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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.
| 201. |
The major organic product formed in the following reaction is A. B. C. D. |
| Answer» Correct Answer - B | |
| 202. |
Aniline on heating with fuming sulphuric acid gives.A. benzene sulphonic acidB. anthranilic acidC. anilineD. sulphanilic acid |
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Answer» Correct Answer - D Anilinium hydrogen sulphate on heating with sulphuric acid at 453-473 K produces sulphanilic acid. |
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| 203. |
Explain why ethylamine is more soluble in water, whereas aniline is not soluble. |
| Answer» Elthyl amine is more soluble in water due to the presence of hydrogen bonding. Where are in case of aniline due to presence of bulky hydrocarbon part, the extent of hydrogen bonding is less and it is not soluble in water. | |
| 204. |
Give one chemical test of distinguish between the following pairs of compounds. Aniline and N.Methylaniline |
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Answer» Aniline `(1^(@) - "amine")` and N.Methyl `(2^(@) - "amine")` aniline are distinguished by carbylamine test (or) isocyanide test. Aniline responds to carbyl amine test to give foul smelling phenyl iso cyanide where as N - methyl aniline does not responds to carbyl amine Test. `C_(6)H_(5)NH_(2)+CHCl_(3)+ak.KOH overset(Delta)to underset("(Foul smelline)") underset("Phenyl isocyamide")(C_(6)H_(5)NC+3KCl+3H_(2)O)` |
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| 205. |
Give one chemical test of distinguish between the following pairs of compounds. Ethylamine and aniline |
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Answer» Ethyl amine (`1^(@)` - aliphatic amine) and aniline (`1^(@)` - aromatic amine) are distinguished by Diazotisation reaction. Aniline under go diazotisation reaction to form benzene diazonium salt where as ethyl amine form highly unstable alkyl diazonium salt `C_(6)H_(5)NH_(2) overset(NaNO_(2)+HCl) to underset("Benzene diazonium chloride")(C_(6)H_(5)N_(2)^(+)Cl^(-)+3NaCl+2H_(2)O)` |
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| 206. |
Gabriel Phthalimide synthesis exclusively forms primary amines only. Explain. |
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Answer» Gabriel Phthalinide synthesis exclusively forms primary amines only. Reason : In this reaction primary amines are formed without the traces of `2^(@)` (or) `3^(@)` amines. |
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| 207. |
How do you carryout the following conversions ? Aniline to Benzene sulphonamide |
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Answer» Conversion of Aniline to Benzene sulphonamide : Aniline reacts with benzene sulphonyl chloride to form N - Phenyl benzene sulphonamide. `C_(6)H_(5)NH_(2)+C_(6)H_(5)SO_(2)Cl to underset("Phenyl benzene sulphonamide")(C_(6)H_(5)NHSO_(2)C_(6)H_(5)+HCl)` |
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| 208. |
How will you convert `Cl-(CH_(2))_(4)-Cl` into hexan-1, 6-diamine ? |
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Answer» `Cl-(CH_(2))_(4)-Cl)` into hexan-1, 6 - diamine `Cl-(CH_(2))_(4)-Cl+2KCN underset(-KCl) overset("Alcohol") (to)CN(CH_(2))_(4)-CN underset(or LiAlH_(4))overset(H_(2)//Pt)(to) underset("Hexan-1, 6-diamine")(NH_(2)CH_(2)(CH_(2))_(4)CH_(2)NH_(2))` |
