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Right option is (b) False

Explanation: AMINES are derivatives of ammonia. Nitrogen of NH3 is sp^3 hybridised with ONE ORBITAL having unshared pairs of electrons as it bonds to only three H atoms. Similarly, amines also have an ELECTRON lone PAIR on nitrogen.

The correct option is (d) N,N-Diphenylaniline

To ELABORATE: It is a secondary aromatic amine that will be FORMED. ONE of the phenyl groups will be considered as a substituent on nitrogen atom (N-phenyl) and the other will ACT as ANILINE with the N atom.

Correct choice is (c) Trigonal pyramidal

Explanation: CH3-NH-CH3 is an amine in which TWO hydrogen atoms are REPLACED by METHYL groups. The HYBRIDISATION of N atom of amines is same as that of ammonia, and is also expected to have a pyramidal geometry, with N at the apex and the groups CH3, H and CH3 occupying the corners of a trigonal base.

Correct CHOICE is (d) Copper chloride

For EXPLANATION I would say: The ANION from the diazonium SALT (Br^–) combines with Cu^+ from copper powder (in HALOGEN acid), to form CuBr. The chlorobenzene formed gets the chlorine from HCl in solution.

The correct choice is (a) less

Best explanation: The -NHCOCH3 group forms two RESONANCE STRUCTURES, where the lone pair of nitrogen INTERACTS with the lone PAIRS on oxygen atom. This makes the lone electron pair on N less AVAILABLE for donation to benzene ring.

The correct OPTION is (d) nitrosoamine

For explanation I would say: SECONDARY amines REACT slowly with nitrous acid at low temperatures to give yellow OILY nitrosoamines. Diethylamine specifically produces N,N-diethylnitrosoamine.

Right answer is (a) True

Easy EXPLANATION: Tertiary amines on reaction with COLD HNO2 remain dissolved, forming amine nitrite SALTS, which DECOMPOSE to nitrosoamines and alcohol on WARMING.

Right choice is (b) N-Phenylethanamide and acetic acid

To explain: The H atom ATTACHED to nitrogen of aniline is replaced by COCH3 GROUP of ETHANOIC anhydride. This forms N-phenylethanamide (or ACETANILIDE). The by PRODUCT (acetic acid) is formed by the attachment of the H atom from aniline with remaining CH3COO part of ethanoic anhydride.

The correct OPTION is (a) Primary amine

To explain: Primary and secondary AMINES react with alkyl halide to form tertiary amines by NUCLEOPHILIC substitution. When a primary or secondary amine ACTS as the nucleophile, a secondary or tertiary amine is generated respectively. Finally, the tertiary amine REACTS with the remaining methyl iodide to form quaternary ammonium iodide salt.

Correct option is (d) 14.22

For explanation I would say: NO2 is an ELECTRON WITHDRAWING group and has a weakening effect on the basicity of anilines. Its presence at para position will be more influencing than at meta, but less weakening than at ortho position. This anomaly is DUE to a complex ortho effect (STEARIC and electronic reasons). Hence o-nitroaniline is a weaker base and has a higher pKb value than p-nitroaniline.

The correct option is (c) N-Ethylethanamine

Easy explanation: Between methyl substituted and ethyl substituted AQUEOUS AMINES, the latter will have relatively higher Kb values due to the ethyl group being larger than methyl group. Of the ethyl substituted amines, the secondary amine will be the most basic because of the combined INDUCTIVE and solvation effect of alkyl group. Thus, being the most basic, it will react faster with HCL.

Correct choice is (b) B > A

Easy EXPLANATION: Butanamine is a LARGER molecule than ethanamine. As the carbon chain increase, the magnitude of van DER Waals forces increases RESULTING in the increase in boiling point.

Right option is (C) C6H5CONH2

Easiest EXPLANATION: Only primary amides undergo this reaction with BROMINE in NaOHto form primary amines. Benzamide undergoes this reaction to form BENZENAMINE or ANILINE.

