Explanation-

Ether synthesis-

The general method for the synthesis of ether is Williamson ether synthesis, involves nucleophilic displacement of a halide ion or other good leaving group by an alkoxide ion.

The name of the reaction was coined after Alexander William Williamson developed it in 1850. Williamson Ether Synthesis is a reaction that uses deprotonated alcohol and an organohalide to form an ether.

• Williamson Ether Synthesis usually takes place as an S_N2 reaction of a primary alkyl halide with a 2° or 3° alkoxide ion. The structure of ethers was proved due to this chemical reaction.
• S_N2 pathway is required for the synthesis of this reaction and is useful only when the alkyl halide is primary or secondary.
• The Ethers produced in this way have more carbon atoms than either of the starting materials and thus are more complex structures.
• 2° and 3° alkyl halides may undergo E2 elimination in the presence of a strong base to form alkenes.

Mechanism of this reaction-

• The nucleophile attacks the alkyl halide forming an ether from the back.
• This response takes place in a single step, which is both cleavage and bond formation.
• If halides are sterically impeded then alkoxide acts as a basis and protons in β-place are accessible.
• The products are derived from a response to elimination.

Explanation:

Given:

⇒ 2A+ Dry Silver Oxide  $\stackrel{\mathrm{\Delta }}{\to }$  R-O-R (ether) + 2AgX 

We know,

An alkyl Halide on heating with dry silver oxide gives ether

⇒ 2 R - X + Ag₂O $\stackrel{\mathrm{\Delta }}{\to }$ R - O - R + 2 AgX

Explanation-

Friedel-Crafts Alkylation-

It refers to the replacement of an aromatic proton with an alkyl group. This is done through an electrophilic attack on the aromatic ring with the help of a carbocation. The Friedel-Crafts alkylation reaction is a method of generating alkylbenzenes by using alkyl halides as reactants.

The Friedel-Crafts alkylation reaction of benzene is illustrated below.

A Lewis acid catalyst such as FeCl₃ or AlCl₃ is employed in this reaction in order to form a carbocation by facilitating the removal of the halide. The resulting carbocation undergoes a rearrangement before proceeding with the alkylation reaction.

Some important limitations of Friedel-Crafts alkylation are listed below-

• Since the carbocations formed by aryl and vinyl halides are extremely unstable, they cannot be used in this reaction.
• The presence of a deactivating group on the aromatic ring (such as an NH₂ group) can lead to the deactivation of the catalyst due to the formation of complexes.
• An excess of the aromatic compound must be used in these reactions in order to avoid polyalkylation (addition of more than one alkyl group to the aromatic compound).
• Aromatic compounds that are less reactive than mono-halobenzenes do not participate in the Friedel-Crafts alkylation reaction.

Friedel craft alkylation will undergo by that aromatic compound, which is electron-rich. As it is the electrophilic substitution reaction.

So both 2 and 4 will undergo  Friedel - Craft's alkylation reaction.

Explanation-

Phenol is an aromatic alcohol. We know aromatic compounds easily give electrophilic substitution reactions.

Electrophilic substitution reactions are the reactions in which an electrophile replaces a functional group generally a proton from the reactant.

Concentrated HNO₃ is concentrated nitric acid. It is a nitrating reagent.

Phenol will undergo a nitration reaction in the presence of concentrated. nitric acid, HNO₃ to give 2, 4, 6-trinitrophenol which is also known as picric acid.

Phenol will undergo nitration in presence of nitric acid also but will form either mono- or di-substituted ortho- or para-nitrophenols.

Phenols are nucleophilic in nature and hence undergo electrophilic substitution reactions like nitration, sulphonation, alkylation, acetylation, and halogenation.

 will be formed on treating phenol with concentrated HNO3 / H2SO4 mixture.

Explanation:

→ The phenoxide ions left from o-hydroxybenzaldehyde and p-hydroxybenzaldehyde are stabilized by the-I and R-effect of the -CHO group.

