91.
What is the product of the following sequence of reactions ?
\[{{\left( C{{H}_{3}} \right)}_{2}}C=CH.C{{H}_{2}}C{{H}_{3}}\] \[\xrightarrow[\left( \text{ii} \right)\,{{H}_{2}}{{O}_{2}},O{{H}^{-}}]{\left( \text{i} \right)\,B{{H}_{3}}/THF}\,\,\xrightarrow[C{{H}_{2}}C{{l}_{2}}]{PCC}\] \[\xrightarrow[\left( \text{ii} \right)\,{{H}_{3}}{{O}^{+}}]{\left( \text{i} \right)\,C{{H}_{3}}MgBr}\]
Solubility of alcohols decreases with increase in size of alkyl group. Boiling points of alcohols increase with increase in the number of carbon atoms and decrease with increase of branching in carbon chain.
Hence, \[C{{H}_{3}}\underset{\begin{smallmatrix}
| \\
\,\,\,C{{H}_{3}}
\end{smallmatrix}}{\overset{\begin{smallmatrix}
\,\,\,\,\,C{{H}_{3}} \\
|
\end{smallmatrix}}{\mathop{-C-}}}\,OH<\] \[C{{H}_{3}}\underset{\begin{smallmatrix}
|\,\,\,\,\, \\
C{{H}_{3}}\,\,
\end{smallmatrix}}{\mathop{-CH-}}\,C{{H}_{2}}OH\] $$ < C{H_3}C{H_2}C{H_2}C{H_2}OH$$
96.
An ether $$\left( A \right),{C_5}{H_{12}}O,$$ when heated with excess of hot concentrated $$HI$$ produced two alkyl halides which when treated with $$NaOH$$ yielded compounds $$(B)$$ and $$(C).$$ Oxidation of $$(B)$$ and $$(C)$$ gave a propanone and an ethanoic acid
respectively. The $$IUPAC$$ name of the ether $$(A)$$ is :
The reaction involves the formation of carbocation as intermediate. Hence more the stability of the carbocation, more will be the rate of reaction. Let us draw the structure of the corresponding carbocation and observe the relative stability of the four benzyl carbocations. The relative stability of the four carbocations is
100.
In the following reaction.
\[{{C}_{2}}{{H}_{5}}O{{C}_{2}}{{H}_{5}}+4H\xrightarrow{\text{Red}\,\,P+HI}2X+{{H}_{2}}O\,;X\] is
Ethers are reduced by red $$P$$ and $$HI$$ to alkanes through alkyl iodides \[{{C}_{2}}{{H}_{5}}O{{C}_{2}}{{H}_{5}}\xrightarrow{\text{red}\,\,P+HI}2{{C}_{2}}{{H}_{5}}I\xrightarrow{\text{red}\,\,P+HI}2{{C}_{2}}{{H}_{6}}\]