Q: Write the equation involved in the following reactions:(i) Reimer-Tiemann reaction (ii) Kolbe's reaction

(i) Reimer-Tiemann reaction \$\rightarrow\$ When phenol is treated with chloroform in the presence of sodium hydroxide, a \$-CHO\$ group is introduced at the ortho position of the benzene ring. This reaction is known as the Reimer-Tiemann reaction. The intermediate is hydrolyzed in presence of alkalis to produce salicylaldehyde.(ii) Kolbe's reaction \$\rightarrow\$ When phenol is treated with sodium hydroxide, sodium phenoxide is produced. This sodium phenoxide when treated with carbon dioxide, followed by acidification, undergoes electrophilic substitution to give ortho-hydroxybenzoic acid as the main product. This reaction is known as Kolbe's reaction.

Q: Explain the preparation of phenol from cumene.

Preparation of phenol from cumene \$\rightarrow\$To prepare phenol, cumene is first oxidised in the presence of air to cumene hydroperoxide.\$\underset{\left( \text{Isoprophyl benzene} \right)}{\text{Cumene}} \xrightarrow[]{{O}_{2}} \text{Cumene hydroperoxide}\$Now, cumene hydroperoxide is treated with dilute acid to prepare phenol and acetone as by-products.\$\text{Cumene hydroperoxide} \xrightarrow[323-363 \; K]{{H}^{+}} \text{Phenol} + \text{Acetone}\$By this method, acetone is obtained in large quantities.

Q: Identify the product in the following reaction.

The Reimer–Tiemann reaction is a chemical reaction used for the ortho-formylation of phenols; with the simplest example being the conversion of phenol to salicylaldehyde

Q: When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place:Give a mechanism for this reaction.(Hint : The secondary carbocation formed in step II rearranges to a more stable tertiary carbocation by a hydride ion shift from 3rd carbon atom.)

The mechanism for the reaction between \$3\$-methylbutan-\$2\$-ol and HBr is shown above.In the first step, \$-OH\$ group is protonated. In the second step, a molecule of water is lost to form secondary carbocation.Third step is the rearrangement of less stable secondary carbocation to more stable tertiary carbocation through \$1,2\$-hydride shift.Final step is the nucelophilic attack of bromide ion on tertiary carbocation to form \$2\$-bromo-\$2\$-methylbutane.

Question 1

Write the mechanism of hydration of ethene to yield ethanol.

Accepted Answer

Three steps are involved in the mechanism of hydration of ethene to yield ethanol.$(1)$ The electrophillic attack of hydronium ion protonates ethene to form carbocation.$(2)$ Water (as a nucleophile) attacks carbocation.$(3)$ Deprotonation gives ethanol.The mechanism is shown above.

Question 2

Write the equation involved in the following reactions:(i) Reimer-Tiemann reaction (ii) Kolbe's reaction

Accepted Answer

(i) Reimer-Tiemann reaction $\rightarrow$ When phenol is treated with chloroform in the presence of sodium hydroxide, a $-CHO$ group is introduced at the ortho position of the benzene ring. This reaction is known as the Reimer-Tiemann reaction. The intermediate is hydrolyzed in presence of alkalis to produce salicylaldehyde.(ii) Kolbe's reaction $\rightarrow$ When phenol is treated with sodium hydroxide, sodium phenoxide is produced. This sodium phenoxide when treated with carbon dioxide, followed by acidification, undergoes electrophilic substitution to give ortho-hydroxybenzoic acid as the main product. This reaction is known as Kolbe's reaction.

Question 3

Explain the preparation of phenol from cumene.

Accepted Answer

Preparation of phenol from cumene $\rightarrow$To prepare phenol, cumene is first oxidised in the presence of air to cumene hydroperoxide.$\underset{\left( \text{Isoprophyl benzene} \right)}{\text{Cumene}} \xrightarrow[]{{O}_{2}} \text{Cumene hydroperoxide}$Now, cumene hydroperoxide is treated with dilute acid to prepare phenol and acetone as by-products.$\text{Cumene hydroperoxide} \xrightarrow[323-363 \; K]{{H}^{+}} \text{Phenol} + \text{Acetone}$By this method, acetone is obtained in large quantities.

