Hydrocarbons

Concept:Buta-1,3-diyne with 2 equivalents of $\mathrm{NaNH}_2$ forms a dianion that undergoes $\mathrm{S_N2}$ substitution with excess trans-$\mathrm{CH_3-CH=CH-CH_2-Br}$ to give product $\mathrm{X}$.Explanation:Buta-1,3-diyne ($\mathrm{HC\equiv C-C\equiv CH}$) reacts with $\mathrm{NaNH}_2$ to remove both terminal hydrogens, yielding the dianion $^{-}\mathrm{C\equiv C-C\equiv C}^{-}$.This dianion reacts with two molecules of trans-$\mathrm{CH_3-CH=CH-CH_2-Br}$ via $\mathrm{S_N2}$ at the allylic carbon.The product $\mathrm{X}$ is: $\mathrm{CH_3-CH=CH-CH_2-C\equiv C-C\equiv C-CH_2-CH=CH-CH_3}$.In $\mathrm{X}$, the central diyne chain ($-\mathrm{C\equiv C-C\equiv C}-$) is linear due to $sp$ hybridization of the triple-bonded carbons.Each terminal $sp$ carbon of the diyne also bonds linearly to the adjacent $\mathrm{CH_2}$ carbon of the allyl group, placing those two $\mathrm{CH_2}$ carbons exactly on the same straight line.Thus the carbons of the $\mathrm{CH_2-C\equiv C-C\equiv C-CH_2}$ fragment are all collinear.That gives a total of 6 carbon atoms in a straight line.The alkene carbons ($\mathrm{CH=CH}$) are not collinear with this line because the $\mathrm{CH_2}$ carbon is $sp^3$ hybridized and the bond to the alkene is at an angle.Hence the maximum number of collinear carbons in $\mathrm{X}$ is 6.Answer:$6$