182.
Which of the following reactions will give the major and minor products?
\[C{{H}_{3}}-C{{H}_{2}}-\underset{\begin{smallmatrix}
|\,\,\,\,\,\,\,\,\,\, \\
Br\,\,\,\,\,\,\,\,
\end{smallmatrix}}{\mathop{CH-}}\,C{{H}_{3}}\xrightarrow[\text{heat}]{alc.\,KOH}\]
$$C{H_3} - \mathop {CH = CH}\limits_{\left( A \right)} - C{H_3} + C{H_3}$$ $$ - \mathop {C{H_2} - CH}\limits_{\left( B \right)} = C{H_2}$$
A
$$(A)$$ is major product and $$(B)$$ is minor product.
B
$$(A)$$ is minor product and $$(B)$$ is major product.
C
Both $$(A)$$ and $$(B)$$ are major products.
D
Only $$(B)$$ is formed and $$(A)$$ is not formed.
Answer :
$$(A)$$ is major product and $$(B)$$ is minor product.
According to Saytzeff's rule, the more substituted product is more stable and is formed as major product. Hence $$(A)$$ is the major product (80%) while $$(B)$$ is the minor (20%) and less stable product.
183.
In a nucleophilic substitution reaction : \[R-Br+C{{l}^{-}}\xrightarrow{DMF}R-Cl+B{{r}^{-}},\] Which one of the following undergoes complete inversion of configuration ?
$${C_6}{H_5}CHC{H_3}Br$$ being an optically active secondary alkyl bromide undergoes $${S_N}2$$ nucleophilic substitution reaction. Hence it undergoes complete inversion of configuration.
184.
Which of the following will give enantiomeric pair on reaction with water due to presence of asymmetric carbon atom?
A
\[{{C}_{2}}{{H}_{5}}\underset{\begin{smallmatrix}
|\,\,\,\,\,\,\,\, \\
{{C}_{2}}{{H}_{5}}\,\,\,
\end{smallmatrix}}{\overset{\begin{smallmatrix}
{{C}_{2}}{{H}_{5}}\, \\
|\,\,\,\,\,\,\,\,
\end{smallmatrix}}{\mathop{-\,\,C-Br}}}\,\]
B
\[{{C}_{2}}{{H}_{5}}\underset{\begin{smallmatrix}
|\,\,\,\,\,\,\,\, \\
C{{H}_{3}}\,\,\,
\end{smallmatrix}}{\overset{\begin{smallmatrix}
{{C}_{2}}{{H}_{5}} \\
|\,\,\,\,\,\,\,
\end{smallmatrix}}{\mathop{-\,\,C-Cl}}}\,\]
C
\[{{C}_{2}}{{H}_{5}}\underset{\begin{smallmatrix}
| \\
\,\,\,\,C{{H}_{3}}
\end{smallmatrix}}{\overset{\begin{smallmatrix}
H \\
|
\end{smallmatrix}}{\mathop{-C-}}}\,I\]
D
\[C{{H}_{3}}\underset{\begin{smallmatrix}
|\,\,\,\,\,\,\,\, \\
{{C}_{2}}{{H}_{5}}\,\,\,
\end{smallmatrix}}{\overset{\begin{smallmatrix}
C{{H}_{3}} \\
|\,\,\,\,\,\,\,
\end{smallmatrix}}{\mathop{-\,\,C-Br}}}\,\]
188.
Pure $$\left( S \right) - C{H_3}C{H_2}CHBrC{H_3}$$ is subjected to monobromination to form several isomers of $${C_4}{H_8}B{r_2}.$$ Which of the following is not produced?