Question 4

An alkene $CH_3CH = CH_2$ is treated with $B_2H_6$ in presence of $H_2O_2$. The final product formed is:

Accepted Answer

Hydroboration-oxidation reaction follows anti-Markovnikov's addition of $H-OH$ across $C=C$ to give alcohol.Thus an alkene $CH_3CH=CH_2$ when treated with $B_2H_6$ in presence of $H_2O_2$ will yield the final product as  $CH_3CH_2CH_2OH$

Question 5

Which of the following is most acidic?

Accepted Answer

Option D is the correct answer. A methoxy group is mesomeric donating (+R) and inductive withdrawing (-I). We know that the +R effect is predominant over the -I effect. But +R effect is only valid when the substituent is at the ortho and para position.We also know that addition of Electron donating groups to the Benzene ring in phenol decreases its acidity while the addition of Electron withdrawing groups increase it.By that logic, m-Methoxyphenol is a stronger acid than phenol as the $-OCH_3$ group at m-position imparts only -I effect to the ring.

Question 6

Which one of the following compounds is most acidic?

Accepted Answer

The nitro group in ortho position withdraws electrons from benzene making it more acidic forming a stable compound after hydrogen is removed. Phenols are more acidic than alcohols due to the stabilisation of phenoxide ion by resonance. Hence the correct option is B.

Question 7

Name the reagents used in the following reactions:(i) Oxidation of a primary alcohol to carboxylic acid.(ii) Oxidation of a primary alcohol to aldehyde.(iii) Bromination of phenol to 2,4,6-tribromophenol.(iv) Benzyl alcohol to benzoic acid.(v) Dehydration of propan-2-ol to propene.(vi) Butan-2-one to butan-2-ol.

Accepted Answer

(i) Oxidation of primary alcohol to carboxylic acid.Acidified $ \displaystyle K_2Cr_2O_7$ or acidified $ \displaystyle KMnO_4$(ii) Oxidation of primary alcohol to aldehyde.Pyridinium chlorochromate (PCC) in $ \displaystyle CH_2Cl_2$ or Pyridinimium dichromate (PDC) in $ \displaystyle CH_2Cl_2$(iii) Bromination of phenol to 2,4,6-tribromophenol.Aqueous bromine ( or bromine water)(iv) Benzyl alcohol to benzoic acid.Acidified or alkaline $ \displaystyle KMnO_4$ followed by hydrolysis with dil $ \displaystyle H_2SO_4$(v) Dehydration of propan-2-ol to propene.85% phosphoric acid(vi) Butan-2-one to butan-2-ol.$ \displaystyle NaBH_4$ or $ \displaystyle Ni / H_2$

Question 8

Write equations of the following reactions:(i) Friedel-Crafts reaction  alkylation of anisole.(ii) Nitration of anisole.(iii) Bromination of anisole in ethanoic acid medium.(iv) Friedel-Crafts acetylation of anisole.

Accepted Answer

The different reactions are shown above.

Question 9

Identify the product in the following reaction.

Accepted Answer

The Reimer&#8211;Tiemann reaction is a chemical reaction used for the ortho-formylation of phenols;  with the simplest example being the conversion of phenol to salicylaldehyde

Question 10

When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place:Give a mechanism for this reaction.(Hint : The secondary carbocation formed in step II rearranges to a more stable tertiary carbocation by a hydride ion shift from 3rd carbon atom.)

Accepted Answer

The mechanism  for the reaction between $3$-methylbutan-$2$-ol and HBr is shown above.In the first step, $-OH$ group is protonated. In the second step, a molecule of water is lost to form secondary carbocation.Third step is the rearrangement of less stable secondary carbocation to more stable tertiary carbocation through $1,2$-hydride shift.Final step is the nucelophilic attack of bromide ion on tertiary carbocation to form $2$-bromo-$2$-methylbutane.