In which of the following compounds the carbon marked with asterisk is most electronegative?

C H_{3} - C H_{2} - ˙ C H_{2} - C H_{3}

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C H_{3} - ˙ C H = C H - C H_{3}

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C H_{3} - C H_{2} - C \equiv ˙ C H

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C H_{3} - C H_{2} - C H = ˙ C H_{2}

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Solution

The correct option is D $C H_{3} - C H_{2} - C \equiv ˙ C H$
Increased s character ( $s p = 50$ %, $s p^{2} = 33$ % and $s p^{3} = 25$ %) implies that the alkyne sp orbital is closer to the nucleus and so there is greater electrostatic stabilisation of the electron pair. Therefore the free radical of the alkyne is the most stable and the most readily formed.

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C H_{3} - C H_{2} - C H_{2} - C H_{2} - C H | C H_{3} - C | C H_{3} - C H_{3} - C H_{2} - C H_{3} | C H - C H_{2} - C H_{2} - C H_{2} - C H_{3}

The IUPAC name of the following compound is?

$C H_{3} - C H_{2} - C H_{2} - C | H C H_{3} - C H_{2} - C H_{3} | C H - C H_{3}$

C H_{3} - C H_{2} - C H_{3} | C H - C H_{2} - C H_{2} - C H_{2} - C H_{2} - C H_{2} - C H_{2} - C H_{2} - C H_{2} - C H_{3} | C H - C H_{3}

{CH}_{3} - {CH}_{2} - CH = {CH}_{3}

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