(i) 

(ii) \(2HCHO\overset { Conc.NaOH }{ \longrightarrow } { CH }_{ 3 }OH+HCOONa\)
(iii) \({ CH }_{ 3 }-CH=N-NH-CO-NH_{ 2 }\)
(iv) In FCH₂ - COOH, fluorine is more electron withdrawing and has stronger -I effect than chlorine in ClCH₂COOH. So, FCH₂COOH is more acidic than CICH₂COOH hence its pK_a value is lesser than ClCH₂COOH.
(v) \(CH_{ 3 }-CH=CH-{ CH }_{ 2 }-CN\overset { DIBAL-H }{ \underset { { H }_{ 2 }O }{ \longrightarrow } } { CH }_{ 3 }-CH=CH-{ CH }_{ 2 }-CHO\)
     Pent-3-enenitrile                                             Pent-3-ene-1-al
(v) Propanal and propanone can be differentiated by Tollen's reagent i.e. propanal will give silver mirror but propanone will not.
\({ CH }_{ 3 }-{ CH }_{ 2 }-CHO+2[Ag({ NH }_{ 3 })_{ 2 }]^{ + }\rightarrow { CH }_{ 3 }-{ CH }_{ 2 }-{ COO }^{ - }+2Ag\downarrow +{ H }_{ 2 }O+4{ NH }_{ 3 }\ (Or\ any\ other\ correct\ test)\)                                                                                    (Silver mirror)

(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 (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

Si (E.N. = 8) is more electropositive than C (E.N. = 2.5), therefore, Me₃Si (trimethylsilyl group) has greater +l-effect than that of Me₃C (r-butyl group). As a result, Me₃Si intesifies the -ve charge on the carboxylate ion relative to r-butyl group and hence Me₃CCH₂COOH is a stronger acid that Me₃SiCH₂COOH.

(a) (i) Cannizzaro's reaction:
When aldehydes which do not have \(\alpha\)-hydrogen gets oxidised as well as reduced in presence of conc. alkali, it is called Cannizzaro's reaction.
\(\underset { Methanal }{ 2HCHO } \overset { 50%NaOH }{ \longrightarrow } \underset { Sodium\quad Methanoate }{ HCOONa } +\underset { Methanol }{ CH_{ 3 }OH } \)
(ii) Decarboxylation : Sodium salts of acids when heated with soda lime, lower alkanes are formed
\(\underset { Sodium\quad acetate }{ CH_{ 3 }COONa } +\underset { Sodalime }{ NaOH } \longrightarrow \underset { Methane }{ CH_{ 4 } } +\underset { sodium\quad carbonate }{ Na_{ 2 }CO_{ 3 } } \)