See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Concept: Carbon hybridization determines bond angles. sp-hybridized carbons have 180° bond angles (linear), sp2-hybridized carbons have 120° bond angles (trigonal planar), and sp3-hybridized carbons have 109.5° bond angles (tetrahedral). Step 1: Identify the hybridization of each carbon in CH3C≡CCH=CH2 (numbered C1 through C5 left to right). - C1 (CH3): sp3 hybridized — tetrahedral, bond angles ~109.5° - C2 (C≡): sp hybridized — linear - C3 (≡C): sp hybridized — linear - C4 (CH=): sp2 hybridized — trigonal planar, bond angle ~120° - C5 (=CH2): sp2 hybridized — trigonal planar Step 2: Determine which consecutive carbons can be collinear. The triple bond (C2≡C3) forces both C2 and C3 to be sp hybridized with 180° bond angles. This means the atoms bonded to C2 and C3 are also arranged linearly through the triple bond axis. - C1–C2≡C3 are linear because the C1–C2 bond lies along the axis of the sp carbon C2, making C1, C2, and C3 collinear. - C2≡C3–C4: C3 is sp hybridized, so C2, C3, and C4 are also collinear. - Therefore, C1, C2, C3, and C4 are all arranged in a straight line. Step 3: Check if C5 can also be included. C4 is sp2 hybridized with bond angles of 120°. The C3–C4=C5 angle is approximately 120°, not 180°, so C5 deviates from the line defined by C1–C2–C3–C4. Thus C5 is NOT collinear with the others. Step 4: Maximum linear arrangement is C1, C2, C3, C4 = 4 carbon atoms. Why other options fail: - (a) 2: Underestimates; the triple bond alone already gives 3 collinear carbons plus C1. - (b) 3: Underestimates; C1 through C4 are all collinear. - (d) 5: Overestimates; C4 is sp2, so C5 is bent away at 120° from the linear chain. Therefore, the correct answer is C.