Question

Which of the following product is not obtain in reaction ?

$C{H}_3-CH=CH-C{H}_3+B{r}_2/H_{2}O\stackrel{aq.Nacl}{⟶}$

• A. $C{H}_3-\stackrel{Br}{\stackrel{|}{C}}H-\stackrel{Br}{\stackrel{|}{C}}H-C{H}_3$
• B. $C{H}_3-\stackrel{OH}{\stackrel{|}{C}}H-\stackrel{Br}{\stackrel{|}{C}}H-C{H}_3$
• C. $C{H}_3-\stackrel{Cl}{\stackrel{|}{C}}H-\stackrel{Br}{\stackrel{|}{C}}H-C{H}_3$
• D. $C{H}_3-\stackrel{OH}{\stackrel{|}{C}}H-\stackrel{OH}{\stackrel{|}{C}}H-C{H}_3$

Answer 1

The correct option is D $C{H}_3-\stackrel{OH}{\stackrel{|}{C}}H-\stackrel{OH}{\stackrel{|}{C}}H-C{H}_3$

${CH}_3-CH=CH-{CH}_3+{Br}_2/H_{2}O\underrightarrow{InNaCl}?$

In this reaction, these are two nucleophiles ${Br}^{-}$ and ${Cl}^{-}$. Firstly, the $\pi$ electrons will attack on ${Br}^{+}$ which is a electrophile obtained from the $Br-Br\left({Br}_2\right)$ bond polarisation. This ${Br}^{+}$ is added to the less substituted carbon, which is in accordance with Markovnikov's rule. This forms a cation / carbocation.

Now the, nucleophile will attack on the carbocation. Therefore, these can be varied products formed depending on the carbocation generated.

$C{H}_3-\stackrel{Br}{\stackrel{|}{C}}H-\stackrel{Br}{\stackrel{|}{C}}H-C{H}_3$ , $C{H}_3-\stackrel{OH}{\stackrel{|}{CH}}-\stackrel{Br}{\stackrel{|}{CH}}-C{H}_3$ , $C{H}_3-\stackrel{Cl}{\stackrel{|}{CH}}-\stackrel{Br}{\stackrel{|}{CH}}-C{H}_3$