Question

give the structures of A, B and C in the following reactions:
i) $C{H_3}Br\xrightarrow{{KCN}}A\xrightarrow{{LiAl{H_4}}}B\xrightarrow[{273K}]{{HN{O_2}}}C$

ii) $C{H_3}COOH\xrightarrow[\Delta ]{{N{H_3}}}A\xrightarrow{{B{r_2} + KOH}}B\xrightarrow{{CHC{l_3} + NaOH}}C$

Answer 1

As we all know that organic chemistry is full of dozens of chemical reactions, so to tackle these reactions, remember the working and functioning of reagents and the reactants to solve for the product. Reagents can be broken into ions to simplify and check whether reduction or oxidation is happening.

Complete step by step answer:
To solve such questions is simple, first look at the pattern like which functional group is involved in the reaction and which reagents do you have to proceed for the further solution.
i) So let us look at the first reaction, we know that when any alkyl halide reacts with $KCN$ it results in the formation of alkyl cyanide because $KCN$ is an ionic compound and easily dissociated into ${K^ + }$ and $C{N^ - }$and now we also know that is an ambident nucleophile which can attack a carbon as well as a nitrogen but because C-C bond is stronger so it will attack carbon atom following the SN2 mechanism and the reaction we can show as the following:
$R - Br + KCN \to R - CN + KBr$
Thus, in the same way the first product A that we will get is $C{H_3}CN$.
Now the next reagent is $LiAl{H_4}$ which is a strong reducing agent and help in the reduction of double and triple bonds and resulting in the conversion of nitriles into primary amines which can be explained using the reaction:
$R - CN + LiAl{H_4} \to R - N{H_2}$
Hence we get our second product B as $C{H_3}N{H_2}$.
Finally the last reagent that is $HN{O_2}$which is a weak acid and is used in making azo dyes by converting amines into diazides and it converts primary amines into alcohol by giving out nitrogen and a colourless gas and this reaction can be shown as:
$R - N{H_2} + HN{O_2} \to R - OH + {N_2} + {H_2}O$
Therefore, our last product C is $C{H_3}OH$ and the overall reaction will be:
\\\ $C{H_3}Br\xrightarrow{{KCN}}C{H_3}CN\xrightarrow{{LiAl{H_4}}}C{H_3}N{H_2}\xrightarrow[{273K}]{{HN{O_2}}}C{H_3}OH$

ii) Similarly in the second part, ammonia will react with ethanoic acid to give an ethanamide giving up a water molecule and then this ethanamide will follow Hoffmann bromamide degradation reaction with reagents bromine and alcoholic KOH, converting the amide into primary amine and finally the primary amine will be converted to alcohol in the presence of chloroform and alcoholic sodium hydroxide which are the reagents of Hoffmann’s carbylamine test. So the overall reaction will be:
\\\
$C{H_3}COOH\xrightarrow[\Delta ]{{N{H_3}}}C{H_3}CON{H_2}\xrightarrow{{B{r_2} + KOH}}C{H_3}N{H_2}\xrightarrow{{CHC{l_3} + NaOH}}C{H_3}OH$

Note:
Always remember that oxidation of alcohols gives aldehydes and oxidation of aldehydes results in carboxylic acids and reverse is the case for reduction that when carboxylic acids are reduced they give aldehydes which on further reduction gives alcohol.