Amines MCQs

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Amines are the compounds that are derived from ammonia (NH3). Amines are classified as the functional groups of the organic compounds of nitrogen that include a basic nitrogen atom along with a lone pair. Amines occur widely in living organisms and many of them are said to be naturally occurring compounds. Amines are mainly used in the production of amino acids and they are also used in pain-relieving medicines. 

Amines are broadly classified into four types: primary, secondary, tertiary, and cyclic according to the number of atoms that are bonded directly to the nitrogen atom. Primary amines consist of one carbon bonded to the nitrogen atom while the secondary amines consist of two carbons bonded to the nitrogen atom and the tertiary atoms consist of three carbons bonded to the nitrogen atom.

Multiple Choice Questions

Question: Which one of the following is the functional group of Amine?

  1. Nitrogen
  2. Oxygen
  3. Barium
  4. Helium

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Answer: 1) Nitrogen

Explanation: An amine is defined as a compound or functional group with a nitrogen atom having a lone pair. Nitrogen has a total valency of 5, which is why it makes a trivalent with a lone pair. When the hydrogen atom of ammonia is replaced by an alkyl or aliphatic group, they are known as alkyl amines and when it is replaced by an aryl or aromatic group, then they are known as aryl amines. Amines have sp3 hybridised N but the shape of the compound is a pyramid due to the presence of a lone pair of electrons. Amines can naturally be obtained from Halogens.

Question: Aniline is less basic than:

  1. Triphenylamine
  2. Benzylamine
  3. Diphenylamine
  4. p-nitroaniline

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Answer: 2) Benzylamine

Explanation: According to Lewis's Theory of acids and bases, a base must have the availability of electrons to donate to an acid. A stronger base is said to be the one whose electrons are easily available for donation. The chemical formula of benzyl amine is C6H5CH2NH2. In this molecule, a lone pair of electrons is present on nitrogen and it is attached to an sp3 hybridised carbon. That is why, Aniline is less basic than benzylamine.

Question: Which one of the following reactions helps in preparing amines from lower amines?

  1. Wurtz reaction
  2. Mannich reaction
  3. Schmidt reaction
  4. Hofmann bromamide reaction

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Answer: 4) Hofmann bromamide reaction

Explanation: The Hofmann bromamide reaction is an organic reaction used to convert a primary amide to a primary amine. An amide is treated with bromine and base (usually NaOH or KOH) in this reaction. Then upon heating that amide, a non-isolated intermediate isocyanate is formed. Then, with the presence of water, the non-isolated isocyanate loses carbon dioxide to produce an amine.

Question: Which one of the following between these compounds has the lowest Boiling rate?

  1. Aniline
  2. Butyl amine
  3. Diethylamine
  4. Propyl amine

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Answer: 4) Propyl amine

Explanation: Amines have higher boiling points than any of their corresponding alkanes only because of Dipole-Dipole interaction and Hydrogen bonding in primary and secondary amines. Thus, the secondary amines are said to have lesser boiling points than primary amines because of their comparable molecular masses. Hence, the boiling point of diethyl amine is less than that of butylamine. Between propylamine and butylamine, the boiling point of propyl amine is less because it has a lower molecular mass than butylamine. Between propylamine and diethylamine, the boiling of propyl amine is less than diethylamine because of its lower molecular weight. Here, between these compounds, Aniline has the highest molecular mass, so it has the highest boiling point. 

Question: Which one of the following is the product that is formed when a primary amine reacts with chloroform in alcoholic KOH?

  1. An isocyanide
  2. An alcohol
  3. An Aldehyde
  4. Cyanide

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Answer: 1) An isocyanide

Explanation: When primary amines are treated with chloroform and ethanol in KOH, it produces isocyanides or carbylamines. The reaction occurs through carbene intermediate. Chloroform reacts with KOH to produce chloro-carbene, whose chemical formula is: CCl2.

The general form of the reaction is: 

R – NH2 + CHCl3 + 3KOH (alc.) ∆ → R – NC + 3KCl + 3H2O

Hence, we now know that when a primary amine reacts with chloroform in alcoholic KOH, the product formed is an isocyanide or a carbylamine.

