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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.
| 51. |
2-Nitropropane on hydrolysis with boiling concentrate solution of HCI gives:A. propaneB. propanalC. propanoneD. propanoic acid |
| Answer» Correct Answer - C | |
| 52. |
Which one is less alkaline?A. B. C. D. All of these |
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Answer» Correct Answer - A Because of presence of electron withdrawing group. |
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| 53. |
Which one of the following forms propanenitrile as the major product ?A. Ethyl bromide + alcoholic KCNB. Propyl bromide + alcoholic KCNC. Propyl bromide + alcoholic AgCND. Ethyl bromide + alcoholic AgCN |
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Answer» Correct Answer - A When ethyl bromide reacts with alcoholic KCN, propane nitrile is obtained as the main product. `C_2H_5Br + KCN(alc) oversetDeltato underset"Propane nitrile"(C_2H_5CN)+KBr` |
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| 54. |
When `C_6H_5N_2Cl` is reduced with `Na_2SnO_2`, the product is:A. B. C. D. |
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Answer» Correct Answer - B Diazo group can be replaced by (H) on reduction with `Na_2SnO_2` or warm with `C_2H_5OH`. |
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| 55. |
Reduction of alkyl nitriles, producesA. secondary amineB. primary amineC. tertiary amineD. amide |
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Answer» Correct Answer - B `underset"Alkyl nitriles "(R-CN+4[H]) overset(LiAlH_4)to underset"Primary amine"(R-CH_2NH_2)` |
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| 56. |
Method by which Aniline can t be prepared is .A. hydrolysis of phenol isocyanide with acidic solutionB. potassium salt of phthalimide treated with chloroebenezene followed by hydrolysis with aquesous NaOH solutionC. reduction of nitrobenzene with `H_(2)//Pd` in ethanolD. degradation of benzamide with bromine in alkaline solution |
| Answer» Correct Answer - B | |
| 57. |
In a reactione a coloured product `C` was obtained The structure of `C` would be A. B. C. D. |
| Answer» Correct Answer - B | |
| 58. |
Mendius reaction converts an alkyl cyanide to"A. a primary amineB. an aldehydeC. a ketoneD. an oxine |
| Answer» Correct Answer - A | |
| 59. |
Write two methods each for the preparation of alkyl cyanide and alkyl isocyanide. |
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Answer» Preparation a) From Alkyl Halides : Alkyl halides with ethanolic potassium cyanide gives cyanides where as with silver cyanide gives alkyl isocyanide. `CH_(3)- underset(Cl)underset(|)(CH_(2)) underset(KCN) overset("ethanolic") (to) CH_(3)-underset("Ethyl cyanide")underset(CN)underset(|)(CH_(2))+KCl` `CH_(3)- underset(Cl)underset(|)(CH_(2)) underset(AgCN) overset("ethanolic") (to) CH_(3) - underset("Ethyl isocyanide")underset(NC)underset(|)(CH_(2))+AgCl` b) From amides and aldoximes : The dehydration of amides (or) oximes with dehydrating agents like `P_(2)O_(5)` (or) with benzene sulphonyl chloride yeild cyanides. `CH_(3)- underset(O)underset(||)(C)-NH_(2)+C_(6)H_(5)SO_(2)Cl underset(70^(@)C)overset("Pysanidine") (to) CH_(3)-CN+C_(6)H_(5)SO_(3)H+HCl` `CH_(3)-CH_(2)-CH=N-OH+(CH_(3)CO)_(2)O to CH_(3)- underset(CN)underset(|)(CH_(2))+2CH_(3)COOH` c) Isocyanides from amines : (Carbyl amine reaction) `R-NH_(2)+CHCl_(2)+3KOH overset("Heat") to R-NC+3KCl+3H_(2)O` |
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| 60. |
