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.
| 301. |
The product C isA. `CH_(3)-CH_(2)CH_(2)-NO_(2)`B. `CH_(3)underset(ONO)underset(|)(CH)-CH_(3)`C. `CH_(3)underset(NO_(2))underset(|)(CH)-CH_(3)`D. `CH_(3)CH_(2)CH_(2)ONO` |
| Answer» Correct Answer - A | |
| 302. |
When benzenediazonium chloride is heated with fluoro boric acid, fluorobenzene is formed. This reaction is calledA. Balz Schiemann reationB. Gattermann reactionC. Sandmeyer reactionD. Hoffmann bromamide reaction |
| Answer» Correct Answer - A | |
| 303. |
By which process aniline can be purified ......... |
|
Answer» Steam distillation |
|
| 304. |
Secondary amines can be prepared by ...... |
|
Answer» Reduction of nitro compounds |
|
| 305. |
Which of the following reactions give the poor yield of amine products?A. `CH_(3)-overset(CH_(3))overset(|)underset(CH_(3))underset(|)C-overset(O)overset(||)C-NH_(2) overset(Br_(2)+NaOH)underset(Delta)to`B. C. `CH_(3)-overset(CH_(3))overset(|)underset(CH_(3))underset(|)C-Br + NH_(3) to `D. |
| Answer» Correct Answer - B::C | |
| 306. |
Which of the following compounds react with `HNO_(2)`?A. B. C. D. |
|
Answer» Correct Answer - B::C::D |
|
| 307. |
Which of the following can give `1^(@)` amine?A. `Ph-overset(CH_(3))overset(|)(CH)-OH overset(NaCN,H^(o+))rarr`B. `Ph-CH=CH-overset(O)overset(||)(C)-NH_(2) overset(NaOCI)underset(CH_(3)OH)rarr`C. `Ph-C-=C-overset(O)overset(||)(C)-NH_(2) overset(NaOBr)rarr`D. `Ph-overset(O)overset(||)(C)-CI overset(NaN_(3))underset(Delta)rarr overset(LiAIH_(4))rarr` |
|
Answer» Correct Answer - B::C::D |
|
| 308. |
`C_(6)H_(5)-CH_(2)-I overset(NaN_(3))underset(Delta)rarr` Products Reaction is assumed to involve nitrene as intermediate, then various possible products are:A. `C_(6)H_(5)CN_(2)NH_(2)`B. `C_(6)H_(5)N=CH_(2)`C. `C_(6)H_(5)CN=NH`D. |
|
Answer» Correct Answer - B::C |
|
| 309. |
Rank the following compounds in order of increasing basic strength. (weakest`to`strongest) A. `4lt2lt1lt3`B. `4lt3lt1lt2`C. `4lt1lt3lt2`D. `2lt1lt3lt4` |
|
Answer» Correct Answer - B (4) `Ph-overset(O)overset(||)C-NH_(2)` is weak base because of resonance. One to -M of `NO_(2)` and ortho effect it is weak base. 3) Strong base lone pair is not involved in resonance. |
|
| 310. |
Which of the following compounds do not give Friedel Crafts reaction?A. B. C. D. |
| Answer» Correct Answer - A::B | |
| 311. |
Which of the following compounds can be converted into amines in the presence of Na and alcohol?(a) Alkyl nitriles(b) Aldoxime (c) Ketoxime (d) All of these |
|
Answer» Option : (d) All of these |
|
| 312. |
Isopropylamine is obtained by the reduction of :(a) acetoxime (b) acetaldoxime (c) formaldoxime (d) aldoxime |
|
Answer» Option : (a) acetoxime |
|
| 313. |
Which of the following compounds does not exist as a zwitter ion ?A. B. `H_(2)NCH_(2)COOH`D. `CH_(3)CH(NH_(2))COOH` |
|
Answer» Correct Answer - C An aromatic `-NH_(2)` group `(K_(a)=10^(-5))` and hence cannot appreciably neutralize the strongly acidic `-SO_(3)H` group as in sulphanilic acid In other words aliphatic `NH_(2)` is sufficiently basic to accept an `H^(+)` from `COOH` The `COOH` is not strong enough to donate `H^(+)` to the weakly basic `ArNH_(2)` but `SO_(3)H` is a sufficiently strong acid to do so . |
|
| 314. |
Which of the following exist as Zwitter ion?(a) Salicylic acid (b) Suphanilic acid (c) p-Aminophenol (d) p-Amino acetophenone |
|
Answer» Option : (b) Suphanilic acid |
|
| 315. |
The reagents that can be used to convert benzenediazonium chloride to benzene are __________.(i) SnCl2/HCl(ii) CH3CH2OH(iii) H3PO2(iv) LiAlH4 |
|
Answer» (ii), (iii) (ii) CH3CH2OH (iii) H3PO2 |
|
| 316. |
Benzene sulphonyl chloride and aqueous NaOH can be used to distinguish three classes of amines such as primary, secondary and tertiary. a) Name the above test.b) How will you distinguish the above amines using this test?c) Give the reactions and justifications, |
|
