AP EAMCET · Chemistry · Hydrocarbons

The following reaction is a/an
\(\begin{aligned} \mathrm{CH}_3-\mathrm{CH}_2- & \mathrm{CH}_2-\mathrm{Br} \xrightarrow{\mathrm{KOH}} \\ & \mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2+\mathrm{KBr}+\mathrm{H}_2 \mathrm{O}\end{aligned}\)

A substitution reaction
B addition reaction
C electrophilic substitution reaction
D elimination reaction

Verified Solution

Answer & Solution

Correct Answer

(D) elimination reaction

Step-by-step Solution

Detailed explanation

\(\begin{aligned} \mathrm{CH}_3-\mathrm{CH}_2- & \mathrm{CH}_2-\mathrm{Br} \xrightarrow{\mathrm{KOH}} \\ & \mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2+\mathrm{KBr}+\mathrm{H}_2 \mathrm{O}\end{aligned}\) In the given reaction alkene is formed from alkane by elimination of…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Explore more questions on app

From AP EAMCET

Explore more questions on app

	$List- I$		$List- II$
$(A)$	Electron-deficient hydride	$(I)$	${CH}_{4}$
$(B)$	Electron-precise hydride	$(II)$	$B_{2} H_{6}$
$(C)$	Electron-rich hydride	$(III)$	${CaH}_{2}$
$(D)$	Saline hydride	$(IV)$	${NiH}_{0.6}$
		$(V)$	${PH}_{3}$

The following reaction is a/an\(\begin{aligned} \mathrm{CH}_3-\mathrm{CH}_2- & \mathrm{CH}_2-\mathrm{Br} \xrightarrow{\mathrm{KOH}} \\ & \mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2+\mathrm{KBr}+\mathrm{H}_2 \mathrm{O}\end{aligned}\)

Step-by-step Solution

The following reaction is a/an
\(\begin{aligned} \mathrm{CH}_3-\mathrm{CH}_2- & \mathrm{CH}_2-\mathrm{Br} \xrightarrow{\mathrm{KOH}} \\ & \mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2+\mathrm{KBr}+\mathrm{H}_2 \mathrm{O}\end{aligned}\)