Question: When ethylamine is treated with methyl magnesium bromide, the product formed is:

  1. Ethane
  2. Methane
  3. Ethanol
  4. Methanol

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Answer: 2) Methane

Explanation: Methyl magnesium bromide, whose chemical formula is: CH3MgX and it falls in the category of Grignard Reagents. In this category, the primary and secondary amines react with Grignard Reagents to form alkanes. For this reaction to occur, there have to be hydrogen atoms that need to be attached to the nitrogen atom. As there are no hydrogen atoms attached to the nitrogen atom of tertiary amines, so they do not undergo the reaction with Grignard reagents. The reaction of ethylamine with methyl magnesium bromide, Methane is formed from the alkyl part of the Grignard Reagents.

Question: Which of the following natural amino acids is optically inactive? 

  1. Glycine
  2. Alanine
  3. Valine
  4. Aspartic acid

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Answer: 1) Glycine

Explanation: An optically inactive substance is defined as a substance that does not have optical activity, i.e. a substance which does not rotate the plane of polarisation of a beam of light that is passed through it. Glycine is said to be the second most common amino acid found in proteins as compared to the other amino acids but almost all the other natural amino acids except glycine are optically active. Glycine is said to be optically inactive because it does not have a chiral carbon in its structure.

Question: The product obtained when methylamine (CH3NH2) is treated with nitrous acid is:

  1. CH3OH
  2. CH3CH2
  3. CH3OCH3
  4. Both 2 & 3

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Answer: 1) CH3OH

Explanation: CH3NH2 is a primary alkyl amine which undergoes a reaction with HNO2. The reaction takes place at a very low temperature of about 0° – 5°C. The intermediate form is methyl diazonium, which upon hydrolysis methanol is formed. Nitrogen is also produced as a side product in this reaction. Hence, we can say that Methyl amine when treated with nitrous acid produces methanol.

Question: Secondary amines can be prepared by:

  1. reduction of nitro compounds
  2. reduction of isonitriles
  3. reduction of nitriles
  4. oxidation of N-substituted amides

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Answer: 2) reduction of isonitriles

Explanation: Secondary amines are only said to be formed when the isonitriles are reduced with hydrogen atoms in the pressure of Pt catalyst.

The reaction is given below:

R – NC 4 H → R – NH – CH3

The product we get after the reduction of isonitriles is always a secondary amine, while the other compounds mentioned giving primary amines which are also shown below:

RNO2 6H → RNH+ 2H2O

RCONH2 4H → RCH2NH+ H2O

RCN 4H → RCH2NH2

Question: Which one of the following is formed in the reaction of an aldehyde with a primary amine?

  1. Carboxylic acid
  2. Aromatic Acid
  3. Schiff’s base
  4. Ketone

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Answer: 3) Schiff’s base

Explanation: Schiff bases are defined as aldehyde or ketone-like compounds in which the carbonyl group gets replaced by the amine group. 

Here, Aldehyde + Primary amines = Schiff’s base

The general reaction is shown below:

reaction

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CBSE CLASS XII Related Questions

  • 1.
    Calculate emf of the following cell at 298 K: \[ \text{Cr(s)} \, | \, \text{Cr}^{3+} (aq) \, (0.1\, M) \; || \; \text{Fe}^{2+} (aq) \, (0.01\, M) \, | \, \text{Fe(s)} \] Given: \[ E^\circ_{\text{Cr}^{3+}/\text{Cr}} = -0.74 \, V, \quad E^\circ_{\text{Fe}^{2+}/\text{Fe}} = -0.44 \, V, \quad \log 10 = 1 \]


      • 2.
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          • A = CH\(_3\)I, B = NaCl
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        • 3.
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            • A
            • B
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          • 4.
            What happens when acidic solution of potassium permanganate is allowed to stand for sometime ? Give the equation involved. What is this type of reaction called ?


              • 5.
                Write IUPAC names of the following coordination compounds:
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                  • 6.
                    For a reaction : \( N_2 + 3H_2 \rightarrow 2NH_3 \), the rate of reaction with respect to \( NH_3 \) is

                      • \( +\frac{1}{3} \frac{\Delta [NH_3]}{\Delta t} \)
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                    CBSE CLASS XII Previous Year Papers

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