What are the IUPAC names of the insecticide DDT and benzene hexachloride? Why is their use banned in India and other countries?
The IUPAC name of DDT is 2,2-bis (4-chlorophenyl)-1,1,1-trichloroethane and that of benzene hexachloride is 1,2,3,4,5,6-hexachlorocyclohexane.
They are banned in india because they are non-biodegradable. Instead, they get deposited and stored in fatty tissues. If this ingestion continues at a steady rate, DDT builds up within the animal over time. This will affects the reproductive system of animals.
If animals including humans are exposed to high levels of benzene hexachloride then it may cause acute poisoning. Apart from that this BHC may affect liver functioning in humans.
Elimination reactions (especially $\beta$-elimination) are as common as the nucleophilic substitution reaction in case of alkyl halides. Specify the reagents used in both cases.
Elimination reactions are as common as the nucleophilic substitution reaction in case of alkyl halides as two reactions occur simultaneously. Generally, at lower temperature and by using weaker base nucleophilic substitution reaction occur while at higher temperature and by using a stronger base elimination reactions (especially $\beta$ - elimination) take place. e.g., If ethyl bromide is treated with aq. KOH , at low temperature it gives ethanol while if it is treated with alc, KOH at high temperature then it gives ethene.
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br} \xrightarrow[373 \mathrm{~K}]{\text { Aq. } \mathrm{KOH}} \underset{\substack{\text { Ethanol }}}{\mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}}$ (Nucleophilic substitution reaction)
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br} \xrightarrow[473-523 \mathrm{~K}]{\mathrm{Alc} \mathrm{KOH}} \underset{\text { Ethene }}{\mathrm{CH}_2=\mathrm{CH}_2}$ (Elimination reaction)
How will you obtain monobromobenzene from aniline?
When aniline, dissolved or suspended in cold aqueous mineral acid, is treated with sodium nitrite, a diazonium salt is formed. This diazo salt on treatment with cuprous bromide gives monobromobenzene.
This reaction is named as Sandmeyer's reaction. If benzene diazonium chloride is treated with copper in HBr then the product formed is bromobenzene. This reaction is known as Gattermann reaction.
Aryl halides are extremely less reactive towards nucleophilic substitution. Predict and explain the order of reactivity of the following compounds towards nucleophilic substitution.
Aryl halides are less reactive towards nucleophilic substitution reaction. Presence of electron withdrawing group at ortho and para position increases the stability of intermediates and hence increases the reactivity of aryl halides towards nucleophilic substitution reaction.
Now, more the number of EWG at ortho and para position, higher will be the reactivity of aryl halide. Compound (III) has three EWG so, it is most reactive and compound (I) has only one EWG so, it is least reactive. So, the order of reactivity is (I)<(II)<(III)
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.