The correct answer is (a) PRIMARY

Best EXPLANATION: MIXED amines are a classification of secondary and tertiary amines. The Hoffmann bromamide degradation is only possible with 1° amides so as to PRODUCE primary amines.

Correct ANSWER is (b) secondary amine

Easiest EXPLANATION: The reduction of isocyanide compounds (where the CN GROUP is attached through N ATOM) with H2/Ni gives secondary amines or N-alkyl amines. For example, phenyl isocyanide gives N-methylaniline.

Correct answer is (a) LiAlH4-dry ether

The explanation is: LiAlH4 MAY be used for the REDUCTION of aliphatic nitro compounds to respective amines. But with AROMATIC nitro compounds (nitrobenzene), it GIVES azobenzene and not primary amines (ANILINE).

Right option is (C) p-aminoazobenzene

To explain I would SAY: Diazonium SALTS undergo coupling with aniline to form p-aminoazobenzene. The coupling PREDOMINANTLY occurs at the para position to the AMINO group.

Right answer is (d) 40°C

For EXPLANATION: When DIAZONIUM fluoroborate is heated with aqueous NANO2 in the PRESENCE of copper metal, the DIAZO group is replaced by NO2 group, to give nitrobenzene.

Right answer is (B) Cu2Cl2/HCl

To EXPLAIN: The Cl^– nucleophile can be INTRODUCED in the benzene ring in the presence of Cu(I) ion which is produced by Cu2Cl2 in hydrochloric acid. This on warming with the DIAZONIUM salt will give chlorobenzene.

The CORRECT answer is (b) PHENOL

For EXPLANATION: Primary aromatic amines react with NITROUS acid at cold temperatures (273-278K) to form diazonium salts. However, if the temperature is more than 278K, phenol is formed along with the EVOLUTION of nitrogen gas.

Correct answer is (a) True

Easy explanation: When the alkyl group is small, there is no stearic hinderance to hydrogen BONDING. But when the alkyl group size or number increases, there will be hinderance to FORMATION of hydrogen BONDS, and this affects the ORDER of basic strength of amines.

Right answer is (C) Tertiary AMINE

For explanation: This is TRIMETHYLAMINE, in which all the three HYDROGEN atoms are replaced by METHYL (CH3) groups. Hence, it is a tertiary or 3° amine.

Right choice is (d) Hexane-1,6-diamine

For explanation I would say: The FORMULA is NH2-(CH2)6-NH2 which is the chain of 6 carbon atoms and two amino GROUPS at the ends of the chain, i.e., FIRST and SIXTH carbon. This type of substitution is named and hexamethylene.

The correct answer is (a) Isopropylamine

To explain: The reaction of Hinsberg’s reagent with Isopropylamine GIVES N-isopropylbenzene sulphonamide. The lone HYDROGEN attached to the N atom is strongly acidic due to the presence of strong electron WITHDRAWING sulphonyl group. Hence, is it soluble in ALKALI like KOH.

The correct option is (b) electrophilic substitution

For EXPLANATION I WOULD say: During a coupling reaction, the DIAZONIUM cation which has a POSITIVE charge on the TERMINAL nitrogen acts as the electrophile, and the electron rich compounds act as nucleophiles.

Right option is (C) CH3CH2OH

The EXPLANATION is: Ethanol is a mild reducing agent which on treatment with a benzenediazonium salt, oxidises to the CORRESPONDING aldehyde, and in the process reducing the DIAZONIUM salt to give BENZENE.

Correct option is (d) 2,4,6-Tribromoaniline

Easy explanation: Since amino group is highly ORTHO and para directing, it substitutes the Br group at the para as well as both ortho positions (2,4,6) to GIVE a white PRECIPITATE of a tribromo substituted ANILINE.

Correct choice is (b) One H atom of NH2

Explanation: Aliphatic and AROMATIC 1° and 2° AMINES react with acid chlorides, anhydrides and esters by NUCLEOPHILIC substitution, where the H atom of NH2 or NH GROUP is replaced by the ACYL group.

