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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.
| 101. |
Which of the following is not a nucleophile?A. `Br^(-)`B. `:NH_(3)`C. `H^(+)`D. `C_(6)H_(6)` (benzene) |
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Answer» Correct Answer - C `H^(+)` is an electrphile, all others are nucleophile. |
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| 102. |
Which of the following does not show electromeric effect ?A. AlkenesB. EthersC. AldehydesD. Ketones. |
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Answer» Correct Answer - B There is no E-effect in ethers . |
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| 103. |
Electromeric effect:A. Comes into play at the demand of attacking reagentB. Involves displacement of electrons in a sigma bondC. Comes into play in the molecule when at least one atom has unshared pair of electronsD. Invovles the distortion of the electron cloud. |
| Answer» Correct Answer - A | |
| 104. |
Assertion: When inductive and electromeric effects operate in oppsite directions, the inductive effect predominates. Reason: Inductive effect is the complete transfer of shared pair of `pi` electrons to one of the atoms.A. If both assertion and reason are true and reason is the correct explanation of assertionB. If both assertion and reason are true but reason is not the correct explanation of assertionC. If assertion is true but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - D When inductive and electromeric effects operate in opposite directions, the electronmeric effect predominates. inductive effect is the polarisation of `sigma` bond caused by the polarisation of adjancent `sigma` bond. |
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| 105. |
Electronegativity of carbon atoms depends upon their state of hybrisation. In which of the following compounds, the carbon marked with asterisk is most electronagative ?A. `CH_(3)-CH_(2)-^(**)CH_(2)-CH_(3)`B. `CH_(3)-^(**)CH=CH-CH_(3)`C. `CH_(3)-CH_(2)-C equiv^(**)CH`D. `CH_(3)-CH_(2)-CH=^(**)CH_(2)` |
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Answer» Correct Answer - C `- C equiv overset(**)(C)H` carbon is the most electronegative since it is sp hydridised and has maximum s-character. The release of `H^(+)` from `C - H` bond becomes easiest. |
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| 106. |
In which of the following, functional group isomerism is not possible ?A. AlcoholsB. AldehydesC. Alkyl halidesD. Cyanides |
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Answer» Correct Answer - C Alcohols and ethers , aldehydes and ketones, cyanides and isocyanides are functional group isomers. Alkyl halides, however, do not show functional group isomerism. |
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| 107. |
In which of the following functional groups, isomerism is not possible ?A. AlcoholsB. AldehydesC. Alkyl halidesD. Cyanides |
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Answer» Correct Answer - C Alkyl halides donot show any functional isomerism. |
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| 108. |
The fragrance of flowers is due to the presence of some steam volatile organic compounds called essential oils. These are generally insoluble in water at room temperature but are miscible with water vapours in vapour phase. A suitable method for the extraction of these oils from the flowers is :A. DistillationB. CrystallisationC. Distillation under reduce pressureD. steam distillation |
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Answer» Correct Answer - D Extraction of oils from flowes is done with the help of steam distillation. |
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| 109. |
During hearing of a court case, the judge suspected that some changes in the documents had been carried out. He asked the forensic department to check the ink used at two different places. According to you which technique can give the best results ?A. Column chromatographyB. Solvent extractionC. DistillationD. Thin layer chromatography |
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Answer» Correct Answer - D The constituents of ink can be identified with the help of thin layer chromatography. |