Answer» a) Hinsberg test. C6H5SO2Cl b) 1° Amine + Hinsberg reagent → a compound soluble in alkali 2° Amine + Hinsberg reagent → A compound which is insoluble in alkali 3° Amine +Hinsberg reagent → no reaction. c) The hydrogen attached to nitrogen in sulphonamide is strongly acidic due to the presence of strong electron withdrawing sulphonyl group. Hence, it is soluble in alkali. Secondary amine react with benzene salphonyl chloride to form N, N-dialkylbenzene- sulphon-amide, which is insoluble in alkali. This is be?cause it does not contain any hydrogen attached to nitrogen atom and is not acidic. Tertiary amines do not react with benzene salphonyl chloride, because they do not possess any replacable hydrogen. |
|
| 317. |
Reduction of benzene diazonium chloride with Zn/HCl gives :(a) phenyl hydrazine (b) hydrazine hydrate (c) aniline (d) ozo benzene |
|
Answer» Option : (c) aniline |
|
| 318. |
Identify ‘Z’ in the sequence:(a) Nitrobenzene(b) Benzene(c) Fluorobenzene(d) Phenol |
|
Answer» (a) Nitrobenzene |
|
| 319. |
When benzene diazonium salt solution is treated with KI ....... is formed. |
|
Answer» lodo benzene (C6H5I) |
|
| 320. |
When benzene diazonium chloride is treated with cuprous chloride and HCl the product formed isA. ChlorobenzeneB. BenzeneC. PhenolD. Chloroazobenzene |
| Answer» Correct Answer - A | |
| 321. |
Hoffmann Bromamide Degradation reaction is shown by __________.(i) ArNH2(ii) ArCONH2(iii) ArNO2(iv) ArCH2NH2 |
|
Answer» (ii) ArCONH2 |
|
| 322. |
The best reagent for converting, 2-phenylpropanamide into 1- phenylethanamine is ____.(i) excess H2/Pt(ii) NaOH/Br2(iii) NaBH4/methanol(iv) LiAlH4/ether |
|
Answer» (ii) NaOH/Br2 |
|
| 323. |
Match column-I (Compound) with coloumn-II (Property) |
| Answer» Correct Answer - A::B::C::D | |
| 324. |
Match the coloumn-I with column-II |
| Answer» Correct Answer - A::B::C::D | |
| 325. |
The gas leaked from a stronge tank of the Union Carbide plant in Bhopal gas tragedy wasA. MethylamineB. AmmoniaC. PhosgeneD. Mehtylisocyanate |
|
Answer» Correct Answer - D Gas leaked was methy1 isocylamine `CH_(3)-N=C=O` . |
|
| 326. |
Write the IUPAC name of benzylamine. |
|
Answer» The IUPAC name is Phenylmethanamine. |
|
| 327. |
The correct increasing order of basic strength for the following compounds is _________.(i) II < III < I(ii) III < I < II(iii) III < II < I(iv) II < I < III |
|
Answer» (iv) II < I < III |
|
| 328. |
Arrange the following compounds in an increasing order of basic strength :Aniline, pnitroaniline, p-toluidine. |
|
Answer» p-nitroaniline < aniline < p-toluidine. |
|
| 329. |
Arrange the following(i) NH3, C2H5NH2 , (C2H5)2NH , (C2H5)3N (Decreasing order of basicity in gaseous phase)(ii) CH3CH2CH3 , CH3CH2NH2 , CH3CH2OH ( Increasing order of Boiling Point) |
|
Answer» (i) (C2H5)3N > (C2H5)2NH > C2H5NH2 > NH3 (ii) CH3CH2CH3 < C2H5NH2 < C2H5O |
|
| 330. |
Write down the IUPAC name of |
| Answer» N, N-Dimethylbenzenamine | |
| 331. |
The correct IUPAC name for `CH_(2)=CHCH_(2)NHCH_(3)` isA. allyl methylamineB. 2-amiono-4-pentaneC. 4-aminopent-1-eneD. N-methylprop-2-ene-1-amine |
|
Answer» Correct Answer - D IUPAC name `overset(3)(CH_(2))=overset(2)(CH)overset(1)(CH_(2) NHCH_(3) ` is N-methylprop-2-ene-1amine Hence, option (d) is correct |
|
| 332. |
The IUPAC name of ethyl dimethyl amine is ……………..(a) 2-amino propane (b) N,N-dimethyl ethanolamine (c) ethyl methanamine (d) propanamine |
|
Answer» Option : (b) N,N-dimethyl ethanolamine |
|
| 333. |
IUPAC name of diethylmethyl amine is ………………(a) methyl amino propane (b) N-Ethyl-N-methylhexanamine (c) methyl diethanamine (d) amino pentane |
|
Answer» Option : (b) N-Ethyl-N-methylhexanamine |
|
| 334. |
Explain the following observation: Electrophilic substitution in case of aromatic amines takes place more readily than benzene. |
| Answer» It is because `-NH_(2)` group is electron releasing which increases electron density on benzene ring and there is `-ve` charge at o- and p- position in resonating structure of aniline. Thus, it activates the benzene ring for electrophilic substitution. | |
| 335. |
`1^(@),2^(@)` and `3^(@)` nitroalkane can be identified by action of:A. `HNO_(3)+NaOH` (aq).B. `CHCI_(3)+NaOH` (aq)C. `HNO_(2)+NaOH` (aq)D. `CHCI_(3)+KOH` (aq.) |