The aniline reaction with…to..yield… as the final product..A. bromine, 2-bromoanilineB. bromine, 2,4,6-tribromoanilineC. chloroform/KOH, phenyl cyanideD. acetyl chloride, benzalide |
| Answer» Correct Answer - B | |
| 61. |
Final product of. hydrolysed alkyl cyanide isA. R COOHB. R `CONH_2`C. `R-undersetunderset(OH)|C=NH`D. `R-C-=overset(o+)NH` |
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Answer» Correct Answer - A `underset"Alkyl cyanide"(R-C-=N) overset"Partial hydrolysis"to underset"Alkyl amide"(R-CONH_2) overset"Complete hydrolysis"to underset"Carboxylic acid ""RCOOH"` |
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| 62. |
`CH_3-CH=CH-CH=N-CH_3overset(LiAlH_4)to` What is final productA. `CH_3-CH_2-CH_2CH_2NH-CH_3`B. `CH_3-CH=CH-CH_2-NH-CH_3`C. `CH_3-CH_2-CH_2-CH_2-N-CH_3`D. `CH_3-CH=CH-CH_2-OH` |
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Answer» Correct Answer - B `CH_3-CH=CH-CH=N-CH_3overset(LiAIH_4)toCH_3-CH=CH-CH_2-NH-CH_3` `LiAIH_4` reduces imine into amine but does not reduces `C=C` double bond. |
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| 63. |
The amine which does not react with acetyl chloride is or which of the following cannot be acetylatedA. `CH_3NH_2`B. `(CH_3)_2NH`C. `(CH_3)_3N`D. none of these |
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Answer» Correct Answer - C `3^@` amine3 cannot be acetylated because replaceable `H-atom` is absent. |
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| 64. |
Aniline when treated acetyl chlroide in presence of alkali, the product formed is:A. acetanilideB. benzoyl chlorideC. acetophenoneD. aniline hydrochloride |
| Answer» Correct Answer - A | |
| 65. |
Isopropylamine with excess of acetyl chloride will give?A. `(CH_3CO)_2N-CH-(CH_3)_3`B. `(CH_3)_2CH-undersetunderset(H)(|)N-COCH_3`C. `(CH_3)_2CHN(COCH_3)_2`D. `CH_3CH_2CH_2-undersetundersetH|N-COCH_3` |
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Answer» Correct Answer - C `underset"(pri amine)"underset"Iso-propyl amine" ((CH_3)_2CHNH_2) + 2CH_3COCl overset(-HCl) to underset"Tert-amine"((CH_3)_2CH-N(COCH_3)_2)` |
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| 66. |
The product of the reaction of alcoholic silver nitrite with ethyl bromide is:A. EthaneB. EtheneC. NitroethaneD. Ethyl alcohol |
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Answer» Correct Answer - C `C_2H_5Brunderset(AgNO_2)overset(alc.)tounderset("Nitroethane")(C_2H_5NO_2)` |
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| 67. |
(A) The main product of reaction of alcoholic silver nitrite and ethyl bromide is nitroethane. (R) Silver nitrite is predominantly covalent compound.A. If both A and R are correct and R is the correct explanation of A.B. If both A and R are correct and R is not the correct explanation of A.C. If A is the correct but R is incorect.D. If A is the incorrect but R is corect. |
| Answer» Correct Answer - A | |
| 68. |
The reaction of chloroform with alcoholic `KOH` and p- toluidine formsA. B. C. D. |
| Answer» Correct Answer - D | |
| 69. |
A mixture of ethyl amine and alcoholic KOH on heating givesA. alkylcyanideB. ethylcyanateC. ethylisocyanideD. ethylisocyanate |
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Answer» Correct Answer - D `underset"Ethyl amine"(C_2H_5NH_2) + CHCl_3 + underset"(Alcoholic)""3KOH" to C_2H_5NC + 3KCl + 2H_2O` It is carbylamine reaction. |
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| 70. |
The end product of the reaction, Ethyl amine `overset(HNO_(2))rarr(A)overset(PCl_(5))rarr(B)overset(KCN)rarr(C )` `(C)` is,A. propane nitrileB. triethylamineC. diethylamineD. propylamine |
