Which amongst the following is a positional isomer of ${CH}_{3} - {CH}_{2} - CH = {CH}_{3}$

{CH}_{3} - {CH}_{2} - {CH}_{2} - {CH}_{3}

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{CH}_{3} - CH = {CH}_{3}

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{CH}_{3} - CH = CH - {CH}_{3}

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{CH}_{3} - C \equiv C - {CH}_{3}

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Solution

The correct option is C ${CH}_{3} - CH = CH - {CH}_{3}$
${CH}_{3} - CH = CH - {CH}_{3}$ and ${CH}_{3} - {CH}_{2} - CH = {CH}_{3}$ are position isomers as they have the same molecular formula $(C_{4} H_{9})$ and the only difference is in the position of unsaturation. The IUPAC name of first compound is but-2-ene while that of the second compound is butene.

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Similar questions

A g N O_{3}

(A) C H_{3} - C H_{2} - C | C O O H H - - C H - C H_{3}

(B) C H_{3} - C | C O O H H - C H_{2} - - C H - C H_{3}

(C) C | C O O H H_{2} - C H_{2} - C H_{2} - - C H - C H_{3}

C H_{3} - C H_{3}, C H_{2} = C H_{2}

C H \equiv C H

S_{N} 2

c o n c . H C l :

(a) C H_{3} - C H_{2} - O H A n h y Z n C l_{2} / △ - -------- \to C H_{3} - C H_{2} - C l

(b) C H_{3} - C H_{3} | C | C H_{3} - C H_{2} O H A n h y Z n C l_{2} / △ - -------- \to C H_{3} - C H_{1} | C | C H_{3} - C H_{2} - C H_{3}

(c) C H_{3} - O H | C | C H_{3} - C H_{3} A n h y Z n C l_{2} / △ - -------- \to C H_{3} - C l | C | C H_{3} - C H_{3}

(d) C H_{2} = C H - C H_{2} - O H A n h y Z n C l_{2} / △ - -------- \to C H_{2} = C H - C H_{2} - C l

A . C H_{3} - C H = C H_{2} + H C l ⟶ C H_{3} - C | C l H - C H_{3}

B . C H_{3} - C H = C H_{2} + H B r ⟶ C H_{3} - C H_{2} - C H_{2} B r

C . C H_{3} - C H = C H_{2} + H B r p e r o x i d e - ---- \to C H_{3} - C H_{2} - C H_{2} - B r

D . C H_{3} - C H = C H_{2} + H I p e r o x i d e - ---- \to C H_{3} - C | I H - C H_{3}

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