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Answer» Correct Answer - B ` 1^@` aromatic amine on diazotisation followed by coupling with ` beta-`naphthol gives azo dye test . |
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| 25. |
Which of the following will yield phenlhydrazine hydrochloride ?A. Benzenamine and hydrazineB. Hydrazine and `HCl`C. Benzenediazonium chlorid and ` SnCl_2//HCl`D. Nitrobenzene and `SnCl_2//HCl` |
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Answer» Correct Answer - C |
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| 26. |
A mixture of two immiscible liquids may be easily separated by using a A. Leibig’s condenser B. fractionisation column C. separating funnel D. none of these. |
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Answer» Correct Answer - C Separating funnel is used to separate two immiscible liquids. |
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| 27. |
A mixture of naphthalene and benzoic acid can be separated byA. extraction with solventB. sublimation C. fractional crystallisation D. distillation. |
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Answer» Correct Answer - A A mixture of naphthalene and benzoic acid can be separated with hot water (extraction with a solvent). |
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| 28. |
A mixture of camphor and benzoic acid can be separated by which of the following techniques A. Chemical methodsB. Sublimation C. Fractiona1 distillationD. Extraction with a solvent. |
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Answer» Correct Answer - A Chemical method using `NaHCO_3` solution . |
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| 29. |
Separation of two substances by fractional crystallisation depends upon their difference inA. Densities B. SolubilitiesC. Meting pointsD. Boilinigipoints. |
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Answer» Correct Answer - B Fectual Question. |
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| 30. |
The purity of an organic compound is determined byA. DensityB. m.pt.C. mixed m.pt.D. molecular weight |
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Answer» Correct Answer - C Factual questions. |
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| 31. |
An organic compound with C =40 % and H= 6.7% will have the empirical formulaA. `CH_(4)`B. `CH_(4)O`C. `C_(3)H_(6)O_(3)`D. `C_(2)H_(4)O_(2)` |
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Answer» Correct Answer - B Ratio of C:H:O `=40/12:(6.7)/(1):(53.33)/(16)=1:2:1` Hence Empirical formula `=CH_O` |
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| 32. |
An organic compound is extracted from its aqueous solution with 100 mL of chloroform. The extraction will be maximum whenA. 100 mL of chloroform is used in one litreB. two 50 mL portions of chloroform are usedC. four 25 mL portions of chloroform are usedD. ten 10 mL portions of chloroform are used. can be removed from commercial |
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Answer» Correct Answer - D See A-Level information |
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| 33. |
The separation of organic compound from its aqueous solution can be done by A. Distillation B. Steam distillation C. Solvent extractionD. Fractional distillation. |
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Answer» Correct Answer - C Since organic solvents are generally in organic solvents than in water, they can beextracted from their aqueous solutions by or- ganic solvents such as `C_6H_6,CHCl_3,CCl_4` etc., which are immiscible with water. |
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| 34. |
An organic compound with C =40 % and H= 13.33 % and N=46.67 % Its empirical empirical formula would beA. CHNB. `C_(2)H_(2)N`C. `CH_(4)N`D. `C_(3)H_(7)N` |
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Answer» Correct Answer - C Ratio of C:H:N `=(40)/(12):(13.33)/(1)=(46.67)/(14)=1:4:1` Empirical formula `=CH_2O` |
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| 35. |
Which of the following organic compounds contains about 52 % carbon ?A. EthanalB. Dismethly etherC. Acetic acidD. Phenol |
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Answer» Correct Answer - B `H_(3)C-OCH_(3) or underset(46 g" "24 g )(C_(2)H_(6)O-=2C` 46 g of compound has C=24 g 100 g of compound has C `=(24)/(46)xx100= 52.3%` |
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| 36. |
An organic substance from its aqueous solution can be separated by A. Distillation B. Steam distillation C. Solvent extraction D. Fractional distillation |
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Answer» Correct Answer - C Factual question |
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| 37. |
An organic compound containing carbon hydrogen and oxygen contains 52 .2 % carbon and 13.04 % hydrogen .Vapour density of the compound is 23 .Its molecular formula will beA. `C_(2)H_(6)O`B. `C_(3)H_(6)O`C. `C_(4)H_(8)O`D. `C_(5)H_(10)O` |
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Answer» Correct Answer - A Precentage of O=100-(52.2+13.04)=34.76 Ratio of atoms C:H:O`=(52.2)/(12):(13.04)/(1):(34.76)/(16)` `=4.35:13:04:2.17=2:6:1.` E.F`=C_2H_6O` E.F mass `=2xx12+6xx1+1xx16=46` Mol.mass `=2xxV.D=2xx23=46` `n=("Mol.mass")/("E.F mass")=46/46=1` `therefore MF=(E.F.)_(n)=(C_2H_6O)_(1)=C_2H_6O` |
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| 38. |
A mixture contains four solid organic compounds containing A, B, C and D. On heating only C changes from solid to vapour state .C can be separated from the rest in the mixture byA. distillationB. sublimation C. Fractional distillationD. crystallisation |
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Answer» Correct Answer - B Sublimation |
