Consider the following reaction sequence

$\stackrel{{\mathrm{HNO}}_3+{\mathrm{H}}_2{\mathrm{SO}}_4}{\to } \stackrel{Sn+HCl}{\to }\underset{\left(6H\right)}{\overset{NaN{O}_2+HCl}{\to }}\stackrel{C{u}_2{\left(CN\right)}_2+HCN}{\to }\stackrel{H_{2}O}{\to }Product$

 

Product is

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

The products R and S are respectively

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

The end product 'Z' of the reaction sequence is

(1)
(2)
(3)
(4)	Both (1) and (2)

Compound P(C₆H₁₀) does not have any geometrical isomer. ON ozonolysis, two products R(C₃H₄O) and Q(C₃H₆O) are formed. R gives negative iodoform test while Q responds positively towards I₂/NaOH solution. S, another isomer of P is an unsyumetrical alkene and on ozonolysis produces T(C₆H10O₂) which also gives a yellow precipitate with I₂/NaOH solution and also gives positive test with Tollen's reagent. Which of the following does not represent any of the molecules amongst P,Q,R,S & T.

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

On halogenation, an alkane gives only one monohalogenated product. The alkane may be :
(1) 2-methyl butane

(2) 2, 2-dimethyl propane

(3) cyclopentane

(4) both (b) and (c)

Which of the following compounds can be best prepared by Wurtz-reaction ?

(1) Iso-butane

(2) n-butane

(3) n-pentane

(4) Iso-pentane

1. 3             
2. 4             
3. 1            
4. 2 

Product (P) in the above-mentioned reaction is:-  

1.		2.
3.		4.	Both (2) & (3)

Determine the double bond equivalent (degree of unsaturation) of the compound 'A':  

(A) $\stackrel{{\mathrm{H}}_2\left(1\mathrm{mole}\right)/\mathrm{Pt}}{\to }$

Given that compound 'A' undergoes hydrogenation with 1 mole of H₂ in the presence of Pt, yielding the depicted product, what is the double bond equivalent of 'A'?
 

1.  1                 

2. 2               

3. 3                 

4. 4