Question 1. Which of the following is a $3^{o}$ amine?
(i) 1-methylcyclohexylamine
(ii) Triethylamine
(iii) tert-butylamine
(iv) N-methyl aniline

Answer:

Option (ii) is the correct answer.

Triethylamine is a tertiary $\left(3^{\circ}\right)$ amine because the nitrogen atom is bonded to three ethyl groups and no hydrogen atoms.

Question 2. The correct IUPAC name for $CH_{2}=CHCH_{2} NHCH_{3}$ is
(i) Allylmethylamine
(ii) 2-amino-4-pentene
(iii) 4-aminopent-1-ene
(iv) N-methylprop-2-en-1-amine

Answer:

Option (iv) is the correct answer.

The amine group is considered a functional group that modifies the parent hydrocarbon chain. In the compound $\mathrm{CH}_2=\mathrm{CHCH}_2 \mathrm{NHCH}_3$, the amine is a substituted primary amine, so the longest chain is numbered from the end closer to the $-\mathrm{NHCH}_3$ group to give it the lowest possible number.

Question 3. Amongst the following, the strongest base in aqueous medium is ____________.
(i) $CH_{3}NH_{2}$
(ii) $NCCH_{2}NH_{2}$
(iii) $(CH_{3})_{2}NH$
(iv) $C_{6}H_{5}NHCH_{3}$

Answer:

Option (iii) is the correct answer.

$(CH_{3})_{2}NH$ is a $2^{o}$ amine and $CH_{3}NH_{2}$ is a $1^{o}$ amine, therefore $(CH_{3})_{2}NH$ is more basic than $CH_{3}NH_{2}$. And because of –I effect of CN group, $NC-CH_{2}NH_{2}$ is less basic than $CH_{3}NH_{2}$. Moreover, $C_{6}H_{5}NHCH_{3}$ is the least basic and less basic than $CH_{3}NH_{2}$ and $(CH_{3})_{2}NH$; it is due to delocalisation of lone pair electrons that are present in nitrogen atom into the benzene ring. Therefore, the decreasing order of amines will be:
$\left ( CH_{3} \right )_{2}NH>CH_{3}NH_{2}>C_{6}H_{5}NHCH_{3}>NC-CH_{2}NH_{2}$

Question 4. Which of the following is the weakest Brönsted base?

Answer:

Option (A) is the correct answer.

The lone pair of electrons on the N-atom has delocalised into the benzene ring; therefore $C_{6}H_{5}NH_{2}$ s the weakest base.

Question 5. Benzylamine may be alkylated, as shown in the following equation:
$C_{6}H_{5}CH_{2}NH_{2} + R-X \rightarrow C_{6}H_{5}CH_{2}NHR$
Which of the following alkyl halides is best suited for this reaction through $S_{N}1$ mechanism?
(i) $CH_{3}Br$
(ii) $C_{6}H_{5}Br$
(iii) $C_{6}H_{5}CH_{2}Br$
(iv) $C_{2}H_{5}Br$

Answer:

Option (iii) is the correct answer.

$S_{N}1$ is a two-step reaction. At first, $R-X$ bond is broken, which produces carbocation. Then nucleophile attacks the carbocation. The more the stability of carbocation, the more will be the rate of reaction. Benzylic halides are highly reactive towards $S_{N}1$ reactions.

Question 6. Which of the following reagents would not be a good choice for reducing an aryl nitro compound to an amine?
(i) $\text {H}_{2}$ (excess)/Pt
(ii) $\text {LiAIH}_{4}$ in ether
(iii) $\text {Fe}$ and $\text {HCl}$
(iv) $\text {Sn}$ and $\text {HCl}$

Answer:

Option (ii) is the correct answer.

$\mathrm{LiAlH}_4$ in ether is not suitable for reducing aryl nitro compounds to amines because it is ineffective in reducing the nitro group directly on an aromatic ring. Catalytic hydrogenation ( $\mathrm{H}_2 / \mathrm{Pt}$ ) or metal-acid reductions ( $\mathrm{Fe} / \mathrm{HCl}$ or $\mathrm{Sn} / \mathrm{HCl}$ ) are more effective for such conversions.