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| 110. |
By mistake, an alcohol (boiling point `68^(@)C`). Suggest a suitable method to separate the two compounds. Explain the reason for your choice. |
| Answer» The difference in boiling point of two liquids is more than `20^(@)C`. Hence, simple distillation can be used and since at the boiling point of low boiling liquid, the vapours would consist entirely of only low boiling liquid without any contamination of vapours of high boiling liquid and vice-versa. Thus, both the liquid can be distilled without any decomposition. | |
| 111. |
By mistake, an alcohol (boiling point `97^(@)C` ) was mixed with a hydrocarbon (boiling point `68^(@)C` ). Suggest a suitable method to separate the two compounds. Explain the reason for your choice. |
| Answer» The separation can be done with the help of simple distillation since the difference in boiling point is more than `20^(@)C`. Low boiling hydrocarbon distils leaving alcohol in the distillation flask. This is possible only if alcohol and hydrocarbon are misicible with each other. | |
| 112. |
For the purification of organic compounds, the latest technique followed is |
| Answer» Chromatography is the most useful and the latest technique of separation and purification of organic compounds. It was first used to separate a mixture of coloured substances | |
| 113. |
The best method to separate a mixture of ortho and para nitrophenol `(1 : 1)` is :A. Steam distillationB. CrystallisationC. vaporisationD. colour spectrum |
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Answer» Correct Answer - A Steam distillation is the most effective because whereas o-nitrophenol is steam volatile, p-nitrophenol is not. |
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| 114. |
Suggest a method to purify : (i) Camphor containing traces of common salt. (ii) Kerosene oil containing water. (iii) A liquid which decomposes at its boiling point. |
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Answer» (i) Sublimation. Camphor sublimes while common salt remains as residue in the china dish. (ii) Since the two liquids are immiscible, the technique of solvent extraction with a separating funnel is used. The mixture is thoroughly shaken and the separating funnel is allowed to stand. Kerosene being lighter tha water forms the upper layer while forms the layer. The lower water layer is run off when kerosene oil is obtained. It is dried over anhydrous `CaCl_(2)` or `MgSO_(4)` and then distilled to give pure kerosene oil. (iii) Distillation under reduced pressure. Since the b.p. of a liquid depends upon the pressure acting on it. Therefore, a liquid which decomposes at its b.p. can be purified safely at a lower temperature if the pressure acting on it reduced. |
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| 115. |
The compound which is not resonance stabilisedA. `CH_(2)=CH-Cl`B. C. `CH_(2)=CH-CH_(2)-Cl`D. |
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Answer» Correct Answer - C There is conjugation in `A,B,D` but not in `C`. |
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| 116. |
Designate each carbon atom as primary, secondary, tertiary and quanternary in the following : (a) `{:(" "CH_(3)),(" |"),(CH_(3)-C-CHCH_(2)CH_(2)CH_(3)),(" |"" |"),(" "CH_(3)""CH_(3)):}` (b) |
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Answer» `{:(" "overset(1^(@))(C)H_(3)" "overset(1^(@))(C)H_(3)),(" |"" |"),((a) overset(1^(@))(C)H_(3)overset(4^(@))(-C)-overset(3^(@))(C)H-overset(2^(@))(C)H_(2)-overset(2^(@))(C)H_(2)-overset(1^(@))(C)H_(3)" "(b)" "overset(1^(@))(C)H_(3)overset(3^(@))(-C)H-overset(2^(@))(C)H_(2)-overset(2^(@))(C)H_(2)-overset(3^(@))(C)H-overset(1^(@))(C)H_(3)),(" |"" |"" |"),(" "underset(1^(@))(C)H_(3)underset(1^(@))(C)H_(3)" ".^(1^(@))CH_(3)):}` `{:(overset(1^(@))(C)H_(3)-overset(3^(@))(C)H-overset(3^(@))(C)H-overset(1^(@))(C)H_(3)),(" |"" |"),(" "underset(1^(@))(C)H_(3)" "underset(1^(@))(C)H_(3)):}` |
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| 117. |
Least stable carbocation among the following isA. B. C. D. |
| Answer» Correct Answer - D | |
