
Identify the product (C) in the given reaction;
(1)
(2)
(3)
(4)
Answer
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Hint:The starting reactant has a carbonyl group, and the name of the functional group is a ketone (acetophenone), with a phenyl group, and if observed the groups are nitrous acid (HONO), acetic anhydride, and the last step is hydrolysis.
Complete step-by-step answer:So, here we have the first reactant is acetophenone, which is reacting with\[{\text{NaN}}{{\text{O}}_{\text{2}}}\]/ ${\text{HCl}}$, this group will give a nitrous acid (HONO), which will attack the acetophenone and form an oxime derivative. Following is the reaction observed in the formation of product ‘A’
${\text{NaN}}{{\text{O}}_{\text{2}}}{\text{ + HCl}} \to {\text{HN}}{{\text{O}}_{\text{2}}}$
${\text{Ph}} - \mathop {\mathop {\text{C}}\limits^{{\text{||}}} }\limits^{\text{O}} - {\text{CH}} = {\text{N}} - {\text{OH}}$ is the product (A) that is formed, now this oxime derivative will react with acetic anhydride and remove ${{\text{H}}_{\text{2}}}{\text{O}}$ molecules from the oxime derivative compound. The following reaction will take place:
Now, in the last step, the nitrile group will get hydrolyzed (${{\text{H}}_3}{{\text{O}}^ + }$) and form a carboxylic acid derivative:
Hence, the correct answer is option(3).
Note:: In the reaction of acetophenone with (HONO), there are two possible products of ‘A’, both have a similar molecular formula, 1st oxime derivative and the other with the tautomerization of oxime (\[{\text{Ph}} - \mathop {\mathop {\text{C}}\limits^{{\text{||}}} }\limits^{\text{O}} - {\text{C}}{{\text{H}}_{\text{2}}} - {\text{N}} = {\text{O}}\]), but we have used the oxime derivative as in the next step of dehydration the oxime derivative will be more easily lose ${{\text{H}}_{\text{2}}}{\text{O}}$ molecules and form the corresponding product with a nitrile group.
Complete step-by-step answer:So, here we have the first reactant is acetophenone, which is reacting with\[{\text{NaN}}{{\text{O}}_{\text{2}}}\]/ ${\text{HCl}}$, this group will give a nitrous acid (HONO), which will attack the acetophenone and form an oxime derivative. Following is the reaction observed in the formation of product ‘A’
${\text{NaN}}{{\text{O}}_{\text{2}}}{\text{ + HCl}} \to {\text{HN}}{{\text{O}}_{\text{2}}}$
${\text{Ph}} - \mathop {\mathop {\text{C}}\limits^{{\text{||}}} }\limits^{\text{O}} - {\text{CH}} = {\text{N}} - {\text{OH}}$ is the product (A) that is formed, now this oxime derivative will react with acetic anhydride and remove ${{\text{H}}_{\text{2}}}{\text{O}}$ molecules from the oxime derivative compound. The following reaction will take place:
Now, in the last step, the nitrile group will get hydrolyzed (${{\text{H}}_3}{{\text{O}}^ + }$) and form a carboxylic acid derivative:
Hence, the correct answer is option(3).
Note:: In the reaction of acetophenone with (HONO), there are two possible products of ‘A’, both have a similar molecular formula, 1st oxime derivative and the other with the tautomerization of oxime (\[{\text{Ph}} - \mathop {\mathop {\text{C}}\limits^{{\text{||}}} }\limits^{\text{O}} - {\text{C}}{{\text{H}}_{\text{2}}} - {\text{N}} = {\text{O}}\]), but we have used the oxime derivative as in the next step of dehydration the oxime derivative will be more easily lose ${{\text{H}}_{\text{2}}}{\text{O}}$ molecules and form the corresponding product with a nitrile group.
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