How many carbon atoms does an alkane (not a c | vedclass

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(D) For an alkane to exist in an enantiomeric form,it must possess at least one chiral carbon atom (a carbon atom bonded to four different groups).Let

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$7$

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(D) For an alkane to exist in an enantiomeric form,it must possess at least one chiral carbon atom (a carbon atom bonded to four different groups).Let the chiral carbon be $C^*$. The four groups attached to $C^*$ must be different.In an alkane,the smallest groups are hydrogen $(-H)$,methyl $(-CH_3)$,ethyl $(-CH_2CH_3)$,and propyl $(-CH_2CH_2CH_3)$.Summing the carbon atoms in these groups: $1$ (for the chiral carbon) $+ 1$ (methyl) $+ 2$ (ethyl) $+ 3$ (propyl) $= 7$ carbon atoms.Therefore,the smallest alkane capable of exhibiting enantiomerism is $3$-methylhexane,which has $7$ carbon atoms.

How many carbon atoms does an alkane (not a cycloalkane) need before it is capable of existing in an enantiomeric form?

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