MCQ
Acetoacetic ester behaves as
- AAn unsaturated hydroxy compound
- BA keto compound
- ✓Both of these ways
- DNone of these
✓
Answer
Correct option: C.
Both of these ways
c
(c)$\underset{\begin{smallmatrix}
\text{Acetoacetic ester} \\
\text{ ( keto form)}
\end{smallmatrix}}{\mathop{C{{H}_{3}}-\overset{O}{\mathop{\overset{||}{\mathop{C}}\,}}\,-C{{H}_{2}}-\overset{O}{\mathop{\overset{||}{\mathop{C}}\,}}\,-O-{{C}_{2}}{{H}_{5}}}}\,$ $ \rightleftharpoons $
$\mathop {C{H_3} - \mathop {\mathop C\limits^| = }\limits^{OH} CH - \mathop {\mathop C\limits^{||} }\limits^O - O - {C_2}{H_5}}\limits_{{\rm{(enolic form)}}} $
(c)$\underset{\begin{smallmatrix}
\text{Acetoacetic ester} \\
\text{ ( keto form)}
\end{smallmatrix}}{\mathop{C{{H}_{3}}-\overset{O}{\mathop{\overset{||}{\mathop{C}}\,}}\,-C{{H}_{2}}-\overset{O}{\mathop{\overset{||}{\mathop{C}}\,}}\,-O-{{C}_{2}}{{H}_{5}}}}\,$ $ \rightleftharpoons $
$\mathop {C{H_3} - \mathop {\mathop C\limits^| = }\limits^{OH} CH - \mathop {\mathop C\limits^{||} }\limits^O - O - {C_2}{H_5}}\limits_{{\rm{(enolic form)}}} $
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Given below are two statements: One is labelled as Assertion $A$ and the other is labelled as Reason $R$
→Assertion $A$: Permanganate titrations are not performed in presence of hydrochloric acid.
Reason $R$ : Chlorine is formed as a consequence of oxidation of hydrochloric acid.
In the light of the above statements, choose the correct answer from the options given below.
Which of the following is Not predominantly acidic or amphoteric oxide?
$C _{7} H _{10} O$ reacts with $CH _{3} MgBr$ to give a compound $C _{8} H _{10} O$ which gives the test with iodoform, than fine out structure of $A$
→(major) '$A$' and '$B$' are respectively
→Which of the following on reaction with $N{H_3}$ gives urinary antiseptic compound
→Major end product of the following sequence of reaction is:
→$CH_3CH_2CH_2CONH_2$ $\mathop {\xrightarrow{{Ca{{(OH)}_2},C{l_2}}}}\limits_\Delta X\xrightarrow{{HN{O_2}}}Z$
Product $(A)$ of the reaction is
→Calculate the molality of $1\,litre$ solution of $ 93\%$ ${H_2}S{O_4}$(weight/volume). The density of the solution is $ 1.84\,g /ml$
→The major reason that phenol is a better Bronsted acid than cyclohexanol is that
Alkane can be iodinated in the presence of: