Question

©C10H140 I wish IH Ovih 21D)* 2-1, 4 HD1 CHS e-CH3

Answer 1

• The formula is C₁₀H₁₄O
• As you have correctly calculated, the degree of unsaturation of the compound will be 4.
• Further 2 protons have around 7.3 ppm shift and 2 protons have 6.8 ppm shift. This suggests an aromatic system and with 4 degree of unsaturation, it means presence of a benzene ring.
• Further, you only have 4 Hydrogens on benzene ring which means that 2 Hydrogen are missing. This suggests two substituents on the benzene ring.
• Moreover, you only are getting 2 peaks corresponding to benzene ring and the peaks appears to be a doublet peak. This is only possible when the substituent are para to each other i.e. 1,4-disubstituted benzene
• Further, there are 9 protons having same nmr shift of around 1.4 ppm. This suggests a tertiary butyl group i.e. (CH₃)₃C-
• Moreover, you see one broad peak around 5 ppm. Moreover, we have one Oxygen atom to account for. Peak in this range suggests a phenolic proton i.e. and -OH group on benzene ring
• Putting together all the pieces, we will obtain following structure-

The compound is 4-(tert-Butyl)phenol.