Product; Product is:

MCPBA $\to$Metachloroperbenzoic acid

  $\underset{\left(2\right) {\mathrm{H}}_2{\mathrm{O}}_2{\mathrm{HO}}^{-}}{\overset{{\mathrm{BH}}_3; \mathrm{THF}}{\to }} \left(\mathrm{A}\right)$ ; Product of the reaction is:

An organic compound C₄H₆ on ozonolysis gives HCHO, CO₂, and CH₃CHO. The compound will be:-

$\underset{{\mathrm{H}}_2\mathrm{O}}{\overset{\mathrm{HCHO},{\mathrm{H}}^{+}}{\to }}$major product of this reaction is:

 $\underset{∆}{\overset{H^{+}}{\to }}\left(A\right) \underset{kMn{O}_4}{\overset{cold dil.}{\to }} \left(B\right) \stackrel{Cr{O}_3}{\to } \left(C\right)$

Product (C) of the reaction is:-

What is the major product expected from the following reaction?

$\stackrel{\mathrm{D}-\mathrm{Cl}}{\to }$ Product   

 $\underset{2.NaB{H}_4}{\overset{1. Hg\left(OAc\right) 2.H_{2}O}{\to }}$ Product; Product is:-

How many stereoisomeric tetrabromides will be formed in the following reaction?

(a) 2

(b) 3

(c) 4

(d) 6

$$\begin{gathered} {\mathrm{CH}}_3-\underset{|}{\mathrm{C}}\mathrm{H}-{\mathrm{CO}}_2\mathrm{K} \\ {\mathrm{CH}}_3-\mathrm{CH}-{\mathrm{CO}}_2\mathrm{K} \end{gathered}$$ $\stackrel{\mathrm{electrolysis}}{\to }$ (A) (Major)

Major product (A) of the above reaction:

(Only one enantiomer is taken)
Which of the following statement is correct about A and B?
(a) A  and B are mixture of diastereomers
(b) A and B are mixture of enantiomers
(c) A and B are optically active
(d) B is racemic mixture

$\stackrel{{\mathrm{HO}}^{⊝}}{\to }$ (X) $\stackrel{\mathrm{CH}2=\mathrm{CH}-\stackrel{\oplus }{\mathrm{P}}\mathrm{Ph}3{\mathrm{Br}}^{⊝}}{\to }$ (Y)

Product (Y) of the above reaction is:

Rank the following in the increasing order of rate of reaction with HBr.

(P) 

(Q)     

(R)    

Which of the following reactions generates the major product? Ignore stereoisomerism

$\left(\mathrm{A}\right)\stackrel{Na/N{H}_3\left(l\right)}{\to }\left(B\right)\stackrel{MCBPA}{\to }\left(C\right)$

Compound (C) in above sequence of reaction is:

$\underset{{\mathrm{R}}_2{\mathrm{O}}_2/\mathrm{hv}}{\overset{\mathrm{HBr}}{\to }}$(A);   

Major product (A) is:

(a) 

(b) 

(c) 

(d) 

In the given reaction 
 
The product P is

In the reaction given below, the product would be:-
$$\begin{gathered} {\mathrm{CH}}_3-\mathrm{CH}=\mathrm{CH}-{\mathrm{CH}}_3 \stackrel{{\mathrm{H}}_3{\mathrm{O}}^{+}}{\to } {\mathrm{CH}}_3-{\mathrm{CH}}_2-\underset{ | \\ \mathrm{OH}}{\mathrm{C}}\mathrm{H}-{\mathrm{CH}}_3 \end{gathered}$$

 $\underset{∆}{\overset{{\mathrm{H}}^{+}}{\to }}$Possible products $\stackrel{{\mathrm{Br}}_2/{\mathrm{CCl}}_4}{\to }$(y) products

The number of possible products for x and y is:

(a) 2,4

(b) 3,5

(c) 3,6

(d) 3,4

$\underset{\mathrm{Liq}. {\mathrm{NH}}_3}{\overset{\mathrm{Na}}{\to }} \left(A\right)\underset{\mathrm{Zn}}{\overset{{\mathrm{O}}_3}{\to }}\left(B\right)\stackrel{{\mathrm{Ph}}_3\mathrm{P}={\mathrm{CH}}_2\left(2\mathrm{mole}\right)}{\to }\left(C\right)$

Product (C) of the above reaction is :

(a) 1,3-Hexadiene
(b) 1,4-Pentadiene
(c) 1,3-Butadiene
(d) 1,3-Heptadiene

To achieve above conversion, the reagents used will be:

$\underset{0°\mathrm{C}}{\overset{{\mathrm{PCl}}_5}{\to }}\underset{70-80\%}{\left(\mathit{A}\right)}\underset{\left(2\right) {\mathrm{H}}^{\oplus }}{\overset{\left(1\right) 3{\mathrm{NaNH}}_2, \mathrm{mineral} \mathrm{oil}}{\to }}\underset{46\%}{\left(B\right)}$; Product (B) is:

Product (K) of the above reaction is:

; This conversion can be achieved by:

Which alkyne will give 3-ethylhexane on catalytic hydrogenation?

$\mathrm{Br}-{\left({\mathrm{CH}}_2\right)}_{12}-\mathrm{C}\equiv \mathrm{CH} \stackrel{{\mathrm{NaNH}}_2}{\to } \left(\mathrm{A}\right) \underset{\mathrm{Catalyst}}{\overset{\mathrm{Lindlar}}{\to }} \left(\mathrm{B}\right)$; Product (B) is :

Ph - C $\equiv \mathrm{CH}\underset{\mathrm{MeOH}}{\overset{{\mathrm{MeO}}^{-}}{\to }}$ Major product of the reaction is:

To carry out above conversion, (A) and (B) respectively, are :

H-C$\equiv$C-Ph   Product; Product obtained in this reaction is: