Question

(a) Identify the compounds ‘A’, ‘B’ and ‘C’ in the following reactions:
(i) C2H5OH + CH3CH2COOH ‘A’ + H2O
(ii) CH3COOC2H5 + NaOH →‘B’ + C2H5OH
(iii) CH3COONa + NaOH(CaO) ‘C’ + Na2CO3
(b) A cyclic compound ‘X’ has molecular formula C6H6. It is unsaturated and burns with a sooty flame. Identify ‘X’ and write its structural formula. Will it decolourise bromine water or not and why?

Answer 1

Answer:

we will be Mr. Why is weak acid and alcohol B will be an Acetic Acid C will be e methane

Explanation:

the first reaction is the Easter synthesis that is the Acetic Acid + alcohol and II reaction of the first question is the reverse of esterification that is that give the initial substrate of esterification that is alcohol and acetic acid and that heard that 80 is Methane and its reaction of decarboxylation in the presence of soda lime

the second question give the molecular formula of benzene the bromine water test is given by a benzene so there are two points which can prove that the molecule is Benzene that is C6 H6 itself

Answer 2

(a) (i) - Ethyl Propionate

(ii) - Sodium Acetate

(iii) - Methane

(b) Benzene

Explanation:

Missing compound:

(i) CH3CH2COOH + C2H5OH = CH3CH2COOC2H5 + H2O

Ethyl propionate

(ii) CH3COOC2H5 + NaOH → CH3COONa + C2H5OH  

Sodium acetate

(iii) CH3COONa + NaOH --> CH4 + Na2CO3

Methane

(b) Molecular formula is C6H6. So the molecule is Benzene. Benzene is considered to be unsaturated as it has double carbon-carbon covalent bonds. Since it is unsaturated, it burns with a sooty flame.  

Benzene molecule is unsaturated but it does not undergo electrophillic addition reaction because saturating the carbon-carbon bonds destroys the delocalized pi-cloud. But Benzene's delocalized pi-structure is very stable. So benzene does not decolourize bromine water.

Please refer the attached picture for the structural formula.