| Answer» Correct Answer - A | |
| 71. |
Examine the following two structures for the anilinium ion and choose the correct statement from the ones given below:A. II is not an acceptable canonical structure because carbocation ions are less stable than ammonium ionsB. II is not an accepted canonical structure because it is non-aromaticC. II is not an acceptable canonical structure because the nitrogen has 10 valence electronsD. II is an acceptable canonical structure |
| Answer» Correct Answer - C | |
| 72. |
`CH_(3)NH_(2)+CHCl_(3)+KOH to` nitrogen containing compound `+KCl+H_(2)O`. Nitrogen containing compound isA. `CH_3NH_2`B. `CH_3-C-=N`C. `CN_3overset(+)(N)-=overset(-)(C)`D. `CH_3-overset(-)N-=overset(+)C` |
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Answer» Correct Answer - C This is an example of carbylamine reaction. `CHCl_3+RNH_2+KOHrarrR-overset(+)N-=overset(-)+3KCl+3K_2O` |
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| 73. |
Statement I: Aniline on reaction with ` NaNO_2HCl` at `0^@C` followed by coupling with `beta`-naphthol gives a dark blue coloured precipitate. Statement II: The colour of the compound formed in the reaction of aniline with `NaNO_2//HCl` at ` 0^@C` followed by coupling with `beta`-naphthol is due to extended conjugation.A. If both A and R are correct and R is the correct explanation of A.B. If both A and R are correct and R is not the correct explanation of A.C. If A is the correct but R is incorect.D. If A is the incorrect but R is corect. |
| Answer» Correct Answer - D | |
| 74. |
Statement I: Aniline on reaction with ` NaNO_2HCl` at `0^@C` followed by coupling with `beta`-naphthol gives a dark blue coloured precipitate. Statement II: The colour of the compound formed in the reaction of aniline with `NaNO_2//HCl` at ` 0^@C` followed by coupling with `beta`-naphthol is due to extended conjugation. |
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Answer» Correct Answer - D Correct assertion: Aniline reacts with `(NaNO_2)/(HCl)` at `0^@C` followed by coupling with `beta`-napthol gives an orange red coloured. |
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| 75. |
The strongest base among the following .A. B. C. D. |
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Answer» Correct Answer - C The lone pair of electrons on nitrogen is not involved in the formation of `pi`-electron cloud of the ring. |
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| 76. |
Among the following the strongest base isA. `C_(6)H_(5)NH_(2)`B. `p-NO_(2)-C_(6)H_(4)NH_(2)`C. `p-CH_(3)-C_(6)H_(4)NH_(2)`D. `C_(6)H_(5)CH_(2)NH_(2)` |
| Answer» Correct Answer - D | |
| 77. |
The strongest base among the following isA. `C_6H_5NH_2`B. `(C_6H_5)_2NH`C. `C_2H_5NH_2`D. `(C_2H_5)_2NH` |
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Answer» Correct Answer - D `(C_2H_5)_2 ddotNH` (`2^@` amine ) is strongest base. Basic nature of amines is due to the presence of lone pair of electron on nitrogen atom which is available for the bond formation with Lewis acid. Due to the +I effect `2^@` amine is better base than `1^@` amine and `NH_3`. In case of aromatic amines the lone pair on nitrogen atom involved in resonance, therefore , not available for bond formation, so aromatic amines are less basic . |
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| 78. |
Which of the following is the strongest base:A. `C_6H_5NH_2`B. `p-NO_2-C_6H_5NH_2`C. `m-NO_2-C_6H_5NH_2`D. `C_6H_5CH_2NH_2` |