Question 7. In order to prepare a 1° amine from an alkyl halide with simultaneous addition of one $\text {CH}_{2}$ group in the carbon chain, the reagent used as source of nitrogen is ___________.
(i) Sodium amide, $\text {NaNH}_{2}$
(ii) Sodium azide, $\text {NaN}_{3}$
(iii) Potassium cyanide, $\text {KCN}$
(iv) Potassium phthalimide, $\text {C}_{6}\text {H}_{4}\left ( \text {CO}_{2} \right )\text N^{-}K^{+}$

Answer:

Option (iii) is the correct answer.

The number of carbon atoms increases with the help of $\text {KCN}$. The alkyl halide reacts with KCN to form a nitrile $(\mathrm{R}-\mathrm{CN})$, which is then reduced to a primary amine.

Question 8. The source of nitrogen in Gabriel synthesis of amines is _____________.
(i) Sodium azide, $\text {NaN}_{3}$
(ii) Sodium nitrite, $\text {NaNO}_{2}$
(iii) Potassium cyanide, $\text {KCN}$
(iv) Potassium phthalimide, $\text {C}_{6}\text {H}_{4}\left ( \text {CO} \right )_{2}\text N^-K^+$

Answer:

Option (iv) is the correct answer.

In Gabriel's synthesis, Potassium Phthalimide is the source of nitrogen.

Steps in Gabriel's synthesis-

1. Potassium phthalimide $+R-X \rightarrow \mathrm{~N}$-alkyl phthalimide
2. N-alkyl phthalimide $+\mathrm{H}_2 \mathrm{O} / \mathrm{H}^{+} \rightarrow$ Primary amine $\left(\mathrm{R}-\mathrm{NH}_2\right)+$ Phthalic acid

Question 9. Amongst the given set of reactants, the most appropriate for preparing 2° amine is _____.
(i) 2° $\text {R-Br+NH}_{3}$
(ii) 2° $\text {R-Br+NaCN}$ followed by $\text {H}_{2}/\text {Pt}$
(iii) 1° $\text {R-NH}_{2}+\text {RCHO}$ followed by $\text {H}_{2}/\text {Pt}$
(iv) 1° $\text {R-Br}\left ( \text {2 mol} \right )+$ potassium phthalimide followed by $\text {H}_{3}\text {O}^{+}$/heat

Answer:

Option (iii) is the correct answer.

A primary amine reacts with an aldehyde (RCHO) to form an imine (Schiff base) which is then reduced (e.g., by $\mathrm{H}_2 / \mathrm{Pt}$ ) to form a secondary amine. This is an efficient method for synthesizing $2^{\circ}$ amines.

Question 10. The best reagent for converting 2–phenylpropanamide into 2-phenylpropanamine is _____.
(i) excess $\text {H}_{2}$
(ii) $\text {Br}_{2}$ in aqueous $\text {NaOH}$
(iii) iodine in the presence of red phosphorus
(iv) $\text {LiAlH}_{4}$ in ether

Answer:

Option (iv) is the correct answer.

$\mathrm{LiAlH}_4$ in ether is a strong reducing agent that converts amides to amines efficiently. So, 2-phenylpropanamide is reduced to 2-phenylpropanamine using this reagent.

Question 11 The best reagent for converting, 2-phenylpropanamide into 1- phenylethanamine is ____.
(i) excess $\text {H}_{2}/\text {Pt}$
(ii) $\text {NaOH}/\text {Br}_{2}$
(iii) $\text {NaBH}_{4}$/methanol
(iv) $\text {LiAlH}_{4}$/ether

Answer:

Option (ii) is the correct answer.

$\mathrm{NaOH} / \mathrm{Br}_2$ (Hofmann bromamide reaction) removes one carbon from the amide and forms a primary amine. So, 2-phenylpropanamide (3 carbon chain) is converted to 1phenylethanamine (2 carbon chain amine).

Question12. Hoffmann Bromamide Degradation reaction is shown by __________.
(i) $\text {ArNH}_{2}$
(ii) $\text {ArCONH}_{2}$
(iii) $\text {ArNO}_{2}$
(iv) $\text {ArCH}_{2}\text {NH}_{2}$

Answer:

Option (ii) is the correct answer.

The Hoffmann Bromamide Degradation is a reaction where a primary amide is converted to a primary amine with one less carbon atom in the presence of Br₂ and a strong base like NaOH or KOH.

$\mathrm{C}_6 \mathrm{H}_5 \mathrm{CONH}_2 \xrightarrow[\text { Heat }]{\mathrm{Br}_2+4 \mathrm{NaOH}} \mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2+\mathrm{Na}_2 \mathrm{CO}_3+2 \mathrm{NaBr}+2 \mathrm{H}_2 \mathrm{O}$

Question13. 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:

Option (iv) is the correct answer.
Explanation: The basic strength is decreased with electron-withdrawing groups, and it is increased with electron releasing groups of aniline.

