Predict the product.
$C H_{3} - C H = C H_{2} + H B r \to$

C H_{3} - C H (B r) - C H_{3}

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C H_{3} - C H (B r) = C H_{3}

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

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none of these

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Solution

The correct option is A $C H_{3} - C H (B r) - C H_{3}$
The addition of hydrogen bromide to propene gives isopropyl bromide as a major product according to Markownikoff's rule.
$C H_{3} - C H = C H_{2} + H - B r \to C H_{3} - C H (B r) - C H_{3}$
Note:
The addition of hydrogen halides to unsymmetrical alkene takes place according to Markovonikov's rule.
When an unsymmetrical reagent adds to an unsymmetrical alkene, the negative part of the reagent gets attached to that unsaturated carbon atom which caries lesser number of hydrogen atoms.

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

-------- the product is

Q. Identify the type of the following reaction of carbon compounds.
a. CH₃ -CH₂ -CH₂-OH → CH₃ -CH₂ -COOH
b. CH₃ -CH₂ -CH₃ → 3CO₂ + 4H₂O
c. CH₃ -CH = CH -CH₃ + Br₂ → CH₃ -CHBr - CHBr -CH₃
d. CH₃ -CH₃ + Cl₂→ CH₃ -CH₂ -Cl + HCl
e. CH₃-CH₂ -CH₂ -CH₂ -OH → CH₃ -CH₂ -CH=CH₂ + H₂O
f. CH₃-CH₂ -COOH + NaOH → CH₃ -CH₂ -COO - Na⁺ + H₂O
g. CH₃ -COOH + CH₃ -OH → CH₃ -COO- CH₃ + H₂O

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. Assertion :When

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

is treated with potassium tert-butoxide

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

is formed. Reason: The above reaction is

E_{2}

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:

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Predict the product.CH3−CH=CH2+HBr→

Predict the product.
$C H_{3} - C H = C H_{2} + H B r \to$