| 118. |
Identify the teritary alcohol and write its IUPAC name. |
| Answer» Compound (III) is tertiary alcohol. Its IUPAC name is : 2-Methylpropan-2-ol. | |
| 119. |
Identify the primary alcohols from the given list. |
| Answer» Compounds (I) and (IV), represent primary alcohols. | |
| 120. |
Which of the following carbocation is most stable?A. `(CH_(3))_(3)Coverset(o+)(C)H_(2)`B. `(CH_(3))_(3)overset(o+)(C)`C. `CH_(3)CH_(2)overset(o+)(C)H_(2)`D. `CH_(3)overset(o+)(C)HCH_(2)CH_(3)`. |
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Answer» Correct Answer - B is the most stable since it is a tertiary carbocation. |
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| 121. |
The best and latest technique for isolation, purification and separation of organic compound isA. CrystallisationB. DistillationC. SublimationD. Chromatography. |
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Answer» Correct Answer - D is the correct answer. |
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| 122. |
The best and latest technique for isolation, purification and separation of organic compound isA. ChromatographyB. Steam distillationC. CrystallisationD. Vaccum distillation. |
| Answer» Correct Answer - A | |
| 123. |
Benzoic acid is an organic compound. Its crude sample can be purified by crystallisation from hot water. What characeteristic difference in the properties of benzoic acid and the impurity make this process of purification suitable ? |
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Answer» Benzoic acid can be purified by hot water because of following characteristics. (i) Benzoic acid is more soluble in hot water and less soluble in cold water. (ii) Impurities present in benzoic acid are either insoluble in water or more soluble in water to such an extent that they remain in solution as the mother liquor upon crystallisation. |
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| 124. |
Give a brief description of the principles of the following techniques taking an example in each case. (a) Crystallisation (b) Distillation (c) Chromatography |
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Answer» (a) Crystallisation Crystallisation is one of the most commonly used techniques for the purification of solid organic compounds. Principle: It is based on the difference in the solubilites of the compound and the impurities in a given solvent. The impure compound gets dissolved in the solvent in which it is sparingly soluble at room temperature, but appreciably soluble at higher temperature. The solution is concentrated to obtain a nearly saturated solution. On cooling the solution, the pure compound crystallises out and is removed by filtration. For example, pure aspirin is obtained by recrystallising crude aspirin. Approximately 2 – 4 g of crude aspirin is dissolved in about 20 mL of ethyl alcohol. The solution is heated (if necessary) to ensure complete dissolution. The solution is then left undisturbed until some crystals start to separate out. The crystals are then filtered and dried. (b) Distillation This method is used to separate volatile liquids from non-volatile impurities or a mixture of those liquids that have a sufficient difference in their boiling points. Principle: It is based on the fact that liquids having different boiling points vapourise at different temperatures. The vapours are then cooled and the liquids so formed are collected separately. For example, a mixture of chloroform (b.p = 334 K) and aniline (b.p = 457 K) can be separated by the method of distillation. The mixture is taken in a round bottom flask fitted with a condenser. It is then heated. Chloroform, being more volatile, vaporizes first and passes into the condenser. In the condenser, the vapours condense and chloroform trickles down. In the round bottom flask, aniline is left behind. (c) Chromatography It is one of the most useful methods for the separation and purification of organic compounds. Principle: It is based on the difference in movement of individual components of a mixture through the stationary phase under the influence of mobile phase. For example, a mixture of red and blue ink can be separated by chromatography. A drop of the mixture is placed on the chromatogram. The component of the ink, which is less adsorbed on the chromatogram, moves with the mobile phase while the less adsorbed component remains almost stationary. |
