tert-Butylbromide reacts with aq. NaOH by $\mathrm{S}_{\mathrm{N}} 1$ mechanism while n -butylbromide reacts by $\mathrm{S}_{\mathrm{N}} 1$ mechanism. Why?
Tert. butyl bromide reacts with aq. NaOH as follows
tert. butyl bromide when treated with aq. NaOH , it forms tert. corbocation which is more stable intermediate. This intermediate is further attacked by ${ }^{-} \mathrm{OH}$ ion. As tert. carbocation is highly stable so tert butylbromide follow $\mathrm{S}_{\mathrm{N}} 1$ mechanism. In case of $n$-nutylbromide, primary carbocation is formed which is least stable so, it does not follow $\mathrm{S}_{\mathrm{N}} 1$ mechanism. Here, stearic hindrance is very less so, it follow $\mathrm{S}_{\mathrm{N}} 2$ mechanism. In $\mathrm{S}_{\mathrm{N}} 2$ mechanism, ${ }^{-} \mathrm{OH}$ will attack from backside and a transition state is formed. The leaving group is then pushed off the eopposite side and the product is formed.
Predict the major product formed when HCl is added to isobutylene, Explain the mechanism involved.
Reaction between the isobutylene added to HCl
Electrophilic addition reaction takes place in accordance with Markownikoff's rule.
We know that $3^{\circ}$ carbocation is more stable than $1^{\circ}$ carbocation because in further step $3^{\circ}$ carbocation is further attacked by $\mathrm{Cl}^{-}$ion.
Discuss the nature of $\mathrm{C}-\mathrm{X}$ bond in the haloarenes.
In haloarenes, carbon of benzene is bonded to halogen. Electronegativity of halogen is more than that of $s p^2$ hybridised carbon of benzene ring. So, $\mathrm{C}-\mathrm{X}$ bond is a polar bond. Apart from this, lone pair of electrons of halogen atom are involved in resonance with benzene ring. So, this $\mathrm{C}-\mathrm{X}$ bond has acquire partial bond character.
This $\mathrm{C}-X$ bond of haloarenes is less polar than $\mathrm{C}-X$ bond of haloalkanes. This is supported by the fact that dipole moment of chlorobenzene ( $\mu=1.69 \mathrm{D}$ ) is little lower than that of $\mathrm{CH}_3 \mathrm{Cl}(\mu=1.83 \mathrm{D})$
How can you obtain iodoethane from ethanol when no other iodine containing reagent except NaI is available in the laboratory?
Ethanol is treated with red phosphorous and bromine mixture and the product formed will be bromoethane. The bromoethane so formed is then treated with Nal to give iodoethane.
$$\begin{gathered} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH} \xrightarrow{\mathrm{Red} \mathrm{P} / \mathrm{Br}_2} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br} \\ \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br} \xrightarrow{\mathrm{Nal}} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{I}+\mathrm{NaBr} \end{gathered}$$
This reaction is known as Finkelstein reaction.
Cyanide ion acts as an ambident nucleophile. From which end it acts as a stronger nucleophile in aqueous medium? Give reason for your answer.
Cyanide ion $(\bar{:}C \equiv \mathrm{~N}$ ) is an ambident nucleophile because it can react either through carbon or through nitrogen. Since, $\mathrm{C}-\mathrm{C}$ bond is stronger than $\mathrm{C}-\mathrm{N}$ bond so, cyanide ion will mainly attack through carbon to form alkyl cyanide.
