Match List - I with List - II.

Choose the correct answer from the options given below:

(A)-(III), (B)-(IV), (C)-(I), (D)-(II)

The correct answer is Option (1) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)

(A) Williamsons Synthesis:

In the Williamson Ether Synthesis, an alkyl halide (or sulfonate, such as a tosylate or mesylate) undergoes nucleophilic substitution \((S_N2)\) by an alkoxide to give an ether. Being an \(S_N2\) reaction, best results are obtained with primary alkyl halides or methyl halides. Tertiary alkyl halides give elimination instead of ethers. Secondary alkyl halides will give a mixture of elimination and substitution. The alkoxide \(RO^–\) can be those of methyl, primary, secondary, or tertiary alcohols. The reaction is often run with a mixture of the alkoxide and its parent alcohol (e.g. \(NaOEt/EtOH\) or \(CH_3ONa/CH_3OH\)). Alternatively, a strong base may be added to the alcohol to give the alkoxide. Sodium hydride \((NaH)\) or potassium hydride \((KH)\) are popular choices. When an alkoxide and alkyl halide are present on the same molecule, an intramolecular reaction may result to give a new ring. This works best for 5- and 6-membered rings. When planning the synthesis of ethers using the Williamson, take care to select the best starting materials for an \(SN_2\) reaction.

Mechanism:

(B) 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 \(C_6H_5OH\) (phenols). When phenol, i.e. \(C_6H_5OH\), is treated with \(CHCl_3\) (chloroform) in the presence of \(NaOH\) (sodium hydroxide), an aldehyde group \((-CHO)\) is introduced at the ortho position of the benzene ring leading to the formation of o-hydroxybenzaldehyde. The reaction is popularly known as the Reimer Tiemann reaction.

Mechanism:

Step I: First, a strongly basic hydroxide solution is used to deprotonate the chloroform. Deprotonation is the elimination of the hydrogen atom. The chloroform carbanion is formed by removing the hydrogen atom. The chloroform carbanion will swiftly undergo alpha elimination, yielding dichlorocarbene \((CCl_2)\), the reaction’s major reactive species.

Step II: Base extracts the hydrogen atom from the \(-OH\) group to generate the phenoxide ion. The phenoxide’s negative charge is delocalized into the aromatic ring, making it more nucleophilic.

Step III: The addition of dichlorocarbene results in the formation of an intermediate i.e., dichloromethyl-substituted phenol. The resulting intermediate is then exposed to basic hydrolysis to yield ortho-hydroxybenzaldehyde.

(C) Dow's Process

In Dow’s process, Phenol is produced from chlorobenzene by fusing it with molten sodium hydroxide at \(350^oC\) to convert it to sodium phenoxide which upon acidification gives phenol. The reaction occurs via elimination-addition mechanism that involves a benzyne intermediate.

Mechanism:

The first step hydroxide ion removes the proton from carbon adjacent to the carbon bonded to chlorine. This leads to the elimination of a water molecule followed by the formation of the short lived benzyne intermediate. In the second step, the hydroxide ion attacks the benzyne intermediate to give the sodium phenoxide which on acidification gives phenol.

(D) Schotten Baumann reaction:

Schotten Baumann reaction mechanism is a convenient method for benzoylation of phenols with benzoyl chloride in presence of cold aqueous sodium hydroxide. The aqueous alkali reacts with the hydrogen halide produced.

Mechanism: