Chemistry Test - 18

Explanation-

Phenyl isocyanide test-

The phenyl isocyanide test is also known as the carbylamine test in which there is a synthesis of an isocyanide with the help of a reaction between chloroform, primary amine, and a base.

As the carbylamine reaction is only effective for primary amines, it can be used as a chemical test for the presence of primary amines. In this context, the reaction is also called Saytzeff's isocyanide test.

In this reaction, the unknown substance is heated up with alcoholic potassium hydroxide and chloroform. If a primary amine is present, the isocyanide (carbylamine) is formed which is indicated by a foul smell.

The carbylamine test gives a negative reaction with secondary and tertiary amines.
The reaction's mechanism involves adding an amine group to the dichlorocarbene, which is simply formed by removing a chlorine atom and a hydrogen atom (overall termed as dehydrohalogenation).

This dichlorocarbene acts as a Lewis acid and receives electrons from the nitrogen atom of the primary amine group. Later there are two base-mediated dehydrochlorination reactions that lead to the formation of isocyanide.

Thus, the correct answer is aromatic primary amines.

Explanation:

Amine is a type of compound that is derived from ammonia (NH₃).

Primary Amine

• Primary amines are formed when one hydrogen atom in ammonia is substituted by an alkyl or aromatic group. Some examples of primary alkyl amines include amino acids and methylamine while primary aromatic amines include aniline.

Secondary amines

• Secondary amines are those amines that have two organic substituents either alkyl or aryl ones or both. They are bound to the nitrogen together with one hydrogen. A common example includes dimethylamine. Likewise, diphenylamine is an example of an aromatic amine.

Methyl isocyanide can be reduced to secondary amine in the presence of Lithium aluminum hydride (LiAIH₄)

CH₃NC → CH₃NHCH₃ ( in the presence of Lithium aluminum hydride)

 Important Points

Tertiary amine-

Tertiary amines are those where the nitrogen consists of three organic substituents. Examples of this type of amine are trimethylamine and EDTA.

Explanation:

• Amine is a type of compound that is derived from ammonia (NH₃).
• The amino group is one of the most important functionalities in organic synthesis and in nature. Amines are a class of organic compounds that contain the functional group -NH₂. The structure of amine is R-NH₂, where R in the case of aliphatic amine is an alkyl group, or for aromatic amines is an aryl group.
• Amines are usually categorized into different types based on their nature.
• They are usually classified as aliphatic amines (having only Hydrogen and alkyl substituents) and aromatic amines (the nitrogen atom is connected to an aromatic ring).

The reaction of Aromatic primary amines with cold HNO₂-

Primary aromatic amines such as aniline react with nitrous acid under ice-cold conditions (273-278 K)  to form Benzene Diazonium salt. The reaction is known as the Diazotization reaction.

Diazo means two nitrogen atoms. when combined with onium, we have diazonium, which means two nitrogen atoms and a positive charge ( N₂⁺). Diazonium ions are produced when an aryl amine reacts with cold nitrous acid.

Nitrous acid is unstable and is prepared just prior to its use by a reaction between sodium nitrate and Hydrochloric acid.

Explanation:

Hofmann bromamide degradation-

• When an amide is treated with bromine in an aqueous or ethanolic solution of sodium hydroxide, degradation of amide takes place leading to the formation of primary amine.
• This reaction involves the degradation of amide and is popularly known as the Hoffmann bromamide degradation reaction.
• The primary amine thus formed contains one carbon less than the number of carbon atoms in that amide.

RCONH₂ +Br₂ + 4NaOH → R-NH₂ + Na₂CO₃ + 2NaBr + 2H₂O

Given data and Analysis-

• In this reaction, we have to find the amide group as all the options groups are different (Hoffmann bromamide reaction causes degradation of amides).
• An amide group is a group in which the carbonyl group is linked to the nitrogen atom that is ArCONH₂.  
• ArCONH2 Group is called an amide group. Among the given options, only option 2 contains this group.

This reaction will be as follows: 

→ ArCONH₂ + Br₂ + 4NaOH → ArNH₂ + Na₂CO₃ + 2NaBr + 2H₂O

Explanation:

Scientists commonly use the boiling point of a liquid as a measure of volatility.

• Volatile liquids have low boiling points.
• A liquid with a low boiling point will begin to boil faster than liquids with higher boiling points.
• Much less energy (in the form of heat) is required to break the intermolecular bonds of a volatile liquid than those of liquids having higher boiling points.
• Once enough energy is supplied to break apart the bonds between molecules, the molecules are free to expand and escape the liquid surface in the form of a gas.

The order of boiling points of isomeric amines is 1⁰ amines > 2⁰  amines > 3⁰ amines.

Because of the absence of H−atom available for hydrogen bonding, 3⁰ amines do not have intermolecular associations.

The intermolecular association is more than 1⁰ amines than in 2⁰ amines as there are two H are also available for H bonding. Hydrocarbons are almost non-polar molecules and possess weak van der Waals forces and hence have the lowest boiling point i.e most volatile.

