At low temperatures, the slow addition of molecular bromine to $C H_{2} = C H - C H_{2} - C \equiv C H$ gives

C H_{2} = C H - C H_{2} - C B r = C H B r

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B r C H_{2} - C H B r - C H_{2} - C \equiv C H

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C H_{2} = C H - C H_{2} - C H_{2} - C B r_{3}

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C H_{3} - C B r_{2} - C H_{2} - C \equiv C H

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Solution

The correct option is A $C H_{2} = C H - C H_{2} - C B r = C H B r$
At low temperatures, the slow addition of molecular bromine produces $C H_{2} = C H - C H_{2} - C B r = C H B r .$

$C H_{2} = C H - C H_{2} - C \equiv C H + B r_{2} \to C H_{2} = C H - C H_{2} - C | B r = C H | B r$

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Similar questions

C H_{3} - C H = C H_{2} + H B r \to

-------- the product is

Q. A hydrocarbon, 'X'

(C_{3} H_{4})

decolourises

B r_{2} / C C l_{4}

solution forming 'Y'. 'X' gave a red precipitate of a compound 'Z' with ammoniacal cuprous chloride.
X, Y, and Z respectively are:

Which are free radical reactions

$I . C H_{3} - C H = C H_{2} + H B r P e r o x i d e - ---- \to C H_{3} - C H_{2} - C H_{2} B r I I . C H_{3} - C H = C H_{2} + H B r \to C H_{3} - C H B r - C H_{3} I I I . C H_{3} - C H = C H_{2} + C l_{2} \to C l C H_{2} - C H = C H_{2} I V . C H_{3} - C H_{3} + C l_{2} \to C l C H_{2} - C H_{3}$

Q. Which will undergo reaction with ammonical

A g N O_{3}

Q. The correct order of C - H bond length in

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

and

C H \equiv C H

is:

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At low temperatures, the slow addition of molecular bromine to CH2=CH−CH2−C≡CH gives

The correct option is A CH2=CH−CH2−CBr=CHBrAt low temperatures, the slow addition of molecular bromine produces CH2=CH−CH2−CBr=CHBr.CH2=CH−CH2−C≡CH+Br2→CH2=CH−CH2−C|Br=CH|Br

At low temperatures, the slow addition of molecular bromine to $C H_{2} = C H - C H_{2} - C \equiv C H$ gives

The correct option is A $C H_{2} = C H - C H_{2} - C B r = C H B r$
At low temperatures, the slow addition of molecular bromine produces $C H_{2} = C H - C H_{2} - C B r = C H B r .$

$C H_{2} = C H - C H_{2} - C \equiv C H + B r_{2} \to C H_{2} = C H - C H_{2} - C | B r = C H | B r$