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Answer» Correct Answer - D Aliphatic amines (in which amino group is attached with alkyl group) are more basic than aromatic amines (in which amino group is bonded directly with benzene nucleus). Hence, `C_6H_5CH_2NH_2` (benzyl amine), being an aliphatic amine, is the most basic among the given compounds. |
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| 79. |
Which of the following is the strongest base:A. B. C. D. |
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Answer» Correct Answer - A In benzylamine lone pair of `._(-1)e_0` can be enter in resonance with benzene ring. |
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| 80. |
An organic compound X having molecular formula `C_4H_11N` reacts with p-toluene sulphonyl chloride to form a compound Ythat is soluble in aqueous KOH. Compound X is optically active and reacts with acetyl chloride to form compound Z. Identify the compound Z.A. `CH_3CH_2CH_2CH_2NHCOCH_3`B. `CH_3CH_2 oversetoverset(CH_3)|C NH HCOCH_3`C. `CH_3 overset(CH_3)overset|C HC H_2NHCOCH_3`D. `CH_3-undersetunderset(CH_3)|oversetoverset(CH_3)|C-NHCOCH_3` |
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Answer» Correct Answer - B Since, `X(C_4H_11N)` reacts with p-toluene sulphonyl chloride (Hinsberg reagent) to form a compound which is soluble in KOH, therefore X must be a primary amine. `undersetX(C_6H_5SO_2Cl+H-oversetoversetH|N-overset(CH_3)overset|undersetHunderset|C-CH_2-CH_3) underset(-HCl)to underset"(Soluble in KOH)"undersetY((C_6H_5SO_2-oversetoverset(H)|N-undersetundersetH|oversetoverset(CH_3)|C-CH_2-CH_3))` `undersetX(CH_3-CH_2-oversetoverset(CH_3)|CH-oversetoversetH|N-H+ClCOCH_3 underset(-HCl)) to undersetZ(CH_3-CH_2-oversetoverset(CH_3)|CH-NHCOCH_3)` |
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| 81. |
Which of the following is the strongest base?A. `C_6H_5NH_2`B. `CH_3NH_2`C. `NH_3`D. `CH_3CONH_2` |
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Answer» Correct Answer - B Basic nature of amine is due to the presence of lone pair of electrons on nitrogen atom which is available for the bond formation with Lewis acid. Due to+ I effect of `-CH_3`group, `CH_3NH_2` is the strongest base. |
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| 82. |
Some meta-directing substituents in aromatic substitution are given which one is the most deactivating?A. `-C-=N`B. `-SO_3H`C. `-COOH`D. `-NO_2` |
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Answer» Correct Answer - D The deactivating tendency of given a groups follows the order `-NO_2 gt -SO_3H gt -C -= N gt - oversetoverset(O)(||)C-OH` Thus, `-NO_2` is the most deactivating group. |
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| 83. |
Reactivity order of the following towards `NaOEt, EtOH` A. `III gt II gt I`B. `II gt I gt III`C. `I gt II gt III`D. `III gt I gt II` |
| Answer» Correct Answer - C | |
| 84. |
Which give only monosubstituted product?A. o-DinitrobenzeneB. m-DinitrobenezeneC. p-DinitrobenzeneD. nitrobenzene |
| Answer» Correct Answer - B | |
| 85. |
A compound with formula `C_5H_(13)` gives a base soluble derivative iin the Hinsberg test `(C_6H_5SO_2Cl` in base). Which of the following best satisfy this condition?A. 2,2-dimethylpropylamineB. IsopropyldimethylamineC. N,N-dimethylpropylamineD. N-methyl-2-methylpropylamine |
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Answer» Correct Answer - A `CH_(3)-overset(CH_(3))overset(|)underset(CH_(3))underset(|)(C)-CH_(2)-NH_(2)+CHCl_(3)+3KOH` `rarr(CH_(3))_(3)C CH_(2)NC+3KCl+3H_(2)O` This is the property of `1^@` amine (i.e., compound having-`NH_2`) group. Thus 2,2- dimethylpropylamine gives this reaction. |
