Duration: 6:32

0 views

0

In the following sequence of reaction a compound
A. (molecular formula \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{2} \) ) with a straight chain structure gives a \( \mathrm{C}_{4} \) carboxylic acid. A is
A \( \frac{\mathrm{LiAlH}_{4}}{\mathrm{H}_{3} \mathrm{O}^{+}} \rightarrow \mathbf{B} \stackrel{\text { Oxidation }}{\longrightarrow} \mathrm{C}_{4} \) - carboxylic acid
(a)
\[
\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{COO}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{3}
\]
(b)
(c) \( \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{COO}-\mathrm{CH}_{2}-\mathrm{CH}_{3} \)
(d) \( \mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{2}-\mathrm{OH} \)
📲PW App Link - https://bit.ly/YTAI_PWAP
🌐PW Website - https://www.pw.live