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The correct option is (b) False

To explain I WOULD say: AMMONOLYSIS is not suitable for the production of ARYL amines, because aryl halides are stable and relatively less REACTIVE TOWARD nucleophilic substitution than alkyl halides. Since chlorobenzene is an aryl halide, it does not undergo ammonolysis to form aniline.

The correct choice is (a) BENZENAMINE

Easy explanation: Since it is an AROMATIC compound, the phenyl group C6H5 has to present with CH4N. Since it is a PRIMARY AMINE, the group NH2 can be SEPARATED leaving us with CH2. So, the formula of the compound is C6H5CH2NH2, which is benzylamine. Benzenamine has the formula C6H5NH2.

The correct answer is (c) N,N-Dimethylmethanamine

To explain I WOULD say: The COMPOUND is a tertiary AMINE with three methyl substituents. One of the methyl groups is considered as the parent chain with the N atom and the other two ACTS as substituents. These are REPRESENTED by N, N- as they are the same groups and are directly attached to the nitrogen atom.

Right option is (b) False

To explain I would say: As the reaction takes place in a basic medium, the COUPLING will take place at the ORTHO position (SINCE para position is ALREADY occupied by amino group) with respect to OH group. This results in the compound 4-amino-2-phenylazophenol.

Correct ANSWER is (b) False

Easy explanation: This reaction yields p-dimethylaminoazobenzene, which is a YELLOW solid commonly known as METHYL yellow. However, in an aqueous solution at low pH, methyl yellow APPEARS red in colour.

The CORRECT option is (B) orange

Explanation: BENZENEDIAZONIUM chloride undergoes coupling with phenol to FORM an azo compound, p-hydroxyazobenzene. This compound is orange in colour.

The correct option is (d) REACTION with CH3CH2Br

The explanation: Both ETHYLAMINE and DIETHYLAMINE on reaction with CH3CH2Br eventually gives quaternary ammonium salt. Hence, the alkylation of primary amines cannot be USED as a distinction METHOD.

Correct OPTION is (b) benzenesulphonyl chloride

For explanation I would say: Benzenesulphonyl chloride (C6H5SO2Cl), or arylsulphonyl chloride, is KNOWN as Hinsberg’s REAGENT. It is an important compound used in the distinction of DIFFERENT classes of AMINES.

Right CHOICE is (b) 3

To explain I would say: Aniline has five resonating structures, out of which three of them have a positive charge on nitrogen. This RESULTS in unavailability of ELECTRON pair for protonation. When aniline accepts a proton, it FORMS anilinium ion which has only two resonating structures and is less stable and more basic than aniline.

The CORRECT answer is (b) False

Easy EXPLANATION: N,N-Dimethylaniline reacts with acetic acid to form a salt, N,N-dimethylanilinium ACETATE. This is due to the BASIC NATURE of amines.

Right answer is (B) False

To elaborate: Only primary and secondary amines form intermolecular hydrogen BONDS as they have INDIVIDUAL at LEAST one hydrogen atom to form bonds with. Tertiary amines do not have intermolecular ASSOCIATION due to absence of hydrogen atom available for hydrogen bonding.

Correct option is (d) Benzenamine > Methanamine > Ethanamine

Best explanation: The HIGHER pKb value means lower basic strength. This means that benzenamine has the lower basic strength among methanamine and ethanamine. This is because of the electron withdrawing nature of the aryl group. Also in primary amines, the increase in SIZE of alkyl group increases +I effect leading to high electron density on N atom and as a RESULT higher basicity.

The correct choice is (c) Secondary and tertiary

The explanation is: When an alcoholic solution of AMMONIA is HEATED with alkyl halides, a mixture of all three types of AMINES, i.e., PRIMARY, secondary and tertiary amines are formed along with a QUATERNARY ammonium salt.

The correct answer is (b) 2

To elaborate: ONE molecule of nitropropane reacts with 3 molecules of hydrogen gas (H2) in the presence of Ni/Pt/Pd catalyst to form one molecule of propanamine ALONG with two molecules of WATER.

Correct CHOICE is (b) 8

The best I can explain: EIGHT isomeric amines with the formula C4H11N are possible. FOUR of them are PRIMARY amines, three are secondary and one of them is a tertiary AMINE.