|
Answer» Correct Answer - C |
|
| 336. |
The final product B obtained in the reaction is: A. B. `(H_(2)C = CH)_(2)`C. `H_(3)C-CH=CH-CH_(2)`D. |
|
Answer» Correct Answer - A |
|
| 337. |
Which of the following reactions belong to electrophilic aromatic substitution?(i) Bromination of acetanilide(ii) Coupling reaction of aryldiazonium salts(iii) Diazotisation of aniline(iv) Acylation of aniline |
|
Answer» (i), (ii) (i) Bromination of acetanilide |
|
| 338. |
For such kind of diazo- coupling reaction the suitable substituents P and S are respectively:A. `-NH_(2)` and `-OCH_(3)`B. `-NO_(2)` and `-overset(O)overset(||)(C)-H`C. `-NH_(2)` and `-NHCH_(3)`D. |
|
Answer» Correct Answer - D |
|
| 339. |
Under which of the following reaction conditions, aniline gives p-nitro derivative as the major product?(i) Acetyl chloride/pyridine followed by reaction with conc. H2SO4 + conc. HNO3.(ii) Acetic anyhdride/pyridine followed by conc. H2SO4 + conc. HNO3.(iii) Dil. HCl followed by reaction with conc. H2SO4 + conc. HNO3.(iv) Reaction with conc. HNO3 + conc.H2SO4. |
|
Answer» (i), (ii) (i) Acetyl chloride/pyridine followed by reaction with conc. H2SO4 + conc. HNO3. (ii) Acetic anyhdride/pyridine followed by conc. H2SO4 + conc. HNO3. |
|
| 340. |
What is the likely product from the following reaction? A. B. C. D. |
|
Answer» Correct Answer - B |
|
| 341. |
Consider the following diazonium ion: The order of reactivity towards diazo coupling with phenol in presence of dil. NaOH:A. `P gt Q gt R gt S`B. `Q gt S gt R gt P`C. `P gt R gt S gt Q`D. `S gt R gt Q gt P` |
|
Answer» Correct Answer - B |
|
| 342. |
Optically active amine having molecular formula `C_(5)H_(13)N` on reaction with `NaNO_(2)+HCI` produces, `3^(@)` optically inative alcohol. Find out structures of amines:A. B. C. D. |
|
Answer» Correct Answer - A::C |
|
| 343. |
Decreasing order of basicity of the three isomers of nitro aniline is:A. p-nitroaniline `gt o`-nitroaniline `gtm`-nitroanilineB. p-nitroaniline `gt m`-nitroaniline `gt o`-nitroalinineC. m-nitroaniline `gt p-`nitroaniline `gto-` nitroanilineD. m-nitroaniline `gt o-` nitroaniline `gt p-` nitroaniline |
|
Answer» Correct Answer - C |
|
| 344. |
Which is the best leaving group ?A. `-overset(o+)(N)-= N`B. `OH^(-)`C. `NH_(2)^(-)`D. `CH_(3)COO^(-)` |
|
Answer» Correct Answer - A |
|
| 345. |
What is the likely product from the following reaction? A. B. C. D. |
|
Answer» Correct Answer - B |
|
| 346. |
When an primary aromatic amine is treated with `NaNO_(2)+HCI` at `0^(@)-5^(@)C`, a diazonium salt is formed and the reaction is called diazo reaction. In this reaction mineral acid must be added to prevent the coulping reaction of diazonium salt with excess of aryl amine. diazonium salt is highly in the synthesis of number of coloured dyes. In the given reaction. The final product isA. B. C. D. |
|
Answer» Correct Answer - A |
|
| 347. |
Strongest base is:A. B. C. D. |
|
Answer» Correct Answer - A |
|
| 348. |
Which compound is the likely from following reaction? A. B. C. D. |
|
Answer» Correct Answer - B |
|
| 349. |
Treatment of primary amines with nitrous acid gives diazonium ions. Aliphatic diazonium ions decompose under the diazotization conditions to give complex mixture of products. Aryldiazonium ions are stable and is a valubale intermediate. They react with activated aromatic compounds to give coupling product. At what pH phenol reacts with benzene diazonium chloride to give coupling product?A. B. C. D. |
| Answer» Correct Answer - A | |
| 350. |
Treatment of primary amines with nitrous acid gives diazonium ions. Aliphatic diazonium ions decompose under the diazotization conditions to give complex mixture of products. Aryldiazonium ions are stable and is a valubale intermediate. They react with activated aromatic compounds to give coupling product. At what pH phenol reacts with benzene diazonium chloride to give coupling product?A. pH=2B. pH=7C. pH=9D. pH=14 |
| Answer» Correct Answer - C | |