Aromatic Aldehydes Ketones and Acids - Result Question 25
####29. The compound $S$ is
(a)
(c)
(b)
(d)
Passage
In the following reactions sequence, the compound $J$ is an intermediate.
$J\left(C _9 H _8 O _2\right)$ gives effervescence on treatment with $NaHCO _3$ and positive Baeyer’s test.
(2012)
Show Answer
Answer:
Correct Answer: 29. (d)
Solution:
- Given (In connection with Q. 17)
$$ P \xrightarrow[\substack{\text { 4. } CHCl _3 / KOH, \Delta \ \text { 5. } H _2 / Pd-C}]{\substack{\text { 1. } H _2 / Pd-C \ \text { 2. } NH _3 / \Delta \ \text { 3. } Br _2 / NaOH}} S $$
So,
(P)
$(S)$
Passage
Sol for (Q. Nos. 30 to 31) The first step of reaction is Perkin’s condensation.
$$ \underbrace{} _I+\left(CH _3 CO\right) _2 O \xrightarrow{C _6 H _5-CH=\underset{J}{CH _3 COONa}} CH-COOH $$
$J$ being a carboxylic acid gives effervescence with $NaHCO _3$. Also, $J$ has olefinic bond, it will decolourise Baeyer’s reagent. In the second step, $J$ on treatment with $H _2 / Pd / C$ undergo hydrogenation at olefinic bond only as :
$$ J+H _2 / Pd \longrightarrow C _6 H _5-CH _2-CH _2-COOH $$
The hydrogenated acid, on treatment with $SOCl _2$ gives acid chloride.
In the final step, acid chloride formed above undergo intramolecular Friedel-Craft acylation as:
