Question

Provide the structure that is consistent with the data below.

C₆H₁₀
IR (cm^-1): 2950, 2230
¹H NMR (d): 2.0 (1H, septet), 1.8 (3H, s), 0.9 (6H, d)
¹³C NMR (d): 78 (s), 72 (s), 45 (d), 18 (q), 15 (q)

Answer 1

Answer – We are given, Molecular formula = C₆H₁₀

Site of unsaturation = C₆H₁₄ – C₆H₁₀ / 2H = 4H/2H = 2

IR = 2950 cm^-1 for the C-H alkane

         2230 cm^-1 for the C triple bond C- alkyne

¹HNMR – 2.0 ppm for the 1 H gives septate, so it is like -CH-(CH₃)₂

1.8 ppm for 3H gives singlet , so it is like -C triple bond C -CH₃

0.9 ppm for the 6H gives doublet , so it is like -CH-(CH₃)₂

So the structure for this data is as follow -