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| 209. |
Acid anhydrides on reaction with primary amine gives…A. amideB. imideC. imineD. `2^(@)` amine |
| Answer» Correct Answer - A | |
| 210. |
Which one of the following is hydrolysed to give secondary amine?A. Alkyl cyanideB. NitroalkanesC. Acid amideD. Dimethyl formamide |
| Answer» Correct Answer - D | |
| 211. |
Arrange the following in increasing order of their basic strength. `C_(2)H_(5)NH_(2),C_(6)H_(5)NH_(2), NH_(3),C_(6)H_(5)CH_(2)NH_(2)and(C_(2)H_(5))_(2)NH` |
| Answer» `C_(6)H_(5)NH_(2)ltNH_(3)ltC_(6)H_(5)CH_(2)NH_(2)ltC_(2)H_(5)NH_(2)lt(C_(2)H_(5))NH` | |
| 212. |
Comparing basic strength of `NH_(3), CH_(3)NH_(2)` and `C_(6)H_(5)NH_(2)` it may be concluded that:A. basic strength remains unaffectedB. basic strength of `NH_(3)` is highestC. basic strength of alkylamine is lowestD. basic strength of arylamine is lowest |
| Answer» Correct Answer - D | |
| 213. |
The correct order of increasing basic nature of the bases `NH_(3),CH_(2)NH_(2) and (CH_(3))_(2)NH` is-A. `CH_(3)NH_(2) gt (CH_(3))_(2)NH gt (CH_(3))_(3)N gt NH_(3)`B. `(CH_(3))_(3)N gt (CH_(3))_(2)NH gt CH_(3)NH_(2) gt NH_(3)`C. `(CH_(3))_(2)NH gt CH_(3)NH_(2) gt (CH_(3))_(3)N gt NH_(3)`D. `NH_(3) gt (CH_(3))_(3)N gt CH_(3)NH_(2) gt (CH_(3))_(2)NH` |
| Answer» Correct Answer - C | |
| 214. |
Arrange the following in increasing order of their basic strength. `CH_(3)NH_(2), (CH_(3))_(2)NH,(CH_(3))_(3)N,C_(6)H_(5)NH_(2),C_(6)H_(5)CH_(2)NH_(2)` |
| Answer» `C_(6)H_(5)NH_(2)ltC_(6)H_(5)CH_(2)NH_(2)lt(CH_(3))_(3)NltCH_(3)NH_(2)lt(CH_(3))_(2)NH` | |
| 215. |
Arrange the following in increasing order of their basic strength. `C_(2)H_(5)NH_(2),(C_(2)H_(5))_(2)NH,(C_(2)H_(5))_(3)N,C_(6)H_(5)NH_(2)` |
| Answer» `C_(6)H_(5)NH_(2)ltC_(2)H_(5)NH_(2)lt(C_(2)H_(5))_(3)Nlt(C_(2)H_(5))_(2)NH` | |
| 216. |
The reagent that is used to distinguish between secondary amine and teritary amine is:A. p-touene sulphonyl chlorideB. `CHCl_(3)` and alc. KOHC. Lucas reagentD. Bromine water |
| Answer» Correct Answer - A | |
| 217. |
Which of these reactions can be used for the preparation of prinmary amines?A. Curtius rearrangementB. Gabriel synthesisC. Schmidt rearrangementD. All of the above |
| Answer» Correct Answer - D | |
| 218. |
N-butylamine (I), diethylamine (II) and N,N-dimethyl ethylamine(III) have the same molar mass. The increasing order of their boiling point is:A. `III lt II lt I`B. `I lt II lt III`C. `II lt III lt I`D. `II lt I lt III` |
| Answer» Correct Answer - A | |
| 219. |
The correct increasing order of basic strength in, `CH_(3)CH_(2)CN, CH_(3)CH_(2)NH_(2), CH_(3)N=CHCH_(3)` isA. `CH_(3)N=CHCH_(3),CH_(3)CH_(2)NH_(2), CH_(3)CH_(2)CN`B. `CH_(3)CH_(2)NH_(2),CH_(3)N=CHCH_(3),CH_(3)CH_(2)CN`C. `CH_(3)CH_(2)CN, CH_(3)N=CHCH_(3), CHCH_(2)NH_(2)`D. `CH_(3)CH_(2)CN, CH_(3)CH_(2)NH_(2), CH_(3)N=CHCH_(3)` |
| Answer» Correct Answer - C | |
| 220. |
In the given reaction `CH_(3)-CH_(2)-underset(NaOH)underset(||)(C)-CH_(3)underset((H_(2)O //H^(+)),Delta)overset(H_(2)SO_(4))to(A)+(B)` A and B are:A. `CH_(3)COOH and C_(2)H_(5)NH_(2)`B. `CH_(3)CH_(2)COOH and CH_(3)NH_(2)`C. `CH_(3)NH_(2) and C_(2)H_(5)NH_(2)`D. `CH_(3)COOH and CH_(3)CH_(2)COOH` |
| Answer» Correct Answer - A | |
| 221. |
