Compound (A) C₆H₁₂O is optically active. Compound (A) give negative Tollens test and positive test with 2-4 dinitro phenyl hydrazine. Identify A.

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

(b) $$\begin{gathered} O \\ \left|\right| \\ C{H}_3—C—CH—C{H}_3 \\ | \\ C{H}_{2}C{H}_3 \end{gathered}$$

(c) $$\begin{gathered} O \\ \left|\right| \\ H —C—C{H}_2—CH—C{H}_3 \\ | \\ C{H}_2—C{H}_3 \end{gathered}$$

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

1. 

2. 

3. 
4. 

$\mathrm{Ph}-{\mathrm{CH}}_3\stackrel{{\mathrm{CrO}}_2{\mathrm{Cl}}_2}{\to }\left(\mathrm{A}\right)\stackrel{\mathrm{conc}.\mathrm{KOH}}{\to }\mathrm{Ph}-{\mathrm{CH}}_2\mathrm{OH} + \left(\mathrm{B}\right)$

Product (B) of above the raction is :

(1) $\mathrm{Ph}-{\mathrm{CO}}_2\mathrm{H}$

(2) $\mathrm{Ph}-{\mathrm{CO}}_2^{-}$

(3) $\mathrm{Ph}-\mathrm{CHO}$

(4) $\mathrm{Ph}-{\mathrm{CH}}_3$

Product (C) of the reaction is :

Product (A) and (C) is :

1. 

2. 

3. 

4. 

$\mathrm{Ph}-\stackrel{\stackrel{\mathrm{OH}}{\mathrm{l}}}{\mathrm{CH}}-{\mathrm{CH}}_3\stackrel{\mathrm{PCC}}{\to }\left(\mathrm{A}\right)\stackrel{{\mathrm{NH}}_2-\mathrm{NH}-\stackrel{\stackrel{\mathrm{O}}{\parallel }}{\mathrm{C}}-{\mathrm{NH}}_2}{\to }\left(\mathrm{B}\right)$

Product (B) is :

1.  $\mathrm{Ph}-\underset{\underset{{\mathrm{CH}}_3}{\mathrm{l}}}{\mathrm{C}}=\mathrm{N}-\mathrm{NH}-\stackrel{\stackrel{\mathrm{O}}{\mathrm{ll}}}{\mathrm{C}}-{\mathrm{NH}}_2$

2. $\mathrm{Ph}-\stackrel{\stackrel{{\mathrm{CH}}_3}{\mathrm{l}}}{\mathrm{C}}=\mathrm{N}-\stackrel{\stackrel{\mathrm{O}}{\mathrm{ll}}}{\mathrm{C}}-\mathrm{NH}-{\mathrm{NH}}_2$

3. $\mathrm{Ph}-\mathrm{CH}=\mathrm{N}-\underset{\underset{{\mathrm{CH}}_3}{\mathrm{l}}}{\mathrm{N}}-\stackrel{\stackrel{\mathrm{O}}{\mathrm{ll}}}{\mathrm{C}}-{\mathrm{NH}}_2$

4. $\mathrm{Ph}-\stackrel{\stackrel{{\mathrm{CH}}_3}{\mathrm{l}}}{\mathrm{CH}}-\mathrm{NH}-\stackrel{\stackrel{\mathrm{O}}{\mathrm{ll}}}{\mathrm{C}}-\mathrm{NH}-{\mathrm{NH}}_2$

(A) and (B) is differentiated by :

1. NaH

2. 2-4 DNP

3. Tollen's reagent 

4. ${\mathrm{NaHSO}}_3$

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

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