Arrange in their increasing order of equilibrium constants for hydraion ?

(1) ${\mathrm{CH}}_3-\underset{\left(\mathrm{A}\right)}{\overset{\stackrel{\mathrm{O}}{\parallel }}{\mathrm{C}}}-{\mathrm{CH}}_3$,     ${\mathrm{CH}}_3-\underset{\left(\mathrm{B}\right)}{\overset{\stackrel{\mathrm{O}}{\parallel }}{\mathrm{C}}}-\mathrm{H}$,   ${\mathrm{CH}}_3-\underset{ \left(\mathrm{C}\right)}{\overset{\stackrel{\mathrm{O}}{\parallel }}{\mathrm{C}}}-{\mathrm{CH}}_2-\mathrm{Cl}, \mathrm{H}-\underset{\left(\mathrm{D}\right)}{\overset{\stackrel{\mathrm{O}}{\parallel }}{\mathrm{C}}}-\mathrm{H}, \mathrm{Cl}-{\mathrm{CH}}_{2\underset{\left(\mathrm{E}\right)}{}}-\stackrel{\stackrel{\mathrm{O}}{\parallel }}{\mathrm{C}}-\mathrm{H}$

(a) A < B < C < D < E

(2) A < C < B < E < D

(3) A < C < E < B < D

(4) C < A < B < E < D

$\mathrm{Ph}-{\mathrm{CH}}_2-\mathrm{CN}\underset{\underset{\left(3\right) {\mathrm{H}}_3{\mathrm{O}}^{\oplus }/∆}{\left(2\right){\mathrm{CH}}_3-\stackrel{\stackrel{\mathrm{O}}{\parallel }}{\mathrm{C}}-\mathrm{Cl}}}{\overset{\left(1\right)\mathrm{EtONA}}{\to }}$ (P); Product (p) of the reaction will be :

(a) $\mathrm{Ph}-{\mathrm{CH}}_2-\stackrel{\stackrel{\mathrm{O}}{\parallel }}{\mathrm{C}}-\mathrm{H}$

(b) $\mathrm{Ph}-{\mathrm{CH}}_2-\stackrel{\stackrel{\mathrm{O}}{\parallel }}{\mathrm{C}}-{\mathrm{CH}}_3$

(c) $\mathrm{Ph}-\mathrm{C}\underset{\stackrel{\text{l}}{{\mathrm{CH}}_3}}{\mathrm{H}}-\stackrel{\stackrel{\mathrm{O}}{\parallel }}{\mathrm{C}}-\mathrm{H}$

(d) $\mathrm{Ph}-\mathrm{C}\underset{\stackrel{\text{l}}{{\mathrm{CH}}_3}}{\mathrm{H}}-\stackrel{\stackrel{\mathrm{O}}{\parallel }}{\mathrm{C}}-{\mathrm{CH}}_3$

1. 

2. 

3. 

4. 

$\underset{\mathrm{Symmetrical} \mathrm{ketone}}{\left(\mathrm{A}\right)} \stackrel{{\mathrm{LiAlH}}_4}{\to }\left(\mathrm{B}\right)\underset{∆}{\overset{{\mathrm{H}}^{\oplus }}{\to }}$Diastereomers Reactent (A) is :

(a) ${\mathrm{CH}}_3-\stackrel{\underset{\parallel }{\mathrm{O}}}{\mathrm{C}}-{\mathrm{CH}}_3$

(b) ${\mathrm{CH}}_3-\stackrel{\underset{\parallel }{\mathrm{O}}}{\mathrm{C}}-{\mathrm{CH}}_2-{\mathrm{CH}}_3$
(c) ${\mathrm{CH}}_3-{\mathrm{CH}}_2-\stackrel{\underset{\parallel }{\mathrm{O}}}{\mathrm{C}}-{\mathrm{CH}}_2-{\mathrm{CH}}_3$

(d) ${\mathrm{CH}}_3-{\mathrm{CH}}_2-{\mathrm{CH}}_2-\mathrm{CH}=\mathrm{O}$

Molecular weight of compound (A) is:

(1) 58

(2) 120

(3) 60

(4) 182

Product (B) is:

(a) $\mathrm{Ph}-\mathrm{NH}-\stackrel{\underset{\parallel }{\mathrm{O}}}{\mathrm{C}}-{\mathrm{CO}}_2\mathrm{H}$

(b) $\mathrm{Ph}-\mathrm{NH}-\stackrel{\underset{\parallel }{\mathrm{O}}}{\mathrm{C}}-{\mathrm{CH}}_2-{\mathrm{CO}}_2\mathrm{H}$

 (c) 

(d) 

This conversion can be achieved by :

(1) Dehydration, Hydrolysis

(2) Retro aldol and further condensation 

(3) Perkin condensation & Clemmensen reduction

(4) Clemmensen and Perkin condensation 

\(\left(CH\right)_{3} - \overset{\overset{O}{\parallel}}{C} - \left(CH\right)_{2} - \left(CH\right)_{2} - \left(CH\right)_{2} - \left(CH\right)_{2} - \overset{\overset{O}{\parallel}}{C} - H \overset{\left(HO\right)^{-} / \Delta}{\rightarrow} (\underset{73 \%}{A})\)
Product (A) is:

$\mathrm{Ph}-\mathrm{CH}=\mathrm{CHCHO} + {\mathrm{CH}}_3\mathrm{CH}=\mathrm{CHCHO}\underset{\mathrm{ErOH}, ∆}{\overset{\mathrm{base}}{\to }}$ (A) 87%; product of this reaction is :

1. $\mathrm{Ph}-{\left(\mathrm{CH}=\mathrm{CH}\right)}_2-\mathrm{CHO}$

2. $\mathrm{Ph}-{\left(\mathrm{CH}=\mathrm{CH}\right)}_3\mathrm{CHO}$

3. $\mathrm{Ph}-{\left(\mathrm{CH}=\mathrm{CH}\right)}_4\mathrm{CHO}$

4. $\mathrm{Ph}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}-{\mathrm{CH}}_3$

CH₃CHO$\underset{5°\mathrm{C}}{\overset{10\%\mathrm{NaOH}}{\to }}\underset{∆}{\to }\underset{\mathrm{Ni}}{\overset{{\mathrm{H}}_2}{\to }}$$$(A) ; Product (A) of the reaction is :

(1) Propanol

(2) ethanol

(3) butanol

(4) pentanol