Identify C in the following sequence of reactions :

(1)
(2)
(3)
(4)

Product (A) obtained is :

(1)
(2)
(3)
(4)

A set of reagents (1 to 8) are successively reacted with the following compound

1. NaHCO₃

2. 2, 4, DNP

3. Na metal

4. AgNO₃ +OH

5. Fehling's solution

6. Cu₂CI₂ + NH₄OH

7. Br₂/H₂O

8. NaNO₂ + HCI

The reagents which give the positive test with the given compound are :

Which of the following alcohol will show positive iodoform test ?

The compound A gives following reactions.

Its structure can be

(1)
(2)
(3)
(4)

In compound A (C₃₀H₆₀O) following tests are observed negatively, A can be :

 ?

The major product of the reaction is:-

Which one of the following is the intermediate in the preparation of a ketone by hydration of an alkyne in the presence of sulfuric acid and mercury (II) sulphate?

 $\stackrel{{\mathrm{RCO}}_2\mathrm{H}}{\to }$ product; Product of this reaction is:

Above conversion can be achieved by:

Product (A) is:

Reagent to carry out above conversion,P,Q,R respectively are:

Above compounds can be differentiated by following reagent:

Product (D) will be :

Product (B) in this reaction is :

$\mathrm{Ph}-\stackrel{\stackrel{\mathrm{O}}{\mathrm{ll}}}{\mathrm{C}}-{\mathrm{CH}}_3\underset{\mathrm{HCl}}{\overset{{\mathrm{NaNO}}_2}{\to }}\left(\mathrm{A}\right)\underset{\mathrm{heat}}{\overset{{\mathrm{AC}}_2\mathrm{O}}{\to }}\left(\mathrm{B}\right)\stackrel{{\mathrm{H}}_3{\mathrm{O}}^{+}}{\to }\left(\mathrm{c}\right)$

Product (C) of the above reaction is :

Product (B) of the reaction is :

Correct order of reactivity of following compounds towards Grignard reagent ?

$$\begin{gathered} {\mathrm{CH}}_3-\stackrel{\stackrel{\mathrm{O}}{\mathrm{ll}}}{\mathrm{C}}-\mathrm{H} \\ \left(\mathrm{I}\right) \end{gathered}$$          $$\begin{gathered} \mathrm{H}-\stackrel{\stackrel{\mathrm{O}}{\mathrm{ll}}}{\mathrm{C}}-\mathrm{H} \\ \left(\mathrm{II}\right) \end{gathered}$$

What reagent and/or reaction conditions would you choose to bring about the following conversion?

${\mathrm{CH}}_3-\mathrm{c}\equiv \mathrm{CH}\underset{\mathrm{dil}. {\mathrm{H}}_2{\mathrm{SO}}_4}{\overset{{\mathrm{HgSO}}_4}{\to }}\left(\mathrm{A}\right)$

${\mathrm{CH}}_3-\mathrm{c}\equiv \mathrm{CH}\underset{\left(2\right) {\mathrm{H}}_2{\mathrm{SO}}_4/{\mathrm{HO}}^{-}}{\overset{\left(1\right){\mathrm{BH}}_3\mathrm{THF}}{\to }}\left(\mathrm{B}\right)$

Product (A) and (B) is differentiated by:

End Product (C) in above reaction is :

(b)

(c)

Compound (X) C₄H₈O, which reacts with 2, 4-DNP derivative and gives negative haloform test is.

Product (B) of the reaction is :

$\mathrm{Ph}-\mathrm{CH}=\mathrm{CH}-\stackrel{\stackrel{\mathrm{O}}{\left|\right|}}{\mathrm{C}}-{\mathrm{CH}}_3\to \mathrm{Ph}-\mathrm{CH}=\mathrm{CH}-{\mathrm{CO}}_2\mathrm{H}$
Above conversion can be achieved by :

(a) ${\mathrm{KMnO}}_4, ∆ \mathrm{followed} \mathrm{by} {\mathrm{H}}^{+}$

(b) ${\mathrm{I}}_2/\mathrm{NaOH} \mathrm{followed} \mathrm{by} {\mathrm{H}}^{+}$

(c) H₂/Pt

(d) LiAlH₄

Product (A) of the reaction is :

$R-\stackrel{\stackrel{\mathrm{O}}{\mathrm{ll}}}{\mathrm{C}}-\mathrm{H}\stackrel{\mathrm{R}-{\mathrm{NH}}_2}{\to }R-\mathrm{CH}=\mathrm{N}-R$. This reaction gives best yield at:

An aromatic compound A of the molecular formula C₈H₁₀O on reaction with iodine and dilute NaOH gives a yellow precipitate. The structure of the compound is expected to be-

In the following sequence :

${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Cl}\stackrel{\mathrm{NaCN}}{\to }\left(\mathrm{i}\right)\stackrel{\mathrm{Ni}/{\mathrm{H}}_2}{\to }\left(\mathrm{ii}\right)\stackrel{\mathrm{acetic} \mathrm{anhydride}}{\to }\left(\mathrm{iii}\right), \mathrm{product} \left(\mathrm{iii}\right) \mathrm{is}$

Carbonyl compounds can generally be converted to hydrocarbons by :

Which statement about the aldol condensation is correct ?