First of all, spectra 1-3 are 13C NMRs. You can tell by the 200 ppm axis and the singlets. [Note: 1H NMR spectra have a scale of 10 ppm and they are more likely to show coupling.] Spectrum 4 is an infrared (IR) spectrum. Almost every single one of you studied IR in the lab course. Moreover, it is not necessary for the solution of the problem. Let's go. Compound A has zero degrees of unsaturation and one oxygen. Thus A cannot be an aldehyde or ketone (one degree of unsaturation per carbonyl group). A must be an acyclic ether or an acyclic alcohol. Compound A contains 7 carbons but the 13C NMR spectrum (#1) shows only 3 singlets, which indicates a high degree of symmetry in the compound. For the IR mavens, the broad absorption at the left-hand side of the spectrum is indicative of a hydroxyl group. In spectrum 1, the signal at ~74 ppm is significantly weaker than the two higher field signals. The carbon whose signal is at 74 ppm is most likely the one attached to the oxygen (lowest field signal) and, owing to its diminished intensity, the carbon most likely bears no hydrogens, i. e., tertiary alcohol (ether?). If indeed this is the case, the remaining 6 carbons in A must be accommodated by the 2 remaining carbon signals in the spectrum at ~ 30 and 8 ppm. This would indicate 3 ethyl groups. The ether with the fewest number of signals (4) in its 13C NMR spectrum that satisfies the formula C7H16O is t-butyl ethyl ether. Already, one can make a tentative assignment of 3-ethyl-3-pentanol for compound A.
The reaction of A with H2SO4 to form B, which subsequently undergoes ozonolysis, tells us that B contains a double bond formed by dehydration of an alcohol (without the IR data, the possibility of an ether may exist if you are not convinced at this point that A is an alcohol). One of the products of ozonolysis, namely compound C, has only 2 singlets in its 13C nmr spectrum (#2). The signal at 200 ppm is indicative of an aldehyde carbonyl carbon. What else can compound C be but acetaldehyde? Compound D (spectrum #3) has 3 singlets. If compound A is an alcohol, then the 7 carbons of A would be in B and their sum would be contained in C plus D. If C contains 2 carbons, then D has 5 carbons. The signal at ~210 ppm (weak intensity) is where the carbon of a ketone carbonyl resonates, which means the remaining 4 carbons of D must be represented by 2 signals. What C5 ketone but 3-pentanone. Thus, the assignment of A is correct and the dehydration of A forms only 3-ethyl-2-pentene, B.
"Elementary, my dear Watson!" - Sherlock Holmes
[ According to one Holmes afficionado, Sir Arthur Conan Doyle never used that phrase in his Holmes series. On the other hand, while Ilsa (Ingrid Bergmann), Rick (Humphrey Bogart), and Laszlo (Paul Henreid) all said "play it" in the film Casablanca (1942), no one ever said "Play it again, Sam". Search the transcript at this site.]