Question 14. Methylamine reacts with $\text {HNO}_{2}$ to form _________.
(i) $\text {CH}_{3}\text {-O-N=O}$
(ii) $\text {CH}_{3}\text {-O-CH}_{3}$
(iii) $\text {CH}_{3}\text {OH}$
(iv) $\text {CH}_{3}\text {CHO}$
Answer:

Option (iii) is the correct answer.
Explaination : $(c)\; \text {CH}_{3}\text {NH}_{2}+\text {HNO}_{3}\overset{0^{o}-5^{o}C}{\rightarrow}\text {CH}_{3}\text {OH}+\text {N}_{2}+\text {H}_{2}\text {O}$

Question 15. The gas evolved when methylamine reacts with nitrous acid is __________.
(i) $\text {NH}_{3}$
(ii) $\text {N}_{2}$
(iii) $\text {H}_{2}$
(iv) $\text {C}_{2}\text {H}_{6}$
Answer:

Option (ii) is the correct answer.
Explanation: The chemical reaction of methylamine with nitrous acid is as follows:

Question 16. In the nitration of benzene using a mixture of conc. $\text {H}_{2}\text {SO}_{4}$ and conc. $\text {HNO}_{3}$, the species which initiates the reaction is __________.
(i) $\text {NO}_{2}$
(ii) $\text {NO}^{+}$
(iii) $\text {NO}_{2}^+$
(iv) $\text {NO}_{2}^{-}$
Answer:

Option (iii) is the correct answer.
Explanation: The process of nitration is initiated by $\text {NO}_{2}^+$ (Nitronium Ion) electrophile. It is then obtained as:
$H_{2}SO_{4}\left ( conc. \right )\rightarrow H^{+}+HSO_{4}^{-}$
$H^{+}+HNO_{3}\rightarrow H_{2}NO_{3}^{+}$
$H_{2}NO_{3}^{+}\rightarrow NO_{2}^{+}+H_{2}O$

Question 17. Reduction of aromatic nitro compounds using $\text {Fe and HCl}$ gives __________.
(i) aromatic oxime
(ii) aromatic hydrocarbon
(iii) aromatic primary amine
(iv) aromatic amide
Answer:

Option (iii) is the correct answer.
Explanation :

Question 18. The most reactive amine towards dilute hydrochloric acid is ___________.

Answer:

Option (ii) is the correct answer.
Explanation: The reactivity towards dilute $HCl$ is more if the strength of the base is more. Therefore,$\left ( CH_{3} \right )_{2}NH$ has the highest basic strength.

Question 19. Acid anhydrides on reaction with primary amines give ____________.
(i) amide
(ii) imide
(iii) secondary amine
(iv) imine
Answer:

Option (i) is the correct answer.
Explanation:
$\\(a)C_{2}H_{5}NH_{2}+\left ( CH_{3}CO \right )_{2}O\rightarrow CH_{3}CONHC_{2}H_{5}+CH_{3}COOH \\$
N-Ethylacetamide

Question 20. The reaction $ArN_{2}^+Cl^{-} \overset{Cu/HCl}{\rightarrow} ArCl + N_2+ CuCl$ is named as _________.
(i) Sandmeyer reaction
(ii) Gatterman reaction
(iii) Claisen reaction
(iv) Carbylamine reaction
Answer:

Option (ii) is the correct answer.
Explanation:

Question 21. The best method for preparing primary amines from alkyl halides without changing the number of carbon atoms in the chain is
(i) Hoffmann Bromamide reaction
(ii) Gabriel phthalimide synthesis
(iii) Sandmeyer reaction
(iv) Reaction with $NH_{3}$
Answer:

Option (ii) is the correct answer.
Explanation: To get primary amines from alkyl halide without changing the number of carbon atoms, Gabriel Phthalimide synthesis is used.

Question 22. Which of the following compound will not undergo an azo coupling reaction with benzene diazonium chloride.
(i) Aniline
(ii) Phenol
(iii) Anisole
(iv) Nitrobenzene
Answer:

Option (iv) is the correct answer.
Explanation: Diazonium Chloride is a very weak electrophile; therefore, it reacts with any electron rich compound which contains electron donating groups; $-OH, -NO_{2}, -OCH_{3}$. The compounds should not contain electron withdrawing groups like: $-NO_{2}$, etc.

