Q19 / 160 · TS EAMCET 2024 (10 May Shift 1) · Chemistry
The major product ' $Y$ ' in the given sequence of reactions is
$\mathrm{C}_3 \mathrm{H}_7 \mathrm{OH} \xrightarrow[443 \mathrm{~K}]{\text { Conc. } \mathrm{H}_2 \mathrm{SO}_4} \mathrm{X} \xrightarrow[\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CO}\right)_2 \mathrm{O}_2]{\mathrm{HBr}} \mathrm{Y}$
$\mathrm{C}_3 \mathrm{H}_7 \mathrm{OH} \xrightarrow[443 \mathrm{~K}]{\text { Conc. } \mathrm{H}_2 \mathrm{SO}_4} \mathrm{X} \xrightarrow[\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CO}\right)_2 \mathrm{O}_2]{\mathrm{HBr}} \mathrm{Y}$
- Correct answerA$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}$
- B$\mathrm{CH}_3 \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_3$
- C$\mathrm{CH}_3 \mathrm{COC}_6 \mathrm{H}_5$
- D$\mathrm{C}_6 \mathrm{H}_5 \mathrm{COBr}$
Solution
$\begin{aligned}
& \mathrm{C}_3 \mathrm{H}_7 \mathrm{OH} \xrightarrow{\mathrm{ConcH}_2 \mathrm{SO}_4} \mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2 \xrightarrow[\left(\mathrm{C}_6 \mathrm{HBr}_5 \mathrm{CO}\right)_2 \mathrm{O}]{ } \\
& \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{Br}
\end{aligned}$
Hence, the correct answer is option: 'A'.