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| 125. |
`CH_(2)=CH^(-)` is more basic than `HC -= C^(-)`. Explain why ? |
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Answer» `CH_(2) = overset(sp^(2))(C)H^(-) " " HC -= overset(sp)(C^(-))` Since, sp-carbon is more electronegative than `sp^(2)`-carbon, therefore, `CH -= C^(-)` is less willing to donate a pair of electrons than `H_(2)C = CH^(-)`. In other words, `H_(2)C = CH^(-)` is more basic than `HC -= C^(-)`. |
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| 126. |
The hybridisation of carbons of C-C single bond of `HC-=C-CH=CH_(2)` isA. `sp^(3)-sp^(3)`B. `sp-sp^(2)`C. `sp^(3)-sp`D. `sp^(2)-sp^(3)` |
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Answer» Correct Answer - B `H_(2)overset(sp^(2))(C)=overset(sp^(2))(CH)-overset(sp)(C)-=overset(sp)(C)H` |
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| 127. |
`CH_(2)=CH^(-)` is a better nucleophile than `HC -= C^(-)`. Explain. |
| Answer» In `H_(2)C = CH^(-)`, the carbon atom carrying the -ve charge is `sp^(2)` hybridized while in `HC -= C^(-)`, the carbon atom carrying the -ve charge is sp-hybridized. Since a `sp^(2)`-hybridized carbon is less electronegative than a sp-hybridized carbon, therefore, `H_(2)C = CH^(-)` is a better nucleophile than `HC -= C^(-)`. | |
| 128. |
Out of `HO^(-)` or `HS^(-)` which one is (a) better nucleophile and (B) a better base ? |
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Answer» (a) S is less electronegative than O, therefore, it can donate a pair of electrons more easily than O. Thus, `HS^(-)` is a better nucleophile than `HO^(-)`. (b) Since O-H bond is stronger than S-H bond, therefore `HO^(-)` has a greater tendency to accept a proton than `HS^(-)` and hence `HO^(-)` is a better base than `HS^(-)`. |
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| 129. |
Write IUPAC names for the following srtuturtes : ( a) `H_3C-underset(OH)underset(|)"CH"-underset(CH_3)underset(|)overset(CH_3)overset(|)C-CH_3` `H_2C-underset(COOH)underset(|)CH-CH_2-CH-overset(Cl)overset(|)CH-CH_3` ( c) `CH_3-underset(CH_3)underset(|)"CH"-underset(CH_3)underset(|)C=CH-CH_3` ( d) `H_3C-underset(CH_(3))underset(|)CH-CH_2-CH=CH_2` (e) `H_3C-underset(CH_3)underset(|)overset(CH_3)overset(|)C-CH_3 ` ( f) `{:(" "C_2H_5),(" "|),(HC-=C-CH_(3)),(" "|),(" "H):}` ( g) `H_3C-CH-CH-overset(CH_3)overset(|)"CH"-CH_3` |
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Answer» (a) 3, 3-Dimethyl 2-butanol (b) 4 Chlero 2-methyl pentanoic acid (c) 3, 4 Dimethyl-2 pentened (d)4 Methyl 1-pentene (e) 2, 2 Dimethy1 propane (f) 3 Methy1 1 pentyne (g) 4-Methyl 2-pentene |
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| 130. |
Freshly prepared solution of sodium nitroprusside is added to the sodium extract. Appearance of a deep violet colour indicates the presence ofA. nitrogenB. sulphurC. both nitrogenn and sulphurD. halogen |
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Answer» Correct Answer - B `Na_(2)S+ underset("Sod nitroprusside")(Na_(2)[Fe(CN)_(5)(NO)])to underset("Sod. Thionitroprusside (Violet)")(Na_(4)[Fe(CN)_(5)(NOS)])` |
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| 131. |
1.216 g of an organic compound was Kjeldahlised and the ammonia evolved was absorbed in 100 mL 1 N `H_(2)SO_(4)`. The remaining acid solution was made upto 500 mL by addition of water. 20 mL of this dilute solution required 32mL of N/10 caustic soda solution for complete neutralisation. Calculate the percentage of nitrogen in the organic compound. |
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Answer» 20 mL of diluted unreacted `H_(2)SO_(4) -= 32 mL` of N/10 NaOH solution `:.` 500 mL of diluted unreacted `H_(2)SO_(4) -= (32)/(20) xx 500 mL` of `(N)/(10) NaOH -= (32)/(20) xx (500)/(10)mL` 1N NaOH But 80 mL 1 N `NaOH -= 80 mL` of `1 N H_(2)SO_(4) :.`Acid left unused = 80 mL 1 `N H_(2)SO_(4)` or Acid used for neutralization of `NH_(3) = (100 - 80) = 20 mL 1 N H_(2)SO_(4)` `:. % N = (14 xx "Normality of the acid" xx "Vol. of acid used")/("Mass of the substance taken") = (1.4 xx 1 xx 20)/(1.216) = 23.026` |
