Question

Name the following halides according to the IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, and tertiary), vinyl or aryl halides:
(i) $(CH_{ 3 })_{ 2 }CHCH(Cl)CH_{ 3 }$
(ii) $CH_{ 3 }CH_{ 2 }CH(CH_{ 3 })CH(C_{ 2 }H_{ 5 })Cl$
(iii) $CH_{ 3 }CH_{ 2 }C(CH_{ 3 })_{ 2 }CH_{ 2 }I$
(iv) $(CH_{ 3 })_{ 3 }CCH_{ 2 }CH(Br)C_{ 6 }H_{ 5 }$
(v) $CH_{ 3 }CH(CH_{ 3 })CH(Br)CH_{ 3 }$
(vi) $CH_{ 3 }C(C_{ 2 }H_{ 5 })_{ 2 }CH_{ 2 }Br$
(vii) $CH_{ 3 }C(Cl)(C_{ 2 }H_{ 5 })CH_{ 2 }CH_{ 3 }$
(viii) $CH_{ 3 }CH=C(Cl)CH_{ 2 }CH(CH_{ 3 })_{ 2 }$
(ix) $CH_{ 3 }CH=CHC(Br)(CH_{ 3 })_{ 2 }$
(x) $p-ClC_{ 6 }H_{ 4 }CH_{ 2 }(CH)(CH_{ 3 })_{ 2 }$
(xi) $m-ClCH_{ 2 }C_{ 6 }H_{ 4 }CH_{ 2 }C(CH_{ 3 })_{ 3 }$
(xii) $o-Br-C_{ 6 }H_{ 4 }CH(CH_{ 3 })CH_{ 2 }CH_{ 3 }$

Answer 1

Hint: As mentioned in the question we just need to make sure they are named properly with the correct suffix and prefix. Then according to the group present in their structure classify them as alkyl, allyl, benzyl (primary, secondary, and tertiary), vinyl, or aryl halides.

Complete step by step solution: Let’s name these compounds one by one -
(i) $(CH_{ 3 })_{ 2 }CHCH(Cl)CH_{ 3 }$ = 2-Chloro-3-methylbutane (secondary alkyl halide)

(ii) $CH_{ 3 }CH_{ 2 }CH(CH_{ 3 })CH(C_{ 2 }H_{ 5 })Cl$ = 3-Chloro-4-methylhexane (secondary alkyl halide)

(iii) $CH_{ 3 }CH_{ 2 }C(CH_{ 3 })_{ 2 }CH_{ 2 }I$ = 1-Iodo-2,2-dimethylbutane (primary alkyl halide)

(iv) $(CH_{ 3 })_{ 3 }CCH_{ 2 }CH(Br)C_{ 6 }H_{ 5 }$ = 1-Bromo-3,3-dimethyl-1-phenylbutane (secondary benzyl halide)

(v) $CH_{ 3 }CH(CH_{ 3 })CH(Br)CH_{ 3 }$ = 2-Bromo-3-methylbutane (secondary alkyl halide)

(vi) $CH_{ 3 }C(C_{ 2 }H_{ 5 })_{ 2 }CH_{ 2 }Br$ = 1-Bromo-2-ethyl-2-methylbutane (primary alkyl halide)

(vii) $CH_{ 3 }C(Cl)(C_{ 2 }H_{ 5 })CH_{ 2 }CH_{ 3 }$ = 3-Chloro-3-methylpentane (tertiary alkyl halide)

(viii) $CH_{ 3 }CH=C(Cl)CH_{ 2 }CH(CH_{ 3 })_{ 2 }$ = 3-Chloro-5-methylhex-2-ene (vinyl halide)

(ix) $CH_{ 3 }CH=CHC(Br)(CH_{ 3 })_{ 2 }$ = 4-Bromo-4-methylpent-2-ene (allyl halide)

(x) $p-ClC_{ 6 }H_{ 4 }CH_{ 2 }(CH)(CH_{ 3 })_{ 2 }$ = 1-Chloro-4-(2-methylpropyl) benzene (aryl halide)

(xi) $m-ClCH_{ 2 }C_{ 6 }H_{ 4 }CH_{ 2 }C(CH_{ 3 })_{ 3 }$ = 1-Chloromethyl-3-(2,2-dimethylpropyl) benzene (primary benzylic halide)

(xii) $o-Br-C_{ 6 }H_{ 4 }CH(CH_{ 3 })CH_{ 2 }CH_{ 3 }$ = 1-Bromo-2-(1-methylpropyl) benzene (aryl halide)

Therefore, we named all the compounds given.

Note: You should not confuse yourself with the structure of the allyl and vinyl group present in these structures.
In chemistry, vinyl or ethenyl (abbreviated as Vi) is the functional group with the formula −CH=$CH_{ 2 }$. It is the ethylene (IUPAC ethene) molecule ($H_{ 2 }$C=C$H_{ 2 }$) with one fewer hydrogen atom.
An allyl group is a substituent with the structural formula $H_{ 2 }$C=CH−C$H_{ 2 }$R, where R is the rest of the molecule.