$C{H}_2—C{H}_2—C{O}_{2}H \underset{BaC{O}_3}{\overset{∆}{\to }} A\stackrel{N{H}_2—N{H}_2}{\to } B\underset{KOH}{\overset{heat }{\to }}\left(C\right),$
$|$
$C{H}_2—C{H}_2—C{O}_{2}H$

Product (C) obtained is :

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

 (b)

  (c) 

 (d)

1. 

2. 

3. 

4. 

${\mathrm{CH}}_3-\stackrel{\stackrel{\mathrm{O}}{\parallel }}{\mathrm{C}}-\mathrm{H}\stackrel{\mathrm{HCN}}{\to }\left(\mathrm{A}\right)\stackrel{{\mathrm{H}}_3{\mathrm{O}}^{\oplus }}{\to }\left(\mathrm{B}\right)\underset{△}{\to }\left(\mathrm{C}\right)\stackrel{{\mathrm{LiAlH}}_4}{\to }\left(\mathrm{D}\right)\stackrel{{\mathrm{HIO}}_4}{\to }\mathrm{HCHO} + \left(\mathrm{E}\right)$

Compound (C) can show geometrical isomerism, Product (E) of the reaction will be :

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

(2) ${\mathrm{CH}}_3-{\mathrm{CH}}_2-\stackrel{\stackrel{\mathrm{O}}{\parallel }}{\mathrm{C}}-\mathrm{H}$

(3) ${\mathrm{CH}}_3-\mathrm{CHO}$

(4) HCHO

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