Question 23. Which of the following compounds is the weakest Brönsted base?

Answer:

Option (iii) is the correct answer.
Explanation: All the amines have a tendency to accept protons, therefore are stronger Bronsted Bases than phenols and alcohols. And since phenol is much more acidic than alcohol, Phenols have fewer tendencies to accept any proton. Hence the weakest.

Question 24. Among the following amines, the strongest Brönsted base is __________.

Answer:

Option (iv) is the correct answer.
Explanation:$NH_{3}$ is a stronger base than aniline because of the delocalisation of the lone pair of electrons at N-atom and into the Benzene Ring. And in Pyrrole, as the lone pair electrons on N-atom donate to aromatic sextet formation, it is not basic at all. Hence, Pyrrolidine accepts the proton readily while also being the strongest base.

Question 25. The correct decreasing order of basic strength of the following species is _______. $H_{2}O, NH_{3}, OH^{-}, NH_{2}^{-}$
$(i)\; NH_{2}^{-}>OH^{-}>NH_{3}>H_{2}O$
$(ii)\; OH^{-}>NH_{2}^{-}>H_{2}O>NH_{3}$
$(iii)\;NH_{3}>H_{2}O>NH_{2}^{-}>OH^{-}$
$(iv)\;H_{2}O>NH_{3}>OH^{-}>NH_{2}^{-}$
Answer:

Option (i) is the correct answer.
Explanation: The electronegativity of $O$ is more than $N$ atom; therefore, $O-H$ bond is more polar than $N-H$ bond. Therefore, $O-H$ is more acidic than the $N-H$ bond. $NH_{2}^-$ and $OH^-$, both have a negative charge, and because of that, they are more basic than $NH_{3}$ and $H_{2}O$.

Question 26. Which of the following should be most volatile?

(i) II
(ii) IV
(iii) I
(iv) III
Answer:

Option (ii) is the correct answer.
Explanation: The boiling points of 1° and 2° amines are higher than 3° amines because of intermolecular H-Bonding. They are also less volatile than 3° amines and hydrocarbons of similar molecular mass.

Question 27. Which of the following methods of preparation of amines will not give the same number of carbon atoms in the chain of amines as in the reactant?
(i) The reaction of nitrite with $LiAlH_{4}$.
(ii) The reaction of the amide with $LiAlH_{4}$ followed by treatment with water.
(iii) Heating alkyl halide with potassium salt of phthalimide followed by hydrolysis.
(iv) Treatment of amide with bromine in the aqueous solution of sodium hydroxide.
Answer:

Option (iv) is the correct answer.

Hoffmann degradation of amides results in the loss of one carbon atom as CO₂. Hence, the amine formed has one carbon less than the reactant.

Question 29. Reduction of nitrobenzene by which of the following reagent gives aniline?
(i) ${Sn}/{HCl}$
(ii) ${Fe}/{HCl}$
(iii) $H_{2}-Pd$
(iv) ${Sn}/{NH_{4}OH}$
Answer:

Option (i), (ii) and (iii) are the correct answers.

$\mathrm{Sn} / \mathrm{HCl}, \mathrm{Fe} / \mathrm{HCl}$, and $\mathrm{H}_2 / \mathrm{Pd}$ effectively reduce nitrobenzene to aniline via complete reduction of the NO₂ group. $\mathrm{Sn} / \mathrm{NH}_4 \mathrm{OH}$ being mildly reducing and in basic medium, does not efficiently convert nitrobenzene to aniline.

Question 30. Which of the following species are involved in the carbylamine test?
(i) $R-NC$
(ii) $CHCl_{3}$
(iii) $COCl_{2}$
(iv) $NaNO_{2}+HCl$
Answer:

Option (i) and (ii) are the correct answers.
Explanation: Amine when reacts with a mixture of $\text {CHCl}_{3}$ and $\text {KOH}$ gives out alkyl isocyanate. This reaction is called a Carbylamine reaction.
Here, $\text {R-NH}_{2}+\text {CHCl}+\text {3KOH}\rightarrow \text {RNC}+\text {3KCl}+\text {3H}_{2}\text {O}$ Only $\text {RNC}$ and $\text {CHCl}_{3}$ are involved in carbylamine reaction.