The best method for preparation of `Me_(3)C-CN` is:A. `"to react" Me_(3)C to OH "with" HCN`B. `"to react" Me_(3)C to Br "with" NaCN`C. `"to react" Me_(3)C to MgBr "with" CICN`D. `"to react" Me_(3)C to Li"with"NH_(2)CN` |
| Answer» Correct Answer - C | |
| 222. |
Which one of the following reagent will convert acetamide to ethanamine?A. `P_(2)O_(5)`B. `SOCl_(2)`C. `LiAlH_(4)`D. `KCN` |
| Answer» Correct Answer - C | |
| 223. |
Amines are basic compounds They act as Lewis base due to the presence of lone pair electrons at nitrogen Amines aho behave as base as well as Bronsted inductive effect, steric hindrance and resosnance. `CH_(3)-NH_(2)+HOH hArr CH_(3)-NH_(3)^(+)+OH^(-)` `CH_(3)-NH_(2)+H^(+)Cl^(-)hArr CH_(3)-NH_(3)^(+)+Cl^(-)` Alkyl groups and electron groups hence these groups increases the electron density at nitrogen as well as the basic amines character of amines. Basic character of teritary amines is reduced due to the steric hindrance of three alkyl groups. Experimentally it is observed that stronger bases have smaller values of `pK_(b)` greater value of `K_(b)`. Which among hte following is the most basic in aqueous medium?A. `NH_(3)`B. `CH_(3)-NH_(2)`C. `(CH_(3))_(2)NH`D. `(CH_(3))_(3)N` |
| Answer» Correct Answer - C | |
| 224. |
Amines behave as:A. aprotic acidB. netural compoundC. Lewis acidD. Lewis base |
| Answer» Correct Answer - D | |
| 225. |
In which of the following compounds is there the least delocalization of the lone pair of electrons on the nitrogen atom?A. B. C. D. |
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Answer» Correct Answer - C In this structure, there is steric inhibition to resonance due to lack of planarity of nitrogen with aromatic ring. This occur due to steric crowding by the three bulky adhacent groups. |
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| 226. |
Amines are basic compounds They act as Lewis base due to the presence of lone pair electrons at nitrogen Amines also behave as base as well as Bronsted inductive effect, steric hindrance and resosnance. `CH_(3)-NH_(2)+HOH hArr CH_(3)-NH_(3)^(+)+OH^(-)` `CH_(3)-NH_(2)+H^(+)Cl^(-)hArr CH_(3)-NH_(3)^(+)+Cl^(-)` Alkyl groups and electron groups hence these groups increases the electron density at nitrogen as well as the basic amines character of amines. Basic character of teritary amines is reduced due to the steric hindrance of three alkyl groups. Experimentally it is observed that stronger bases have smaller values of `pK_(b)` greater value of `K_(b)`. Which among the following is the most basic in aqueous medium?A. `NH_(3)`B. `CH_(3)-NH_(2)`C. `(CH_(3))_(2)NH`D. `(CH_(3))_(3)N` |
| Answer» Correct Answer - D | |
| 227. |
p-Amino-N,N-dimethylaniline is added to a strongly acidic solution of X. The resulting solution is treated with a few drops of aqueous solution of Y to yield blue colouration due to the formation of methylene blue. Treatment of aqueous solution of Y with reagent potassium hexacyanoferrate (II) leads to the formation of an intense blue precipitate. The precipitate dissolves on excess addition of the reagent. Similarly, the treatment of the solution of Y with the solution of potassium hexacyanoferrate (III) leads to a brown colouration due to the formation of Z. Q. Compound X isA. `NaNO_(3)`B. `NaCl`C. `Na_(2)SO_(4)`D. `Na_(2)S` |
| Answer» Correct Answer - D | |
| 228. |
