$\left(\mathrm{a}\right) 2-$methyl-$1, 3$-cyclohexanedione is more acidic than cyclohexanone - explain with reason

$\left(\mathrm{b}\right)$ Explain why $\mathrm{HCN}$ will add to the double bond in ${\mathrm{CH}}_2=\mathrm{CHCOOH}$ but not in $\mathrm{RCH}=\mathrm{CHR}$

$\left(\mathrm{a}\right)$ Convert

$\mathrm{PhCHO}$ into $\mathrm{PhCH}=\mathrm{CHCOPh}$

$\left(\mathrm{b}\right)$ Identify $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and $\mathrm{D}$ in the following reaction.

Complete the following reactions:

$\left(\mathrm{i}\right)$

Write the structures of $\mathrm{A}$ to $\mathrm{D}$ and give the $\mathrm{IUPAC}$ name of $\left(\mathrm{D}\right).$

$\left(\mathrm{ii}\right)$ 

Give reasons for the following :

$\left(\mathrm{a}\right)$ Ketones are less electrophilic than aldehydes.

$\left(\mathrm{b}\right)$ Aldehydes are reducing agents and ketones are not.

$\left(\mathrm{c}\right)$ $\mathrm{HBr}$ fails to give addition products with carbonyl compounds.

Identify $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}$ and $\mathrm{E}$ in the following sequence of reactions:

When the compound shown was heated in refluxing hydrochloric acid, a compound with the molecular formula $\mathrm{A}\left({\mathrm{C}}_5{\mathrm{H}}_6{\mathrm{O}}_3\right)$ was isolated. Identify this product. Along with this product, three other carbon-containing substances are formed. What are they ?

What happens when $\mathrm{A}\left({\mathrm{C}}_5{\mathrm{H}}_6{\mathrm{O}}_3\right)$ reacts with

$\left(\mathrm{A}\right) \mathrm{HCN}$ follwed by hydrolysis

$\left(\mathrm{B}\right)$ soda $\frac{\mathrm{lime}}{\mathrm{\Delta }}$

$\left(\mathrm{C}\right) \frac{{\mathrm{NH}}_2{\mathrm{NH}}_2}{\mathrm{glycol}}, {\mathrm{OH}}^{-}$

$\left(\mathrm{D}\right) \frac{\mathrm{P}}{{\mathrm{Br}}_2}$

$\left(\mathrm{E}\right) {\mathrm{N}}_3\mathrm{H}$

What are the product in each case?

$\left(\mathrm{A}\right) {\mathrm{NH}}_3$

$\left(\mathrm{B}\right) {\mathrm{LiAlH}}_4$

$\left(\mathrm{C}\right) \mathrm{EtOH}, {\mathrm{H}}^{+}$

$\left(\mathrm{D}\right) \frac{{\mathrm{NaBH}}_4}{\mathrm{EtOH}}$