Question 31. The reagents that can be used to convert benzene diazonium chloride to benzene are __________.
(i) ${SnCl_{2}}/{HCl}$
(ii) $CH_{3}CH_{2}OH$
(iii) $H_{3}PO_{2}$
(iv) $LiAlH_{4}$
Answer:

Option (ii) and (iii) are the correct answers.
Explanation :

Question 32. The product of the following reaction is __________.

Answer:

Option (A) and (B) is the correct answer.
Explanation :

Question 33. Arenium ion involved in the bromination of aniline is __________.

Answer:

Option (i), (ii) and (iii) are the correct answers.
Explanation: Bromination of aniline involves Arenium is as follows:

Question 34. Which of the following amines can be prepared by Gabriel synthesis.
$\\(i) Isobutyl amine\\ (ii) 2-Phenylethylamine\\ (iii) N-methyl benzylamine\\ (iv) Aniline$
Answer:

Option (i) and (ii) are the correct answers.
Explanation:$(CH_{3})_{2}CH-CH_{2}NH_{2}$ and $C_{6}H_{5}CH_{2}NH_{2}$ are the primary aliphatic amines that can be prepared by Gabriel synthesis. $2^{o}$ amines are $C_{6}H_{5}CH_{2}NHCH_{3} (C)$ and $1^{o}$ amine, $C_{6}H_{5}NH_{2}$ cannot be prepared by this process.

Question 35. Which of the following reactions are correct ?

Answer:

Options (i) and (iii) are the correct answers.
Explanation : $CH_{3}CH_{2}NH_{2}+NH_{4}Cl$

Question 36. Under which of the following reaction conditions, aniline gives p-nitro derivative as the major product?
(i) Acetyl chloride/pyridine followed by reaction with $conc. H_{2}SO_{4} + conc. HNO_{3}$
(ii) Acetic anhydride/pyridine followed by $conc. H_{2}SO_{4} + conc. HNO_{3}$
(iii) Dil. HCl followed by reaction with $conc. H_{2}SO_{4} + conc. HNO_{3}$
(iv) Reaction with $conc. HNO_{3}+conc. H_{2}SO_{4}$
Answer:

Option (i) and (ii) are the correct answers.
Explanation: When aniline reacts with acetyl chloride or acetic anhydride in the presence of pyridine produces N-acetyl aniline. This is a ortho, para directing group which produces p-nitroaniline on reaction with nitrating mixtures. As shown below:

Question 37. 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:

Option (i) and (ii) are the correct answers.
Explanation: Nucleophilic substitution reaction where $-NH_{2} \; and\; H$$�NH_{2} \; and\; H$ atoms are replaced by acyl groups is known as Acylation. Diazotisation is also a type of nucleophilic substitution reaction.

Question 39. Why is $\text {NH}_{2}$ group of aniline acetylated before carrying out nitration?
Answer:

The $\text {NH}_{2}$ group of aniline is acetylated before carrying out the nitration for controlling the nitration reaction, carry oxidation products and nitro derivative products formation.
As the acetyl group is an electron-withdrawing group, it attracts the lone pair of the electrons in the N atom towards the carbonyl group.

Question 40. What is the product when $\text {C}_{6}\text {H}_{5}\text {CH}_{2}\text {NH}_{2}$ reacts with $\text {HNO}_{2}$ ?
Answer:

$\text {HNO}_{2}$ is reacted $\text {C}_{6}\text {H}_{5}\text {CH}_{2}\text {NH}_{2}$ which forms unstable diazonium salt, and in turn, alcohol is given out. This is the reaction:
$\text {C}_{6}\text {H}_{5}\text {CH}_{2}\text {NH}_{2}+\text {HNO}_{2}\rightarrow \text {C}_{6}\text {H}_{5}\text {CH}_{2}\text {OH}+\text {N}_{2}+\text {H}_{2}\text {O}$

Question 41. What is the best reagent to convert nitrile to primary amine?

Answer:

$\text {LiAlH}_{4}$ and Sodium/Alcohol are the best reagents for the conversion of nitrile to a primary amine. And nitriles can be converted into a corresponding primary amine using reduction.

Question 42. Give the structure of 'A' in the following reaction.

Answer:

3-Methylnitrobenzene is the product formed in the above reaction.

Question 43. What is Hinsberg reagent?

Answer:

$\text {C}_{6}\text {H}_{5}\text {SOCl}$ aka Benzene sulphonylchloride is known as Hinsberg's reagent. It is used to differentiate between primary, secondary and tertiary amines.