Amines are basic compounds They act as Lewis base due to the presence of lone pair electrons at nitrogen Amines aho behave as base as well as Bronsted inductive effect, steric hindrance and resosnance. `CH_(3)-NH_(2)+HOH hArr CH_(3)-NH_(3)^(+)+OH^(-)` `CH_(3)-NH_(2)+H^(+)Cl^(-)hArr CH_(3)-NH_(3)^(+)+Cl^(-)` Alkyl groups and electron groups hence these groups increases the electron density at nitrogen as well as the basic amines character of amines. Basic character of teritary amines is reduced due to the steric hindrance of three alkyl groups. Experimentally it is observed that stronger bases have smaller values of `pK_(b)` greater value of `K_(b)`. Which of the following is the most basic?A. `CH_(3)-CH_(2)`B. `C_(2)H_(5)-NH_(2)`C. `C_(3)H_(7)-NH_(2)`D. |
| Answer» Correct Answer - C | |
| 229. |
The correct order of basicity of amines in water is:A. `(CH_3)_2NHgt(CH_3)_3NgtCH_2NH_2`B. `CH+3NH_2gt(CH_3)_3NHgt(CH_3)_3N`C. `(CH_3)_3Ngt(CH_3)_2NHgt(CH_3NH_2`D. `(CH_3)_3NgtCH_3NH_2gt(CH_3)_2NH` |
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Answer» Correct Answer - A Basicity of amines increase with increase in number of `-CH_3` groups (or any group which cause `+I` effect), due to increase in electron density on N atom. As a rule, th basicity of t-amine should be more than that of s-amine, but actually it is found to be lesser than s-amines, this is due to stearic hindrance of bulkier alkyl groups, which decreases the availability of line pair of electron on the N atom of the amino group. Hence the correct order of basicity is `(CH_3)_2NHgt(CH_3)_3NgtCH_3NH_2` |
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| 230. |
Among the following which one does not act as an intermediate in hofmann rearrangementA. `RNCO`B. `RCOoverset(..)N`C. `RCOoverset(..)(N)HBr`D. `RNC` |
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Answer» Correct Answer - D (i). `Roverset(O)overset(||)(C)NH_(2)+Br_(2)+KOHtoRCONHBr+KBr+H_(2)O` (ii). `RCONHBr+KOHtoRNCO+KBr+H_(2)O` `underline((iii)" "RNCO+2KOHtoRNH_(2)+K_(2)CO_(3))` `RCONH_(2)+Br_(2)+4KOHtoRNH_(2)+2KBr+K_(2)CO_(3)+2H_(2)O` |
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| 231. |
Which of the following is more basic than aniline? .A. p-NitroanilineB. BenzylamineC. DiphenylamineD. Triphenylamine |
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Answer» Correct Answer - B `C_6H_5CH_2NH_2` has least negative inductive effect and thus shows more basic nature. |
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| 232. |
`C_2H_5NH_2 overset(HNO_2)to A overset(PCl_5)to B overset(NH_3)to C ` The compound C is identified asA. acetamideB. ethylamineC. ethyl cyanideD. methyl amine |
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Answer» Correct Answer - B `C_2H_5NH_2 overset(HNO_2)to C_2H_5OH overset(PCl_5)to C_2H_5Cl overset(NH_3)to underset"Ethyl amine"(C_2H_5NH_2)` |
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| 233. |
Aniline in a set of reactions yielded a product `D` `C_(6)H_(5)NH_(2)underset(HCI)overset(NaNO_(2))rarrAoverset(CuCN)rarrBunderset(Ni)overset(H_(2))rarrCoverset(HNO_(2))rarrD` The structure of the product `D` would be .A. `C_6H_5NHCH_2CH_3`B. `C_6H_5CH_2NH_2`C. `C_6H_5CH_2OH`D. `C_6H_5NHOH` |