Right answer is (a) C6H5NH2

Easiest explanation: CH3, OCH3 and NH2 are all electron donating groups. When these are substituted at para position in aniline, it stabilizes the respective anilinium ion FORMED and thus increases the basic strength. THEREFORE, aniline is the weakest BASE from the given compounds and will have the highest pKb value.

The CORRECT OPTION is (a) 1° aryl

For explanation: Only one hydrogen atom is replaced by an AROMATIC GROUP, in which the nitrogen is directly linked to the sp^2 hybridised carbon of benzene ring. Hence, it is a primary aromatic aryl amine.

Right CHOICE is (c) N-Ethyl-N-methylbezenamine

The best EXPLANATION: There are THREE substituents viz., methyl group, ethyl group and PHENYL group. Since the phenyl group is the largest, it is considered as the main group with CH3 and C2H5 acting as substituents. Also, ethyl is named before methyl due to alphabetical REASONS.

The correct choice is (d) Mixed amine

Explanation: In the given amine, there are TWO substituents, ONE METHYL GROUP and one phenyl group. Hence, it is a secondary aryl amine and SINCE both the substituent groups are different, it is a mixed amine.

Correct ANSWER is (a) True

The best explanation: In GATTERMAN REACTION, there is an additional by product along with chlorobenzene and nitrogen gas, which is a copper salt (with the anion of diazonium salt). The contribution towards FORMATION of this product results in the yield on chlorobenzene to be slightly less.

Right choice is (c) gas

For EXPLANATION I would say: Nitrogen gas is liberated when diazonium salts are treated with SUITABLE REAGENTS so as to DISPLACE the N2^+X^– (X=anion) with an anionic GROUP. This may or may not involve the production of another by-product.

Right ANSWER is (a) Benzenediazonium hydrogensulphate

The best I can explain: Diazonium salts are named by suffixing diazonium to the parent hydrocarbon (benzene) from which they are FORMED, followed by the NAME of the ANION (HSO4^–).

Right answer is (c) N,N-Dimethylethanamine

For EXPLANATION I would SAY: For benzoylation to take PLACE, there should be a hydrogen ATOM PRESENT at the nitrogen of the amine for the C6H5CO^– group to replace. This is not the case in tertiary amines like N,N-dimethylethanamine and hence they do not undergo benzoylation.

Correct choice is (c) white solid

To EXPLAIN I would say: METHYLAMINE and hydrogen iodide are allowed to MIX at very cold temperatures for about 2 HOURS. The resulting product is allowed to evaporate and methanaminium iodide (CH3NH3I) is obtained in the form of a white POWDER.

Correct option is (a) Aniline > Diphenylaniline > Triphenylaniline

To explain: When HYDROGEN atoms of amino groups are replaced by electron withdrawing groups (in this CASE, phenyl GROUP), the BASIC character of the resulting aryl amine decreases.

Correct option is (d) PRESENCE of an aromatic ring

Easy EXPLANATION: Since only basicity of alkylamines is in question the presence of aromatic ring is not considered. As the size and NUMBER of alkyl groups increases, the stability of ammonium ion (formed from the amine) increases due to DISPERSAL of more positive charge by the +I effect of alkyl groups. The physical state, GASEOUS or aqueous, also determines the basicity as the hydration effect comes into play.

The correct OPTION is (a) Only A

The explanation is: Propanamine has three carbon atoms and so does the PARENT amide. So, the reduction of CO to CH2 will give the required compound. Whereas in path B, the amide undergoes HOFFMANN bromamide degradation to give ETHANAMINE.

Right option is (c) N,N-Dimethylphenylmethanamine

The EXPLANATION is: BENZYL CHLORIDE reacts with ethanolic ammonia to form benzylamine (A) which reacts with ONE molecule of CH3Cl to give N-methyphenylmethanamine, which further reacts with one more molecule of CH3Cl to give N, N-dimethylphenylmethanamine.