I am not that quite sure of the answer of this question ??

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2 Answers
Truong-Son N.
May 2, 2018

You've chosen #"C"_6"H"_12# (cis-dimethylcyclobutane) and #"C"_6"H"_14# (2-ethylbutane), respectively. The choices are cis-dimethylcyclobutane, a five-coordinate carbon molecule (makes no sense), 2,3-dimethylbutene, and 2-ethylbutane.

What you want is #"C"_6"H"_12# for all chosen answers...

Simply count the number of carbon atoms (all of these have six), and then draw out the implicit hydrogens. Any given carbon atom in MANY cases has at most 4 bonds.

  • cis-dimethylcyclobutane is valid. Four carbon ring atoms, two being #"CH"_2# and two being #"CH"#, and two external carbons each being #"CH"_3#.
  • Any five-coordinate hydrocarbon is invalid by virtue of not existing...
  • 2,3-dimethylbutene is valid. Two double-bond carbon atoms with no #"H"#, and four external carbons each being #"CH"_3#.
  • 2-ethylbutane is invalid. The main chain has four carbon atoms, two of which are #"CH"_3#, one is #"CH"_2#, and one is #"CH"#. The ethyl adds #"CH"_2# and #"CH"_3#, giving #14# hydrogen atoms, not #12#.
anor277
May 2, 2018

Well, a formula of #C_6H_12# has ONE degree of saturation...

Explanation:

And what's that? Alkanes have formula of #C_nH_(2n+2)#...each two hydrogens LESS than this formula specifies a double bond, or a ring junction.

The cyclobutane derivative, #"option a."#, thus has formula #C_6H_12#. #"Option c."#, the olefin, also has this formula. #"Option b."# is NOT a real molecule. And #"Option d."#, #C_6H_14# is an alkane....

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