An alkene C6H12 gives only propanone on ozonolysis followed by Zn/H2O. The number of signals in the ^13C NMR spectrum of this alkene is

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An alkene $C X_{6} H X_{12}$ gives only propanone on ozonolysis followed by $Zn / H X_{2} O$ . The number of signals in the $^{13}$ C NMR spectrum of this alkene is

Only propanone,

(CH X_{3}) X_{2} C = O

, is formed, so the alkene is

(CH X_{3}) X_{2} C = C (CH X_{3}) X_{2}

(2,3-dimethylbut-2-ene). The four methyl carbons are equivalent and the two sp

^{2}

carbons are equivalent, giving 2 signals.