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| 39. |
In Lassaigne’s test, the organic compound is fused with a piece of sodium metal in order toA. ncrease the ionisation of the compound B. decrease the melting point of the compound C. increase the reactivity of the compound D. convert the covalent compound into a mixture of ionic compounds. |
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Answer» Correct Answer - D To convert covalent compounds into ionic com- pounds such as NaCN, `Na_(2)S`, NaX etc. |
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| 40. |
When an organic compound containing 1 phosphorus is fused with fusion mixture, it gives isA. `Na_(2)HPO_(4)`B. `Na_(3)PO_(4)`C. `NaH_(2)PO_(4)`D. `Na_(3)PO_(3)` |
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Answer» Correct Answer - A Fusion of organic compound with fussion mix- ture `(Na_2CO_3+K_2CO_3) " changes P to " Na_3PO_4`. |
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| 41. |
The density of a gaseous organic compound is `1.3xx10^(-3) g mL^(-1)` at N.T.P. Its vapour densitA. 0.00065B. 0.65C. 14.4816D. 14.56 |
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Answer» Correct Answer - D Mass of 1 mL of `H_(2) " at NTP "=2/(22400)g` `therefore` Density of `H_(2)` at N.T.P. `=2/(22400) g mL ^(-1)` vapour density `=("Density of given gas /Vapour")/("Density of Hydrogen (at same T and P")` `(1.3xx 10^(-3)xx 22400)/(2)=1.3 xx 11.2 =14.56` |
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| 42. |
Distillation under reduced pressure is employed for A. `C_(6) H_(6)`B. PetrolC. `CH_(2)OHCHOHCH_(2)OH`D. Organic compounds used in medicine. |
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Answer» Correct Answer - C Distillation under reduced pressure is employed for the compound which deconpose near their boiling point . |
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| 43. |
In the Lassaigne’s test, the blood red colouration is due to the formation of A. `Fe(CNS)_(2)`B. NaCNSC. `NH_(4)CNS`D. `Fe(CNS)_(3)` |
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Answer» Correct Answer - D Blood red colour is due to ferric thiocyanate (or sulphocyanide) which is blood red in colour. |
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| 44. |
The element X (atomic weight = 75) and Y (atomic weight = 16) combine to give a compound containing 75.8% X. The molecular formula of the compound isA. XYB. `X_(2)Y`C. `X_(2)Y_(2)`D. `X_(2)Y_(3)` |
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Answer» Correct Answer - D Similar to Q.59 |
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| 45. |
If `0.24 g` of a volatile liquid upon vaporization gives `45 ml` of vapors at `NTP`, what will be the vapor density of the substance? (Density of `H_(2) = 0.089 f L^(-1)`)A. 95.39B. 39.95C. 99.53D. 59.93 |
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Answer» Correct Answer - D Density of vapours of volatile liquid at NTP `=0.24/45 g mL ^(-1)` `=0.24/45 g mL ^(-1)` `=0.24/45xx10^3 gL ^(-1)= 240/45 gL^(-1)` =Vapour density `("Dendity of gas / vapour ")/("Density of H_(2)(Similar T & P))` `=(240//45)/(0.089)=(240000)/(89 xx 45)=59.93` |
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| 46. |
An organic compound contains, `C, H` and `S`. The minimum molecular weight of the compound containing `8%` sulphur is : `("atomic weight of" S=32 "amu")`A. 200B. 400C. 800D. 600 |
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Answer» Correct Answer - C Minimum molecular mass is obtained when it is supposed that one molecule of the compound contains only one atom of the given element under investigation. `therefore` Min .mol.mass `= 100xx At. ("Mass of element under investigation ")/("% of the element ")` `=(100xx32 ) /(4)=800` |
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| 47. |
Acetamide is reated separately with the following reagents . Which one of these would give methylamine ?A. ` PCl_5`B. `Soda lime`C. ` NaOH +Br_2`D. `Hot conc . H_2SO_4` |
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Answer» Correct Answer - C `underset("Acetamide")(Me-overset(O)overset(||)C-NH_(2))overset(NaOH+Br_(2))underset("Hoffmann bromamide")rarr underset("Methylamine")(Me-NH_(2))` |
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| 48. |
The compound which on reaction with aqueous nitrous acid at low tremperature produces an oily nitrosoamine is :A. MethylamineB. EthylamineC. DiethylamineD. Triethgylamin |
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Answer» Correct Answer - C `2^@` amines react with `HNO_2` at low temperature to give ily nitrosamine. ` Et_2 NH + NHO_2 rarr underset (N-"Nitroso N-ethyl ethanamine") (Et_2 N-N=O + H_2O)`. |
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| 49. |
0.44 gm of organic compound `C_(x) H_(y)` O which occupied 224 ml at NTP and on combustion gave 0.88 gm `CO_(2)`. The ratio of X to Y in the compound 1sA. `1:1`B. `1:2`C. `1:3`D. `1:4` |
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Answer» Correct Answer - B 224 mL at NTP of O.C weighs = `0*44` g `therefore` 22400 ml at NTP of O.C. weighs = `(0*44)/(224) xx 22400 = 44 g ` Wt. of carbon 1 mol (44 g) of the O.C. = `(12)/(44) xx 0*88 xx (44)/(0*44) = 24 `g = 2g atom of carbon . `therefore` X = 2 As the molecular mass of the compound is 44 . Therefore molecular formula can be `C_(2)H_(4)O` Hence Y = 4 , `therefore X : Y = 2 : 4 = 1 : 2 ` |
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| 50. |
If on adding `FeCl_(3)` solution to acidified Lassaigne solution ,a blood red colouraton is produced ofA. SB. NC. N and SD. S and Cl |
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Answer» Correct Answer - C See Comprehensive Review. |
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