Predict the correct intermediate and product in the following reaction: 



1. \(A :\) \(H_{3} C - \underset{OH}{\underset{\left|\right.}{C}} = CH_{2}\) \(\) \(;\) \(B :\) \(\) \(H_{3} C - \underset{HSO_{4}}{\underset{\left|\right.}{C}} = CH_{2}\)

2. \(A :\) \(H_{3} C - \underset{O}{\underset{\left|\right. \left|\right.}{C}} - CH_{3}\) \(\) \(;\) \(B :\) \(\) \(H_{3} C - C \equiv CH\)
3. \(A :\) \(H_{3} C - \overset{OH}{\overset{\left|\right.}{C}} = CH_{2}\) \(\) \(;\) \(B :\) \(\) \(H_{3} C - \overset{O}{\overset{\left|\right. \left|\right.}{C}} - CH_{3}\)

4. \(A :\) \(H_{3} C - \underset{HS O_{4}}{\underset{\left|\right.}{C}} = CH_{2}\) \(\) \(;\) \(B :\) \(\) \(H_{3} C - \underset{O}{\underset{\left|\right. \left|\right.}{C}} - CH_{3}\)${\mathrm{}}_{}$

The compound that will react most readily with gaseous bromine has the formula:

Hydrocarbon (A) reacts with bromine by substitution reaction to form an alkyl bromide B.
B undergoes the Wurtz reaction to give a gaseous hydrocarbon containing less than four carbon atoms.

The formula of (A) is:

1. $\mathrm{CH}\equiv \mathrm{CH}$
2. ${\mathrm{CH}}_2={\mathrm{CH}}_2$ 
3. ${\mathrm{CH}}_3-{\mathrm{CH}}_3$
4. ${\mathrm{CH}}_4$

The compound ${\mathrm{C}}_7{\mathrm{H}}_8$ undergoes the following reactions

${\mathrm{C}}_7{\mathrm{H}}_8\stackrel{3{\mathrm{Cl}}_2/\mathrm{hv}}{\to }\mathrm{A}\stackrel{{\mathrm{Br}}_2/\mathrm{Fe}}{\to }\mathrm{B}\stackrel{\mathrm{Zn}/\mathrm{HCl}}{\to }$ C

The product 'C' is-

1. m–Bromotoluene

2. o–Bromotoluene

3. 3–Bromo–2,4,6–trichlorotoluene

4. p–Bromotoluene

 
P, Q and R in the above-mentioned sequence of reactions are respectively:

1.
2.
3.
4.

The molecule among the following that has the hybridization ${\mathrm{sp}}^2, {\mathrm{sp}}^2, \mathrm{sp},\; and \mathrm{sp}$ from left to right atoms is:

1. $\mathrm{HC}\equiv \mathrm{C}-\mathrm{C}\equiv \mathrm{H}$

2. ${\mathrm{CH}}_2=\mathrm{CH}-\mathrm{C}\equiv \mathrm{CH}$

3. ${\mathrm{CH}}_2=\mathrm{CH}-\mathrm{CH}={\mathrm{CH}}_2$ 

4. ${\mathrm{CH}}_2-\mathrm{CH}=\mathrm{CH}-{\mathrm{CH}}_3$

The correct statement regarding the comparison of staggered and eclipsed conformations of ethane is:

X and Y in the above-mentioned reaction are respectively:
1. X = 2–Butyne; Y = 3–Hexyne
2. X = 2-Butyne; Y = 2-Hexyne
3. X = 1-Butyne; Y = 2-Hexyne
4. X = 1-Butyne; Y = 3–Hexyne

HCl with an alkene X reacts in accordance with Markovnikov’s rule to give 1-Chloro-1-methylcyclohexane. The structure of alkene (X) is:

1.		2.
3.	(1) and (2)	4.

The reaction of C₆H₅CH =CHCH₃ with HBr produces:

1. \(C_{6} H_{5} \underset{Br}{\underset{\left|\right.}{CH}} CH_{2} CH_{3} \)

2. \(C_{6} H_{5} CH_{2} \underset{Br}{\underset{\left|\right.}{CH}} CH_{3} \) 

3. \(C_{6} H_{5} CH_{2} CH_{2} CH_{2} Br \)

4.