Predict the product in the following: CH 3 - CH = CH 2 → H 2 O / H +

The hydrolysis of propene in the presence of dilute sulphuric acid forms 2-propanol. Let’s see by reaction that the double bond gets converted into a single bond and an alcoholic group gets introduced to the carbon atom having lesser number of hydrogen around double bond. Thus the addition takes place according to Markonikov rule. CH 3 - CH = CH 2 → H 2 O / H + CH 3 - CH ( OH ) - CH 3 ( Major ) + CH 3 - CH 2 - CH 2 - OH ( Minor )

How will you bring about the formaldehyde to methyl alcohol transformation?

i Formaldehyde when treated with Na - Hg gets reduced to methanol. In this reaction, sodium amalgam in ethanol brings out the reduction of formaldehyde. HCHO → Na - Hg CH 3 OH ii Formaldehyde on reduction by using lithium aluminum hydride followed by acid hydrolysis gives methanol. HCHO → ii . Hydrolysis i . LiAlH 4 CH 3 OH

Identify ' X ' , ' Y ' and ' Z ' in the following: (a) C 6 H 5 OH → Zn / Heat ' X ' → Anhyd AlCl 3 CH 3 COCl ' P ' (b) CH 3 CH = CH 2 → H 2 O / H + ' Q ' → K 2 Cr 2 O 7 / H + ' Y ' (c) C 6 H 5 OH + Na → Z

(a) Distillation of phenol with zinc-dust gives benzene and ZnO as a side-product. The reaction between benzene and an acyl chloride in the presence of an aluminium chloride catalyst gives acetophenone. Therefore, X is benzene and P is acetophenone. C 6 H 5 OH → Zn / Heat C 6 H 6 → Anhyd AlCl 3 CH 3 COCl C 6 H 5 COCH 3 (b) Propene on hydration gives propan-2-ol in the presence of a dilute strong acid. Propan-2-ol is secondary alcohol which is oxidised to give acetone with sodium or potassium dichromate(VI) solution acidified with dilute sulphuric acid. Therefore, Q is propan-2-ol and Y is acetone. CH 3 CH = CH 2 → H 2 O / H + CH 3 CH ( OH ) CH 3 → K 2 Cr 2 O 7 / H + CH 3 COCH 3 (c) Phenol reacts with sodium and form sodium phenoxide and hydrogen gas. This is an acid - metal reaction. Therefore, Z is sodium phenoxide. C 6 H 5 OH + Na → C 6 H 5 O - Na + + 1 2 H 2

Predict the product: CH 3 3 C - OC 2 H 5 → HI

( CH 3 ) 3 C − O − CH 3 on reaction with HI gives ( CH 3 ) 3 C − I and CH 3 OH as the main products. CH 3 3 C - OC 2 H 5 → HI ( CH 3 ) 3 C − I + CH 3 OH The above reaction has proceeded through nucleophilic substitution of first order. In the SN 1 mechanism, the rate of reaction is fastest for tertiary carbocation. Hence, the formation of tertiary carbocation is favourable and forms the basis of the final product.

Identify 'X' and 'Y' in the following: CH 3 - CH 2 - OH → conc . H 2 SO 4 / 443 K ' X ' → Alkaline KMnO 4 H 2 O + O Y

Ethanol on reaction with concentrated sulphuric acid form ethene which reacts with alkaline KMnO 4 to form the ethylene glycol. The first reaction is dehydration and the second one is oxidation. CH 3 - CH 2 - OH → conc . H 2 SO 4 / 443 K CH 2 = CH 2 → Alkaline KMnO 4 H 2 O + O CH 2 ( OH ) - CH 2 ( OH )

Predict the product: CH 3 - CH ( OH ) - CH 3 → 440 K Conc . H 3 PO 3

Propan - 2 - ol can be dehydrated to give propene by heating it with an excess of concentrated phosphoric acid at about 440 K . Concentrated sulphuric acid, H 2 SO 4 , can be used instead. CH 3 - CH ( OH ) - CH 3 → 440 K Conc . H 3 PO 3 CH 3 CH = CH 2 This reaction is an example of dehydration reaction.

Complete the following: C 2 H 5 Cl + C 2 H 5 ONa →

Chloroethane reacts with sodium ethoxide to form diethyl ether and sodium chloride. This reaction is example of Williamson synthesis. It is a substitution reaction. C 2 H 5 Cl + C 2 H 5 ONa → C 2 H 5 OC 2 H 5 + NaCl

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY>Chapter -1 - Alcohols, Phenols and Ethers>Jharkhand Board-2018>Q 1

MEDIUM

12th Jharkhand Board

IMPORTANT

Earn 100

Ethanoyl chloride $+$ Ethanol $\to$ A

'A' is

50% studentsanswered this correctly

Chapter-Wise 3D Chemistry Videos from Embibe Experts EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY

04 min 43 sec

Preparation of Alcohols

Alcohol is a psychosomatic drug and it is bad for health if consumed. However, it is a useful organic solvent in chemistry labs. Tune in to learn more about...

04 min 06 sec

Preparation of Alcohols from Alkenes

The alcohols are important organic compounds that can be prepared by the hydration of alkenes. Do you know how the reaction proceeds? Let’s watch the video t...

04 min 00 sec

Laboratory Preparation of Alcohols

Alcohols can be prepared from different organic compounds such as alkenes, aldehydes and ketones. Do you know what reagents are involved in the reactions? Le...