A. Write the product(s) in the following reactions:

(i)

(ii)

(iii)

(B) Give simple tests to distinguish between the following pairs of compounds :

(i) Ethanol and Phenol

(ii) Propanol and 2-methylpropan-2-ol

OR

(a) Write the formula of reagents used in the following reactions :

(i) Bromination of phenol to 2,4,6-tribromophenol

(ii) Hydroboration of propene and then oxidation to propanol.

(b) Arrange the following compound groups in the increasing order of their property indicated :

(i) p-nitrophenol, ethanol, phenol (acidic character)

(ii) Propanol, Propane, Propanal (boiling point)

(c) Write the mechanism (using curved arrow notation) of the following reaction :

(A) (i)

This is salicylic acid. Acetylation of salicylic acid produces aspirin and CH₃COOH

(ii)

If R-O-R’ is the ether which reacts with HX ( hydrogen halides, X = F,Cl,Br,I)

Then, RX and R’OH are formed.

So, here products are

(iii) (PCC), pyridinium chlorochromate is a complex of chromium trioxide with pyridine and HCl. It yields CH₃ – CH = CH – CHO but–2-ene-1-al. It oxidises primary alcohols to aldehydes.

(B) (i) Ethanol and Phenol – bromine water tests

(ii) Propanol and 2-methylpropan-2-ol – lucas test

1. Bromine water test - orange-brown coloured bromine water, can help test the presence of unsaturation. When we test ethanol and phenol with this, we see no change in the ethanol and formation of white precipitate in the case of phenol. Only phenols give this reaction.

2. Propanol is a primary alcohol and 2-methylpropan-2-ol is a tertiary alcohol. CH₃-CH₂-CH₂-OH is propanol. 2-methylpropan-2-ol is –

Alcohols are soluble in Lucas reagent which is conc. HCl and ZnCl₂ but their halides are immiscible and produce turbidity in solution. So, these alcohols are first reacted with hydrogen halides to form alkyl halides. In case of tertiary alcohols, turbidity is produced immediately as they form the halides easily. Primary alcohols do not produce turbidity at room temperature. This Lucas test distinguishes them.

ROH + HX → R–X + H₂O

OR

A. (i) Bromination of phenol to 2,4,6- tribromophenol

Bromine water (Br ) is used in this reaction.

(ii) Hydroboration of propene and then oxidation to propanol

Diborane (BH₃)₂ reacts with propene to give tripropyl boranes as addition product.

This is oxidized to propanol by hydrogen peroxide ( H₂O₂) in the presence of aqueous sodium hydroxide (NaOH)

B. (i) p-niotrophenol, ethanol, phenol (acidic character)

The presence of electron withdrawing groups such as nitro group, enhances the acidic strength of phenol, especially when group is at para ortho positions.

The hydroxyl group, in phenol is directly attached to the sp²hybridised carbon of benzene ring which acts as an electron withdrawing group, making it more acidic then alcohols

So, acidic character – ethanol < phenol < p-nitrophenol

(ii) Propanol, Propane, Propanal (boiling point)

Taking hydrogen bonding into consideration, more the bonding, more is the boiling point.

Alcohols display more hydrogen bonding than aldehydes. Alkanes do not show hydrogen bonding. So, Boiling point – propane < propanal < propanol

(C) In given reaction, alcohol acts as nucleophile and attacks on carbocation. It is the intermediate step, involved in the preparation of ethers.