Which of the following compounds contains car | vedclass

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(D) $1$. In $C_4H_{10}$ (butane),all carbon atoms are $sp^3$ hybridized.$2$. In $CH_3-CH=CH-CH_3$ (but$-2-$ene),the terminal carbons are $sp^3$ hybrid

Vedclass · Accepted Answer

Both $CH_3-CH=CH-CH_3$ and $CH_2=CH-C\equiv CH$

Vedclass · Answer

(D) $1$. In $C_4H_{10}$ (butane),all carbon atoms are $sp^3$ hybridized.$2$. In $CH_3-CH=CH-CH_3$ (but$-2-$ene),the terminal carbons are $sp^3$ hybridized,while the carbons involved in the double bond are $sp^2$ hybridized.$3$. In $CH_2=CH-C\equiv CH$ (but$-1-$en$-3-$yne),the terminal $CH_2$ carbon is $sp^2$,the middle $CH$ carbon is $sp^2$,and the carbons in the triple bond are $sp$ hybridized.$4$. Since both $CH_3-CH=CH-CH_3$ and $CH_2=CH-C\equiv CH$ contain carbon atoms with different types of hybridization,the correct answer is $D$.

Which of the following compounds contains carbon atoms with more than one type of hybridization?

Similar Questions

In the following compound,the number of $sp$-hybridised carbons is:

What is the type of hybridisation of each carbon atom in the following compounds?
$(a)$ $CH_{3}Cl$
$(b)$ $(CH_{3})_{2}CO$
$(c)$ $CH_{3}CN$
$(d)$ $HCONH_{2}$
$(e)$ $CH_{3}CH=CHCN$

The pair of electrons in the given carbanion,$CH_{3}C\equiv C^{-},$ is present in which of the following orbitals?

In the molecule $CH_2=C=CH-CH_3$ (where the atoms are numbered $1, 2, 3,$ and $4$ respectively),the hybridization of carbon $1, 2, 3,$ and $4$ are:

Which one of the following does not have $sp^2$ hybridised carbon?

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