(i) Since compound (A) with M.F. C₂H₄O reduces Tollens' reagent, it must be an aldehyde, i.e., acetaldehyde (CH₃HO).
\(\underset{Acetaldehyde}{CH_3CHO}+\underset{Tollens\ reagent}{2[Ag(NH_3)_2]^+}+3OH^-\longrightarrow CH_3COO^-+2Ag\downarrow+4NH_3+2H_2O\)
(ii) In presence of Al(OC₂H₅)₃ aldehydes undergo Tischenko reaction to give esters. Thus, when two moles of acetaldehyde (CH₃CHO) react in presence of AI(OC₂H₅)₃ ethyl acetate (B) with M.F. C₄H₈O₂ is produced
\(\underset{Acetaldehyde(A)\\(Two\ moles)}{CH_3CHO+OHCCH_3}\xrightarrow[(Tischenko reaction)]{Al(OC_2H_5)_3}\underset{Ethyl\ acetate\\ M/F.\ C_4H_8O_2}{CH_3COOCH_2CH_3}\)
(iii) The structure of ethyl acetate (B) is confirmed by the observation that on treatment with NH₃, it gives one molecule of an alcohol, i.e., ethyl alcohol, CH₃CH₂OH (C) and one molecule of an amide, i.e., acetamide, CH₃CONH₂(D)
\(\underset{Ethyl\ acetate(B)}{CH_3COOCH_2CH_3}\xrightarrow{NH_3}\underset{Ethyl alcohol (C)\\ M.F. C_2H_6O}{CH_3CH_2OH}+\underset{Acetamide (D)\\ M.F. C_2H_5NO}{CH_3CONH_2}\)

(i) Since compound A on oxidation gives compound 8 with M.F. C₂H₄O₂, therefore, compound 8 may be acetic acid, CH₃COOH and A may be acetaldehyde, CH₃CHO.
(ii) Since compound A, i.e., acetaldehyde reacts with dil. NaOH, therefore, it undergoes aldol condensation to afford an aldol.Further since this aldol on heating gives compound (C), therefore, (C) must be an α, β- unsaturated aldehyde, i.e., but-2-en-I-al (crotonaldehyde).
(iii) Since compound (C) on catalytic hydrogenation gives compound D, therefore, D may be either I-butanal or I-butanol depending upon the extent of hydrogenation.
AIl the reactions involved in this question are explained below:

(i) Since compound (E) with molecular formula, C₃H₆0 does not reduce Tollens' reagent and Fehling's solution but forms 2, 4-dinitrophenylhydrazone, it must be a ketone. But the only possible ketone having the molecular formula, C₃H₆O is acetone or propanone. Thus, compound (E) is acetone or (propanone) CH₃COCH3·
(ii) Since acetone (E) is obtained by heating compound (8) with Ca(OH)₂ therefore, (B) must be acetic acid (ethanoic acid), CH₃COOH.
(iii) Since (D) on oxidation with KMn04 gives acetic acid (8), therefore, (D) must be ethyl alcohol (ethanol), CH₃CH₂OH.
(iv) Since acetic acid (8) and ethyl alcohol (D) are obtained by hydrolysis of (C) under acidic conditions, therefore, (C) must be ethyl acetate (ethyl ethanoate), CH₃COOC₂H₅
(v) Since ethyl acetate (C) and acetic acid (8) are obtained by treatment of compound (A) with ethyl alcohol, therefore, compound (A) must be acetic anhydride (ethanoic anhydride), (CH₃COO)₂O.
(vi) All the reactions involved in this problem can now be explained as follows

(i) Since ketone A (C₄H₈O) undergoes haloforrn reaction, it must contain the grouping CH}CO. Therefore, ketone A (C₄H₈O) must be butanone (CH₃COCH₂CH₃).
(ii) Since ketone A, i.e., butanone gives compound B on reduction, therefore, B must be 2-butanol (CH₃CHOHCH₂CH₃).
(iii) Since B, i.e., 2-butanol on heating with H₂SO₄ gives compound C which forms a mono-ozonide therefore, compound C must be an alkene.
(iv) Since alkene C forms a mono-ozonide D which on hydrolysis in presence of zinc dust (z.e., reductive ozonolysis) gives only acetaldehyde E, therefore, C must be a symmetrical alkene, i.e., 2-butene.
(v) All the reactions involved in the problem can now be explained as follows:

During acetylation, one H-atom (at. mass = Iamu) of the OH group is replaced by an acetyl group, i.e., CH₃CO (mol. mass = 43 amu). .
-OH + (CH₃CO)₂O ⇾ -O-COCH₃ + CH₃COOH
In other words, acetylation of each OH group increases the mass by 43 - I = 42 amu. Now the mol. mass of C₄H₁₀O = 106 amu while that of the acetylated product is 190 amu, therefore, the number of OH groups III the compound\(={190-106\over 42}=2\)