The correct option is (b) OCH3

The BEST explanation: p-ANISIDINE is better known as 4-methoxybenzenamine, which has a methoxy GROUP at PARA position to the amino group. The common name anisidine is DERIVED from the common names anisole and aniline.

The CORRECT answer is (c) 4-Bromoaniline

Best explanation: In the COMMON SYSTEM, the aromatic amines have special names derived from the basic compound aniline and depending upon the type of SUBSTITUENT group. The prefixes o-, m- and p- are used to depict the position of the substituent in common system. Whereas, the number position of carbon is used in the IUPAC system.

The correct option is (a) N-Propylamine

The best I can explain: The SHOWN COMPOUND is a primary amine and hence the locant N cannot be used as it is INVALID. The amino group is attached to the second carbon of the PROPYL group (which is also known as isopropyl substituent).

Right choice is (b) 107°

For explanation: The bond angle for a STANDARD TETRAHEDRAL GEOMETRY is 109.5°. But the ammonia molecule contains a lone electron pair on nitrogen. As it is closer to the N atom than the other orbitals, it creates a repulsive EFFECT and PUSHES the Hydrogen atoms close to each other, thus reducing the bond angle to 107°.

Right option is (b) halogen<BR>
Best explanation: Diazonium salts are IMPORTANT intermediates for the introduction of -F, -Cl, -Br, -I, etc. GROUPS into an aromatic ring. CERTAIN compounds like aryl fluorides and aryl iodides cannot be prepared by direct halogenation and are prepared from diazonium salts.

Correct choice is (b) Dye

To elaborate: COUPLING REACTIONS of diazonium salts produce azo compounds which are COLOURED in nature. This is USED as an application in the manufacturing of dyes.

The CORRECT answer is (B) 3

To explain: The three reactions are as FOLLOWS:

\(\Rightarrow\) Reaction 1: Aniline on treatment with bromine water gives 2,4,6-tribromoaniline.

\(\Rightarrow\) Reaction 2: This on diazotisation with HNO2 gives 2,4,6-tribromobenzenediazonium CHLORIDE.

\(\Rightarrow\) Reaction 3: This on reaction with H3PO2, reduces the diazo group to hydrogen, to give 1,3,5-tribromobenzene.

Correct answer is (B) deamination

For explanation: The complete process of diazotisation of aniline followed by the REDUCTION of DIAZONIUM salt or replacement of the diazo group by hydrogen is CALLED deamination.

The correct answer is (d) Benzenediazonium fluoroborate

The explanation is: Arenediazonium fluoroborate is obtained as a precipitate when arenediazonium chloride is TREATED with HBF4. This SALT on heating DECOMPOSES to give ARYL fluoride.

Correct choice is (a) Bromobenzene

The explanation is: Gatterman reaction is a modification of Sandmeyer reaction. It is used for introducing CHLORINE or bromine in the benzene ring by treating diazonium SALT (aqueous) with CORRESPONDING HALOGEN acid (HCl or HBr) in the PRESENCE of Cu powder.

The correct CHOICE is (a) HCl

Easy explanation: Amine SALTS on treatment with a base LIKE NaOH gives BACK the PARENT amine along with a water molecule and a salt. For example, CH3NH3Cl with NaOH gives methanamine , water and sodium chloride.

Right answer is (d) Sn, HCl

Explanation: Nitriles on reduction with LITHIUM aluminium hydride or CATALYTIC hydrogenation or with Na(HG)-ethanol PRODUCE primary amines.

Correct option is (a) Ethylenediamine

The best I can explain: In this compound, more than one amino GROUP is PRESENT at different positions in the PARENT chain. Both the amino groups are given equal importance and NONE of them is considered a substituent on the other. The prefix di- is used before amine, and since it is a disubstituted ethane, it is referred to as ethylene.

The correct choice is (b) SECONDARY

To ELABORATE: Amines are classified as secondary or 2° when two of the hydrogen atoms of ammonia are replaced by an ALKYL group and the third H remains attached as it is.