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| 132. |
Name the first organic compound to be prepared in the laboratory. |
| Answer» Correct Answer - Urea | |
| 133. |
What is the molecular formula for the alkane of smallest molecular weight which possesses a stereogenic centre?A. `C_(7)H_(16)`B. `C_(6)H_(14)`C. `C_(5)H_(12)`D. `C_(8)H_(18)` |
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Answer» Correct Answer - A `CH_(3)-CH_(2)-CH_(2)-underset(CH_(3))underset(|)overset(H)overset(|)(C)-CH_(2)CH_(3)` Smallest alkane with a chiral carbon |
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| 134. |
0.6723 g of an organic compound gave on combustion 1.530 g of carbon dioxide and 0.625g of water. Find the percentage of carbon and hydrogen in the compound. |
| Answer» Correct Answer - C = 62.07%, H = 10.33% | |
| 135. |
Ammonia produced when 0.75 g of a substance was kjeldahlised neutralised `30 cm^(3)` of 0.25 `NH_(2)SO_(4)`. Calculate the persentage of nitrogen in the compound. |
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Answer» Mass of the organic substance =0.75 g Volume of `H_(2)SO_(4)` used up `=30 cm^(3)` Normality of sulphuric acid `= 0.25 N` `30 cm^(3)` of `H_(2)SO_(4)` of normality `0.25 N equiv 30 cm^(3)` of `NH_(3)` solution of normality 0.25 N But `1000 cm^(3)` of ammonia solution of normality 1 contain 14 g of nitrogen. `:.` Therefore, `30 cm^(3)` of 0.25 N ammonia solution will contain nitrogen `=14/1000xx30xx0.25` `:.` Percentage of nitrogen `=("Mass of nitrogen")/("Mass of substance")xx100=14/1000xx(30xx0.25)/(0.75)xx100=14.00`. |
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| 136. |
Two immiscible liquid present in a bottle can be separated by:A. Separating funnelB. Steam distillationC. Fractional distillationD. Chromatography |
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Answer» Correct Answer - A Separating funnel is used to separate two immiscible liquids |
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| 137. |
Paper chromatography has following mobile and stationary phases respectivelyA. liquid, solidB. solid, liquidC. gas, liquidD. liquid, liquid |
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Answer» Correct Answer - D It is the correct answer. |
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| 138. |
Which of the following compounds possesses the C-H bond with the lowest bond dissociation energy ?A. TolueneB. BenzeneC. n-PentaneD. 2, 2-Dimethylpropane |
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Answer» Correct Answer - A More stable the free radical formed upon homolytic fission of C-H bond, lesser is the bond dissociation energy. Since the stability of the radicals formed from toluene, benzene, n-pentane, 2, 2-dimethylpropane follows the order : `C_(6)H_(5)overset(*)(C)H_(2) gt (CH_(3))_(3)overset(*)(C) gt CH_(3)overset(*)(C)HCH_(2)CH_(2)CH_(3) gt overset(*)(C_(6))H_(5)`, therefore, the C-H bond of toluene has the lowest bond dissociation energy. |
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| 139. |
The increasing order of stability of the following free radicals isA. `(CH_(3))_(2)overset(*)(C)H lt (CH_(3))_(3)overset(*)(C) lt (C_(6)H_(5))_(2) overset(*)(C)H lt (C_(6)H_(5))_(3)overset(*)(C)`B. `(C_(6)H_(5))_(3)overset(*)(C) lt (C_(6)H_(5))_(2)overset(*)(C)H lt (CH_(3))_(3)overset(*)(C) lt (CH_(3))_(2)overset(*)(C)H`C. `(C_(6)H_(5))_(2)overset(*)(C)H lt (C_(6)H_(5))_(3)overset(*)(C) lt (CH_(3))_(3)overset(*)(C) lt (CH_(3))_(2)overset(*)(C)H`D. `(CH_(3))_(2)overset(*)(C)H lt (CH_(3))_(3)overset(*)(C) lt (C_(6)H_(5))_(3)overset(*)(C) lt (C_(6)H_(5))_(2)overset(*)(C)H` |
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Answer» Correct Answer - A More the number of phenyl groups, greater is the stabilization due to resonance, therefore, `(C_(6)H_(5))_(3)C` is more stable than `(C_(6)H_(5))_(2)overset(*)(C)H`. Further, more the number of methyl groups, more is the number of hyperconjugation structures and hence more stable is the radical, i.e, `(CH_(3))_(3)overset(*)(C)` is more stable than `(CH_(3))_(2)overset(*)(C)H`. Thus, option (a) is correct. |
