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Question of 147

Q.Write IUPAC names of the following:

(i) CH3−CH=CH−CH(CH3)BrCH_3-CH=CH-CH(CH_3)Br
(ii) (CH3)3C−CH2Br(CH_3)_3C-CH_2Br
(iii) CH3CH2CH2−CH(−C(CH3)3)−CH(I)−CH2CH3CH_3CH_2CH_2-CH(-C(CH_3)_3)-CH(I)-CH_2CH_3
(iv) (CH3)2CBrCH2CH3(CH_3)_2CBrCH_2CH_3
(v) (CH3)3CCl(CH_3)_3CCl OR Explain with reasons:
(i) Sulphuric acid is not used in the reaction of alcohol and KI.
(ii) Haloalkanes form alkyl cyanide as chief product on reaction with KCN, while isocyanide as chief product on reaction with AgCN.
(iii) Although chlorine is an electron withdrawing group, even then it is ortho- and para-directing in aromatic electrophilic substitution reaction.
Uttar Pradesh UpmspUP Board (UPMSP) Intermediate 2023Subjective· 5mImportance★★★★★
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IUPAC names of the five halides (main alternative of this OR question): (i) 4-bromopent-2-ene, (ii) 1-bromo-2,2-dimethylpropane, (iii) 4-tert-butyl-3-iodoheptane, (iv) 2-bromo-2-methylbutane, (v) 2-chloro-2-methylpropane.

(i) CH3−CH=CH−CH(CH3)BrCH_3-CH=CH-CH(CH_3)Br: The longest chain including the double bond is 5 carbons — CH3−CH=CH−CH(Br)−CH3CH_3-CH=CH-CH(Br)-CH_3. Numbering to give the double bond the lower locant: double bond at C-2, Br at C-4 ⇒\Rightarrow 4-bromopent-2-ene.

(ii) (CH3)3C−CH2Br(CH_3)_3C-CH_2Br: Longest chain 3 C (propane); central C carries two methyls and C-1 bears Br ⇒\Rightarrow 1-bromo-2,2-dimethylpropane.

(iii) CH3CH2CH2−CH[C(CH3)3]−CH(I)−CH2CH3CH_3CH_2CH_2-CH[C(CH_3)_3]-CH(I)-CH_2CH_3: Longest chain = 7 C (heptane) with a C(CH3)3C(CH_3)_3 (tert-butyl) branch and iodo. Numbering to give the substituent set the lowest locants: iodo at C-3, tert-butyl at C-4 ⇒\Rightarrow 4-tert-butyl-3-iodoheptane (IUPAC substitutive: 3-iodo-4-(1,1-dimethylethyl)heptane).

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