Right answer is (d) PYRAMIDAL

To elaborate: The nitrogen atom of AMMONIA undergoes hybridisation to form 4 sp^3 ORBITALS. Three of these overlap with s orbital of H forming three N-H bonds. This results in a pyramidal geometry with N atom at the apex and three H atoms at the CORNERS of a TRIANGLE base.

Correct choice is (a) True

The EXPLANATION is: If the temperature of benzenediazonium salt solutions is ALLOWED to RISE upto 283K, the diazonium group is replaced by OH group, and therefore the salt get hydrolysed to phenol.

Correct choice is (a) YELLOW

For explanation: DIAZONIUM salts on reaction with aniline form azo COMPOUNDS which are coloured. This is because of the complex system involving the N-N double bond between two aromatic RINGS. In this case, compound A is p-aminobenzene, which is yellow coloured.

Correct choice is (c) ACETIC acid

To explain: Ethanol is a mild reducing agent that reduces benzenediazonium bromide to benzene, and itself gets oxidized to an aldehyde, ETHANAL (CH3CHO).

Right answer is (c) BENZENE

To explain: When benzenediazonium salts are TREATED with mild REDUCING AGENTS like HYPOPHOSPHOROUS acid, the complete diazo and anionic group is replaced by hydrogen atom to form benzene.

Correct option is (d) N-Phenylbenzamide

Explanation: AMINES REACT with benzoyl CHLORIDE (C6H5COCl) in the PRESENCE of a base to form benzoyl DERIVATIVES in which the C6H5CO^– group is introduced. This is called benzoylation reaction.

Right OPTION is (d) nucleophilic substitution

For explanation I would say: The amines itself acts as nucleophiles and attack the alkyl HALIDE at the carbon-halogen BOND to substitute the halogen atom by itself. This occurs by production of small amount of acid at each step, which MAY prevent the reaction by proceeding by protonating the amine.

Correct answer is (b) N and N-H

To explain I WOULD SAY: Primary and SECONDARY amines are involved in intermolecular molecular HYDROGEN bonding between nitrogen of one and hydrogen of another molecule.

The correct choice is (a) Benzylamine

Easiest EXPLANATION: Since Gabriel phthalimide synthesis is a nucleophilic substitution reaction, ARYL HALIDES cannot be prepared form this. This is because of the sp^2 hybridised aryl carbon and STABLE benzene ring.

Right choice is (c) N,N-Diethylbutan-1-amine

For EXPLANATION: This is a tertiary amine with two identical alkyl substituents (ethyl group) and one butyl group which acts as the major chain. The prefix N,N-Di is used for the ethyl GROUPS and this collective group is attached at the first CARBON of the butyl group.

The correct choice is (B) False

Explanation: p-tert-Butyl aniline is a DERIVATIVE of aniline (which is a primary amine) with a tert-Butyl substituent ((CH3)3C) at PARA position. Hence, it is a primary amine.

Correct option is (d) Pentane-1,5-diamine

Easiest explanation: The PARENT carbon chain CONSISTS of five atoms with one AMINO group at each end, i.e., first and fifth carbon. The prefix di is used to indicate the two amino groups and the letter ‘e’ of the SUFFIX part of pentane is retained. Hence, the name pentane-1,5-diamine.

Right answer is (B) NaBF4

For explanation I would say: The diazonium salt dissociates in solution GIVING a free fluoroborate ANION (BF4^–). This combines with the Na^+ ions in solution to FORM NaBF4.

Right choice is (b) 4

To elaborate: The coupling reaction with phenol OCCURS in a basic medium of pH value approximately 9 to 10. This is to counter the acidic nature of phenol and as a result produce water (by the COMBINATION of OH^– and H^+) during the reaction.

Correct option is (a) True

Easiest explanation: The NITROGEN from the DIAZO group of the diazonium SALT is liberated as FREE nitrogen molecule (N2) is the form of a gas, as a by-product during the REACTION.

Right answer is (d) H3PO2

Explanation: In the reaction, benzenediazonium chloride is reduced to benzene. This is possible in the PRESENCE of a reducing agent. Since, one of the by products is H3PO3, it must be FORMED from the oxidation of the reducing agent, and HENCE must either have less O ATOMS or more H atoms than in H3PO3. Therefore, it is H3PO2.