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| 140. |
Assertion (A) : Compounds of a mixture of red and blue inks can be separated by distributing the compounds between stationary and mobile phases in paper chromatography. Reason (R) : The coloured components of inks migrate at different rates because paper selectively retains different components according to the difference in their partition between the two phases.A. Both A and R are correct and R is the correct explanation of A.B. Both A and R are correct but R is not the correct explanation of A.C. Both A and R are not correctD. A is not correct but R is correct |
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Answer» Correct Answer - A Reason is the correct explanation for Assertion |
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| 141. |
A misture of camphor and benzoic acid can be separated by :A. SublimationB. Chemical methodC. Fractional CrystallisationD. Extractin with solvent |
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Answer» Correct Answer - B The mixture is reacted with aqueous `NaOH` solution when benzoic acid becomes water soluble as sodium salt and can be separated. It can be regenerated by reacting the solution with dilute `HCl`. `underset("(Soluble)")(C_(6)H_(5)COOH+NaOH rarr C_(6)H_(5)COONa+H_(2)O)` `C_(6)H_(5)COONa+HCl rarr underset("(Regenerated)")(C_(6)H_(5)COOH)+HCl` |
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| 142. |
Without using column chromatography, how will you separate a mixture of camphor and benzoic acid? |
| Answer» Sublimation cannot be used since both camphor and benzoic acid sublime on heating. Therefore, a chemical method using `NaHCO_(3)` solution is used when benzoic acid dissolves leaving camphor behind. The filtrate is cooled and then acidified with dil. HCl to get benzoic acid. | |
| 143. |
In sodium fusion test of organic compounds, the nitrogen of an organic compound is converted toA. Soda limeB. Sodium cyanideC. Sodium nitriteD. Sodium nitrate. |
| Answer» Correct Answer - B | |
| 144. |
A mixture of camphor and benzoic acid can be separated byA. SublimationB. Chemical methodC. Fractional crystallisationD. Extraction with solvent. |
| Answer» Correct Answer - B | |
| 145. |
The purity of an organic compound is determined byA. DensityB. Melting pointC. Mixed melting pointD. Molecular mass |
| Answer» Correct Answer - B | |
| 146. |
The purity of an organic solid is determined by :A. densityB. melting pointC. mixed melting pointD. molecular mass |
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Answer» Correct Answer - C Mixed melting point is used to test the purity of organic solid. |
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| 147. |
Assertion (A) : Sulphur present in an organic compound can be estimated quantitatively by Carius method. Reason (R) : Sulphur is separated easily from other atoms in the molecule and gets precipitated as light yellow solid.A. Both A and R are correct and R is the correct explanation of A.B. Both A and R are correct but R is not the correct explanation of A.C. Both A and R are not correct.D. A is not correct but R is correct. |
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Answer» Correct Answer - D Correct (R). Sulphur is oxidised to `H_(2)SO_(4)` in Carius method and then estimated as barium sulphate. |
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| 148. |
An organic substance from its aqueous solution can be separated byA. distillationB. steam distillationC. solvent distillationD. fractional distillation |
| Answer» Correct Answer - C | |
| 149. |
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. DistitationB. Kinetic resolutionC. CrystalizationD. Sublimation. |
| Answer» Correct Answer - D | |
| 150. |
How many of the following cannot show tautomerism ? acetophenone, acetaldehyde, cyclohexanone, acetylacetone, benzoquinone, acetone , benzaldehyde, butanone, ethyl acetoacetate. |
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Answer» Correct Answer - 2 Two, i.e., benzoquinone and benzaldehyde. |
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