Answer: 4-allylanisole
Explanation: The doublets behind the 7 ppm belongs to the
para-substituted benzene ring. The three single-proton multi-plets around 5−6 ppm predicts that there has to be a single subsituted alkene group
A single plus a doublet around 3-4 ppm belongs to CH3 and CH2 Groups as they could be attached to the subsituted alkene group.
Moreover the interpretation of the NMR that there is no peak with a higher intensity for >180 ppm represents an absence of Carbonyl group.
The Predicted Number is attached from a chemical database along with their peaks information
Answer:
C is the answer to your question
This is called the pedigree chart.
Answer:

Explanation:
1. Write the balanced chemical equation.

2. Calculate the moles of HCOOH

3. Calculate the moles of NaOH.

4. Calculate the volume of NaOH

HNO3 + Ca —-> H2O + N20 + Ca(NO3)2