Given below are two statements:Statement I: 3 | vedclass

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(A) Statement $I$: $3$-phenylpropene $(C_6H_5CH_2CH=CH_2)$ reacts with $HBr$ via electrophilic addition following Markovnikov's rule.The intermediate

Vedclass · Accepted Answer

Both Statement $I$ and Statement $II$ are true

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(A) Statement $I$: $3$-phenylpropene $(C_6H_5CH_2CH=CH_2)$ reacts with $HBr$ via electrophilic addition following Markovnikov's rule.The intermediate carbocation formed at $C-2$ $(C_6H_5CH_2CH^+CH_3)$ is more stable due to the inductive effect and hyperconjugation compared to the primary carbocation at $C-1$.The major product is $2$-bromo-$3$-phenylpropane $(C_6H_5CH_2CH(Br)CH_3)$,which contains a chiral center at $C-2$.Thus,Statement $I$ is true.Statement $II$: The Sandmeyer reaction uses $Cu_2Cl_2/HCl$ or $Cu_2(CN)_2/KCN$ to convert diazonium salts to aryl chlorides or aryl cyanides.The Gattermann reaction is a modification of the Sandmeyer reaction that uses copper powder $(Cu)$ in the presence of the corresponding acid ($HCl$ or $HCN$).Both reactions are standard methods for the preparation of aryl halides and cyanides.Thus,Statement $II$ is true.

Column-$A$	Column-$B$
$(1)$ $CH_3-CH_2-Cl + KOH_{(aq)} \rightarrow$	$(A)$ $1,2-\text{dichloroethane}$
$(2)$ $CH_3-CH_2-CH_2-CH_3 \xrightarrow{AlCl_3/HCl}$	$(B)$ $\text{Chloromethane}$
$(3)$ $CH_3-CH_2-Br + \text{alc. } KOH \rightarrow$	$(C)$ $\text{But-2-ene}$
$(4)$ $CH_2=CH_2 + Cl_2 \rightarrow$	$(D)$ $\text{Ethanol}$
	$(E)$ $\text{Chloroethane}$
	$(F)$ $\text{Ethene}$
	$(G)$ $\text{Isobutane}$

Given below are two statements:
Statement $I$: $3$-phenylpropene reacts with $HBr$ and gives secondary alkyl bromide having a chiral carbon atom as the major product.
Statement $II$: Aryl chlorides and aryl cyanides can be prepared by Sandmeyer reaction as well as Gattermann reaction.

Similar Questions

Which of the following is not formed in the given reaction?
$C_6H_5CH_2Cl + CH_3Cl \xrightarrow{Na, \text{dry ether}}$

Consider the above sequence of reactions. $151 \ g$ of $2-$bromopentane is made to react. The yield of the major product $P$ is $80 \%$ whereas the yield of $Q$ is $100 \%$. The mass of product $Q$ obtained is $........ \ g$. (Given molar mass in $g \ mol^{-1}$: $H: 1, C: 12, Br: 80$)

Which product would be expected to predominate in the given reaction?

Identify the correct set of reagents or reaction conditions '$X$' and '$Y$' in the following set of transformation.

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Given below are two statements:Statement $I$: $3$-phenylpropene reacts with $HBr$ and gives secondary alkyl bromide having a chiral carbon atom as the major product.Statement $II$: Aryl chlorides and aryl cyanides can be prepared by Sandmeyer reaction as well as Gattermann reaction.