→ But due to chelation in the o-isomer, it is difficult to remove the H-atom.

Hence, p-hydroxybenzaldehyde is the strongest acid.

Important Points→ Acidity of a compound depends on the nature of groups presented on that compound. As we can say –I (Inductive effect) and +R (Resonance effect) increase the acidity whereas the +I effect increases the basicity. 

Explanation:

→ o - Isomer has a maximum dipole moment.

Additional Information→ Due to intramolecular hydrogen bonding, derivatives of phenol, i.e. o-nitrophenol has a lower solubility in water, and lower boiling point.

→ Due to this, association with different molecules decreases hence, comparatively volatile in nature. On the other hand, m and p- isomer have intermolecular hydrogen bonding.

Thus, they have an appreciable value of boiling point and are soluble in water.

Explanation-

Liebermann's reaction -

When phenol is reacted with NaNO₂ and concentrated H₂SO₄, it provides a deep green or blue color which changes to red on dilution with water. Generated substance in presence of NaOH / KOH restores the original green or blue color. This reaction is termed as Liebermann's nitroso reaction.

Other Important Points-

Reimer - Tiemann reaction-

Reimer Tiemann reaction is a type of substitution reaction, named after chemists Karl Reimer and Ferdinand Tiemann. The reaction is used for the ortho-formylation of C₆H₅OH (phenols).

Schotten - Baumann reaction-

Schotten Baumann reaction refers to the method of chemically synthesizing amides from acyl chlorides and amines. This organic chemical reaction is named after the German chemist Carl Schotten and Eugen Baumann, who discovered this method of synthesizing amides.

Explanation:

1.  IUPAC name is the name that follows the chemistry rules for naming a compound.
2. In IUPAC naming, the organic compounds are named after the parent hydrocarbon by using certain prefixes or suffixes to indicate the presence of a functional group

In the above molecule, it is an ether ROR′. The substituent name for the  RO−  function is alkoxy, and it is correct to name  R−O−R′  compounds as alkoxy derivatives of hydrocarbons.

The longest chain of carbons contains 3 carbons so it becomes prop and single bond so ane. The methoxy group is in 2nd position.

Thus its name is: 2 - methoxypropane

Explanation:

Diethyl sulphate in the presence of NaOH acts as an alkylating agent, it causes the alkylation of phenol to give ethyl phenyl ether which is also called phenetole.

→ C₆H₅OH + NaOH → C₆ H₅ O⁻ Na⁺ + H₂O

→ C6 H5 O− Na⁺ + C₂H₅SO₄ (diethyl sulphate) →  C₆H₅OC₂H₅ (phenetole) + C₂H₅NaSO₄

Explanation-

The reaction of ether with the strong hydriodic acid (HI), causes the cleavage of the ether C-O bond, following the nucleophilic substitution reaction.

This cleavage further depends on the degree of alkyl group present in the ether as the primary and the secondary alkyl ethers have less steric hindrance. Thus, it shows the S_N2 mechanism.

In the primary and the secondary alkyl ethers, following the S_N2 mechanism, due to less hindrance, the iodide attaches to the smaller alkyl group, whereas in the tertiary alkyl ether following the S_N1 mechanism, the high stability of the tertiary carbocation the iodide attaches to it, and the smaller alkyl forms the alcohol.

Option 1-

So, in  CH3 - CH2 - CH2 - CH2 - O - CH3, through the S_N2 mechanism, having 1^∘ alkyl group, on reaction with HI, it forms butanol and methyl iodide.

Option 2-

In   , having a 2^∘ alkyl group, on reaction it forms methyl iodide and isobutyl alcohol.

Option 3-

In the case of the tertiary alkyl ether due to steric hindrance, it follows the S_N1 mechanism through the formation of the stable carbocation intermediate. So, in the compound given in option 3, on reaction we get tertiary- butyl iodide and methanol. As follows:

Option 4-

 In reaction with HI, it forms 2-methyl propanol and methyl iodide.

So option 3 is the correct answer.