Question 44. Why is benzene diazonium chloride not stored and is used immediately after its preparation?
Answer:

Benzene Diazonium is very stable at low temperatures and highly soluble in water at high temperatures. It is recommended to use it immediately after its preparation as it is unstable.

Question 45. Why does the acetylation of $\text {-NH}_{2}$ group of aniline reduce its activating effect?
Answer:

The activating effect of $\text {-NH}_{2}$ group of aniline is reduced by its acetylation. It happens because of the lone pair electrons on the nitrogen of acetanilide react with oxygen atom because of resonance.

Question 46. Explain why $\text {MeNH}_{2}$ is a stronger base than $\text {MeOH}$ ?
Answer:

$\text {MeOH}$ is a weaker base than $\text {MeNH}_{2}$ because $\text {MeNH}_{2}$ has lower electronegativity and has lone pair electrons present on the nitrogen atom.

Question 47. What is the role of pyridine in the acylation reaction of amines?
Answer:

By acylation of $\text {-NH}_{2}$ group with acetic anhydride in the presence of pyridine and after that carrying substitution followed with hydrolysis of the substituted amide with the substituted amine, the activating effect of $\text {-NH}_{2}$ group can be controlled. $\text {HCl}$ is a side product, which is removed with the help of pyridine as a base.

Question 48. Under what reaction conditions (acidic/basic), the coupling reaction of aryldiazonium chloride with aniline is carried out?
Answer:

The reaction is performed in a mild basic medium and known as electrophilic substitution reaction. Aniline and Aryldiazonium chloride react to form a yellow dye of p-Aminoazobenzene.

Question 49. Predict the product of the reaction of aniline with bromine in a non-polar solvent such as $\text {CS}_{2}$.
Answer:

When aniline and bromine react in a non-polar solvent, such as $\text {CS}_{2}$, non-polar products are formed that are 4-Bromoaniline and 2-Bromoaniline in which 4-Bromoaniline is present in the majority.

Question 50. Arrange the following compounds in increasing order of dipole moment.
$\text {CH}_{3}\text {CH}_{2}\text {CH}_{3},\text {CH}_{3}\text {CH}_{2}\text {NH}_{2},\text {CH}_{3}\text {CH}_{2}\text {OH}$
Answer:

$\text {CH}_{3}\text {CH}_{2}\text {CH}_{3}<\text {CH}_{3}\text {CH}_{2}\text {NH}_{2}<\text {CH}_{3}\text {CH}_{2}\text {OH}$
$\text {CH}_{3}\text {CH}_{2}\text {OH}$ has dipole moment greater than $\text {CH}_{3}\text {CH}_{2}\text {NH}_{2}$. Whereas,$\text {CH}_{3}\text {CH}_{2}\text {CH}_{3}$ has the least dipole moment amongst the three compounds. $\text {CH}_{3}\text {CH}_{2}\text {CH}_{3}$ is an almost non-polar molecule.

Question 51. What is the structure and IUPAC name of the compound, allyl amine?
Answer:

The structure of allyl amine is and its IUPAC name is prop-2-en-1-amine.

Question 52. Write down the IUPAC name of

Answer:

IUPAC name of N, N-Dimethylbenzenamine

Question 53. A compound Z with molecular formula $\text {C}_{3}\text {H}_{9}\text {N}$ reacts with $\text {C}_{6}\text {H}_{5}\text {SO}_{2}\text {Cl}$ to give a solid, insoluble in alkali, identify Z.
Answer:

Solid, insoluble alkali is given off when compound Z with molecular formula $\text {C}_{3}\text {H}_{9}\text {N}$ and is an aliphatic amine on treatment with $\text {C}_{6}\text {H}_{5}\text {SO}_{2}\text {Cl}$. And hence, the product does not have any replaceable hydrogen on its nitrogen atom. The amine (Z) should be a secondary amine which means Z is ethyl methylamine $(\text {C}_{2}\text {H}_{5}\text {NHCH}_{3})$.