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Answer» Correct Answer - C `C_6H_5NH_2underset(HCl)overset((NaNO_2)/(HCl))toC_6H_5N_2Cloverset(CuCN)toC_6H_5CN` `Boverset(H_2)toC_6H_5CH_2NH_2overset(HNO_2)toC_6H_5CN_2OH` |
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| 234. |
What is the end product in the following sequence of operations? `C_2 H_5 NH_2 overset(HNO_2) to A overset(PCl_5) to B overset ("alc." NH_3) to C`A. `CH_3NH_2`B. `C_2H_5NH_2`C. `CH_3CH=NH`D. `(CH_3)_2NH` |
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Answer» Correct Answer - B `underset"Ethyl amine"(C_2H_5NH_2) overset(HNO_2)to undersetAunderset"Ethanol"(C_2H_5OH) underset(-POCl_3 -HCl)overset(PCl_5)to undersetBunderset"Ethyl chloride"(C_2H_5Cl) underset"-HCl"overset(NH_3) to undersetC underset"Ethyl amine"(C_2H_5NH_2)` |
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| 235. |
In the reaction, `C_(6)H_(5)NH_(2)underset(0-5^(@)C)overset(NaNO_(2)+HCl)rarr(A)underset(KCN)overset(CuCN)rarr(B)overset(H^(+)//H_(2)O)rarr(C)` the product `( C)` isA. `C_(6)H_(5)CH_(2)NH_(2)`B. `C_(6)H_(5)COOH`C. `C_(6)H_(5)OH`D. none of these |
| Answer» Correct Answer - B | |
| 236. |
`CH_3-CH_2-Br overset(Alc. KCN)to CH_3CH_2CN overset(HOH)to X`, then X isA. acetic acidB. propionic acidC. butyric acidD. formic acid |
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Answer» Correct Answer - B `CH_3CH_2Br overset"Alc. KCN"to CH_3CH_2CN` `CH_3CH_2CN overset"HOH"to CH_3C H_2 oversetOoverset(||)COH` |
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| 237. |
`CH_(3)Br+KCN(Alco.)to X underset(Na+C_(2)H_(5)OH)overset("Reduction")to Y. ` What is Y in the series ?A. `CH_3CN`B. `C_2H_5CN`C. `C_2H_5NH_2`D. `CH_3NH_2` |
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Answer» Correct Answer - C `underset"Methyl bromide"(CH_3Br)overset"KCN (alc.)"to underset"Methyl cyanide"(CH_3CN) overset(Na-C_2H_5OH)to undersetYunderset"Ethyl amine"(CH_3CH_2NH_2)` |
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| 238. |
An aromatic amine (X) was treated with alcoholic potash and another compound (Y) when foul smelling gas was formed `C_(6)H_(5)NC`. The compound (Y) was formed by reacting a compound (Z) with `Cl_(2)` in the presence of slaked lime. The compound (Z) is:A. `CHCl_(3)`B. `CH_(3)COCH_(3)`C. `CH_(3)OH`D. `C_(6)H_(5)NH_(2)` |
| Answer» Correct Answer - B | |
| 239. |
In this molecuar `C-N-C` bond angle is `108^@`. The hydridization and shape of the above molecule areA. `sp^3` pyramidalB. `sp^3` tetrahedralC. `sp^2` angularD. `sp^2` pyramidal |
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Answer» Correct Answer - A In this molecule, there are 3 bond pairs and a lone pair, so hybridization is `sp^3` because of the involvement of 4 pairs of electrons. Howerver. Presence ofa lone pair, changes its geometry from tetrahedral to pyramidal |
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| 240. |
A nitrogenous compound is treated with `HNO_(2)` and the product so formed is further treated with NaOH solutoin which produces blue colouration. The nitrogenous compound is:A. `CH_(3)CH_(2)NH_(2)`B. `CH_(3)CH_(2)NO_(2)`C. `CH_(3)CHONO`D. `CH_(3)underset(CH_(3))underset(|)(CH)NO_(2)` |
| Answer» Correct Answer - D | |
| 241. |
Write chemical equations for the following conversions : `C_(6)H_(5)-CH_(2)-Cl " into " C_(6)H_(5)-CH_(2)-CH_(2)-NH_(2)` |
