Hydrocarbons - Result Question 22

####24. In the following reactions,

[2011]

(i)

(ii)

$ (\begin{matrix} \text{ Major } \\ \text{ product } \end{matrix} )^{+} \underset{(\begin{matrix} \text{ Minor } \\ \text{ product } \end{matrix} )}{D} $

the major products $(A)$ and $(C)$ are respectively:

(a)C=C(C)CCCCC(C)CC

(b)CCC(C)(Br)C(C)(C)C(C)C(C)C

(c)CC=C(C)CCC(C)C(C)Br

(d)C=C(C)CCCCC(C)(C)Br

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Solution:

(b) (i)CC(C)C(C)OC=CC(C)C(C)CC(C)C(C)C

(A)

(B)

In this case, dehydration is governed by Saytzeff’s rule, according to which hydrogen is preferentially eliminated from the carbon atom with fewer number of hydrogen atoms i.e., poor becomes poorer. Thus, 2-methyl but-2-ene is the major product.

(ii)CC=C(C)C(C)BrCCC(C)(C)Br

This reaction is governed by Markownikoff’s rule according to which when an unsymmetrical reagent e.g. $HBr$ adds to an unsymmetrical alkene, then the negative part of the reagent is added to that carbon atom of the double bond which bears the least number of hydrogen atom. Thus, in above case 2-methyl-2-bromobutane will be the major product.

Peroxide effect is shown by $HBr$ only because $H-Cl$ bond is stronger than $H-Br$ and is not cleaved by the free radicals, whereas $H-I$ bond is weaker and Iodine free radical combine to form iodine molecule.



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