Which of the following compounds is optically | vedclass

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Question

(A) compound is optically inactive if it does not contain a chiral center (an asymmetric carbon atom bonded to four different groups).$1.$ $2-$Chloro-

Vedclass · Accepted Answer

$2-$Chloro-$2-$methylbutane

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(A) compound is optically inactive if it does not contain a chiral center (an asymmetric carbon atom bonded to four different groups).$1.$ $2-$Chloro-$2-$methylbutane: The $C-2$ carbon is bonded to two identical methyl groups,so it is not chiral. Thus,it is optically inactive.$2.$ $3-$Chlorohexane: The $C-3$ carbon is bonded to $-H$,$-Cl$,$-CH_2CH_3$,and $-CH_2CH_2CH_3$. Since all four groups are different,it is chiral and optically active.$3.$ $2-$Chloro-$3-$methylbutane: The $C-2$ carbon is bonded to $-H$,$-Cl$,$-CH_3$,and $-CH(CH_3)_2$. Since all four groups are different,it is chiral and optically active.$4.$ $2-$Chloropentane: The $C-2$ carbon is bonded to $-H$,$-Cl$,$-CH_3$,and $-CH_2CH_2CH_3$. Since all four groups are different,it is chiral and optically active.Therefore,$2-$Chloro-$2-$methylbutane is the optically inactive compound.

List-$I$ (Reagent)	List-$II$ (Product)
$A$. $KOH$ (alc)	$I$. Nitrile
$B$. $KCN$ (alc)	$II$. Ester
$C$. $AgNO_2$	$III$. Alkene
$D$. $CH_3COOAg$	$IV$. Nitroalkane

Which of the following compounds is optically inactive?

Similar Questions

Which of the following compounds is most reactive in $S_N 1$ reaction?

What are nucleophiles and nucleophilic substitution reactions? Provide a basic overview of the nucleophilic substitution reaction of haloalkanes.

Match List-$I$ with List-$II$.
$1$-Bromopropane is reacted with reagents in List-$I$ to give products in List-$II$.

List-$I$ (Reagent) List-$II$ (Product)

$A$. $KOH$ (alc) $I$. Nitrile

$B$. $KCN$ (alc) $II$. Ester

$C$. $AgNO_2$ $III$. Alkene

$D$. $CH_3COOAg$ $IV$. Nitroalkane

What is $E$?
Cyclohexanone $\xrightarrow{Br_2 + CH_3COOH} A$ $\xrightarrow{Alc. KOH} B$ (Product with six-membered ring) $\xrightarrow[(2) H_2O]{(1) CH_3MgBr} C$ $\xrightarrow{H^{\oplus} + \Delta} D$ $\xrightarrow{HBr (1 \text{ mole})} E$

The following reaction belongs to $(CH_3)_3CBr \xrightarrow{H_2O} (CH_3)_3COH$.

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