Consider the simplest alkane having only one `2^@C` one `3^@C`, and one `4^@C`, how may structural isomers are possible?

Options

A. only one
B. Three
C. Two
D. four

Answer

B

Explanation

The three kinds of C atoms `[H-underset(|)overset(|)(C)-H,-underset(|)overset(|)(C)-H`, and
`underset(|)overset(|)(C)-]` can be joined in three different ways,
(i). `2^(@),3^(@),4^(@)`
(ii). `2^(@),4^(@),3^(@)`
(iii). `4^(@)-2^(@)-3^(@)`
`Me-underset(H)underset(|)overset(H)overset(|)(C)-underset(H)underset(|)overset(Me)overset(|)(C)-underset(Me)underset(|)overset(Me)overset(|)(C)-Me" "Me-underset(H)underset(|)overset(H)overset(|)(C)-underset(Me)underset(|)overset(Me)overset(|)(C)-underset(H)underset(|)overset(Me)overset(|)(C)-Me`
`Me-underset(Me)underset(|)overset(Me)overset(|)(C)-underset(H)underset(|)overset(H)overset(|)(C)-underset(H)underset(|)overset(Me)overset(|)(C)-Me`
Here e, stands for methyl, that the reverse orde3r
`3^(@)-2^(@)-4^(@)` gives the same isomer.