Complete the following reaction and name the products: 2 CH 3 OH → mild O 2 A → ∆ 50 % KOH CH 3 OH + B

Methanol on oxidation with a mild oxidising agent like CrO 3 forms an aldehyde, methanal ( A ). Methanal on heating with potassium hydroxide undergoes self oxidation and reduction to give methanol and potassium formate ( B ). This reaction is called Cannizzaro's reaction. 2 CH 3 OH → mild O 2 HCHO A → ∆ 50 % KOH CH 3 OH + HCOOK B

Write the product of the following reaction with mechanism : C 2 H 5 OH → 443 K H 2 SO 4 ?

Ethyl alcohol undergoes dehydration in the presence of concentrated sulphuric acid at 443 K to form ethene. C 2 H 5 OH → 443 K H 2 SO 4 CH 2 = CH 2 + H 2 O The mechanism involves the following steps: (i) Due to the presence of lone pair of electrons on oxygen atoms, alcohol acts as a weak base, therefore, it reacts with acid to form protonated alcohol. (ii) The protonated ethanol eliminates a molecule of water to form ethyl carbocation: (iii) The carbocation loses a proton to form ethene.

Write the product of the following reaction:

When benzyloxybenzene is heated with hydrogen iodide, benzyl iodide and phenol are formed. In this reaction, the cleavage of C 6 H 5 CH 2 - O bond takes place, since aryl-oxygen bond is partially double due to resonance. The reaction can be written as given below: C 6 H 5 CH 2 - O - C 6 H 5 + HI → ∆ C 6 H 5 CH 2 I + C 6 H 5 OH

Arrange the following compounds in increasing order of their acid strength: H 2 O , CH 3 OH , C 2 H 5 OH and C 6 H 5 OH

The hydroxyl group in phenol is directly attached to sp 2 -hybridised carbon atom of benzene ring, therefore, the electron density on oxygen atom of hydroxyl group decreases, which in turn increases the acidity of the molecule. Therefore, phenol is stronger acid than alcohols and water. Alcohols are, however weaker acid than water. Thus, the increasing order of acid strength of the given compounds are given below: CH 3 CH 2 OH < CH 3 OH < H 2 O < C 6 H 5 OH

Give the equation with conditions for the preparation of diethylether from ethanol.

When excess of ethyl alcohol is heated with concentrated sulphuric acid, at 140 ° C , the symmetrical ether diethyl ether is formed. The reaction can be written as given below: 2 CH 3 CH 2 OH → 140 ° C con . H 2 SO 4 C 2 H 5 - O - C 2 H 5 + H 2 O

How will you distinguish between Pentan- 2 -ol and Pentan- 3 -ol?

Any secondary alcohol having the hydroxyl group on C - 2 will test positively to the iodoform reaction, just like a methyl ketone.Therefore, Pentan- 2 -ol on heating with iodine and sodium hydroxide produces a yellow precipitate of iodoform ( CHI 3 ), but Pentan- 3 -ol does not give this reaction. CH 3 CH 2 CH 2 CH OH CH 3 pentan - 2 - ol → I 2 + NaOH CH 3 CH 2 CH 2 COOH + CHI 3 iodoform

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EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter -1 - Alcohols, Phenols and Ethers>Meghalaya Board-2019>Q 6

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