What is the hybridization shown by C_1 and C_ | vedclass

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(A) The structure of the compound is $OHC-CH=CH-CH_2-COOCH_3$. Numbering the carbons as indicated in the structure: $C_1$ is the carbonyl carbon of th

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$sp^2$ and $sp^3$

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(A) The structure of the compound is $OHC-CH=CH-CH_2-COOCH_3$. Numbering the carbons as indicated in the structure: $C_1$ is the carbonyl carbon of the ester group $(-COOCH_3)$,which is bonded to one oxygen by a double bond and two other groups by single bonds. It has $3$ sigma bonds and $0$ lone pairs,so it is $sp^2$ hybridized. $C_2$ is the methylene carbon $(-CH_2-)$ bonded to four atoms by single bonds. It has $4$ sigma bonds and $0$ lone pairs,so it is $sp^3$ hybridized. Therefore,$C_1$ is $sp^2$ and $C_2$ is $sp^3$ hybridized.

What is the hybridization shown by $C_1$ and $C_2$ carbons,respectively in the given compound? $OHC-CH=CH-CH_2-COOCH_3$

Similar Questions

In the following molecule,the two carbon atoms marked by asterisk $()$ possess the following type of hybridized orbitals: $CH_3-C^ \equiv C^*-CH_3$

In the molecule $CH \equiv C - CH = CH_2$,the hybridisation of the $C - C$ single bond is:

Which of the following molecules represents the order of hybridisation $sp^2, sp^2, sp, sp$ from left to right for the carbon atoms?

Amongst the given options,the compound$(s)$ in which all the atoms are in one plane in all the possible conformations (if any),is (are):
$(A)$ $CH_2=CH-CH=CH_2$
$(B)$ $HC\equiv C-CH=CH_2$
$(C)$ $H_2C=C=O$
$(D)$ $H_2C=C=CH_2$

In the molecule $HC^{1} \equiv C^{2} - CH^{3} = CH^{4} - CH_{3}^{5}$,which carbon atom exhibits the lowest electronegativity?

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