10.16 Write the equations for the preparation of 1-iodobutane from (i) 1-Butanol (ii) 1-Chlorobutane (iii) But-1-ene. 

(a) 1-butanol is treated with KI in the presence of H₃PO₄ where the OH is being replaced by the iodine and gives H₂O and KH₂PO₄ as the by-product with 1-iodobutane as the final

(b) The conversion of 1-chlorobutane to 1-iodobutane simply takes place by treating the reactant with KI in the presence of acetone. The iodine from KI normally replaces the chlorine from the reactant and gives 1-iodobutane as the final product with KCl as the by product.

(c) The conversion of but-1-ene to 1-iodobutane takes place according to anti-markovnikoff (positive charged ion H⁺ goes to the carbon which has less number of hydrogens and negative part Br-

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(a) 1-butanol is treated with KI in the presence of H₃PO₄ where the OH is being replaced by the iodine and gives H₂O and KH₂PO₄ as the by-product with 1-iodobutane as the final

(b) The conversion of 1-chlorobutane to 1-iodobutane simply takes place by treating the reactant with KI in the presence of acetone. The iodine from KI normally replaces the chlorine from the reactant and gives 1-iodobutane as the final product with KCl as the by product.

(c) The conversion of but-1-ene to 1-iodobutane takes place according to anti-markovnikoff (positive charged ion H⁺ goes to the carbon which has less number of hydrogens and negative part Br- goes to the carbon which has more number of hydrogens.) i.e. the attack of HBr on but-1-ene in the presence of peroxide gives 1-bromobutane as the intermediate which on treatment with NaI+Acetone gives 1- iodobutane as the final product and NaBr as the by-product.

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<p><strong> (a)</strong> 1-butanol is treated with KI in the presence of H₃PO₄ where the OH is being replaced by the iodine and gives H₂O and KH₂PO₄ as the by-product with 1-iodobutane as the final</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1690439046phpqgcVNV_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1690439046phpqgcVNV.jpeg" alt="" width="233" height="94"></picture></div></div><p><strong> (b) </strong>The conversion of 1-chlorobutane to 1-iodobutane simply takes place by treating the reactant with KI in the presence of acetone. The iodine from KI normally replaces the chlorine from the reactant and gives 1-iodobutane as the final product with KCl as the by product.</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1690439081phpa7YFLC_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1690439081phpa7YFLC.jpeg" alt="" width="315" height="63"></picture></div></div><p><strong> (c)</strong> The conversion of but-1-ene to 1-iodobutane takes place according to anti-markovnikoff (positive charged ion H⁺ goes to the carbon which has less number of hydrogens and negative part Br- goes to the carbon which has more number of hydrogens.) i.e. the attack of HBr on but-1-ene in the presence of peroxide gives 1-bromobutane as the intermediate which on treatment with NaI+Acetone gives 1- iodobutane as the final product and NaBr as the by-product.</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1690439112phpukNuPI_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1690439112phpukNuPI.jpeg" alt="" width="" height=""></picture></div></div>

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