| Answer» `underset("(Benzyl chloride)") underset("Chlorophenylmethane") (C_(6)H_(5)-CH_(2)-Cl) overset("EthanolicNaCN")(to) underset("(Benzyl cyanide)") underset("Phenylethanenitrile")(C_(6)H_(5)-CH_(2)-C-=N) overset(H_(2)//Ni)(to) underset("2-Phenylethanamine")(C_(6)H_(5)-CH_(2)-CH_(2)-NH_(2))` | |
| 242. |
Complete the following conversions. `? 2H_(2)O to CH_(3)NH_(2)+HCOOH` |
| Answer» `underset("Methyl isocyanide")(CH_(3)NC+2H_(2)O) to underset("Methyl amine Formic acid")(CH_(3)-NH_(2)+HCOOH)` | |
| 243. |
Explain the following name reactions : Sandmeyer reaction |
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Answer» Sandmeyer reaction : Formation of chloro benzene, Brono benzene (or) cyano benzene from benzene diazonium salts with reagents `Cu_(2)Cl_(2)//HCl, Cu_(2)Br_(2)//HBr, CuCN//KCN` is called sandmayers reaction. `C_(6)H_(5)N_(2).^(+)X^(-) overset(Cu_(2)Cl_(2)//HCl) to C_(6)H_(5)Cl+N_(2)` `C_(6)H_(5)N_(2).^(+)X^(-) overset(Cu_(2)Br_(2)//HBr) to C_(6)H_(5)Br+N_(2)` `C_(6)H_(5)N_(2).^(+)X^(-) overset(CuCN//KCN) to C_(6)H_(5)CN+N_(2)` |
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| 244. |
Complete the following conversions. `CH_(3)NC+HgO to ?` |
| Answer» `underset("Methyl isocyanide")(CH_(3)-NC+HgO) to underset("Methyl isocyanate")(CH_(3)-N=C=O+Hg` | |
| 245. |
Explain the following name reactions : Gatterman reaction |
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Answer» Gatterman reaction : Formation of chloro benzene, Bromobenzene from benzene diazonium salts with reagents `Cu//Hcl, Cu//HBr` is referred as gatterman reaction. `C_(6)H_(5)N_(2).^(+)X^(-) overset(Cu//HCl) to C_(6)H_(5)Cl+N_(2)+CuX` `C_(6)H_(5)N_(2).^(+)X^(-) overset(Cu//HBr) to C_(6)H_(5)Br+N_(2)+CuX` |
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| 246. |
Leakage of which gas was responsible for the Bhopal tragedy in 1984A. `CH_3-N=C=O`B. `CH_3-CH_2-N=S`C. `CHCl_3`D. `C_6H_5COCl` |
| Answer» Correct Answer - A | |
| 247. |
Consider the nitratio of benzene using mixed conc. `H_(2)SO_(4)` and `HNO_(3)`. If a large amount of `KHSO_(4)` is added to the mixture, the rate of nitration will be `:`A. unchargedB. doubledC. fasterD. slower |
| Answer» Correct Answer - D | |
| 248. |
Identify the product Z in the series `CH_3CNrarroverset(Na+C_2H_5OH)toXoverset(HNO_2)toYunderset(H_2SO_4)overset(K_2Cr_2O_7)toZ`A. `CH_3CHO`B. `CH_3CONH_2`C. `CH_3COOH`D. `CH_3CH_2NHOH` |
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Answer» Correct Answer - C `CH_3-Crarrunderset(reduction)overset(Na+EtOH)rarrCH_3-CH_2-NH_2overset(HNO_2)toCH_3CH_2-OHunderset(H_2SO_4)overset(K_2Cr_7)toCH_3COOH` |
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| 249. |
The correct order for boiling point of isomeric alkyl amines isA. `1^@ gt 2^@ gt 3^@`B. `1^@ gt 2^@ lt 3^@`C. `1^@ lt 2^@ lt 3^@`D. `1^@ lt 2^@ gt 3^@` |
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Answer» Correct Answer - A The order of boiling points of the isomeric amines is as follows : Primary aminesgt secondary aminesgt tertiary amines. `(1^@ gt 2^@ gt 3^@)` |
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| 250. |
By reduction of mitrosobenzene which f the following is not obtainedA. B. C. D. |
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Answer» Correct Answer - D Nitrobanzene is not obtained by reduction of nitrosobazene. |
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