14CH2-CH-CH3 NBS- Mg-either- D2O- ProductsProduct s is-are -

The correct option is C Both of theseReaction: ^14CH_2=CH CH_3NBS→ #160;^14CH_2=CH CH_2Br #160;Mg/either→ #160; #160; ^14CH_2=CH CH_2MgBr #1 ...

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Introduction to Electrophilic Substitution

14CH2=CH-CH3 ...

Question

$^{14} C H_{2} = C H - C H_{3}$ $N B S \to$ $M g / e i t h e r \to$ $D_{2} O \to$ Product(s)
Product (s) is/are :

^{14} C H_{2} = C H - C H_{2} - D

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C H_{2} = C H -^{14} C H_{2} - D

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Both of these

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none of these

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Solution

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Similar questions

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

C H \equiv C H

C H_{2} = C H - . . N H - C H_{3}

C H_{2} = C H - . . O . . - C H_{3}

C H_{2} = C H - . . O . . - C H = C H_{2}

C H_{2} = C H - N H_{2}

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)

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(I) (I I)

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

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

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

A g N O_{3}

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