Question

Match the reactions given in Column I with the types of reactions given in Column II.

	Column I	Column II
(i)		(a) Nucleophilic aromatic substitution
(ii)	\[\begin{array}{cc} \ce{CH3 - CH = CH2 + HBr -> CH3 - CH - CH3}\\ \phantom{............................}|\phantom{}\\ \phantom{.............................}\ce{Br}\phantom{} \end{array}\]	(b) Electrophilic aromatic substitution
(iii)		(c) Saytzeff elimination
(iv)		(d) Electrophilic addition
(v)	\[\begin{array}{cc} \ce{CH3  CH2 CH CH3 ->[alc.KOH] CH3  CH = CH CH3}\\ \phantom{}|\phantom{..........................}\\ \phantom{}\ce{Br}\phantom{........................} \end{array}\]	(e) Nucleophilic substitution (SN1)

Answer 1

	Column I	Column II
(i)		(b) Electrophilic aromatic substitution
(ii)	\[\begin{array}{cc} \ce{CH3 - CH = CH2 + HBr -> CH3 - CH - CH3}\\ \phantom{............................}|\phantom{}\\ \phantom{.............................}\ce{Br}\phantom{} \end{array}\]	(d) Electrophilic addition
(iii)		(e) Nucleophilic substitution (SN1)
(iv)		(a) Nucleophilic aromatic substitution
(v)	\[\begin{array}{cc} \ce{CH3  CH2 CH CH3 ->[alc.KOH] CH3  CH = CH CH3}\\ \phantom{}|\phantom{..........................}\\ \phantom{}\ce{Br}\phantom{........................} \end{array}\]	(c) Saytzeff elimination

Explanation:

(i) In this reaction, an electrophile CF attacks on to the benzene ring and substitution takes place.

(ii) In this reaction, addition of \[\ce{HBr}\] takes place on to the doubly bonded carbons of propene in accordance with Markownikoff’s rule and electrophilic addition takes place.

(iii) In this reaction, the reactant is secondary halide. Here, halogen atom is substituted by hydroxyl ion. As it is secondary halide so it follows S_N1 mechanism.

(iv) In this reaction, halogen atom is directly bonded to aromatic ring. So, it is nucleophilic aromatic substitution as OH^– group has substituted halogen of given compound.

(v) It is an elimination reaction. It follows Saytzeff elimination rule.