Question 54. A primary amine, $\text {RNH}_{2}$ can be reacted with $\text {CH}_{3}-\text {X}$ to get secondary amine, $\text {RNHCH}_{3}$, but the only disadvantage is that $3^{o}$ amine and quaternary ammonium salts are also obtained as side products. Can you suggest a method where $\text {RNH}_{2}$ forms only $2^{o}$ amine?
Answer:

$\begin{array}{lcc}\mathrm{R}-\mathrm{NH}_2+\mathrm{CHCl}_3+3 \mathrm{KOH} \rightarrow \mathrm{R}-\mathrm{NC} \xrightarrow[\text { (Reduction) }]{\mathrm{H}_2 / \mathrm{Pd}} \mathrm{R}-\mathrm{NH}-\mathrm{CH}_3 \\ 1^{\circ} \text { Amine } \quad \text { Alkyl isocyanide } & 2^{\circ} \text { Amine }\end{array}$
Carbylamine reaction is shown by $1^{o}$ amine only which result in the replacement of two hydrogen atoms attached to N atom of $-\text {NH}_{2}$ group by one carbon atom to form isocyanide. On catalytic reduction, the isocyanide will give a secondary amine with one methyl group.

Question 55. Complete the following reaction.

Answer:

Question 56. Why is aniline soluble in aqueous $\text {HCl}$ ?
Answer:

Anilinium chloride salt is formed when aniline reacts with aqueous $\text {HCl}$, and it is soluble in water.

Question 57. Suggest a route by which the following conversion can be accomplished.

Answer:

Question 58. Identify A and B in the following reaction.

Answer:

Question 59. How will you carry out the following conversions?
(i) toluene $\rightarrow$ p-toIuidine
(ii) p-toluidine diazonium chloride $\rightarrow$ p-toluic acid
Answer:

Question 60. Write the following conversions:
(i) nitrobenzene $\rightarrow$ acetanilide
(ii) acetanilide $\rightarrow$ p-nitroaniline
Answer:

Question 61. A solution contains 1 g mol each of p-toluene diazonium chloride and p-nitrophenyl diazonium chloride. To this 1 g mol of an alkaline solution of phenol is added. Predict the major product. Explain your answer.
Answer:

This reaction is an example of electrophilic aromatic substitution. Phenol forms phenoxide ion in an alkaline medium, which is more electron-rich than phenol and therefore more reactive in the electrophilic attack. Arydiazonium cation is electrophile in this reaction, and p-Nitrophenyldiazonium cation is a stronger electrophile than p-toluene diazonium cation. Hence, it pairs preferentially with phenol.

Question 62. How will you bring out the following conversion?

Answer:

Question 63. How will you carry out the following conversion?

Answer:

Question 64. How will you carry out the following conversion?

Answer:

Question 65. How will you carry out the following conversion?

Answer:

Question 67. Match the compounds given in Column I with the items given in Column II.

Answer:

(i $\rightarrow$ b), (ii $\rightarrow$ a), (iii $\rightarrow$ d), (iv $\rightarrow$ c)

Benzene sulphonyl chloride $\rightarrow$ Hinsberg reagent because it is used in the Hinsberg test to distinguish amines.

Sulphanilic acid $\rightarrow$ Zwitter ion, Exists as $\mathrm{H}_3 \mathrm{~N}^{+}-\mathrm{C}_6 \mathrm{H}_4-\mathrm{SO}_3^{-}$

Alkyl diazonium salts $\rightarrow$ Conversion to alcohols because it is Very unstable

Aryl diazonium salts $\rightarrow$ Dyes, Undergo azo coupling to form brightly coloured azo dyes.

Question 69. In the following question, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices:
Assertion (A): Hoffmann's bromamide reaction is given by primary amines.
Reason (R): Primary amines are more basic than secondary amines.
(i) Both assertion and reason are wrong.
(ii) Both assertion and reason are correct statements but reason is not correct explanation of assertion.
(iii) Assertion is correct statement but reason is wrong statement.
(iv) Both assertion and reason are correct statements and reason is correct explanation of assertion.
(v) Assertion is wrong statement but reason is correct statement.
Answer:

(iii) Assertion is correct statement but reason is wrong statement.

Hofmann’s bromamide reaction is given by amides, not by amines. Moreover, primary amines are basic than secondary amines.

Question 70. In the following question, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices:
Assertion (A): N-Ethylbenzene sulphonamide is soluble in alkali.
Reason (R): Hydrogen attached to nitrogen in sulphonamide is strongly acidic.
(i) Both assertion and reason are wrong.
(ii) Both assertion and reason are correct statements but reason is not correct explanation of assertion.
(iii) Assertion is correct statement but reason is wrong statement.
(iv) Both assertion and reason are correct statements and reason is correct explanation of assertion.
(v) Assertion is wrong statement but reason is correct statement.
Answer:

Ethylbenzene sulphonamide is soluble in alkali because it has acidic hydrogen.
Hence (iv) Both assertion and reason are correct statements and reason is correct explanation of assertion.

