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Question: Given the following sequence of reaction \(C{{H}_{3}}C{{H}_{2}}I\xrightarrow{NaCN}A\xrightarrow[part...

Given the following sequence of reaction CH3CH2I→NaCNA→partial hydrolysisOH−B→Br2/NaOHCC{{H}_{3}}C{{H}_{2}}I\xrightarrow{NaCN}A\xrightarrow[partial\,hydrolysis]{O{{H}^{-}}}B\xrightarrow{B{{r}_{2}}/NaOH}C,
the major product C is:
A.CH3CH2NH2C{{H}_{3}}C{{H}_{2}}N{{H}_{2}}
B. CH3−CH2−COONH4C{{H}_{3}}-C{{H}_{2}}-COON{{H}_{4}}
C. CH3CH2−COONH4C{{H}_{3}}C{{H}_{2}}-COON{{H}_{4}}
D. CH3−CH2CO−NBr2C{{H}_{3}}-C{{H}_{2}}CO-NB{{r}_{2}}

Explanation

Solution

Haloalkanes undergo nucleophilic substitution (SN) reactions, to obtain a compound in which the halo group is replaced by the nucleophile.

Complete answer: A reaction sequence is given in which iodo ethane, reacts with sodium cyanide to form a product A. This product A on partial hydrolysis gives a product B, which in reaction with Br2B{{r}_{2}}and sodium hydroxide, gives product C. We have to find the product C.
As we know the reaction of haloalkanes is through nucleophilic substitution 2, SN2{{S}_{N}}^{2} reaction. This means that the leaving group from ethyl iodide is replaced by nucleophile CN from sodium cyanide, so A is ethane nitrile. Therefore the reaction is,
CH3CH2I→NaCNCH3CH2CN→partial hydrolysisOH−B→Br2/NaOHCC{{H}_{3}}C{{H}_{2}}I\xrightarrow{NaCN}C{{H}_{3}}C{{H}_{2}}CN\xrightarrow[partial\,hydrolysis]{O{{H}^{-}}}B\xrightarrow{B{{r}_{2}}/NaOH}C
The partial hydrolysis of any cyanide, yields amide, as the CN group gets oxygen and hydrogen attached with it, so product B, formed by partial hydrolysis of ethane nitrile will be ethanamide, the reaction will be,
CH3CH2I→NaCNCH3CH2CN→partial hydrolysisOH−CH3CH2CONH2→Br2/NaOHCC{{H}_{3}}C{{H}_{2}}I\xrightarrow{NaCN}C{{H}_{3}}C{{H}_{2}}CN\xrightarrow[partial\,hydrolysis]{O{{H}^{-}}}C{{H}_{3}}C{{H}_{2}}CON{{H}_{2}}\xrightarrow{B{{r}_{2}}/NaOH}C
Now, this product B, which is ethanamide, reacts with Di-bromine and sodium hydroxide, this reaction is named as Hoffmann Bromamide reaction. This reaction yields aliphatic primary amines, when amides react with di-bromine and sodium hydroxide. So, the product C is obtained as a primary amine in the reaction,
CH3CH2I→NaCNCH3CH2CN→partial hydrolysisOH−CH3CH2CONH2→Br2/NaOHCH3CH2NH2C{{H}_{3}}C{{H}_{2}}I\xrightarrow{NaCN}C{{H}_{3}}C{{H}_{2}}CN\xrightarrow[partial\,hydrolysis]{O{{H}^{-}}}C{{H}_{3}}C{{H}_{2}}CON{{H}_{2}}\xrightarrow{B{{r}_{2}}/NaOH}C{{H}_{3}}C{{H}_{2}}N{{H}_{2}}
Hence, the major product, C formed is ethanamine, with the formula, CH3CH2NH2C{{H}_{3}}C{{H}_{2}}N{{H}_{2}}.

So, option A is correct.

Note: The reaction of an amide in Hoffmann bromamide reaction, will yield a primary amine, so, all other options are wrong, as they do not consist of primary amines.