So CH3CH2CH3 is the most volatile.​

Explanation:

E2 reaction-

• In an E2 mechanism which refers to bimolecula Given data and Analysis

  E2​ reaction involves a one-step mechanism in which carbon-hydrogen and carbon halogen bonds break to form a double bond.

  As the given compound is basic in nature so it is not easy to break carbon and hydrogen bonds. So it does not undergo an E2 Elimination reaction.r elimination is basically a one-step mechanism.
• Here, the carbon-hydrogen and carbon-halogen bonds mostly break off to form a new double bond.
• However, in the E2 mechanism, a base is part of the rate-determining step and it has a huge influence on the mechanism.
• The reaction rate is mostly proportional to the concentrations of both the eliminating agent and the substrate.
• It exhibits second-order kinetics.

Given data and Analysis

E2​ reaction involves a one-step mechanism in which carbon-hydrogen and carbon halogen bonds break to form a double bond.

As the given compound is basic in nature so it is not easy to break carbon and hydrogen bonds. So it does not undergo an E2 Elimination reaction.

Explanation-

Amine is a type of compound that is derived from ammonia (NH3).

Primary Amine-

Primary amines are formed when one hydrogen atom in ammonia is substituted by an alkyl or aromatic group. Some examples of primary alkyl amines include amino acids and methylamine while primary aromatic amines include aniline.

Secondary amines-

Secondary amines are those amines that have two organic substituents either alkyl or aryl ones or both. They are bound to the nitrogen together with one hydrogen. A common example includes dimethylamine. Likewise, diphenylamine is an example of an aromatic amine.

\(CH_3NH_2 \ \xrightarrow[\Delta]{CHCI_3/KOH}\) CH₃ - NH - CH₃

\(CH_3NH_2 \ \xrightarrow[\Delta]{CHCl_3/KOH}\)  \(\ \xrightarrow[]{Sn/HCl}\) CH₃ - NH - CH3

This method is the best method for preparing 2^O amine.

So option 3 is the correct answer.

Explanation-

Amine salts are water-soluble but insoluble in organic solvents such as alcohol and ether. This reaction is the basis for the separation of amines from non-basic organic compounds which are insoluble in water.

The reaction between amines and mineral acids forms ammonium salts which clearly depicts the basic nature of amines

The greater the basicity of amines the higher is its reactivity.

Order of basicity if the −R group is −CH₃​ group: secondary amine  >primary amine  > tertiary amine> Ammonia > Aryl compounds. 

In an aqueous solution, the most basic is Dimethyl amine so it will have the highest reactivity.

, is a secondary amine is the most reactive amine towards dilute hydrochloric acid.

Explanation-

Pyridine

Pyridine is a heterocyclic compound that is colorless to yellow liquid with a chemical formula C₅H₅N.

It is also known as Azine. The structure is like benzene, with one methine group replaced by a nitrogen atom. It has a sour, putrid, and fish-like odor.

Pyridine can be synthesized from ammonia, formaldehyde, and acetaldehyde or it can be made from crude coal tar.

Pyridine is the hydrogen derivative of this ring, it is benzene in which one CH- or methine group is replaced by a nitrogen atom.

The structure of pyridine is completely analogous to that of benzene, being related by the replacement of CH with N.

Structure of Pyridine -

Pyridine is much less basic than typical aliphatic amines, but for a very different reason: the unshared pair is in an sp², which is much lower in energy than the electron pair of aliphatic amines, which is in an sp³.

Therefore, pyridine is less easily protonated than typical aliphatic amines such as triethylamine.

Concept:

Grignard Reagent:

• The Grignard reagent is given as RMgX where X is a halogen, and R is an alkyl or aryl (based on a benzene ring) group.
• They react with a variety of compounds such as acids, aldehydes, ketones, alkynes to form addition products.
• Grignard reagent usually attacks the nucleophilic center.
• The polarity of Grignard reagents is opposite to that expected because here the carbon atom bears a negative charge and acts as nucleophiles.
• This effect is called the Umpolang Effect or reversal of polarity.

• Methyl magnesium bromide is CH3MgX and falls in the category of Grignard Reagents.
• The methyl group bears a partial negative charge and the metal bears a partial positive charge.

Explanation:

• Primary and secondary amines react with Grignard reagents to form alkanes.
• For the reaction to occur, there have to be hydrogen atoms attached to the nitrogen atom.
• The hydrogen is extracted by the negatively charged alkyl group of the Grignard reagents leading to the formation of alkanes.
• As there are no hydrogens attached to the Nitrogen atom of tertiary amines, 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 Reagent.
• The reaction can be written as:

• When ethyl amine will be treated with ethyl magnesium bromide, it will give ethane. The reaction will be:

C₂H₅NH₂ + C₂H₅MgBr → C₂H₆

Hence, when ethylamine is treated with methyl magnesium bromide, the product formed is methane.

Mistake Points 

• Ethylamine when treated with NaNO₂ + HCl, gives ethyl alcohol as the product.

• Similarly, methylamine will yield methanol.