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Introduction to Alcohols

Two optically...

Question

Two optically active acyclic compounds $X$ and $Y$ (molecular formula $C_{5} H_{9} B r$ ) give the given reactions
Which of the following is/are correct?
$X H_{2} (1 m o l e) - ----- \to W (a c h i r a l); Y H_{2} (1 m o l e) - ----- \to S (c h i r a l)$

W = C H_{3} - C H_{2} - C H | B r - C H_{2} - C H_{3}

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Y = C H_{3} - C H | B r - C H_{2} - C H = C H_{2}

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X = C H_{3} - C H_{2} - C | B r = C H - C H_{3}

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S = C H_{3} - C H | B r - C H_{2} - C H_{2} - C H_{3}

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Solution

The correct options are
A $W = C H_{3} - C H_{2} - C H | B r - C H_{2} - C H_{3}$
B $Y = C H_{3} - C H | B r - C H_{2} - C H = C H_{2}$
C $X = C H_{3} - C H_{2} - C | B r = C H - C H_{3}$
D $S = C H_{3} - C H | B r - C H_{2} - C H_{2} - C H_{3}$

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Similar questions

S_{N} 2

K I

C H_{3} - C H_{2} - C H_{2} - C H_{2} - C l a n d C H_{3} - C H ∣ C H_{3} - C H_{2} - C l

(I) (I I)

C H_{3} - C H_{2} - C H_{2} - C l a n d C H_{3} - C H_{2} - C H_{2} - B r

(I) (I I)

C H_{3} - C H ∣ C H_{3} - C H_{2} - C H_{2} - C l a n d C H_{3} - C H_{3} ∣ C ∣ C H_{3} - C H_{2} C l

(I) (I I)

C H_{3} - C H_{2} - C H_{2} - B r ∣ C H - C H_{3} a n d C H_{3} - C H_{3} ∣ C H - C H_{2} - B r ∣ C H - C H_{3}

(I) (I I)

C H_{3} - C H_{2} - \oplus C H_{2}

C H_{3} - C H | F - \oplus C H_{2}

C H_{3} - C H | B r - \oplus C H_{2}

C H_{3} - C H | C l - \oplus C H_{2}

(A) C H_{3} - C H_{2} - C | B r H - + C H - C H_{3}

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

C H_{3} - C H_{2} - C H_{2} - C H_{2} - C H | C H_{3} - C | C H_{3} - C H_{3} - C H_{2} - C H_{3} | C H - C H_{2} - C H_{2} - C H_{2} - C H_{3}

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

C H \equiv C H

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