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| 86. |
A nitrogen containg compound dissolves is `10%` aqueous sulphuric acid. The Hinsberg test `(C_6H_5SO_2Cl` in base) gives a solid product that is not soluble in `10%` aqueous `NaOH`. Which of the following would best fit in these fact?A. N,N-dimethylanilineB. N-methylbenzamideC. BenzylamineD. N-methylaniline |
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Answer» Correct Answer - D These are the cracteristics of secondary amines and `C_6H_5NHCH_3` is a secondary amine. |
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| 87. |
The product of mustard oil reaction isA. Alkyl isothiocyanateB. Dithio carbonamideC. Dithio ethylacetateD. Thioether |
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Answer» Correct Answer - A Mustard oil reaction. `CH_3-underset("Ethyl amine")(CH_2-NH_2)+CS_2overset(HgCl_2)to` `underset("ethyl isothiocyanate")(CH_3-CH_2-N=)S+H_2S` |
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| 88. |
Hoffmann Mustard Oil ReactionA. `(Na)/(C_2H_5OH)`B. `(Sn)/(HCl)`C. `CS_2`D. `(K_2Cr_2O_7)/(H_2SO_4)` |
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Answer» Correct Answer - C Amine is treated with `CS_2` in reaction. |
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| 89. |
Observe the following reactions, `A+C_6H_5SO_2Cloverset(-OH)tooverset(KOH)to` clear solution `overset(H^(+))to` substance is soluble in aicdic medium `B+C_6H_5SO_2Cloverset(-OH)to` Insoluble substance in `(H^(+))/(OH^(-))` `C+C_6H_5SO_2Cloverset(-OH)to` Insoluble in base but soluble in `H^(+)overset(H^(+))to` Clear solution.A. `1^@`,`3^@` and `2^@` aminesB. `1^@`, `2^@` and `3^@` aminesC. `1^@`, amiline and `2^@` amineD. `1^@` amine, `2^@` amine and aniline |
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Answer» Correct Answer - B `1^@` amine forms `N-`alkyl benzene sulphonamide with `C_6H_5SO_2Cl`, which if soluble is alkali. `2^@` amine forms N, N-dialkyl benzene sulphonamide, which is insoluble in alkali as well as acid. `3^@` amines do not react with `C_6H_5SO_2Cl` and hence are soluble in acid. Thus, A,B and C here represent `1^@,2^@` and `3^@`. |
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| 90. |
How do you distinguish cyanides and isocyanides by hydrolysis and reducation. |
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Answer» Hydrolysis : Cyanides on hydrolysis give carboxylic acids and ammmonia where as isocynanides on hydrolysis give primary amines and formic acid. `R-CN+2H_(2)O overset(H^(+)//OH^(-)) to R-COOH+NH_(3)` `R-CN+2H_(2)O overset(H^(+)//OH^(-)) to R-NH_(2)+HCOOH` ii) Reduction : Reducation of nitriles give primary amines where as reduction of isocyanides yield secondary amines. `R-CN+4(H) underset(Na .Alcohol)overset(LiA//H_(4)(or)) to R-CH_(2)-NH_(2)" ""Primary amine"` `underset("alkyl isocyanide")(R-NC+4(H)) overset(H_(2)//Ni) to R-NH-CH_(3)" ""Secondary amine"` |
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| 91. |
A student tried to synthesis 4-nitro-N, N-dimethylaniline from N, N-dimethylaniline via electrophilic aromati substitution using nitric acid `(HNO_3)` and sulphuric acid `(H_2SO_4)`. To his surprice, the major product he obtained was 3-nitro-N, N-dimethylaniline. Why the disigned reaction failed to provide the desired product, 4-nitro-N, N-dimethylaniline?A. The amine group is meta directing.B. Protonation at amine group makes it meta directing.C. The amine group is strong electron withdrawing group by -I effect.D. Steric inhibition to resonance is responsible for the above result. |