Question 71. In the following question, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices:
Assertion (A): N, N-diethyl benzene sulphonamide is insoluble in alkali.
Reason (R): Sulphonyl group attached to the nitrogen atom is a strong electron-withdrawing group.
(i) Both assertion and reason are wrong.
(ii) Both assertion and reason are correct statements but reason is not correct explanation of assertion.
(iii) Assertion is correct statement but reason is wrong statement.
(iv) Both assertion and reason are correct statements and reason is correct explanation of assertion.
(v) Assertion is wrong statement but reason is correct statement.
Answer:

N, N-diethyl benzene sulphonamide is not soluble in alkali as it there is no acidic hydrogen present on this. Sulphonyl group is an electron-withdrawing group which is attached to a nitrogen atom.
Hence, (ii) Both assertion and reason are correct statements but reason is not correct explanation of assertion.

Question 72. In the following question, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices:
Assertion (A): Only a small amount of $\text {HCl}$ is required in the reduction of nitro compounds with iron scrap and $\text {HCl}$ in the presence of steam.
Reason (R):$\text {FeCl}_{2}$ formed gets hydrolysed to release $\text {HCl}$ during the reaction.
(i) Both assertion and reason are wrong.
(ii) Both assertion and reason are correct statements but reason is not correct explanation of assertion.
(iii) Assertion is correct statement but reason is wrong statement.
(iv) Both assertion and reason are correct statements and reason is correct explanation of assertion.
(v) Assertion is wrong statement but reason is correct statement.
Answer:

$\text {Fe}+\text {2HCl}\rightarrow \text {FeCl}_{2}+\text {2[H]}$
Nacent hydrogen is reduced into nitro compounds.
$\text {FeCl}_{2}+\text {H}_{2}\text {O}\text {(g)}\rightarrow \text {FeO}+\text {2HCl}$
Hence, (iv) Both assertion and reason are correct statements and reason is correct explanation of assertion.

Question 73. In the following question, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices:
Assertion (A): Aromatic $1^{o}$ amines can be prepared by Gabriel phthalimide synthesis.
Reason (R): Aryl halides undergo nucleophilic substitution with anion formed by phthalimide.
(i) Both assertion and reason are wrong.
(ii) Both assertion and reason are correct statements but reason is not correct explanation of assertion.
(iii) Assertion is correct statement but reason is wrong statement.
(iv) Both assertion and reason are correct statements and reason is correct explanation of assertion.
(v) Assertion is wrong statement but reason is correct statement.
Answer:

(i) Both assertion and reason are wrong

Aromatic amines are never obtained by Gabriel phthalimide synthesis.

Aryl halides are resistant to nucleophilic substitution under normal Gabriel synthesis conditions.

Question 74. In the following question, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct answer out of the following choices:
Assertion (A): Acetanilide is less basic than aniline.
Reason (R): Acetylation of aniline results in the decrease of electron density on nitrogen.
(i) Both assertion and reason are wrong.
(ii) Both assertion and reason are correct statements but reason is not correct explanation of assertion.
(iii) Assertion is correct statement but reason is wrong statement.
(iv) Both assertion and reason are correct statements and reason is correct explanation of assertion.
(v) Assertion is wrong statement but reason is correct statement.
Answer:

(iv) Both assertion and reason are correct statements and reason is correct explanation of assertion. Both are correct acetylation withdraws nitrogen’s lone pair via resonance, hence acetanilide is less basic than aniline.

Question 76. A colourless substance 'A' $\left ( \text {C}_{6}\text {H}_{7}\text {N} \right )$ is sparingly soluble in water and gives a water-soluble compound 'B' on treating with a mineral acid. On reacting with $\text {CHCl}_{3}$ and alcoholic potash 'A' produces an obnoxious smell due to the formation of compound 'C'. The reaction of 'A' with benzene sulphonyl chloride gives compound 'D' which is soluble in alkali. With $\text {NaNO}_{2}$ and $\text {HCl}$, 'A' forms compound 'E' which reacts with phenol in alkaline medium to give an orange dye 'F' Identify compounds' A' to 'F'.

Answer:

Question 77. Predict the reagent or the product in the following reaction sequence.

Answer: