In which of the following compounds is the un | vedclass

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(B) To determine the hybridization of a carbon atom,we count the number of sigma $(\sigma)$ bonds and lone pairs attached to it. $A$ carbon atom with

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$CH_3\underline{C}H_2OH$

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(B) To determine the hybridization of a carbon atom,we count the number of sigma $(\sigma)$ bonds and lone pairs attached to it. $A$ carbon atom with $4$ sigma bonds is $sp^3$ hybridized.$A$. In $CH_3\underline{C}OOH$,the underlined carbon is part of a carbonyl group $(C=O)$,so it has $3$ sigma bonds and is $sp^2$ hybridized.$B$. In $CH_3\underline{C}H_2OH$,the underlined carbon is bonded to one $CH_3$ group,two $H$ atoms,and one $OH$ group. It forms $4$ sigma bonds,so it is $sp^3$ hybridized.$C$. In $CH_3\underline{C}OCH_3$,the underlined carbon is part of an ether linkage $(C-O-C)$,but the carbon itself is bonded to an oxygen and another carbon. Specifically,in $CH_3-O-CH_3$,the carbons are bonded to $3$ hydrogens and $1$ oxygen,forming $4$ sigma bonds,making them $sp^3$ hybridized. However,looking at the structure $CH_3\underline{C}OCH_3$ (which is likely $CH_3-O-CH_3$),the carbon is $sp^3$. Let's re-evaluate the options. Option $B$ is clearly $sp^3$.$D$. In $CH_2 = \underline{C}H - CH_3$,the underlined carbon is part of a double bond,so it has $3$ sigma bonds and is $sp^2$ hybridized.Therefore,the correct option is $B$.

In which of the following compounds is the underlined carbon atom $sp^3$ hybridized?

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