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Answer» Correct Answer - B This is because of the protonation of amine group which makes it meta directing. |
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| 92. |
Write the equations showing the conversion of aniline diazoniumchloride to a) chlorobenzene, b) Iodobenzene and c) Bromobenzene |
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Answer» a) Conversion of aniline diazonium chloride to chloro benzene `C_(6)H_(5)N_(2)^(+)Cl^(-) overset(CH_(2)Cl_(2)//HCl) to C_(6)H_(5)Cl+N_(2)` b) Conversion of aniline diazonium chloride to Iodo benzene `C_(6)H_(5)N_(2)^(+)Cl^(-)+KI to C_(6)H_(5)I+N_(2)+KCl` c) Conversion of aniline diazonium chloride to Bromo benzene `C_(6)H_(5)N_(2)^(+)Cl^(-) overset(Cu_(2)Br_(2)//HBr) to C_(6)H_(5)Br+N_(2)` |
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| 93. |
Complete the following conversions : Aniline to Fluorobenzene |
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Answer» Aniline to Fluorobenzene `underset("Aniline")(C_(6)H_(5)NH_(2)) underset(0-5^(@)C)overset(NaNO_(2)+HCl) to underset("diazonium chloride") underset("Benzene")(C_(6)H_(5)N_(2)^(+)Cl^(-)) underset(Delta) overset(HBF_(4)) to underset("Fluorobenzene")(C_(6)H_(5)-F+BF_(3)+N_(2))` |
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| 94. |
Complete the following conversions : Aniline to Cyanobenzene |
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Answer» Aniline to cyano benzene `underset("Aniline")(C_(6)H_(5)NH_(2)) underset(0-5^(@)C)overset(NaNO_(2)+HCl) to underset("diazonium chloride") underset("Benzene")(C_(6)H_(5)N_(2)^(+)Cl^(-)) overset(CuCN//KCN) to underset("Cyano Benzene")(C_(6)H_(5)-CN+N_(2))` |
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| 95. |
Complete the following conversions : Aniline to Phenol |
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Answer» Aniline to phenol `underset("Aniline")(C_(6)H_(5)NH_(2)) underset(0-5^(@)C)overset(NaNO_(2)+HCl) to underset("diazonium chloride") underset("Benzene")(C_(6)H_(5)N_(2)^(+)Cl^(-)) overset(H_(2)O) to underset("Phenol")(C_(6)H_(5)OH+N_(2)+HCl)` |
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| 96. |
Which of the following reaction will not give primary amine ?A. `CH_3CONH_2 overset(Br_2,KOH)to`B. `CH_3CN overset(LiAlH_4)to`C. `CH_3NC overset(LiAlH_4)to`D. `CH_3CONH_2 overset(LiAlH_4)to ` |
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Answer» Correct Answer - C `CH_3CONH_2 overset(Br_2, KOH)to underset((1^@ "amine"))(CH_3CH_2)` `CH_3CN overset(LiAlH_4)to underset((1^@ "amine"))(CH_3CH_2NH_2)` `CH_3NC overset(LiAlH_4)to underset((2^@ "amine"))(CH_3NHCH_3)` `CH_3CONH_2 overset(LiAlH_4)to underset((1^@ "amine")) (CH_3CH_2NH_2)` |
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| 97. |
By distilling glucine with braium hydroxide, it gives:A. ethylamineB. methylamineC. amino acidD. acetic acid |
| Answer» Correct Answer - B | |
| 98. |
In the reaction `A` is .A. `H_(3)PO_(2) and H_(2)O`B. `Cu_(2)Cl_(2)`C. `HgSO_(3)//H_(2)SO_(4)`D. `H^(+)//H_(2)O` |
| Answer» Correct Answer - A | |
| 99. |
In order to distinguish between `C_(2)H_(5)NH_(2) and C_(6)H_(5)NH_(2)`, Which of the following reagents(s) is useful?A. Hinsberg reagentB. `beta-"napthol"`C. `CH_(3)Cl_(3)//KOH`D. NaOH |
| Answer» Correct Answer - B | |
| 100. |
In order to distinguish between `C_(2)H_(5)NH_(2) and C_(6)H_(5)NH_(2)`, Which of the following reagents(s) is useful?A. Hinsberg reagentB. `beta-"napthol"`C. `CHCl_(3)//KOH`D. NaOH |
| Answer» Correct Answer - B | |